Chapter 4 – Chemical Composition
4.1
(a) mole; (b) Avogadro’s number; (c) empirical formula; (d) solute; (e) molarity; (f) concentrated solution
4.2
(a) molar mass; (b) percent composition by mass; (c) solvent; (d) concentration; (e) dilute solution; (f)
dilution
4.3
The percent composition of calcium in calcium carbonate is set up as:
% Ca =
4.4
grams Ca
9.88 g Ca
 100% 
 100% = 40.0 % Ca
grams CaCO3
24.7 g CaCO3
The percent composition of oxygen in calcium carbonate is set up as:
%O=
grams O
1.80 g O
 100% 
 100% = 48.0 % O
grams CaCO3
3.75 g CaCO3
The percent composition of carbon in calcium carbonate is set up as:
%C=
4.5
4.6
4.7
grams O
0.450 g O
 100% 
 100% = 12.0 % C
grams CaCO3
3.75 g CaCO3
Mass percent is defined as the mass of the part divided by the mass of the whole times 100%. It does not
matter what mass units are used, just so long as they are the same for the part and the whole. That allows
the units to cancel properly.
Mass of Part
Mass % =
100%
Mass of Whole
30.0 μg
Mass %C =
100% = 60.0% C
50.0 μg
Mass percent is defined as the mass of the part divided by the mass of the whole times 100%. It does not
matter what mass units are used, just so long as they are the same for the part and the whole. That allows
the units to cancel properly.
Mass of Part
Mass % =
100%
Mass of Whole
3.54 mg
Mass %N =
100% = 17.7% N
20.0 mg
To calculate the actual mass of a substance in a sample, we rearrange the percent equation in the following
way:
Mass of Part
Mass %  Mass of Whole
Mass % =
100% becomes Mass of Part =
Mass of Whole
100%
Mass of lithium =
4.8
1.2 g 18.8%
= 0.226 g lithium
100%
To calculate the actual mass of a substance in a sample, we rearrange the percent equation in the following
way:
Mass of Part
Mass %  Mass of Whole
Mass % =
100% becomes Mass of Part =
Mass of Whole
100%
41
Mass of silicon =
348 g  46.7%
= 163 g silicon
100%
4.9
Chemical formulas must have whole number subscripts. (a) H2S; (b) A ratio of 1.5 oxygen atoms to 1
nitrogen atom means that the number of each atom needs to be doubled to give whole-number coefficients
for both atoms. Doubling the values works because it converts 1.5 to 3 (a whole number). This gives a
ratio of 3 oxygen atoms to 2 nitrogen atoms (note that both quantities are doubled). The chemical formula
is N2O3. (c) A ratio of one-half calcium ion to one chloride ion means that the number of each ion needs to
be doubled to give whole-number coefficients for both ions. This gives a ratio of 1 calcium ion to 2
chloride ions. The chemical formula for calcium chloride is CaCl2.
4.10
Chemical formulas must have whole number subscripts. (a) PH3; (b) A ratio of 2.5 oxygen atoms to 1 atom
means that the number of each atom needs to be doubled to give whole-number coefficients for both atoms.
This works because it converts 2.5 to 5 (whole number). This gives a ratio of 5 oxygen atoms to 2 nitrogen
atoms. The chemical formula is N2O5. (c) A ratio of 1/3 aluminum ion to one chlorate ion implies that the
number of each ion must be tripled to give whole-number coefficients for both atoms. This gives a ration
of one aluminum ion to three chlorate ions. The chemical formula is Al(ClO3)3.
4.11
The formulas are derived by counting the atoms of each element in the figures. (a) H2SO4; (b) SCl4; (c)
C2H4
4.12
The formulas are derived by counting the atoms of each element in the figures. (a) H2SO3; (b) SCl6; (c)
C2H6
4.13
Each molecule pictured contains 1 carbon atom and 2 oxygen atoms. The formula is CO2.
4.14
Each molecule pictured shows 2 nitrogen and four hydrogen atoms. The formula is N 2H4.
4.15
(a) The formula unit for an ionic compound is described by its formula, which shows the ratio of ions in
lowest possible whole numbers. Since the formula of sodium chloride is NaCl, one formula unit is
NaCl. From the image, you can see a one-to-one correspondence of sodium ions and chloride ions in
the nearest atoms (you should be able to count 14 chloride and 14 sodium ions). This also gives the
formula NaCl (1:1 ratio of Na+ and Cl).
(b) For molecular compounds consisting of discrete molecules, the formula unit is the same as its
molecular formula. In this case there are two Cl atoms in each molecule, so the formula unit is Cl2.
(c) For methane, another molecular compound, each molecule contains one carbon atom and four
hydrogens so the molecular formula and the formula unit is CH4.
(d) The image of silicon dioxide is more difficult to analyze because it is not composed of discrete
molecules. As with ionic compounds, the formula unit is a ratio with lowest possible whole numbers.
On average each silicon atom shares four oxygen atoms with its neighbors, so the formula unit is one
silicon and two oxygens (½ × 4 oxygens); SiO2. However, as with ionic compounds and molecules the
formula unit can be derived from its name, silicon dioxide.
4.16
(a) The formula unit for an ionic compound is described by its formula, which shows the ratio of ions in
lowest possible whole numbers. Since the formula of cesium chloride is CsCl, one formula unit is
CsCl. From the image, you can see a one-to-one correspondence of cesium ions and chloride ions.
This also gives the formula CsCl (1:1 ratio of Cs+ and Cl).
(b) For molecular compounds consisting of discrete molecules, the formula unit is the same as its
molecular formula. In this case there are two oxygen atoms in each molecule, so the formula unit is
O2.
(c) For sulfur dioxide, there is one sulfur and two oxygens in each molecule, so the formula is SO2.
(d) For sodium, a metal, the formula unit is simply Na.
42
4.17
One mole of any object contains 6.022  1023 of those objects. One mole of NH3 contains 6.022  1023
NH3 molecules. One-half mole of NH3 contains half of that:
Molecules NH3 = 0.5 mol NH3 
6.022×1023 NH3 molecules
 3.0×1023 NH3 molecules
1 mole NH3
Since there is one nitrogen atom for each NH3, we can expect that there are 3.0  1023 N atoms as well. We
can express that as follows:
1 N atom
N Atoms = 3.0  1023 NH3 
= 3.0  1023 N atoms
1 NH3
In a similar fashion we can see that there are three hydrogen atoms in each NH 3.
3 H atoms
H atoms = 3.0  1023 NH3 
= 9.0  1023 H atoms
1 NH3
4.18
One mole of any object contains 6.022  1023 of those objects. One mole of SO2 molecules contains
6.022  1023 SO2. If we have 1.75 moles then we have 1.75 times that number:
Molecules SO2 = 1.75 mol SO2 
6.022×1023 SO2 molecules
 1.05×1024 SO2 molecules
1 mole SO2
Since there is one sulfur for each sulfur dioxide, we can expect that there are 1.05  1024 S atoms as well.
We can express that as follows:
S Atoms = 1.051024 SO2 
1 S atom
 = 1.05  1024 S atoms
1 SO2
The molecular formula also shows us that there are two oxygen atoms in each SO2 molecule, so there are
twice as many O atoms than SO2 molecules:
O Atoms = 1.05381024 SO2 
4.19
2 O atom
 = 2.11  1024 O atoms
1 SO2
One mole of Cu2S contains 6.022  1023 Cu2S formula units. One-half mole of Cu2S contains half that
value:
Formula units of Cu 2S = 0.50 mol Cu 2S 
4.20
6.022×1023 Cu 2S formula units
 3.0×1023 Cu 2S formula units
1 mole Cu 2S
One mole of CuSO4 contains 6.022  1023 CuSO4 formula units. 1.5 moles of CuSO4 contains 1.5 times
that many formula units:
Formula units of CuSO4 = 1.50 mol CuSO4 
4.21
6.022×1023 CuSO4 formula units
 9.0×1023 CuSO4 formula units
1 mole CuSO4
To calculate the number of S atoms in 0.2 mol of SO2, calculate the number of molecules in 0.2 mol and
then multiply by the number of S atoms in each SO2 molecule (1 S per molecule) as mapped below:
NA
formula ratio
Map: Moles 
 Number of molecules 
 Number of atoms
Atoms S = 0.2 mol 
6.022  1023 molecule
1 S atom

= 1  1023 S atoms
mol
1 molecule
43
4.22
To calculate the number of O atoms in 0.2 mol of SO2, calculate the number of molecules in 0.2 mol and
then multiply by the number of O atoms in each SO2 molecule (2 O atoms per molecule) as mapped below:
NA
formula ratio
Map: Moles 
 Number of molecules 
 Number of atoms
Atoms O = 0.2 mol 
4.23
6.022  1023 molecule
2 O atom
= 2  1023 O atoms

mol
1 molecule
To calculate the number of calcium ions in 1 mol of CaCl 2, calculate the number of formula units in 1 mol
and then multiply by the number of calcium ions per formula unit (1 Ca2+ per CaCl2) as mapped below:
NA
formula ratio
Map: Moles 
 Number of formula units 
 Number of ions
Ca2+ ions = 1 mol 
4.24
6.022  1023 formula unit
mol

1 Ca 2 
= 6  1023 Ca2+ ions
1 formula unit
To calculate the number of chloride ions in 2 mol of CaCl2, calculate the number of formula units in 2 mol
and then multiply by the number of chloride ions per formula unit (2 Cl  per CaCl2) as mapped below:
NA
formula ratio
Map: Moles 
 Number of formula units 
 Number of ions
Cl ions = 2 mol 
4.25
6.022  1023 formula unit
mol

2 Cl
= 2  1024 Cl ions
1 formula unit
The molar mass is the sum of the masses of the component elements in the chemical formula. The masses
for one mole of each element are found on the periodic table.
NaCl
Mass of 1 mol of Na
= 1 mol  22.99 g/mol =
Mass of 1 mol of Cl
= 1 mol  35.45 g/mol =
Mass of 1 mol of NaCl =
The molar mass is 58.44 g/mol
22.99 g
35.45 g
58.44 g
Cl2
Mass of 2 mol of Cl = 2 mol  35.45 g/mol = 70.90 g
The molar mass is 70.90 g/mol
CH4
Mass of 1 mol of C = 1 mol  12.01 g/mol =
Mass of 4 mol of H = 4 mol  1.008 g/mol =
Mass of 1 mol of CH4
=
The molar mass is 16.04 g/mol
12.01 g
4.03 g
16.04 g
SiO2
Mass of 1 mol of Si= 1 mol  28.09 g/mol = 28.09 g
Mass of 2 mol of O= 2 mol  16.00 g/mol = 32.00 g
Mass of 1 mol of SiO2
= 60.09 g
The molar mass is 60.09 g/mol
44
4.26
The molar mass is the sum of the masses of the component elements in the chemical formula. The masses
for one mole of each element are found on the periodic table.
CsCl
Mass of 1 mol of Cs
= 1 mol  132.9 g/mol = 132.9 g
Mass of 1 mol of Cl
= 1 mol  35.45 g/mol = 35.45 g
Mass of 1 mol of CsCl =
168.4 g
The molar mass is 168.4 g/mol
O2
Mass of 2 mol of O= 2 mol  16.00 g/mol = 32.00 g
The molar mass is 32.00 g/mol
SO2
Mass of 1 mol of S
= 1 mol  32.07 g/mol = 32.07 g
Mass of 2 mol of O
= 2 mol  32.00 g/mol = 32.00 g
Mass of 1 mol of SO2
= 64.07 g
The molar mass is 64.07 g/mol
Na
Mass of 1 mol of Na =
The molar mass is 22.99 g/mol
4.27
1 mol  22.99 g/mol
=
22.99 g
The molar mass is the sum of the masses of the component elements in the chemical formula. The masses
for one mole of each element are found on the periodic table.
(a) Hg2Cl2
Mass of 2 mol of Hg
= 2 mol  200.6 g/mol = 401.2 g
Mass of 2 mol of Cl
= 2 mol  35.45 g/mol = 70.90 g
Mass of 1 mol of Hg2Cl2
472.1 g
Molar mass = 472.1 g/mol
(b) CaSO4•2H2O Note that the mass of 1 water molecule was calculated separately as 18.016 g/mol.
Mass of 1 mol of Ca = 1 mol  40.08 g/mol = 40.08 g
Mass of 1 mol of S
= 1 mol  32.06 g/mol = 32.06 g
Mass of 4 mol of O
= 4 mol  16.00 g/mol = 64.00 g
Mass of 2 mol of H2O = 2 mol  18.016 g/mol = 36.032 g
Mass of 1 mol of CaSO4•2H2O
= 172.17 g
Molar mass = 172.17 g/mol
(c) Cl2O5
Mass of 2 mol of Cl
= 2 mol  35.45 g/mol = 70.90 g
Mass of 5 mol of O
= 5 mol  16.00 g/mol = 80.00 g
Mass of 1 mol of Cl2O5
150.90 g
Molar mass = 150.90 g/mol
(d) NaHSO4
Mass of 1 mol of Na = 1 mol  22.99 g/mol
Mass of 1 mol of H
= 1 mol  1.008 g/mol
Mass of 1 mol of S
= 1 mol  32.06 g/mol
Mass of 4 mol of O
= 4 mol  16.00 g/mol
Mass of 1 mol of NaHSO4
Molar mass = 120.06 g/mol
45
= 22.99 g
=
1.008 g
= 32.06 g
= 64.00 g
= 120.06 g
4.28
The molar mass is the sum of the masses of the component elements in the chemical formula. The masses
for one mole of each element are found on the periodic table.
(a) K2SO4
Mass of 2 mol of K
= 2 mol  39.10 g/mol = 78.20 g
Mass of 1 mol of S
= 1 mol  32.06 g/mol = 32.06 g
Mass of 4 mol of O
= 4 mol  16.00 g/mol = 64.00 g
Mass of 1 mol of K2SO4
174.26 g
Molar mass = 174.26 g/mol
(b) NiCl2•6H2O; Note that the mass of 1 water molecule was calculated separately as 18.016 g/mol (four
significant figures).
Mass of 1 mol of Ni = 1 mol  58.69 g/mol = 58.69 g
Mass of 2 mol of Cl
= 2 mol  35.45 g/mol = 70.90 g
Mass of 6 mol of H2O = 6 mol  18.016 g/mol = 108.108 g
Mass of 1 mol of NiCl2•6H2O
= 237.69 g
Molar mass = 237.69 g/mol
(c) C2H4Cl2
Mass of 2 mol of C
= 2 mol  12.01 g/mol = 24.02 g
Mass of 4 mol of H
= 4 mol  1.008 g/mol = 4.032 g
Mass of 2 mol of Cl
= 2 mol  35.45 g/mol = 70.90 g
Mass of 1 mol of C2H4Cl2
98.95 g
Molar mass = 98.95 g/mol
(d) Mg(NO3)2
Mass of 1 mol of Mg = 1 mol  24.31 g/mol
Mass of 2 mol of N
= 1 mol  14.01 g/mol
Mass of 6 mol of O
= 6 mol  16.00 g/mol
Mass of 1 mol of Mg(NO3)2
Molar mass = 148.33 g/mol
4.29
The molar mass is the sum of the masses of the component elements in the chemical formula. The masses
for one mole of each element are found on the periodic table.
(a) I2
Mass of 2 mol of I
= 2 mol 126.9 g/mol = 253.8 g
Molar mass = 253.8 g/mol
(b) CrCl3
Mass of 1 mol of Cr = 1 mol  52.00 g/mol
Mass of 3 mol of Cl
= 3 mol  35.45 g/mol
Mass of 1 mol of CrCl3
Molar mass = 158.4 g/mol
(c) C4H8
Mass of 4 mol of C
= 4 mol  12.01 g/mol
Mass of 8 mol of H
= 8 mol  1.008 g/mol
Mass of 1 mol of C4H8
Molar mass = 56.10 g/mol
4.30
= 24.31 g
= 28.02 g
= 96.00 g
= 148.33 g
= 52.00 g
= 106.4 g
= 158.4 g
=
=
48.04 g
8.064 g
56.10 g
The molar mass is the sum of the masses of the component elements in the chemical formula. The masses
for one mole of each element are found on the periodic table.
46
(a) P4
Mass of 4 mol of P
= 4 mol 30.97 g/mol = 123.9 g
Molar mass = 123.9 g/mol
(b) CrO2Cl2
Mass of 1 mol of Cr = 1 mol  52.00 g/mol
Mass of 2 mol of O
= 2 mol  16.00 g/mol
Mass of 2 mol of Cl
= 2 mol  35.45 g/mol
Mass of 1 mol of CrO2Cl2
Molar mass = 154.90 g/mol
(c) CaF2
Mass of 1 mol of Ca = 1 mol  40.08 g/mol
Mass of 2 mol of F
= 2 mol  19.00 g/mol
Mass of 1 mol of CaF2
Molar mass = 78.08 g/mol
= 52.00 g
= 32.00 g
= 70.90 g
= 154.90 g
=
=
40.08 g
38.00 g
78.08 g
4.31
To measure out a useful number of atoms by counting would not be possible because atoms are too small
for us to see and manipulate. If you could count 1000 water molecules per second for slightly over 20
billion years, you would have counted the molecules in one small drop of water (about 0.020 mL). Instead,
because we know the mass of one mole of any substance, we can carefully weigh out an amount of any
substance and determine the number of atoms our sample contains.
4.32
One mole is 6.022  1023 of any object. Scientists use the mole because atoms, ions, and molecules are too
small to work with individually. Large collections of these units are more reasonable to deal with.
4.33
The molar mass and atomic mass have the same numerical value but different units. If the molar mass of
LiCl is 42.39 g/mol, the mass of one formula unit of LiCl is 42.39 amu.
4.34
The molar mass and atomic mass have the same numerical value but different units. If the mass of one
formula unit of CuCl2 is 134.5 amu, then the mass of 1.00 mol of CuCl2 is 134.5 g (i.e. the molar mass is
134.5 g/mol).
4.35
The units for molar mass are grams per mole:
g
Molar mass =
mol
This means that if you have the numbers of grams and moles, you can calculate the molar mass by dividing
the number of grams by the number of moles of substance this mass represents. In this case, you are given
grams (12.0) and molecules of substance.
NA
Map: Number of molecules 
 Moles
Molecules = 2.01  1023 molecules 
Molar mass =
4.36
mole
6.022  1023 molecules
= 0.334 mol
12.0 g
= 35.9 g/mol
0.334 mol
In this problem you are asked to convert from molecules to mass. You can make this conversion by first
converting from molecules to moles; then use the molar mass to calculate mass of the sample.
NA
MM  Mass in grams
 Moles 
Map: Number of molecules 
47
Mass = 3.01  1023 molecules 
4.37
mole
6.022  10 molecules
23

