Example 2.6 Converting between Number of Moles and Number of Atoms
Calculate the number of copper atoms in 2.45 mol of copper (Cu).
SOLUTION
SORT You are given the amount of copper in moles
and asked to find the number of copper atoms.
GIVEN 2.45 mol Cu
FIND Cu atoms
STRATEGIZE Convert between number of moles
and number of atoms by using Avogadro’s number as
a conversion factor.
CONCEPTUAL PLAN
RELATIONSHIPS USED
6.022 1023 = 1 mol (Avogadro’s number)
SOLVE Follow the conceptual plan to solve the
problem. Begin with 2.45 mol Cu and multiply by
Avogadro’s number to get to Cu atoms.
SOLUTION
CHECK Since atoms are small, it makes sense that the answer is large. The number of moles of copper is
almost 2.5, so the number of atoms is almost 2.5 times Avogadro’s number.
Example 2.6 Converting between Number of Moles and Number of Atoms
Continued
For Practice 2.6
A pure silver ring contains 2.80
1022 silver atoms. How many moles of silver atoms does it contain?
Example 2.7 Converting between Mass and Amount (Number of Moles)
Calculate the amount of carbon (in moles) contained in a 0.0265 g pencil “lead.” (Assume that the pencil lead is
made of pure graphite, a form of carbon.)
SOLUTION
SORT You are given the mass of carbon and asked
to find the amount of carbon in moles.
GIVEN 0.0265 g C
FIND mol C
STRATEGIZE Convert between mass and amount
(in moles) of an element by using the molar mass of
the element.
CONCEPTUAL PLAN
RELATIONSHIPS USED
12.01 g C = 1 mol C (carbon molar mass)
SOLVE Follow the conceptual plan to solve the
problem.
SOLUTION
CHECK The given mass of carbon is much less than the molar mass of carbon. Therefore the answer (the
amount in moles) is much less than 1 mol of carbon.
Example 2.7 Converting between Mass and Amount (Number of Moles)
Continued
For Practice 2.7
Calculate the amount of copper (in moles) in a 35.8 g pure copper sheet.
For More Practice 2.7
Calculate the mass (in grams) of 0.473 mol of titanium.
Example 2.8 The Mole Concept—Converting between Mass and Number of Atoms
How many copper atoms are in a copper penny with a mass of 3.10 g? (Assume that the penny is composed of
pure copper.)
SOLUTION
SORT You are given the mass of copper and asked
to find the number of copper atoms.
GIVEN 3.10 g Cu
FIND Cu atoms
STRATEGIZE Convert between the mass of an
element in grams and the number of atoms of the
element by first converting to moles (using the molar
mass of the element) and then to number of atoms
(using Avogadro’s number).
CONCEPTUAL PLAN
RELATIONSHIPS USED
63.55 g Cu = 1 mol Cu (molar mass of copper)
6.022 1023 = 1 mol (Avogadro’s number)
SOLVE Follow the conceptual plan to solve the
problem. Begin with 3.10 g Cu and multiply by the
appropriate conversion factors to arrive at the number
of Cu atoms.
SOLUTION
CHECK The answer (the number of copper atoms) is less than 6.022 1023 (one mole). This is consistent with
the given mass of copper atoms, which is less than the molar mass of copper.
Example 2.8 The Mole Concept—Converting between Mass and Number of Atoms
Continued
For Practice 2.8
How many carbon atoms are there in a 1.3-carat diamond? Diamonds are a form of pure carbon. (1 carat =
0.20 grams)
For More Practice 2.8
Calculate the mass of 2.25
1022 tungsten atoms.
Example 2.9 The Mole Concept
An aluminum sphere contains 8.55
density of aluminum is 2.70 g/cm3.
1022 aluminum atoms. What is the radius of the sphere in centimeters? The
SOLUTION
SORT You are given the number of aluminum atoms
in a sphere and the density of aluminum. You are
asked to find the radius of the sphere.
GIVEN 8.55 1022 Al atoms
d = 2.70 g/cm3
FIND radius (r) of sphere
STRATEGIZE The heart of this problem is density,
which relates mass to volume, and though you aren’t
given the mass directly, you are given the number of
atoms, which you can use to find mass.