98.09 g
= 49.0 g
mole
The conversion from mass to number of moles uses the following problem solving map (MM is the molar
mass):
MM  Moles
Map: Mass in grams 
(a) The molar mass of KHCO3 is 100.12 g/mol (calculated by adding the masses of the component
elements of the chemical formula).
Moles KHCO3 = 10.0 g KHCO3 
1 mol KHCO3
100.12 g KHCO3
= 0.0999 mol KHCO3
(b) The molar mass of H2S is 34.08 g/mol.
Moles H2S = 10.0 g H2S 
1 mol H2S
= 0.293 mol H2S
34.08 g H2S
(c) The molar mass of Se is 78.96.
Moles Se = 10.0 g Se 
1 mol Se
= 0.127 mol Se
78.96 g Se
(d) The molar mass of MgSO4 is 120.37 g/mol.
Moles MgSO4 = 10.0 g MgSO4 
4.38
1 mol MgSO4
= 0.0831 mol MgSO4
120.37 g MgSO4
The conversion from mass to number of moles uses the following problem solving map (MM is the molar
mass):
MM  Moles
Map: Mass in grams 
(a) The molar mass of SO2 is 64.06 g/mol.
Moles SO2 = 100.0 g SO2 
1 mol SO2
= 1.561 mol SO2
64.06 g SO2
(b) The molar mass of Na2SO4 is 142.05 g/mol.
Moles Na2SO4 = 100.0 g Na 2SO4 
1 mol Na 2SO4
= 0.7040 mol Na2SO4
142.05 g Na 2SO4
(c) The molar mass of BaSO4 is 233.4 g/mol.
Moles BaSO4 = 100.0 g BaSO4 
1 mol BaSO4
= 0.4284 mol BaSO4
233.4 g BaSO4
(d) The molar mass of KAl(SO4)2 is 258.2 g/mol.
48
Moles KAl(SO4)2 = 100.0 g KAl(SO4 ) 2 
4.39
1 mol KAl(SO4 )2
= 0.3873 mol KAl(SO4)2
258.2 g KAl(SO4 ) 2
The conversion from mass to number of moles uses the following problem solving map (MM is the molar
mass). In these problems, if the mass is not given in grams, it must first be converted using an appropriate
problem solving map:
MM  Moles
Map: Mass in grams 
(a) The molar mass of NaCl is 58.44 g/mol.
Moles NaCl = 32.5 g NaCl 
1 mol NaCl
= 0.556 mol NaCl
58.44 g NaCl
(b) The molar mass of C9H8O4 is 180.15 g/mol. The mass is converted to grams using 1 mg = 103 g.
Moles C9H8O4 = 250.0 mg C9 H8 O4 
103 g C9 H8 O4
1 mg C9 H8 O4

1 mol C9 H8 O4
= 1.388  103 mol C9H8O4
180.15 g C9 H8 O4
(c) The molar mass of CaCO3 is 100.09 g/mol. The mass is converted to grams using 1 kg = 103 g.
Moles CaCO3 = 73.4 kg CaCO3 
103 g CaCO3
1 kg CaCO3

1 mol CaCO3
= 733 mol CaCO3
100.09 g CaCO3
(d) The molar mass of CuS is 95.61 g/mol. The mass is converted to grams using 1 g = 106 g.
Moles CuS = 5.47 μg CuS 
4.40
106 g CuS
1 μg CuS

1 mol CuS
= 5.72  10–8 mol CuS
95.61 g CuS
The conversion from mass to number of moles uses the following problem solving map (MM is the molar
mass). In these problems, if the mass is not given in grams, it must first be converted using an appropriate
problem solving map:
MM  Moles
Map: Mass in grams 
(a) The molar mass of K2SO4 is 174.26 g/mol.
Moles K2SO4 = 72.2 g K 2SO4 
1 mol K 2SO4
174.26 g K 2SO4
= 0.414 mol K2SO4
(b) The molar mass of C18H21NO4 is 315.36 g/mol. The mass is converted to grams using 1 mg = 10 3 g.
Moles C18H21NO4 = 160.0 mg C18 H21 NO4 
103 g C18 H 21 NO4
1 mg C18 H 21 NO4

1 mol C18 H 21 NO4
315.36 g C18 H21 NO4
4
= 5.074  10 mol C18H21NO4
(c) The molar mass of Fe3O4 is 231.55 g/mol. The mass is converted to grams using 1 kg = 10 3 g.
49
Moles Fe3O4 = 2.82 kg Fe3O4 
103 g Fe3O4
1 kg Fe3O4

1 mol Fe3O4
= 12.2 mol Fe3O4
231.55 g Fe3O4
(d) The molar mass of C12H22O11 is 342.30 g/mol. The mass is converted to grams using 1 g = 106 g.
Moles C12H22O11 = 5.00 μg C12 H 22 O11 
106 g C12 H 22 O11
1 μg C12 H 22 O11

1 mol C12 H22 O11
342.30 g C12 H22 O11
–8
= 1.46  10 mol C12H22O11
4.41
If the molar mass of a substance is relatively small, it will take more moles of that substance to equal 1
gram than it would take of a substance with a larger molar mass. This means that carbon, which has the
smallest molar mass of the substances given, contains the most moles of atoms in a 1.0-g sample.
4.42
If the molar mass of a substance is relatively large, it will take fewer moles of that substance to equal 5.0 g
than it would take of a substance with a smaller molar mass. This means that silver, which has the highest
molar mass of the substances given, contains the least moles of atoms in a 5.0-g sample.
4.43
To convert moles to grams we use the following problem solving map (MM = molar mass)
MM  Mass in grams
Map: Moles 
(a) The molar mass of Ba(OH)2 is 171.32 g/mol.
171.32 g BaSO4
Mass Ba(OH)2 = 2.50 mol BaSO4 
1 mol BaSO4
(b) The molar mass of Cl2 is 70.90 g/mol.
70.90 g Cl2
Mass Cl2 = 2.50 mol Cl2 
1 mol Cl2
= 428 g Ba(OH)2
= 177 g Cl2
(c) The molar mass of K2SO4 is 174.26 g/mol.
174.26 g K 2SO4
Mass K2SO4 = 2.50 mol K 2SO4 
1 mol K 2SO4
= 436 g K2SO4
(c) The molar mass of PF3 is 87.97 g/mol.
87.97 g PF3
Mass K2SO4 = 2.50 mol PF3 
= 220. g K2SO4
1 mol PF3
4.44
To convert moles to grams we use the following problem solving map (MM = molar mass)
MM  Mass in grams
Map: Moles 
(a) The molar mass of I2 is 253.80 g/mol.
253.80 g I2
Mass I2 = 0.750 mol I2 
= 190. g I2
1 mol I2
(b) The molar mass of Mg(NO3)2 is 148.33 g/mol.
148.33 g Mg  NO3 2
Mass Mg(NO3)2 = 0.750 mol Mg  NO3 2 
1 mol Mg  NO3 2
(c) The molar mass of SiO2 is 60.09 g/mol.
60.09 g SiO2
Mass SiO2 = 0.750 mol SiO2 
= 45.1 g SiO2
1 mol SiO2
410
= 111 g Mg(NO3)2
(c) The molar mass of Na3PO4 is 163.94 g/mol.
163.94 g Na 3 PO4
Mass Na3PO4= 0.750 mol Na3 PO4 
1 mol Na3 PO4
4.45
= 123 g Na3PO4
The conversion of 2.7 moles of Zn(CH3CO2)2 to an equivalent number of grams requires the molar mass
(MM). We use the following problem solving map:
MM  Mass in grams
Map: Moles 
The molar mass of Zn(CH3CO2)2 is 183.47 g.
Grams Zn(CH3CO2)2 = 2.7 mol Zn(CH3 CO2 )2 
4.46
183.47 g Zn(CH3 CO2 )2
mol Zn(CH3 CO2 )2
= 5.0  102 g Zn(CH3CO2)2
The conversion of 3.4 moles of Cu(HCO3)2 to an equivalent number of grams requires the molar mass
(MM). We use the following problem solving map:
MM  Mass in grams
Map: Moles 
The molar mass of Cu(HCO3)2 is 185.59 g.
Grams Cu(HCO3)2 = 3.4 mol Cu(HCO3 ) 2 
4.47
185.59 g Cu(HCO3 ) 2
mol Cu(HCO3 )2
= 6.3  102 g Cu(HCO3)2
(a) The conversion of 30.0 g NH3 to an equivalent number of moles requires the molar mass (MM). We
use the following problem solving map:
MM  Moles
Map: Mass in grams 
The molar mass of NH3 is 17.03 g/mol.
Moles NH3 = 30.0 g NH3 
mol NH3
17.03 g NH3
= 1.76 mol NH3
(b) To calculate the number of NH3 molecules in the sample, we use Avogadro’s number (NA = 6.022 
1023) and the following problem solving map:
NA
Map: Moles 
 Number of molecules
Molecules NH3 = 1.7612 mol NH3 
6.022  1023 molecules
= 1.06  1024 molecules NH3
mol NH3
(c) To calculate the number of nitrogen atoms in the sample, we look at the ratio of the elements in the
chemical formula. The ratio for nitrogen in ammonia is 1 atom N/1 molecule NH 3. Therefore, the
number of N atoms is the same as the number of NH3 molecules, 1.06  1024 N atoms.
(d) From the chemical formula, we know that there are 3 moles of H for each mole of NH 3. This is the
formula ratio needed to calculate the moles of H in the sample of NH 3:
formula ratio
Map: Moles NH3 
 Moles H
411
Moles H = 1.7612 mol NH3 
4.48
3 mol H
= 5.28 mol H
1 mol NH3
(a) The conversion of 15.0 g Na2CO3 to an equivalent number of moles requires the molar mass (MM). We
use the following problem solving map:
MM  Moles
Map: Mass in grams 
The molar mass of Na2CO3 is 105.99 g/mol.
Moles Na2CO3 = 15.0 g Na 2 CO3 
mol Na 2 CO3
105.99 g Na 2 CO3
= 0.142 mol Na2CO3
(b) To calculate the number of formula units in the sample, we use Avogadro’s number (NA = 6.022 
1023) and the following problem solving map:
NA
Map: Moles 
 Number of formula units
Formula units Na2CO3 = 0.14152 mol Na 2 CO3 
6.022  1023 formula units
mol Na 2 CO3
= 8.52  1022 formula units Na2CO3
(c) To calculate the number of sodium ions in the sample, we look at the ratio of the elements in the
chemical formula. The ratio for sodium in Na2CO3 is 2 Na+ ions/1 formula unit.
formula ratio
Map: Number of formula units 
 Number of Na ions
Na+ ions = 8.5223  1022 formula units Na 2 CO3 
2 Na  ions
= 1.70  1023 Na+ ions
formula units Na 2 CO3
(d) To calculate the number of moles carbonate ions in the sample, we look at the ratio of the carbonate
ions to formula units of Na2CO3. This ratio is 1 mol CO32 per mol of Na2CO3. Therefore, the number
of carbonate ions is the same as the number of moles of Na2CO3: 0.142 mol CO32.
4.49
The number of atoms per mole of substance depends on the number of atoms in the chemical formula.
H2SO4 has the most atoms per mole because it has more atoms per molecule. Na has the least atoms per
mole.
4.50
The number of atoms per mole of substance depends on the number of atoms in the chemical formula.
C2H6 has the most atoms per mole of substance, and Fe has the least atoms per mole of substance.
4.51
To calculate the number of molecules in 0.050 g of water, we need to find the appropriate conversion
factors. This can be done using the following conversion map:
NA
MM  Moles 
Map: Mass in grams 
 Number of molecules
Note that the units of molar mass (g/mol) serve as a connection between mass and moles, and the units of
Avogadro’s number (molecules/mole) serve as a connection between molecules and moles.
The molar mass of water is 18.02 g/mol.
412
= 0.050 g H2 O 
Molecules H2O
mol H2 O
18.02 g H2 O

6.022  1023 molecules H 2 O
mol H 2 O
= 1.7  1021 molecules H2O
4.52
To calculate the number of SiO2 formula units in 1 grain of sand (7.7  104 g) we need to find the
appropriate conversion factors. Assuming that sand is 100% SiO 2, we use the following conversion map:
NA
MM  Moles 
Map: Mass in grams 
 Number of formula units
Note that the units of molar mass (g/mol) serve as a connection between mass and moles, and the units of
Avogadro’s number (formula units/mole) serve as a connection between formula units and moles.
The molar mass of SiO2 is 60.09 g/mol.
mol SiO2
Formula units SiO2 = 7.7  104 g SiO2 
60.09 g SiO2

6.022  1023 formula units SiO2
mol SiO2
= 7.7  1018 formula units SiO2
4.53
To calculate the number of formula units we use the following conversion map:
NA
MM  Moles 
Map: Mass in grams 
 Number of formula units
(a) Br2 = 159.80 g/mol
mol Br2
6.022  1023 formula units Br2

Formula units Br2 = 250.0 g Br2 
159.80 g Br2
mol Br2
= 9.421  1023 formula units Br2
(b) MgCl2 = 95.21 g/mol
Formula units MgCl2 = 250.0 g MgCl2 
mol MgCl2
159.80 g MgCl2

6.022  1023 formula units MgCl2
mol MgCl2
= 1.581  10 formula units MgCl2
24
(c) H2O = 18.02 g/mol
Formula units H2O = 250.0 g H2 O 
mol H2 O
18.02 g H2 O

6.022  1023 formula units H 2 O
mol H 2 O
= 8.355  10 formula units H2O
24
(d) Fe = 55.85 g/mol
Formula units Fe = 250.0 g Fe 
mol Fe
58.85 g Fe

6.022  1023 formula units Fe
mol Fe
= 2.696  10 formula units Fe
24
4.54
To calculate the number of formula units we use the following conversion map:
NA
MM  Moles 
Map: Mass in grams 
 Number of formula units
(a) Cu = 63.55 g/mol
413
Formula units Cu = 375.0 g Cu 
mol Cu

63.55 g Cu
6.022  1023 formula units Cu
mol Cu
= 3.554  10 formula units Cu
24
(b) NaBr = 102.89 g/mol
Formula units NaBr = 375.0 g NaBr 
mol NaBr
102.89 g NaBr

6.022  1023 formula units NaBr
mol NaBr
= 2.195  10 formula units NaBr
24
(c) SO2 = 64.06 g/mol
Formula units SO2 = 375.0 g SO2 
mol SO2
64.06 g SO2

6.022  1023 formula units SO2
mol SO2
= 3.525  10 formula units SO2
24
(d) NH4Cl = 53.49 g/mol
Formula units NH4Cl = 375.0 g NH4 Cl 
mol NH4 Cl

53.49 g NH4 Cl
6.022  1023 formula units NH 4 Cl
mol NH 4 Cl
= 4.222  10 formula units NH4Cl
24
4.55
To calculate the number of atoms or ions of each element in 140.0 g of each substance, the best strategy is
to first calculate the number of formula units. Once you have determined this, use the formula ratios to
calculate the number of atoms or ions. In part (b), for example, you will calculate the formula units of
Ca(NO3)2 and then use formula ratios to calculate the number of Ca2+ and NO3– ions. Two problem solving
maps are applied:
Number of formula units:
NA
MM  Moles 
Mass in grams 
 Number of formula units
Number of ions or atoms:
formula ratio
Number of formula units 
 ions or atoms
(a) Since there we are only looking for atoms of one element, we can combine the two problem solving
maps. The appropriate conversions are: H2 = 2.016 g/mol; 2 H atom = 1 H2 molecule
Atoms = 140.0 g H2 
mol H2
2.016 g H2
6.022  1023 H 2

mol H 2

2 H atom
= 8.364  1025 H atoms
H2
(b) Ca(NO3)2 = 164.10 g/mol
Formula units Ca(NO3)2 =
140.0 g Ca  NO3 2 
mol Ca  NO3 2
164.10 Ca  NO3 2

6.022  1023 Ca  NO3 2
mol Ca  NO3 2
= 5.138  1023 Ca(NO3)2
Ca2+ ions: 1 Ca2+ ion = 1 Ca(NO3)2
Ca2+ ions = 5.138  1023 Ca  NO3 2 
1 Ca 2 
1Ca  NO3 2
NO3 ions: 2 NO3 ions = 1 Ca(NO3)2
414
= 5.138  1023 Ca2+ ions
NO3 ions = 5.138  1023 Ca  NO3 2 
2 NO3
1Ca  NO3 2
= 1.028 1024 NO3 ions
(c) N2O2 = 60.02 g/mol
Formula units N2O2 = 140.0 g N 2 O2 
mol N 2 O2
60.02 g N 2 O2

6.022  1023 N 2 O2
= 1.405  1024 N2O2
mol N 2 O2
N atoms = 1.405  1024 N 2 O2 
2N
= 2.809  1024 N atoms
1 N 2 O2
O atoms = 1.405  1024 N 2 O2 
2O
= 2.809  1024 O atoms
1 N 2 O2
(d) K2SO4 = 174.26 g/mol
Formula units = 140.0 g K 2SO4 
mol K 2SO4
174.26 g K 2SO4

6.022  1023 K 2SO4
= 4.838  1023 K2SO4
mol K 2SO4
K+ ions: 2 K+ ions = 1 K2SO4
K+ ions = 4.838  1023 K 2SO4 
2 K
= 9.676  1023 K+ ions
1K 2SO4
SO42 ions: 1 SO42 ion = 1 K2SO4
SO42 ions = 4.838  1023 K 2SO4 
4.56
1 SO42 
= 4.838  1023 SO42 ions
1K 2SO4
To calculate the number of atoms or ions of each element in 140.0 g of each substance, the best strategy is
to first calculate the number of formula units. Once you have determined this, use the formula ratios to
calculate the number of atoms or ions. In part (a), for example, you will calculate the formula units of
BaSO4 and then use formula ratios to calculate the number of Ba2+ and SO42– ions. Two problem solving
maps are applied:
Number of formula units:
NA
MM  Moles 
Mass in grams 
 Number of formula units
Number of ions or atoms:
formula ratio
Number of formula units 
 ions or atoms
(a) BaSO4 = 233.36g/mol
Formula units = 140.0 g BaSO4 
mol BaSO4
233.36 g BaSO4
Ba2+ ions = 3.613  1023 BaSO4 
SO42– ions = 3.613  1023 BaSO4 

6.022  1023 BaSO4
= 3.613  1023 BaSO4
mol BaSO4
1 Ba 2 
= 3.613  1023 Ba2+ ions
1BaSO4
1 SO42 
= 3.613  1023 SO42– ions
1BaSO4
Total ions = (3.613  1023 Ba2+ ions) + (3.613  1023 SO42– ions) = 7.225  1023 ions
You can also calulate the number of sulfur and oxygen atoms:
415
S atoms = 3.613  1023 BaSO4 
1S
= 3.613  1023 S atoms
1BaSO4
O atoms = 3.613  1023 BaSO4 
4O
= 1.445  1024 O atoms
1BaSO4
(b) Mg3(PO4)2 = 262.87 g/mol;
Formula units
= 140.0 g Mg3  PO4 2 
1mol Mg3  PO4 2
262.87 g Mg3  PO4 2

6.022  1023 Mg3  PO4 2
mol Mg3  PO4 2
= 3.207  1023 Mg3(PO4)2
Mg2+ ions = 3.207  1023 Mg3  PO4 2 
PO43– ions = 3.207  1023 Mg3 (PO4 ) 2 
3 Mg 2 
= 9.622  1023 Mg2+ ions
1Mg3  PO4 2
2 PO43 
= 6.414  1023 PO43– ions
1Mg3 (PO4 ) 2
Total ions = (9.622  1023 Mg2+ ions) + (6.414  1023 PO43– ions) = 1.604  1024 ions
You can also calulate the number of phosphorus and oxygen atoms:
2P
P atoms = 3.207  1023 Mg3  PO4 2 
= 6.414  1023 P atoms
1Mg3  PO4 2
O atoms = 3.207  1023 Mg3  PO4 2 
8O
1Mg3  PO4 2
= 2.566  1024 O atoms
(c) Since we are only looking for one type of atom, we can combine the two problem solving maps. The
appropriate conversions are: O2 = 32.00 g/mol; 2 O atoms = 1 O2 formula unit
O atoms = 140.0 g O2 
mol O2
32.00 g O2

6.022  1023 O2
mol O2

2 O atom
= 5.269  1024 O atoms
O2
(d) KBr = 119.00 g/mol;
Formula units = 140.0 g KBr 
mol KBr
119.00 g KBr
K+ ions = 7.085  1023 KBr 

6.022  1023 KBr
= 7.085  1023 KBr
mol KBr
1 K
= 7.085  1023 K+ ions
1KBr
1 Br 
= 7.085  1023 Br ions
1KBr
Total ions = (7.085  1023 K+ ions + (7.085  1023 Br ions) = 1.417  1024 ions
Br ions = 7.085  1023 KBr 
4.57
To calculate the mass of 6.4  1022 molecules of SO2 we use the following conversion map:
NA
MM  Mass in grams
 Moles 
Map: Number of molecules 
The molar mass of SO2 is 64.06 g/mol.
416
Mass SO2 = 6.4  1022 molecules SO2 
4.58
1 mol SO2
6.022  10
23
moleculesSO2

64.06 g SO2
= 6.8 g SO2
1 mol SO2
To calculate the mass of 1.8  1021 molecules of H2SO4 we use the following conversion map:
NA
MM  Mass in grams
Map: Number of molecules 
 Moles 
The molar mass of H2SO4 is 98.08 g/mol.
Mass H2SO4 = 1.8  1021 molecules H 2SO4 
1 mol H 2SO4
6.022  10
23
molecules H 2SO4

98.08 g H 2SO4
1 mol H 2SO4
= 0.29 g H2SO4
4.59
The substance that has the most nitrogen atoms per 25.0-g sample will also have the highest number of
moles of nitrogen atoms. We use the following conversion map:
formula ratio
MM  Moles 
Map: Mass in grams 

 Moles of N atoms
The molar mass of NH3 is 17.03 g/mol and there is 1 mol N atoms per mol NH3.
Moles N atoms = 25.0 g NH3 
mol NH3
17.03 g NH3

1 mol N atoms
= 1.47 mol N atoms
mol NH3
The molar mass of NH4Cl is 53.49 g/mol and there is 1 mol N atoms per mole NH4Cl.
Moles N atoms = 25.0 g NH 4 Cl 
mol NH 4 Cl
1 mol N atoms
= 0.467 mol N atoms

mol NH 4 Cl
53.49 g NH 4 Cl
The molar mass of NO2 is 46.01 g/mol and there is 1 mol N atoms per mole NO 2.
Moles N atoms = 25.0 g NO2 
mol NO2
1 mol N atoms
= 0.543 mol N atoms

mol NO2
46.01 g NO2
The molar mass of N2O3 is 76.02 g/mol and there are 2 mol N atoms per mole N 2O3.
Moles N atoms = 25.0 g N 2 O3 
mol N 2 O3
76.02 g N 2 O3

2 mol N atoms
= 0.658 mol N atoms
mol N 2 O3
Because NH3 has the highest number of moles of nitrogen atoms, we know that it has the most nitrogen
atoms per 25.0-g sample.
4.60
The substance that has the most chlorine atoms per 100.0-g sample will also have the highest number of
moles of chlorine atoms. We use the following conversion map:
formula ratio
MM  Moles 
Map: Mass in grams 

 Moles of Cl atoms
The molar mass of NaCl is 58.44 g/mol and there is 1 mol Cl atoms per mol NaCl.
417
Moles Cl atoms = 100.0 g NaCl 
mol NaCl
1 mol Cl atoms
= 1.711 mol Cl atoms

mol NaCl
58.44 g NaCl
The molar mass of PCl3 is 137.3 g/mol and there are 3 mol Cl atoms per mole PCl 3.
Moles Cl atoms = 100.0 g PCl3 
mol PCl3
137.3 g PCl3

3 mol Cl atoms
= 2.185 mol Cl atoms
mol PCl3
The molar mass of CaCl2 is 110.98 g/mol and there are 2 mol Cl atoms per mole CaCl2.
Moles Cl atoms = 100.0 g CaCl2 
mol CaCl2
2 mol Cl atoms
= 1.802 mol Cl atoms

mol CaCl2
110.98 g CaCl2
The molar mass of HClO2 is 68.46 g/mol and there is 1 mol Cl atoms per mole HClO2.
Moles Cl atoms = 100.0 g HClO2 
mol HClO2
1 mol Cl atoms
= 1.461 mol Cl atoms

mol HClO2
68.46 g HClO2
Because PCl3 has the highest number of moles of chlorine atoms, we know that it has the most chlorine
atoms per 100.0-g sample.
4.61
No. Molecules of the same substance have the same percent compositions.
4.62
Percent composition can be used to compare the composition of different substances. Copper ore, for
example, is analyzed for percent copper composition, and fertilizers are analyzed for their percent nitrogen
content. Percent composition can also be used to determine the empirical formula of a compound.
4.63
The empirical formula shows the relative amounts of each atom in a compound, expressed as small whole
numbers. The molecular formula shows the exact numbers of each atom present in one molecule of that
compound.
4.64
Chemical formulas are much more compact and give the same information (i.e. the relative composition of
substances). For example, we could report that water is 88.81% oxygen and 11.19% hydrogen by mass or
simply report that its formula is H2O.
4.65
The empirical and molecular formulas are different if the subscripts in the molecular formula are all
divisible by a common factor other than 1. For example, in the formula H 2O2, both subscripts are divisible
by 2, so the empirical formula (HO) is different than the molecular formula. Of the substances listed, those
with different empirical and molecular formulas are:
Molecular
Empirical
H2O2
HO
N2O4
NO2
4.66
The empirical and molecular formulas are the same if the subscripts in the molecular formula are not
divisible by a common factor other than 1. For example, in the formula N2O3, the subscripts have no
common factors other than 1, so the empirical formula is the same as the molecular formula. Of the
substances listed, those with the same empirical and molecular formulas are N2O3 and NaCl.
4.67
The empirical and molecular formulas are the same if the subscripts in the molecular formula are not
divisible by a common factor other than 1. Often, it is desirable to simplify the molecular formula in order
418
to determine if the molecular and empirical formulas are different. For part (d) HO2CC4H8CO2H can be
simplified to C6H10O4.
Molecular Common
Empirical
Formula
Factor
Formula
(a) P4O10
2
P2O5
(b) Cl2O5
none
same as molecular
(c) PbCl4
none
same as molecular
(d) C6H10O4
2
C3H5O2
4.68
The empirical and molecular formulas are the same if the subscripts in the molecular formula are not
divisible by a common factor other than 1.
Molecular Common
Empirical
Formula
Factor
Formula
(a) As4O6
2
As2O3
(b) H2S2
2
HS
(c) CaCl2
none
same as molecular
(d) C3H6
3
CH2
4.69
The empirical and molecular formulas are the same if the subscripts in the molecular formula are not
divisible by a common factor other than 1.
Molecular Common
Empirical
Formula
Factor
Formula
(a) C6H4Cl2
2
C3H2Cl
(b) C6H5Cl
none
same as molecular
(c) N2O5
none
same as molecular
4.70
The empirical and molecular formulas are the same if the subscripts in the molecular formula are not
divisible by a common factor other than 1.
Molecular Common
Empirical
Formula
Factor
Formula
(a) N2O4
2
NO2
(b) H2C2O4
2
HCO2
(c) C2H4O2*
2
CH2O
*It is easier to find the empirical formula of CH3CO2H if the molecular formula is simplified to C2H4O2.
4.71
NO2 and N2O4 have the identical empirical formulas (NO2). The empirical formulas of the other
compounds are different than NO2.
Molecular Common
Empirical
Formula
Factor
Formula
N2O
none
N2O
NO
none
NO
NO2
none
NO2
N2O3
none
N2O3
N2O4
2
NO2
N2O5
none
N2O5
4.72
C2H4 and C3H6 have identical empirical formulas (CH2). The empirical formulas of the other compounds
are different than CH2.
Molecular Common
Empirical
Formula
Factor
Formula
CH4
none
CH4
C2H4
2
CH2
C3H6
3
CH2
C4H12
4
CH3
C6H6
6
CH
419
4.73
The empirical formula shows the whole-number ratio of moles of each element in a compound. To
determine the empirical formula of a compound, we calculate the number of moles of each element in the
sample, and then determine the mole ratios. When presented with percent composition data, it is easiest to
assume that we have exactly 100 grams of the substance.
(a) The percent composition of our sample is 72.36% Fe and 27.64%O. If we have a 100-gram sample, it
will contain 72.36 g Fe and 27.64 g O. We can convert these masses to the equivalent number of
moles as follows:
Moles Fe = 72.36 g Fe 
Moles O = 27.64 g O 
1 mol Fe
= 1.296 mol Fe
55.85 g Fe
1 mol O
= 1.728 mol O
16.00 g O
Next divide the number of moles of each element by the smallest of the molar amounts (i.e. 1.296 mol
Fe).
moles Fe 1.296 mol Fe 1 mol Fe


moles Fe 1.296 mol Fe 1 mol Fe
moles O
1.728 mol O 1.333 mol O


moles Fe 1.296 mol Fe
1 mol Fe
Notice that one of the ratios is not a whole number. Since the subscripts of chemical formulas must be
whole numbers (i.e. you can’t have fractions of atoms), we multiply each ratio by 3 to convert the ratio
into a whole number. We find that the empirical formula has 4 moles of oxygen for every 3 moles of
iron. The empirical formula is Fe3O4.
(b) The percent composition of our sample is 58.53% C, 4.09% H, 11.38% N, and 25.99% O. If we have a
100 gram sample, it will contain 58.53 g C, 4.09 g H, 11.38 g N, and 25.99 g O. We can convert these
masses to the equivalent number of moles as follows:
Moles C = 58.53 g C 
Moles H = 4.09 g H 
1 mol C
= 4.873 mol C
12.01 g C
1 mol H
= 4.06 mol H
1.008 g H
Moles N = 11.38 g N 
1 mol N
= 0.8123 mol N
14.01 g N
Moles O = 25.99 g O 
1 mol O
= 1.624 mol O
16.00 g O
Next divide the number of moles of each element by the smallest of the molar amounts (i.e. 0.8123
mol N).
moles C
4.873 mol C
6.000 mol C


moles N 0.8123 mol N
1 mol N
420
moles H
4.06 mol H
5.00 mol H


moles N 0.8123 mol N
1 mol N
moles N 0.8123 mol N 1 mol N


moles N 0.8123 mol N 1 mol N
moles O
1.624 mol O
2.000 mol O


moles N 0.8123 mol N
1 mol N
The empirical formula has 6 moles of carbon, 5 moles of hydrogen, and 2 moles of oxygen for every
1 mole of nitrogen. The empirical formula is C6H5NO2.
4.74
The empirical formula shows the whole-number ratio of moles of each element in a compound. To
determine empirical formula of a compound, we calculate the number of moles of each element in the
sample and then determine the mole ratios. When presented with percent composition data, it is easiest to
assume that we have exactly 100 grams of the substance.
(a) The percent composition of our sample is 85.62% C and 14.38%H. If we have a 100 gram sample, it
will contain 85.62 g C and 14.38 g H. We can convert these masses to the equivalent number of moles
as follows:
Moles C = 85.62 g C 
1 mol C
= 7.129 mol C
12.01 g C
Moles H = 14.38 g H 
1 mol H
= 14.27 mol H
1.008 g H
Next divide the number of moles of each element by the smallest of the molar amounts (i.e. 7.129 mol
C).
moles C 7.129 mol C 1 mol C


moles C 7.129 mol C 1 mol C
moles H 14.27 mol H 2.001 mol H


moles C 7.129 mol C
1 mol C
Rounding the mole ratios to nearest whole number we find that for each mole of carbon there are 2
moles of hydrogen. The empirical formula is CH2.
(b) The percent composition of our sample is 63.15% C, 5.30% H, and 31.55% O. If we have a 100-gram
sample, it will contain 63.15 g C, 5.30 g H, and 31.55 g O. We convert these masses to the equivalent
number of moles as follows:
Moles C = 63.15 g C 
Moles H = 5.30 g H 
1 mol C
= 5.258 mol C
12.01 g C
1 mol H
= 5.26 mol H
1.008 g H
421
Moles O = 31.55 g O 
1 mol O
= 1.972 mol O
16.00 g O
Next divide the number of moles of each element by the smallest of the molar amounts (i.e. 1.972 mol
O).
moles C 5.258 mol C 2.667 mol C


moles O 1.972 mol O
1 mol O
moles H
5.26 mol H
2.67 mol H


moles O 1.972 mol O
1 mol O
moles O 1.972 mol O 1 mol O


moles O 1.972 mol O 1 mol O
The ratios are not all whole numbers and must be converted to whole numbers by multiplying by an
appropriate factor. In this case, the fractional portion (0.667) represents 2/3, so we should be able to
convert these values to whole numbers by multiplying each ratio by 3. Thus for every 3 moles of
oxygen we have 8 moles of carbon and 8 moles of hydrogen. The empirical formula is C 8H8O3.
4.75
The percent composition of our sample is 73.19% C, 19.49% O, and 7.37% H. If we have a 100-gram
sample, it will contain 73.19 g C, 19.49 g O, and 7.37 g H. To determine the empirical formula, we convert
the masses of each element to the equivalent number of moles, and then determine the relative number of
moles of each element in the substance.
Moles C = 73.19 g C 
1 mol C
= 6.094 mol C
12.01 g C
Moles O = 19.49 g O 
1 mol O
= 1.218 mol O
16.00 g O
Moles H = 7.37 g H 
1 mol H
= 7.31 mol H
1.008 g H
Next divide the number of moles of each element by the smallest of the molar amounts (i.e. 1.218 mol O).
moles C 6.094 mol C 5.003 mol C


moles O 1.218 mol O
1 mol O
moles O 1.218 mol O 1 mol O


moles O 1.218 mol O 1 mol O
moles H
7.31 mol H
6.00 mol H


moles O 1.218 mol O
1 mol O
Rounding the mole ratios to the nearest whole numbers, we find that chemical formula has 5 moles of
carbon, 6 moles of hydrogen, and 1 mole of oxygen. The empirical formula for eugenol is C5H6O.
4.76
The percent composition of our sample is 52.66% Ca, 12.30% Si, and 35.04% O. If we have a 100 gram
sample, then it will contain 52.66 g Ca, 12.30 g Si, and 35.04 g O. To determine the empirical formula, we
422
convert the masses of each element to the equivalent number of moles, and then determine the relative
number of moles of each element in the compound.
Moles Ca = 52.66 g Ca 
1 mol Ca
= 1.314 mol Ca
40.08 g Ca
Moles Si = 12.30 g Si 
1 mol Si
= 0.4379 mol Si
28.09 g Si
Moles O = 35.04 g O 
1 mol O
= 2.190 mol O
16.00 g O
Next divide the number of moles of each element by the smallest of the molar amounts (i.e. 0.4379 mol Si).
moles Ca
1.314 mol Ca
3.001 mol Ca


moles Si
0.4379 mol Si
1 mol Si
moles Si 0.4379 mol Si 1 mol Si


moles Si 0.4379 mol Si 1 mol Si
moles O
2.19 mol O
5.001 mol O


moles Si 0.4379 mol Si
1 mol Si
Rounding the mole ratios to the nearest whole number, we find that chemical formula has 3 moles of
calcium and 5 moles of oxygen for every 1 mole of silicon. The empirical formula is Ca3SiO5.
4.77
The percent composition of our sample is 37.01% C, 2.22% H, 18.50% N, and 42.27% O. Assuming a 100
gram sample, it will contain 37.01 g C, 2.22 g H, 18.50 g N, and 42.27 g O. To determine the empirical
formula, we convert the masses of each element to the equivalent number of moles, and then determine the
relative number of moles of each element in the substance.
Moles C = 37.01 g C 
Moles H = 2.22 g H 
1 mol C
= 3.082 mol C
12.01 g C
1 mol H
= 2.202 mol H
1.008 g H
Moles N = 18.50 g N 
1 mol N
= 1.320 mol N
14.01 g N
Moles O = 42.27 g O 
1 mol O
= 2.642 mol O
16.00 g O
Next divide the number of moles of each element by the smallest of the molar amounts (i.e. 1.320 mol N).
moles C
3.082 mol C 2.335 mol C


moles N 1.320 mol N
1 mol N
423
moles H 2.202 mol H 1.668 mol H


moles N 1.320 mol N
1 mol N
moles N 1.320 mol N 1 mol N


moles N 1.320 mol N 1 mol N
moles O 2.642 mol O 2.002 mol O


moles N 1.320 mol N
1 mol N
Notice that two of the ratios are not whole numbers. The fractional portions (0.333 and 0.668) represent
one-third (1/3) and two-thirds (2/3). We can convert these values to whole numbers by multiplying each
ratio by 3. The empirical formula has 7 moles of carbon, 5 moles of hydrogen, and 6 moles of oxygen for
every 3 moles of nitrogen. The empirical formula is C7H5N3O6.
4.78
The percent composition of strychnine is 75.42% C, 6.63% H, 8.38% N, and 9.57% O. A 100-gram sample
will contain 75.42 g C, 6.63 g H, 8.38 g N, and 9.57 g O. To determine the empirical formula, we convert
the masses of each element to the equivalent number of moles, and determine the relative number of moles
of each element in the substance.
Moles C = 75.42 g C 
1 mol C
= 6.280 mol C
12.01 g C
Moles H = 6.63 g H 
1 mol H
= 6.58 mol H
1.008 g H
Moles N = 8.38 g N 
1 mol N
= 0.598 mol N
14.01 g N
Moles O = 9.57 g O 
1 mol O
= 0.598 mol O
16.00 g O
Next divide the number of moles of each element by the smallest of the molar amounts (i.e. 0.598 mol O or
0.598 mol N).
moles C 6.280 mol C 10.5 mol C


moles O 0.598 mol O
1 mol O
moles H
6.58 mol H
11.0 mol H


moles O 0.598 mol O
1 mol O
moles N 0.598 mol N 1 mol N


moles O
0.598 mol O 1 mol O
moles O 0.598 mol O 1 mol O


moles O 0.598 mol O 1 mol O
Notice that ratio of carbon to oxygen is not a whole number. The fractional portion (0.5) represents onehalf (1/2). We can convert this value to a whole number by multiplying each ratio by 2. The empirical
424
formula for strychnine has 21 moles of carbon, 22 moles of hydrogen, 2 moles of oxygen, and 2 moles of
nitrogen. The empirical formula is C21H22N2O2.
4.79
Convert the percentages to moles of each element, assuming that we have a 100.0 g sample:
1 mol C
mol C = 45.42 g C 
= 3.782 mol
12.01 g C
mol H = 2.720 g H 
1 mol H
= 2.698 mol
1.008 g H
mol O = 51.86 g C 
1 mol O
= 3.241 mol
16.00 g O
Divide each by the smallest number:
mol C
3.782 mol

= 1.402
mol H
2.698 mol
mol H
2.698 mol

= 1.000
mol H
2.698 mol
mol O
3.241 mol

= 1.201
mol H
2.698 mol
Although we did not get whole numbers for the relative amounts of carbon and oxygen, these amounts can
be converted to whole numbers by multiplying by 5:
1.402 mol C × 5 = 7.010 mol C = 7 mol C
1.000 mol H × 5 = 5.000 mol H = 5 mol H
1.201 mol O × 5 = 6.005 mol O = 6 mol O
Now that we have whole number element amounts, we can write an empirical formula: C 7H5O6.
4.80
Convert the percentages to moles of each element, assuming that we have a 100.0 g sample:
1 mol P
mol P = 35.57 g P 
= 1.149 mol
30.97 g P
mol S = 64.43 g S 
1 mol S
= 2.010 mol
32.06 g S
Divide by the smallest number:
mol S
2.010 mol

= 1.749
mol P
1.149 mol
mol P
1.149 mol

= 1.000
mol P
1.149 mol
Although we did not get a whole number for the relative amount of sulfur, these amounts can be converted
to whole numbers by multiplying by 4:
1.749 mol S × 4 = 6.996 mol S = 7 mol S
1.000 mol P × 4 = 4.000 mol P = 4 mol P
Now that we have whole number element amounts, we can write an empirical formula: P 4S7.
4.81
The percent composition by mass and the molar mass are needed to determine the molecular formula.
4.82
A molecular formula represents the exact numbers of atoms of each element in one molecule of the
compound while an empirical formula only represents the whole-number molar ratios of each element in
the compound.
4.83
The ratio of the molar mass to the mass calculated from the empirical formula gives the information
necessary to determine the molecular formula. The mass of the empirical formula, CH 2O is:
425
Mass of 1 mol of C
= 1 mol  12.01 g/mol = 12.01 g
Mass of 2 mol of H
= 2 mol  1.008 g/mol = 2.016 g
Mass of 1 mol of O
= 1 mol  16.00 g/mol = 16.00 g
Mass of 1 mol of CH2O
30.03 g
The ratio of the molar mass to empirical formula mass is:
Molar mass ratio =
90 g / mol
=3
30.03 g / mol
Multiplying each of the subscripts in CH2O by three, we obtain the molecular formula C3H6O3.
4.84
The ratio of the molar mass to the mass calculated from the empirical formula gives the information
necessary to determine the molecular formula. The mass of the empirical formula, NO2 is:
Mass of 1 mol of N
= 1 mol  14.01 g/mol = 14.01 g
Mass of 1 mol of O
= 2 mol  16.00 g/mol = 32.00 g
Mass of 1 mol of NO2
46.01 g
The ratio of the molar mass to empirical formula mass is:
Molar mass ratio =
92 g / mol
= 2.0
46.01 g / mol
Multiplying each of the subscripts in NO2 by 2, we obtain the molecular formula N2O4.
4.85
The strategy for determining the molecular formula from percent composition and molar mass involves two
key steps. First, determine the empirical formula and the molar mass of the empirical formula from the
percent composition data. Next, use the ratio of the molar mass of the compound to the empirical formula
mass to determine the molecular formula.
The percent composition of our sample is 40.00% C, 6.72% H, and 53.29% O. A 100-gram sample of the
compound will contain 40.00 g C, 6.72 g H, and 53.29 g O. To determine the empirical formula, we
convert the masses of each element to the equivalent number of moles, and then determine the relative
number of moles of each element in the compound.
Moles C = 40.00 g C 
Moles H = 6.72 g H 
1 mol C
= 3.331 mol C
12.01 g C
1 mol H
= 6.67 mol H
1.008 g H
Moles O = 53.29 g O 
1 mol O
= 3.331 mol O
16.00 g O
Next divide the number of moles of each element by the smallest of the molar amounts (i.e. 3.331 mol C or
3.331 mol O).
moles C 3.331 mol C 1 mol C


moles O 3.331 mol O 1 mol O
moles H
6.67 mol H
2.00 mol H


moles O 3.331 mol O
1 mol O
426
moles O 3.331 mol O 1 mol O


moles O 3.331 mol O 1 mol O
The empirical formula has 2 moles of hydrogen for each mole of oxygen and carbon. The empirical
formula is CH2O. The mass of the empirical formula is calculated as:
Mass of 1 mol of C
= 1 mol  12.01 g/mol = 12.01 g
Mass of 2 mol of H
= 2 mol  1.008 g/mol = 2.016 g
Mass of 1 mol of O
= 1 mol  16.00 g/mol = 16.00 g
Mass of 1 mol of CH2O
30.03 g
The ratio of the molar mass to the empirical formula molar mass is:
Molar mass ratio =
4.86
180 g / mol
= 6.0
30.03 g / mol
Multiplying the subscripts in CH2O by 6, we obtain the molecular formula C6H12O6.
The strategy for determining the molecular formula from percent composition and molar mass involves two
key steps. First, determine the empirical formula and the molar mass of the empirical formula from the
percent composition data. Next, use the ratio of the molar mass of the compound to the empirical formula
mass to determine the molecular formula.
The percent composition of our sample is 50.7% C, 9.9% H, and 39.4% N. A 100-gram sample of the
compound will contain 50.7 g C, 9.9 g H, and 39.4 g N. To determine the empirical formula, we convert
the masses of each element to the equivalent numbers of moles, and then determine the relative number of
moles of each element in the compound.
Moles C = 50.7 g C 
Moles H = 9.9 g H 
1 mol C
= 4.22 mol C
12.01 g C
1 mol H
= 9.8 mol H
1.008 g H
Moles N = 39.4 g N 
1 mol N
= 2.81 mol N
14.01 g N
Next divide the number of moles of each element by the smallest of the molar amounts (i.e.2.81 mol N).
moles C
4.22 mol C 1.50 mol C


moles N 2.81 mol N
1 mol N
moles H 9.82 mol H 3.49 mol H


moles N 2.81 mol N
1 mol N
moles N 2.81 mol N 1 mol N


moles N 2.81 mol N 1 mol N
Notice that two of the ratios are not whole numbers. The fractional portions (0.5) represent one-half (1/2).
Therefore, we can multiply the ratios by 2 to produce whole numbers. The empirical formula contains 3
moles carbon and 7 moles hydrogen for every 2 moles of nitrogen. The empirical formula is C3H7N2.
427
The mass of the empirical formula is calculated as:
Mass of 3 mol of C
= 3 mol  12.01 g/mol = 36.03 g
Mass of 7 mol of H
= 7 mol  1.008 g/mol = 7.056 g
Mass of 2 mol of N
= 2 mol  14.01 g/mol = 28.02 g
Mass of 1 mol of C3H7N2
71.11 g
The ratio of the molar mass to the empirical formula mass:
Molar mass ratio =
142 g / mol
= 2.00
71.11 g / mol
Multiplying the subscripts in C3H7N2 by 2, we obtain the molecular formula: C6H14N2.
4.87
To calculate the percent composition of a compound from the chemical formula, assume a sample size of 1
mole. Calculate the mass of each element in 1 mole of the compound, divide by the molar mass of the
compound and multiply by 100.
(a) Percent composition of SO2
Mass of 1 mol of S
= 1 mol  32.06 g/mol = 32.06 g
Mass of 2 mol of O
= 2 mol  16.00 g/mol = 32.00 g
Mass of 1 mol of SO2
= 64.06 g
%S=
32.06 g
 100% = 50.05% S
64.06 g
%O=
32.00 g
 100% = 49.95% O
64.06 g
(b) Percent composition of CuCl2
Mass of 1 mol of Cu
= 1 mol  63.55 g/mol = 63.55 g
Mass of 2 mol of Cl
= 2 mol  35.45 g/mol = 70.90 g
Mass of 1 mol of CuCl2
134.45 g
% Cu =
63.55 g
 100% = 47.27% Cu
134.45 g
% Cl =
70.90 g
 100% = 52.73% Cl
134.45 g
(c) Percent composition of Na3PO4
Mass of 3 mol of Na
= 3 mol  22.99 g/mol = 68.97 g
Mass of 1 mol of P
= 1 mol  30.97 g/mol = 30.97 g
Mass of 4 mol of O
= 4 mol  16.00 g/mol = 64.00 g
Mass of 1 mol of Na3PO4
163.94 g
% Na =
68.97 g
 100% = 42.07% Na
163.94 g
428
%P =
30.97 g
 100% = 18.89% P
163.94 g
%O=
64.00 g
 100% = 39.04% O
163.94 g
(d) Percent composition of Mg(NO3)2
Mass of 1 mol of Mg
= 1 mol  24.31 g/mol = 24.31 g
Mass of 2 mol of N
= 2 mol  14.01 g/mol = 28.02 g
Mass of 6 mol of O
= 6 mol  16.00 g/mol = 96.00 g
Mass of 1 mol of Mg(NO3)2
148.33 g
% Mg =
4.88
24.31 g
 100% = 16.39 % Mg
148.33 g
%N=
28.02 g
 100% = 18.89 % N
148.33 g
%O=
96.00 g
 100% = 64.72 % O
148.33 g
To calculate the percent nitrogen of a compound from the chemical formula, assume a sample size of 1
mole. Calculate the mass of nitrogen in 1 mole of the compound, divide by the molar mass of the
compound and multiply by 100.
(a) Percent nitrogen in NaNO3
Mass of 1 mol of Na
= 1 mol  22.99 g/mol = 22.99 g
Mass of 1 mol of N
= 1 mol  14.01 g/mol = 14.01 g
Mass of 3 mol of O
= 3 mol  16.00 g/mol = 48.00g
Mass of 1 mol of NaNO3
85.00 g
%N=
14.01 g
 100% = 16.48% N
85.00 g
(b) Percent nitrogen in NH4Cl
Mass of 1 mol of N
= 1 mol  14.01 g/mol = 14.01 g
Mass of 4 mol of H
= 4 mol  1.008 g/mol = 4.032 g
Mass of 1 mol of Cl
= 1 mol  35.45 g/mol = 35.45 g
Mass of 1 mol of NH4Cl
53.49 g
%N=
14.01 g
 100% = 26.19% N
53.49 g
(c) Percent nitrogen in N2H4
Mass of 2 mol of N
= 2 mol  14.01 g/mol = 28.02 g
Mass of 4 mol of H
= 4 mol  1.008 g/mol = 4.032 g
Mass of 1 mol of N2H4
32.05 g
28.02 g
 100% = 87.43% N
%N=
32.05 g
(d) Percent nitrogen in N2O
429
Mass of 2 mol of N
= 2 mol  14.01 g/mol = 28.02 g
Mass of 1 mol of O
= 1 mol  16.00 g/mol = 16.00 g
Mass of 1 mol of N2O
44.02 g
28.02 g
%N=
 100% = 63.65% N
44.02 g
4.89
A comparison of the percent composition of each mineral shows that cuprite, CuO, has the highest
percentage of copper (79.89% Cu) with chalcocite, Cu2S, an extremely close second at 79.84% Cu.
(a) Cu2S
Mass of 2 mol of Cu
= 2 mol  63.55 g/mol = 127.1 g
Mass of 1 mol of S
= 1 mol  32.06 g/mol = 32.06 g
Mass of 1 mol of Cu2S
= 159.2 g
%Cu =
127.1 g
 100% = 79.84% Cu
159.2 g
(b) Cu2(CO3)(OH)2
Mass of 2 mol of Cu
= 2 mol  63.55 g/mol
Mass of 1 mol of C
= 1 mol  12.01 g/mol
Mass of 5 mol of O
= 5 mol  16.00 g/mol
Mass of 2 mol of H
= 2 mol  1.008 g/mol
Mass of 1 mol of Cu2(CO3)(OH)2
%Cu =
127.1 g
 100% = 57.49% Cu
221.1 g
(c) CuO
Mass of 1 mol of Cu
Mass of 1 mol of O
Mass of 1 mol of CuO
%Cu =
= 1 mol  63.55 g/mol = 63.55 g
= 1 mol  16.00 g/mol = 16.00 g
79.55 g
63.55 g
 100% = 79.89% Cu
79.55 g
(d) Cu3(CO3)2(OH)2
Mass of 3 mol of Cu
= 3 mol  63.55 g/mol
Mass of 2 mol of C
= 2 mol  12.01 g/mol
Mass of 8 mol of O
= 8 mol  16.00 g/mol
Mass of 2 mol of H
= 2 mol  1.008 g/mol
Mass of 1 mol of Cu3(CO3)2(OH)2
%Cu =
4.90
= 127.1 g
= 12.01 g
= 80.00 g
= 2.016 g
221.1 g
= 190.65 g
= 24.02 g
= 128.0 g
= 2.016 g
344.7 g
190.7 g
 100% = 55.32% Cu
344.7 g
A comparison of the percent composition of each mineral shows that wustite, FeO, has the highest
percentage of iron (77.73% Fe).
FeO
Mass of 1 mol of Fe
= 1 mol  55.85 g/mol = 55.85 g
430
Mass of 1 mol of O
Mass of 1 mol of FeO
%Fe =
= 1 mol  16.00 g/mol = 16.00 g
71.85 g
55.85 g
 100% = 77.73% Fe
71.85 g
Fe2O3
Mass of 2 mol of Fe
= 2 mol  55.85 g/mol = 111.7 g
Mass of 3 mol of O
= 3 mol  16.00 g/mol = 48.00 g
Mass of 1 mol of Fe2O3
159.7 g
%Fe =
111.7 g
 100% = 69.94% Fe
159.7 g
Fe3O4
Mass of 3 mol of Fe
= 3 mol  55.85 g/mol = 167.55 g
Mass of 4 mol of O
= 4 mol  16.00 g/mol = 64.00 g
Mass of 1 mol of Fe3O4
231.55 g
%Fe =
167.55 g
 100% = 72.36%
231.55 g
FeCO3
Mass of 1 mol of Fe
= 1 mol  55.85 g/mol = 55.85 g
Mass of 1 mol of C
= 1 mol  12.01 g/mol = 12.01 g
Mass of 3 mol of O
= 3 mol  16.00 g/mol = 48.00 g
Mass of 1 mol of FeCO3
115.86 g
%Fe =
55.85 g
 100% = 48.20% Fe
115.86 g
4.91
A solution is a homogenous mixture of two or more substances. Some common solutions: clear drinks
(coffee and tea), window cleaner, soapy water, tap water, air, brass (a homogenous mixture of copper and
zinc).
4.92
The solute is the substance that is dissolved and is usually present in lesser amount. The solvent is the
substance doing the dissolving and is usually present in greater amount. In the case of solutions formed
from solids and liquids, however, the solid is considered the solute and the liquid is the solvent. Most
products available in the grocery store contain many different solutes.
Solution
Solute
Soda
Sugar (C12H22O11)
Soy Sauce
Salt (NaCl)
Vinegar
Acetic acid (CH3COOH)
Window Cleaner
Ammonia (NH3)
Bleach
Sodium hypochlorite (NaClO)
4.93
Water is the solvent because it is present in the larger amount. Calcium chloride, CaCl2, is the solute
because it is present in the smaller amount. Pure calcium chloride is a solid, but when it is added to water,
it dissociates into its ions as it is dissolving (as shown in the figure).
431
4.94
Water is the solvent because it is present in the larger amount. Sodium chloride is the solute because it is
present in the smaller amount. Pure sodium chloride is a solid, but when it is added to water, it dissociates
into its ions as it is dissolving (as shown in the figure).
4.95
Dilute and concentrated are relative terms. A concentrated solution has a relatively high solute
concentration while a dilute solution has a lower solute concentration.
4.96
Brine is a very concentrated solution of sodium chloride and is used to pickle different kinds of foods.
Saline eye drops are also solutions of sodium chloride, but they are very dilute. There is a much lower
concentration of sodium chloride in eye drops than there is in brine. Liquid pesticides can also be dilute or
concentrated. If you purchase a bottle that is “ready to use,” it is a dilute solution; if the instructions say to
mix with water, then the solution is concentrated and must be diluted before use. Liquid detergent would
also be a good example of a concentrated solution. You add it to water to make it into a dilute solution
before you use it.
4.97
Concentration describes the relationship between the quantities of solute and solvent in a solution.
Molarity (number of moles of solute per liter of solution) and percent concentration (number of grams of
solute per gram of solution) are quantitative ways of expressing concentration.
4.98
Molarity is defined as the number of moles of solute dissolved in one liter of solution.
moles of solute
 mol 
Molarity 
  liters of solution
L


4.99
Solution A is more concentrated. The volume shown in both figures is the same, but solution A shows
more solute particles.
4.100
Solution A is more dilute. The volume shown in both figures is the same, but solution A has fewer solute
particles.
4.101
The molarity of a solution is calculated from the number of moles of solute and volume of solution
(measured in liters). The volumes are given, but the solute quantities are given in grams. In each case, we
must convert the mass of solute into moles using the molar mass. The overall process can be described by
the following map:
MM
volume
Map: Grams of solute  Moles of solute  Molarity
(a) CH3CO2H (60.05 g/mol)
1 mol CH3CO2 H
Moles of CH3CO2H = 122 g CH3CO2 H 
= 2.03 mol CH3CO2H
60.05 g CH3CO2 H
Molarity =
2.03 mol CH3CO2 H
= 2.03 M CH3CO2H
1.00 L solution
(b) C12H22O11 (342.3 g/mol)
Moles of C12H22O11 = 185 g C12 H 22 O11 
Molarity =
1 mol C12 H 22 O11
= 0.540 mol C12H22O11
342.3 g C12 H22 O11
0.540 mol C12 H 22 O11
= 0.540 M C12H22O11
1.00 L solution
(c) HCl (36.46 g/mol)
Moles of HCl = 70.0 g HCl 
1 mol HCl
= 1.92 mol HCl
36.46 g HCl
432
Molarity =
1.92 mol HCl
= 3.20 M HCl
0.600 L solution
(d) This problem is slightly different than the others because the volume is not given in liters.
KOH (56.11 g/mol)
1 mol KOH
Moles of KOH = 45.0 g KOH 
= 0.802 mol KOH
56.11 g KOH
Convert the volume to liters:
Volume (L) = 250.0 mL 
Molarity =
4.102
1L
= 0.2500 L
1000 mL
0.802 mol KOH
= 3.21 M KOH
0.2500 L solution
The molarity of a solution is calculated from the number of moles of solute and volume of solution
(measured in liters). To calculate molarity we must convert the mass of solute into moles using the molar
mass, and convert the volumes from milliliters to liters. The process can be described by the following
map:
MM
volume
Map: Grams of solute  Moles of solute  Molarity
(a) HNO3 (63.02 g/mol)
1L
Volume (L) = 255 mL 
= 0.255 L
1000 mL
Moles of HNO3 = 6.30 g HNO3 
Molarity =
1 mol HNO3
63.02 g HNO3
= 0.100 mol HNO3
0.100 mol HNO3
= 0.392 M HNO3
0.255 L solution
(b) H2SO4 (98.08 g/mol)
Volume (L) = 125 mL 
1L
= 0.125 L
1000 mL
Moles of H2SO4 = 49.0 g H 2SO4 
Molarity =
1 mol H 2SO4
= 0.500 mol H2SO4
98.08 g H 2SO4
0.500 mol HNO3
= 4.00 M H2SO4
0.125 L solution
(c) KOH (56.11 g/mol)
Volume (L) = 525 mL 
1L
= 0.525 L
1000 mL
Moles of KOH = 2.80 g KOH 
1 mol KOH
= 0.0499 mol KOH
56.11 g KOH
433
Molarity =
0.0499 mol KOH
= .0951 M KOH
0.525 L solution
(d) Ca(OH)2 (74.10 g/mol)
Volume (L) = 200.0 mL 
1L
= 0.2000 L
1000 mL
Moles of Ca(OH)2 = 7.40 g KOH 
Molarity =
4.103
1 mol KOH
= 0.0999 mol Ca(OH)2
74.10 g KOH
0.0999 mol Ca(OH) 2
= 0.499 M Ca(OH)2
0.2200 L solution
To calculate moles of substance, it helps to remember that molarity is the conversion between moles and
volume (L). Since you are given molarity and volume, you can calculate moles of a substance. The moles
of ions are calculated using the formula ratios of ions per formula unit.
M
Volume 
 Moles of solute
Moles Na2SO4 = 150.0 mL Na 2SO4 
103 L Na 2SO4
1mL Na 2SO4

0.124 mol Na 2SO4
= 0.0186 mol Na2SO4
L Na 2SO4
2 mol Na   1 mol Na 2SO4
Moles of Na 2SO4 
mol Na 
Mol Na+ = 0.0186 mol Na 2SO4 
2 mol Na 
= 0.0372 mol Na+
1mol Na 2SO4
1 mol SO24   1 mol Na 2SO4
Moles of Na 2SO4 
 mol SO42 
Mol SO42 = 0.0186 mol Na 2SO 4 
4.104
1 mol SO42
1mol
Na 2SO24 
= 0.0186 mol SO42
To calculate moles of substance, it helps to remember that molarity is the conversion between moles and
volume (L). Since you are given molarity and volume, you can calculate moles of a substance. The moles
of ions are calculated using the formula ratios of ions per formula unit.
M
Volume 
 Moles of solute
Moles Mg(NO3)2 = 225.0 mL Mg(NO3 ) 2 
103 L Mg(NO3 )2
1mL Mg(NO3 )2

1.20 mol Mg(NO3 )2
L Mg(NO3 )2
= 0.270 mol Mg(NO3)2
1 mol Mg 2   1 mol Mg(NO3 )2
Moles of Mg(NO3 )2 
 mol Mg 2 
Mol Mg2+ = 0.270 mol Mg(NO3 ) 2 
1 mol Mg 2 
= 0.270 mol Mg2+
1mol Mg(NO3 ) 2
2 mol NO3  1 mol Mg(NO3 )2
Moles of Mg(NO3 )2 
 mol NO3
434
Mol NO3 = 0.270 mol Mg(NO3 ) 2 
4.105
2 mol NO3
1mol Mg(NO3 ) 2
= 0.540
Multiplying the volume of solution (in liters) by the molarity gives you the moles of solute:
molarity
Map: Volume of solution 
 Moles of solute
Make sure to convert the volume to liters. Then use the molar mass of the substance to convert the moles
of solute to mass:
MM
Map: Moles of solute  Grams of solute
(a) 250.0 mL of 1.50 M KCl (MM = 74.55 g/mol)
1L
Volume of solution = 250.0 mL 
= 0.2500 L
1000 mL
Moles KCl = 0.2500 L 
1.50 mol KCl
= 0.375 mol KCl
1L
Grams KCl = 0.375 mol KCl 
74.55 g KCl
= 28.0 g KCl
1 mol KCl
(b) 250.0 mL of 2.05 M Na2SO4 (MM = 142.04 g/mol)
1L
Volume of solution = 250.0 mL 
= 0.2500 L
1000 mL
Moles Na2SO4 = 0.2500 L 
2.05 mol Na 2SO4
= 0.512 mol Na2SO4
1L
Grams Na2SO4 = 0.5125 mol Na 2SO4 
4.106
142.04 g Na 2SO4
= 72.8 g Na2SO4
1 mol Na 2SO4
Multiplying the volume of solution (in liters) by the molarity gives you the moles of solute:
molarity
Map: Volume of solution 
 Moles of solute
Make sure to convert the volume to liters. Then use the molar mass of the substance to convert the moles
of solute to mass:
MM
Map: Moles of solute  Grams of solute
(a) 150.0 mL of 0.245 M CaCl2 (MM = 110.98 g/mol)
1L
Volume of solution = 150.0 mL 
= 0.1500 L
1000 mL
Moles CaCl2 = 0.1500 L 
0.245 mol CaCl2
= 0.0368 mol CaCl2
L
Grams CaCl2= 0.0368 mol CaCl2 
110.98 g CaCl2
= 4.08 g CaCl2
1 mol CaCl2
(b) 1450 mL of 0.00187 M H2SO4 (MM = 98.08 g/mol)
435
Volume of solution = 1450 mL 
Moles H2SO4 = 1.45 L 
1L
= 1.45 L
1000 mL
0.00187 mol H 2SO4
= 0.00271 mol H2SO4
1L
Grams H2SO4 = 0.00271mol H 2SO4 
4.107
98.08 g H 2SO4
= 0.266 g H2SO4
1 mol H 2SO4
We convert moles of solute to volume of solution using molarity:
molarity
Map: Moles of solute 
 Volume of solution
Make sure you check to determine that your answer makes sense. If the number of moles of solute you
need is greater than the molarity of the solution, you will need more than one liter of solution. Conversely,
if the number of moles of solute you need is smaller than the molarity of the solution, you will need less
than one liter of solution.
1L
(a) Volume of solution = 0.250 mol AlCl3 
= 1.00 L
0.250 mol AlCl3
(b) Volume of solution = 0.250 mol HCl 
4.108
1L
= 0.0833 L
3.00 mol HCl
We convert moles of solute to volume using molarity:
molarity
Map: Moles of solute 
 Volume of solution
Make sure you check to determine that your answer makes sense. If the number of moles of solute you
need is greater than the molarity of the solution, you will need more than one liter of solution. Conversely,
if the number of moles of solute you need is smaller than the molarity of the solution, you will need less
than one liter of solution.
1L
(a) Volume of solution = 0.250 mol H 2SO4 
= 0.167 L
1.50 mol H 2SO4
(b) Volume of solution = 0.250 mol NaCl 
1L
= 0.333 L
0.750 mol NaCl
4.109
Count the copper ions in the image before and after the addition of water. Before the addition of water,
there are 10 copper ions. After the addition of water, there are two copper ions shown in the same volume.
When diluting solutions, the concentration is inversely proportional to volume. Since the concentration
decreased by a factor of five (i.e. 2/10 = 1/5) the volume must have increased by five (i.e.10/2 = 5).
Starting with a volume of 10.0 mL, the final volume is 50.0 mL. 40.0 mL of water was added to the
original solution.
4.110
Count the H2S molecules in the image before and after the addition of water. Before the addition of water,
there are 10 H2S molecules. After the addition of water, there are six H2S molecules shown in the same
volume. When diluting solutions, the concentration is inversely proportional to volume. Since the
concentration decreased by a ratio of 6/10, the volume must have increased by the inverse proportion
(10/6). Starting with a volume of 50.0 mL, the final volume is 83.3 mL. 33.3 mL of water was added to
the original solution.
436
4.111
The relationship between the concentration and molarity of the dilute and concentrated solutions is given
as:
M conVcon  M dilVdil
To calculate the volume added, you need to first determine the volume of the dilute solution, and then
calculate the increase in volume.
Vdil 
M conVcon
M dil
To help you solve this problem it is important to identify the appropriate variables:
Mcon = 0.1074 M
Vcon = 935.0 mL
Mdil = 0.1000 M
Since the volume of the concentrated solution is given in milliliters, we first convert the volume to units of
liters:
1L
Vcon = 935.0 mL 
= 0.9350 L
1000 mL
Vdil =
 0.1074 M   0.9350 L
0.1000 M
= 1.004 L
The final volume required is 1.004 L. A volume of 0.0692 L should be added. You should note that this
answer is reasonable since the volume of the dilute solution should always be larger than the more
concentrated solution.
4.112
The relationship between the concentration and molarity of the dilute and concentrated solutions is given
as:
M conVcon  M dilVdil
To calculate the volume added, you need to first determine the volume of the dilute solution, and then
calculate the increase in volume.
Vdil 
M conVcon
M dil
To help you solve this problem correctly it is important to identify the appropriate variables in the problem:
Mcon = 3.0 M
Vcon = 25 mL
Mdil = 0.055M
Since the volume of the concentrated solution is given in milliliters, we first convert the volume to units of
liters:
1L
Vcon = 25 mL 
= 0.025 L
1000 mL
Vdil =
3.0 M   0.025 L = 1.4 L
0.055 M
The volume of the dilute solution is 1.4 L. Note that this answer is reasonable because the volume of the
dilute solution should always be larger than the more concentrated solution.
437
4.113
The relationship between the concentration and molarity of the dilute and concentrated solutions is given
as:
M conVcon  M dilVdil
To calculate the molarity of the dilute solution, we solve the equation for Mdil.
M dil 
M conVcon
Vdil
(a) To help you solve this problem correctly it is important to identify the appropriate variables in the
problem:
Mcon = 0.1832 M
Vcon = 24.75 mL
Vdil = 250.0 mL
At this point, there are two ways to calculate the molarity of the dilute solution. Since the volumes are
given in milliliters, it is not necessary to convert the units to liters. Notice how the mL units cancel in
the following calculation:
Mdil =
 0.1832 M   24.75 mL 
250.0 mL
= 0.01814 M
Alternatively, you could convert the volumes to units of liters and carry out the same calculation:
Mdil =
 0.1832 M   0.02475 L 
= 0.01814 M
0.2500 L
(b) Mcon = 1.187 M
Vcon = 125 mL = 0.125 L (The volumes need to be converted to the same units.)
Vdil = 0.500 L
Mdil =
1.187 M   0.125 L 
0.500 L
= 0.297 M
(c) Mcon = 0.2010 M
Vcon = 10.00 mL
Vdil = 50.00 mL
Mdil =
4.114
 0.2010 M  10.00 mL 
50.00 mL
= 0.04020 M
The relationship between the concentration and molarity of the dilute and concentrated solutions is given
as:
M conVcon  M dilVdil
To calculate the molarity of the dilute solution, we solve the equation for Mdil.
M dil 
M conVcon
Vdil
(a) To help you solve this problem correctly it is important to identify the appropriate variables in the
problem:
438
Mcon = 6.00 M
Vcon = 35.45 mL
Vdil = 150.0 mL
At this point, there are two ways to calculate the molarity of the dilute solution. Since the volumes are
given in milliliters, it is not necessary to convert the units to liters. Notice how the mL units cancel in
the following calculation:
Mdil =
 6.00 M   35.45 mL 
150.0 mL
= 1.42 M
Alternatively, you could convert the volumes to units of liters and carry out the same calculation:
Mdil =
 6.00 M   0.03545 L 
0.1500 L
= 1.42 M
(b) Mcon = 0.00102 M
Vcon = 250.0 mL
Vdil = 500.0 mL
Mdil =
 0.00102 M   250.0 mL 
500.0 mL
= 5.10  104 M
(c) Mcon = 0.8045 M
Vcon = 5.00 mL
Vdil = 250.0 mL
Mdil =
4.115
 0.8045 M   5.00 mL 
250.0 mL
= 0.0161 M
To calculate the mass of 0.100 mol of Cu(OH)2 we use the molar mass:
MM
Map: Moles  Mass in grams
Cu(OH)2
Mass of 1 mol of Cu
= 1 mol  63.55 g/mol = 63.55 g
Mass of 2 mol of O
= 2 mol  16.00 g/mol = 32.00 g
Mass of 2 mol of H
= 2 mol  1.008 g/mol = 2.016 g
Mass of 1 mol of Cu(OH)2
97.57 g
Mass Cu(OH)2 = 0.100 mol Cu(OH)2 
4.116
97.57 g
= 9.76 g Cu(OH)2
1 mol Cu(OH)2
(a) To calculate the number of moles of V2O5 in 52.5 g of V2O5, we use the molar mass and the mole ratio
of vanadium in the compound:
MM
formula ratio
 Moles V
Map: Mass in grams  Moles 
Molar Mass of V2O5
Mass of 2 mol of V
= 2 mol  50.94 g/mol = 101.88 g
Mass of 5 mol of O
= 5 mol  16.00 g/mol = 80.00 g
Mass of 1 mol of V2O5
181.88 g
439
Moles V= 52.5 g V2 O5 
1 mol V2 O5
181.88 g V2 O5

2 mol V
= 0.577 mol V
1 mol V2 O5
(b) The mass of vanadium is calculated from the moles using the molar mass of vanadium (50.94 g/mol):
MM
Map: Moles  Mass in grams
Mass V = 0.577 mol V 
50.94 g V
= 29.4 g V
1 mol V
(c) The number of atoms of vanadium is calculated from the number of moles of vanadium, using
Avogadro’s number:
NA
Map: Moles 
 Number of atoms
Atoms V = 0.577 mol V 
6.022  1023 atoms V
= 3.48  1023 atoms V
1 mol V
4.117
The average mass of one argon atom, in amu, is the value given on the periodic table, 39.95 amu. This is
also the mass of one mole of average argon atoms (i.e. 39.95 g/mol). To calculate the average mass of a
single Ar atom we look for units of grams per atom. We can derive this result using Avogadro’s number to
convert g/mol to g/atom:
1 mol Ar
39.95 g Ar

Average Ar atom mass (g) =
= 6.634  1023 g Ar/atom
23
1 mol Ar
6.022  10 atom
4.118
To calculate the number of molecules in the sample we must first determine the number of moles of
C4H9OH that are present, using the molar mass:
NA
MM  Moles 
Map: Mass in grams 
 Number of molecules
C4H9OH
Mass of 4 mol of C
= 4 mol  12.01 g/mol = 48.04 g
Mass of 10 mol of H
= 10 mol  1.008 g/mol = 10.08 g
Mass of 1 mol of O
= 1 mol  16.00 g/mol = 16.00 g
Mass of 1 mol of C4H9OH
74.12 g
Molecules C4H9OH = 15.43 g 
4.119
1 mol
6.022  1023 molecules
= 1.254  1023 molecules

74.12 g
1 mol
(a) To calculate the number of moles of argon, use the molar mass:
Moles Ar = 36.1 g 
1 mol Ar
= 0.904 mol Ar
39.95 g
Calculate the number of atoms from the number of moles, using Avogadro’s number:
Atoms Ar = 0.904 mol Ar 
6.022  1023 atom
= 5.44  1023 atoms
1 mol Ar
440
(b) Use the conversion 1 carat = 0.200 g to calculate the mass of carbon in the Hope diamond. Then
convert this mass to moles and to number of atoms, using the following conversion map:
0.200 g = 1 carat
NA
MM  Moles 
Map: Mass in carats 
 Mass in grams 
 Number of atoms
Moles C = 44.5 carat 
0.200 g C
1 mol C
= 0.741 mol C

1 carat
12.01 g C
Atoms C = 0.741mol C 
6.022  1023 atom
= 4.46  1023 atoms C
1 mol C
(c) We can use the density of mercury to convert from volume to mass. Then we can convert from mass
to numbers of moles and atoms using the conversion pathway:
MM
NA
Map: Mass in grams  Moles 
 Number of atoms
Mass Hg = 2.50 mL 
13.6 g Hg
= 34.0 g
mL
Moles Hg = 34.0 g Hg 
1 mol Hg
= 0.169 mol Hg
200.6 g Hg
Atoms Hg = 0.169 mol Hg 
4.120
6.022  1023 atom Hg
= 1.02  1023 atoms
1 mol Hg
The mass of one mole of iodine atoms is 126.9 g. The mass of one mole of bromine atoms is 79.90 g. If
you have 50.0 g of iodine atoms, the same number of bromine atoms has a mass of 179 g. This problem
can also be solved using mole conversions:
MM I
1 mol I  1 mol Br
MM Br
Map: Mass I 
 Moles I  Moles Br 
 Mass Br
The 1 mol I = 1 mol Br conversion satisfies the requirement that there are equal numbers of atoms (or
moles of atoms) in the samples:
Mass Br = 50.0 g I 
4.121
1 mol I
126.9 g I

1 mol Br
1 mol I
79.90 g Br
= 31.5 g Br
1 mol Br
To calculate moles, first convert the mass to grams and then use the molar mass of calcium carbonate
(100.09 g/mol) to determine the number of moles
1 mg  103 g
1 mol  100.09 g
 Mass g CaCO3 
 Moles CaCO3
Map: Mass mg CaCO3 
Moles CaCO3 = 750.0 mg CaCO3 
4.122

103 g
1 mg

1 mol
= 7.493  103 mol CaCO3
100.09 g
The percent composition of our sample is 92.3% C and 7.7% H. A 100-gram sample contains 92.3 g C and
7.7 g H. We convert these masses to the equivalent numbers of moles:
441
Moles C = 92.3 g C 
1 mol C
= 7.69 mol C
12.01 g C
Moles H = 7.7 g H 
1 mol H
= 7.6 mol H
1.008 g H
Since the mole ratios are essentially the same, we conclude that the empirical formula must be CH. The
empirical formula mass of CH is 13.02 g/mol. From the molar mass of the compound, 78.1 g/mol, we
calculate the ratio of the molar mass of the compound to the empirical formula mass:
Molar mass ratio =
78.1 g / mol
= 6.00
13.02 g / mol
Multiplying the subscripts of the empirical formula by 6, we determine that the molecular formula for the
compound is C6H6.
4.123
The percent composition of tear gas is 40.25% C, 6.19% H, 8.94% O, and 44.62% Br. A 100-gram sample
contains 40.25 g C, 6.19 g H, 8.94 g O, and 44.62 g Br. To determine the empirical formula, we convert
the masses of each element to the equivalent number of moles and determine the relative number of moles
of each element in the substance.
1 mol C
Moles C = 40.25 g C 
= 3.351 mol C
12.01 g C
Moles H = 6.19 g H 
1 mol H
= 6.14 mol H
1.008 g H
Moles O = 8.94 g O 
1 mol O
= 0.559 mol O
16.00 g O
Moles Br = 44.62 g Br 
1 mol Br
= 0.5584 mol Br
79.90 g Br
Next divide the number of moles of each element by the smallest of the molar amounts (i.e. 0.5584 mol
Br).
Moles C
3.351 mol C
6.001 mol C


Moles Br 0.5584 mol Br
1 mol Br
Moles H
6.14 mol H
11.0 mol H


Moles Br 0.5584 mol Br
1 mol Br
Moles I
0.559 mol O
1.00 mol O


Moles Br 0.5584 mol Br
1 mol Br
The empirical formula has 6 moles of carbon and 11 moles of hydrogen for each mole of O and Br. The
empirical formula is C6H11OBr.
4.124
The molar mass of CO2 is 44.01 g/mol. The mass of dry ice needed is:
442
Mass CO2 (g) = 3.54 mol CO2 
4.125
44.01 g CO2
= 156 g CO2
1 mol CO2
To calculate the number of oxygen atoms, first we calculate the number of moles of H3PO4. Then we use
the mole ratio and Avogadro’s number to calculate the number of oxygen atoms. The molar mass of H3PO4
is 97.99 g/mol.
NA
MM
mole ratio
Map: Mass in grams  Moles  Moles O 
 Number of O atoms
Oxygen Atoms = 10.00 g H3 PO4 
1 mol H3 PO4
97.99 g H3 PO4

4 mol O
1 mol H3 PO4

6.022  1023 atoms O
1 mol O
= 2.458  1023 atoms O
4.126
Percent composition (or mass percent) Na2CO3
Mass of 2 mol of Na
= 2 mol  22.99 g/mol = 45.98 g
Mass of 1 mol of C
= 1 mol  12.01 g/mol = 12.01 g
Mass of 3 mol of O
= 3 mol  16.00 g/mol = 48.00 g
Mass of 1 mol of Na2CO3
105.99 g
% Na =
4.127
45.98 g
 100% = 43.38% Na
105.99 g
%C=
12.01 g
 100% = 11.33% C
105.99 g
%O=
48.00 g
 100% = 45.29% O
105.99 g
The molar mass of CO2 is 44.01 g/mol.
Moles CO2 = 960 g CO2 
4.128
1 mol CO2
44.01 g CO2

6.022  1023 molecules CO2
To calculate moles, first convert the mass to grams and then use the molar mass of ibuprofen, C13H18O2
(206.27 g/mol) to determine the number of moles
1 mg  103 g
1 mol  206.27 g
Map: Mass mg C13H18O2 
 Mass g C13H18O2 
 Moles C13H18O2
Moles C13H18O2 = 200.0 mg C13 H18 O2 ×
4.129
1 mol CO2
= 1.3 × 1025 molecules CO2
10-3 g
1 mg
×
1 mol
= 9.696  104 mol C13H18O2
206.27 g
(a) The percent composition of vanillin is 63.15% C, 5.30% H, and 31.55% O. A 100-gram sample will
contain 63.15 g C, 5.30 g H, and 31.55 g O. To determine the empirical formula, we convert the mass
of each element to the equivalent number of moles, and determine the relative number of moles of each
element in the substance.
Moles C = 63.15 g C 
1 mol C
= 5.258 mol C
12.01 g C
443
Moles H = 5.30 g H 
1 mol H
= 5.258 mol H
1.008 g H
Moles O = 31.55 g O 
1 mol O
= 1.972 mol O
16.00 g O
Next divide the number of moles of each element by the smallest of the molar amounts (i.e., 1.972 mol
O).
Moles C 5.258 mol C 2.667 mol C


Moles O 1.972 mol O
1 mol O
Moles H 5.258 mol H 2.667 mol H


Moles O 1.972 mol O
1 mol O
Moles O 1.972 mol O 1 mol O


Moles O 1.972 mol O 1 mol O
The fractional portion of the ratios (0.667) indicates two thirds (i.e., 2/3) and can be converted to
whole numbers by multiplying by 3. The empirical formula has 8 moles of carbon, 8 moles of
hydrogen for every 3 moles of oxygen. The empirical formula for vanillin is C 8H8O3. The molar
mass of the empirical formula is:
Mass of 8 mol of C
= 8 mol  12.01 g/mol = 96.08 g
Mass of 8 mol of H
= 8 mol  1.008 g/mol = 8.064 g
Mass of 3 mol of O
= 3 mol  16.00 g/mol = 48.00 g
Mass of 1 mol of Na2CO3
152.14 g
(b) Since the molar mass of the empirical formula and the vanillin molecule are the same, the molecular
formula is also C8H8O3.
4.130
Percent composition of monosodium glutamate, NaC5H8NO4
Mass of 1 mol of Na
= 1 mol  22.99 g/mol = 22.99 g
Mass of 5 mol of C
= 5 mol  12.01 g/mol = 60.05 g
Mass of 8 mol of H
= 8 mol  1.008 g/mol = 8.064 g
Mass of 1 mol of N
= 1 mol  14.01 g/mol = 14.01 g
Mass of 4 mol of O
= 4 mol  16.00 g/mol = 64.00 g
Mass of 1 mol of NaNO3
169.11 g
% Na =
22.99 g
 100% = 13.59% Na
169.11 g
%C=
60.05 g
 100% = 35.51% C
169.11 g
%H=
8.064 g
 100% = 4.768% H
169.11 g
444
4.131
%N=
14.01 g
 100% = 8.284% N
169.11 g
%O=
64.00 g
 100% = 37.85% O
169.11 g
To calculate the empirical formula, we need to know the mass of each element in the compound. Since
5.00 g of aluminum produced 9.45 g of aluminum oxide, the principle of conservation of mass tells us that
4.45 g of oxygen has been incorporated into the aluminum oxide (i.e. 9.45 g  5.00 g = 4.45 g). Starting
with those masses, we convert them to moles, and then determine the mole ratio of the elements:
Moles Al = 5.00 g Al 
Moles O = 4.45 g O 
1 mol Al
= 0.185 mol Al
26.98 g Al
1 mol O
= 0.278 mol O
16.00 g O
The mole ratio of aluminum to oxygen (the substance present in smaller number of moles) is:
moles O
0.278 mol O 1.50 mol O


moles Al 0.185 mol Al
1 mol Al
The fractional portion of the number indicates one half (1/2). Multiplying both parts of the ratio by 2
will produce a whole number. The empirical formula has 2 moles of aluminum for every 3 moles of
oxygen. The empirical formula is Al2O3.
4.132
To calculate the empirical formula, we need to know the mass of each element in the compound. Since
4.32 g of copper produced 5.41 g of chalcocite, we know that all 1.09 grams of sulfur reacted (i.e. 4.32 g +
1.09 g = 5.41 g). We convert the masses of copper and sulfur in chalcocite to the equivalent numbers of
moles, and determine the mole ratio of the elements:
Moles Cu = 4.32 g Cu 
Moles S = 1.09 g S 
1 mol Cu
= 0.0680 mol Cu
63.55 g Cu
1 mol S
= 0.0340 mol S
32.06 g S
The mole ratio of copper to sulfur (the substance present in smaller number of moles) is:
moles Cu 0.0680 mol Cu 2 mol Cu


moles S
0.0340 mol S
1 mol S
The empirical formula has 2 moles of copper for every 1 mole of sulfur.
The empirical formula is Cu2S.
4.133
First we must calculate the molar mass of calcium nitrate so that we can use it to convert between moles
and grams:
MM Ca(NO3 )2  40.08 g/mol + 2(14.01 g/mol) + 6(16.00 g/mol) = 164.10 g/mol
445
Then we multiply molar mass by moles to get grams:
164.10 g Ca(NO3 )2
MassCa(NO3 )2  0.742 mol 
= 122 g
1 mol Ca(NO3 )2
4.134
(a) CH4 has a greater mass percent carbon because four hydrogen atoms have a lower mass than two
oxygen atoms attached to carbon in CO2.
(b) This comparison is easier if we compare empirical formulas (CH 4 and CH3). We can do this because
empirical formulas have the same mass percent composition as their molecular formulas. The
empirical formula CH3 has the greater mass percent carbon because it has fewer hydrogen atoms and
the same number of carbon atoms, so the compound C 2H6 must have a greater mass percent carbon
than CH4.
(c) These two compounds have the same ratio of carbon to hydrogen so they have the same empirical
formula, CH2. Therefore they have the same mass percent carbon.
4.135
Both compounds have the same ratio of calcium to anion, so we can compare the molar masses of the
anions to determine relative mass percent calcium.
MM PO 3–  30.97 g/mol + 4(16.00 g/mol) = 94.97 g/mol
4
MM C H O 3–  6(12.01 g/mol) + 5(1.008 g/mol) + 7(16.00 g/mol) = 189.1 g/mol
6
5
7
The phosphate ion contributes a lower mass to its calcium compound compared to the citrate ion, so
Ca3(PO4)2 has the greater mass percent calcium.
4.136
The number of Ca3(C6H5O7)2 formula units is related to moles by Avogadro’s number:
Avogadro ' s Number
Moles 
 Number of Formula Units
Number of formula units 
5.00 mol Ca 3 (C6 H5 O7 ) 2 
6.022 1023 formula units
1 mol Ca 3 (C6 H5 O7 ) 2
= 3.011024 Ca 3 (C6 H5 O7 )2 formula units
4.137
This is similar to the question: Which box contains more balls, a 10-pound box of baseballs or a 10-pound
box of ping-pong balls? Many more of the ping-pong balls are needed to make a total mass of 10 pounds
than the heavier baseballs. Similarly more of the lighter-weight F2 molecules are needed to make a 1.0
gram sample than the heavier-weight SF2 and CF4 molecules. We can also reason this out mathematically:
Moles are proportional to number of molecules, so if we determine which sample has the greater number of
moles, which is also the sample with the greater number of molecules. To calculate moles of each
substance, we divide 1.0 g by the molar mass of each substance. The larger the molar mass the smaller the
calculated number of moles. The molar mass of F2 is the smallest, so a 1.0 gram sample of F2 has the
greatest number of moles and molecules.
4.138
(a) The samples are each 1.0 mole so each contains the same number of molecules: 6.0221023 molecules.
(b) While each sample contains the same number of molecules, the substance with the greatest number of
atoms per molecule will have the greatest number of atoms in the sample. A CF4 molecule has the
most atoms (5) compared to SF2 (3 atoms) and F2 (2 atoms) so a 1.0 mol sample of CF4 will have the
greatest number of atoms.
4.139
(a) 98% of 20.0 g is: 0.98  20.0 g = 19.6 g
446
(b) To determine the mass of Na in 19.6 g NaCl, we must first determine the mass percent Na in the
sample:
% Na =
Mass of 1 mol Na
22.99 g Na
100 
100  39.34% Na
Mass of 1 mol NaCl
58.44 g NaCl
Mass Na in 19.6 g NaCl = 0.3934  19.6 g = 7.71 g Na
(c) Since sodium and chlorine are the only elements in the compound, we can calculate the mass of
chlorine by subtracting the mass of sodium from the total mass of the compound:
Mass Cl in 19.6 g NaCl = 19.6 g NaCl – 7.71 g Na = 11.9 g Cl
4.140
(a) Molarity is a ratio of moles to liters: M =
mol
.
VL
Rearranging to solve for moles we get: mol = M  VL
Before we can substitute volume into this equation we must convert volume units from mL to L. The two
step process is: VmL
1 L = 1000 mL
Molarity

 VL 
 moles
1L
= 0.0100 L
1000 mL
0.30 mol
Moles citric acid = M  VL =
× 0.0100 L = 0.0030 mol H3C6H5O7
1 L
Volume in liters = 10.0 mL 
(b) We can convert from moles of H3C6H5O7 to grams using molar mass. First we must calculate the
molar mass of H3C6H5O7:
MM H3C6H5O7  8(1.008 g/mol) + 6(12.01 g/mol) + 7(16.00 g/mol) = 192.12 g/mol
Mass H3C6 H5 O7  0.00300 mol H3C6 H5 O7 
4.141
192.1 g H3C6 H5 O7
= 0.576 g H3C6 H5 O7
1 mol H3 C6 H5 O7
In each case we will calculate the number of formula units using Avogadro’s number, and then we will
multiply by the number of Cl ions per formula unit to determine the number of Cl ions in each quantity.
(a) Number of Cl ions =
6.022 1023 AlCl3 formula units
3 Cl ions
1.0 mol AlCl3 

 1.8 1024 Cl ions
1 mol AlCl3
1 AlCl3 formula unit
(b) Number of Cl ions =
6.022 1023 AlCl3 formula units
3 Cl ions
0.25 mol AlCl3 

 4.5 1023 Cl ions
1 mol
1 AlCl3 formula unit
(c) Number of Cl ions =
6.022 1023 MgCl2 formula units
2 Cl ions
0.25 mol MgCl2 

 3.0 1023 Cl ions
1 MgCl2 formula unit
1 mol MgCl 2
4.142
The definition of molarity ( M =
mol
) tells us that 1.0 L of a 2.5 M Mg(NO3)2 solution contains 2.5 mol of
VL
Mg(NO3)2. We also know that each formula unit of Mg(NO3)2 consists of 1 Mg2+ ion and 2 NO3 ions, so
447
when Mg(NO3)2 dissolves, the concentration of NO3 is twice that of Mg2+ ion. We have 2.5 mol of
Mg(NO3)2, so it dissolves to form 2.5 mol Mg2+ ion and 5.0 mol of NO3 ion.
CONCEPT REVIEW
4.143
Answer: D; the percent composition will be the same for compounds that have the same ratio of the carbon
to hydrogen atoms. In this case 2/4 = 0.5 for C2H4 and 3/6 = 0.5 for C3H6.
In each case, the compound with the lower carbon/ hydrogen ratio would have the lower percent carbon.
A. The carbon/ hydrogen ratios are 3/4 = 0.75 for C3H4 and 3/6 = 0.5 for C3H6. C3H6 has the lower
percent carbon by mass.
B. The carbon/ hydrogen ratios are 2/4 = 0.5 for C2H4 and 3/4 = 0.75 for C3H4. C2H4 has the lower
percent carbon by mass.
C. The carbon/ hydrogen ratios are 2/4 = 0.5 for C2H4 and 4/2 = 2 for C4H2. C2H4 has the lower percent
carbon by mass.
E. The carbon/ hydrogen ratios are 4/8 = 0.5 for C4H8 and 3/8 = 0.375 for C3H8. C3H8 has the lower
percent carbon by mass.
4.144
Answer: A and E; the molar mass of C3H8 is 44.0 g/mol, so this represents 1 mol; 1 mol of propane
contains 3 mol of carbon, or 36.0 g.
B. 44.0 g of C3H8 would contain 8 mol or 8 .0 g of hydrogen: the amount of propane that contains 8.0 g of
hydrogen
C. The 44.0 g of propane represents 1 mol of propane, not 44.0 mol: the amount of propane that contains
6.02 × 1023 molecules of propane
D. The 44.0 g of propane represents 1 mol of propane, not 8 mol: the amount of propane that contains
6.02 × 1023 molecules of propane
4.145
Answer: D; the number of molecules is:
molecules As 4 = 1 g 
1 mol
6.022  1023 molecules

= 2  1021 molecules As 4
299.68 g
1 mol
The number of molecules of the other substances can be obtained by a similar equation, but using the
appropriate molar mass of the substance. Since the other molar masses are smaller, the number of
molecules would be larger. It is only necessary to determine the molar mass of each substance to compare
the number of molecules in the same mass of each. The molar masses are 123.88 g/mol P 4, 70.90 g/mol
Cl2, 28.02 g/mol N2, 299.68 g/mol As4, and 256.48 g/mol S8. The larger the molar mass, the smaller the
number of molecules in 1 g, so the order will be As 4 < S8 < P4 < Cl2 < N2.
4.146
Answer: C; the number of atoms will equal:
atoms of N = 1 gN2 
1 mol N 2
28.02 g N2

6.022  1023 molecules N 2
1 mol N 2

2 atoms N
1 molecule N 2
= 4.3  1023 N atoms
The number of atoms present in the other substances can be calculated in a similar way. However, the
substance with the smallest product of 1/molar mass and number of atoms per molecule will contain the
largest number of atoms. The products of 1/molar mass and number of atoms/molecule are:
A. P4: 4/123.88 g/mol = 0.032
B. Cl2: 2/70.90 g/mol = 0.028
C. N2: 2/28.02 g/mol = 0.071
D. As4: 4/299.68 g/mol = 0.013
E. S8: 8/256.48 g/mol = 0.031
The order of increasing number of atoms in 1 g is As4 < Cl2 < S8 < P4 < N2.
448
4.147
Answer: C; The number of moles of oxygen atoms in this sample of O 2 is:
2 mol O atoms
mol of O atoms = 0.60 mol O2 
= 1.2 mol O atoms
1 mol O2
1 mol O atoms
= 1.0 mol O atoms
1 mol H 2 O
A.
mol of O atoms = 1.0 mol H 2 O 
B.
mol of O atoms = 0.20 mol N 2O4 
D.
mol of O atoms = 0.11 mol As 4 O10 
E.
mol of O atoms = 0.30 mol O3 
4 mol O atoms
= 0.80 mol O atoms
1 mol N 2 O4
10 mol O atoms
= 1.1 mol O atoms
1 mol As4 O10
3 mol O atoms
= 0.90 mol O atoms
1 mol O3
The order of increasing number of oxygen atoms is B < E < A < D < C.
4.148
Answer: E; none of the subscripts can be simplified to smaller numbers.
A.
B.
C.
D.
4.149
Answer: C; the molecular formula is either the same as the empirical formula (a multiple of 1) or a whole
number multiple of the empirical formula.
A.
B.
D.
E.
4.150
Both subscripts can be divided by 2, giving an empirical formula of NF.
Both subscripts can be divided by 2, giving an empirical formula of NF 2.
Both subscripts can be divided by 2, giving an empirical formula of HC.
All three subscripts can be divided by 2, giving an empirical formula of HNO.
H2C2O4 and HCO2 are of equal complexities.
H2O is both the molecular and the empirical formula of water.
Not all empirical formulas are the same as the molecular formula: CH and C2H2.
The molecular formula is a multiple, not a fraction of the empirical formula. The empirical formula
H2O cannot be divided to give smaller whole number subscripts.
Answer: C; molarity =
0.10 mol NaCl
= 0.20 M NaCl
0.5000 L
A. Dissolving 0.040 mol NaCl in water to make 0.20 L of solution
0.20 mol NaCl
mol NaCl = 2.00 L 
= 0.040 mol NaCl
1 L
B. Dissolving 0.32 mol NaCl in water to make 1.6 L of solution
0.20 mol NaCl
mol NaCl = 1.60 L 
= 0.32 mol NaCl
1 L
D. Dissolving 0.020 mol NaCl in water to make 100.0 mL of solution
0.20 mol NaCl
mol NaCl = 0.1000 L 
= 0.020 mol NaCl
1 L
E. Dissolving 0.0020 mol NaCl in water to make 10.0 mL of solution
0.20 mol NaCl
mol NaCl = 0.0100 L 
= 0.0020 mol NaCl
1 L
4.151
Answer: A; the solution contains:
449
mol Ca 3 (PO4 )2 = 2.00 L 
0.100 mol Ca 3 (PO4 )2
1 L
= 0.200 mol Ca 3 (PO4 )2
B. The solution contains 1.60 mol of oxygen atoms.
0.100 mol Ca 3 (PO4 )2
8 mol O atoms
mol O atoms = 2.00 L 

= 1.60 mol O atoms
1 L
1 mol Ca 3 (PO4 )2
C. 1.00 L of this solution contains 0.300 mol Ca2+ ions.
0.100 mol Ca 3 (PO4 )2
3 mol Ca 2+ ions
mol Ca 2+ ions = 1.00 L 

= 0.300 mol Ca 2+ ions
1 L
1 mol Ca 3 (PO4 )2
D. There 6.02 × 1022 P atoms in 500.0 mL of this solution.
P atoms = 0.5000 L 
0.100 mol Ca 3 (PO4 )2
1 L

2 mol P atoms
6.022  1023 P atoms

= 6.02  1022 P atoms
1 mol Ca 3 (PO4 )2
1 mol P atoms
E. This solution contains 0.600 mol of Ca2+.
0.100 mol Ca 3 (PO4 )2
3 mol Ca 2+ ions
mol Ca 2+ ions = 2.00 L 

= 0.600 mol Ca 2+ ions
1 L
1 mol Ca 3 (PO4 )2
4.152
Answer: D; this solution contains
mol Cl = 0.300 L 
0.200 mol NaCl
1 mol Cl

= 0.600 mol Cl
1 L
1 mol NaCl
A. mol Cl = 0.100 L 
0.200 mol NaCl
1 mol Cl

= 0.0200 mol Cl
1 L
1 mol NaCl
B. mol Cl = 0.200 L 
0.100 mol CaCl2
C. mol Cl = 0.150 L 
0.100 mol FeCl3
E. mol Cl = 0.100 L 
0.050 mol CaCl2
1 L
1 L
1 L

2 mol Cl
= 0.0400 mol Cl
1 mol CaCl2

3 mol Cl
= 0.0450 mol Cl
1 mol FeCl3

2 mol Cl
= 0.010 mol Cl
1 mol CaCl2
The order of decreasing moles chloride ion is D > C > B > A > E.
450