CONCEPTUAL PLAN
1. Convert from number of atoms to number of
moles using Avogadro’s number as a conversion
factor.
2. Convert from number of moles to mass using
molar mass as a conversion factor.
3. Convert from mass to volume (in cm3) using
density as a conversion factor.
4. Once you calculate the volume, find the radius
from the volume using the formula for the volume
of a sphere.
RELATIONSHIPS AND EQUATIONS USED
6.022 1023 = 1 mol (Avogadro’s number)
26.98 g Al = 1 mol Al (molar mass of aluminum)
2.70 g Al = 1 cm3 Al (density of aluminum)
(volume of a sphere)
Example 2.9 The Mole Concept
Continued
SOLUTION
SOLVE Follow the conceptual plan to solve the
problem. Begin with 8.55 1022 Al atoms and
multiply by the appropriate conversion factors to
arrive at volume in cm3.
SOLUTION
Then solve the equation for the volume of a sphere
for r and substitute the volume to compute r.
CHECK The units of the answer (cm) are correct. The magnitude cannot be estimated accurately, but a radius
of about one-half of a centimeter is reasonable for just over one-tenth of a mole of aluminum atoms.
Example 2.9 The Mole Concept
Continued
For Practice 2.9
A titanium cube contains 2.86
4.50 g/cm3.
1023 atoms. What is the edge length of the cube? The density of titanium is
For More Practice 2.9
Find the number of atoms in a copper rod with a length of 9.85 cm and a radius of 1.05 cm. The density of copper
is 8.96 g/cm3.
Example 3.1 Molecular and Empirical Formulas
Write empirical formulas for the compounds represented by the molecular formulas.
(a) C4H8
(b) B2H6 (c) CCl4
SOLUTION
To determine the empirical formula from a molecular formula, divide the subscripts by the greatest common
factor (the largest number that divides exactly into all of the subscripts).
(a) For C4H8, the greatest common factor is 4. The empirical formula is therefore CH2.
(b) For B2H6, the greatest common factor is 2. The empirical formula is therefore BH3.
(c) For CCl4, the only common factor is 1, so the empirical formula and the molecular formula are identical.
For Practice 3.1
Write the empirical formula for the compounds represented by the molecular formulas.
(a) C5H12
(b) Hg2Cl2 (c) C2H4O2
Example 3.11 Calculating Formula Mass
Calculate the formula mass of glucose, C6H12O6.
SOLUTION
To find the formula mass, we sum the atomic masses of each atom in the chemical formula:
Formula mass = 6 × (atomic mass C) + 12 × (atomic mass H) + 6 × (atomic mass O)
= 6(12.01 amu)
+ 12(1.008 amu)
+ 6(16.00 amu)
= 180.16 amu
For Practice 3.11
Calculate the formula mass of calcium nitrate.
Example 3.12 The Mole Concept—Converting between Mass and Number
of Molecules
An aspirin tablet contains 325 mg of acetylsalicylic acid (C9H8O4). How many acetylsalicylic acid molecules does it
contain?
SOLUTION
SORT You are given the mass of acetylsalicylic
acid and asked to find the number of molecules.
GIVEN 325 mg C9H8O4
FIND number of C9H8O4 molecules
STRATEGIZE Convert between mass and the
number of molecules of a compound by first
converting to moles (using the molar mass of the
compound) and then to the number of molecules
(using Avogadro’s number). You need both the molar
mass of acetylsalicylic acid and Avogadro’s number
as conversion factors. You also need the conversion
factor between g and mg.
CONCEPTUAL PLAN
RELATIONSHIPS USED
1 mg = 10–3 g
C9H8O4 molar mass = 9(12.01) + 8(1.008) + 4(16.00)
= 180.15 g/mol
6.022
1023 = 1 mol
Example 3.12 The Mole Concept—Converting between Mass and
Number of Molecules
Continued
SOLUTION
SOLUTION
SOLVE Follow the conceptual plan to
solve the problem.
CHECK The units of the answer, C9H8O4 molecules, are correct. The magnitude seems appropriate because it is
smaller than Avogadro’s number, as expected, since we have less than one mole of acetylsalicylic acid.
For Practice 3.12
Find the number of ibuprofen molecules in a tablet containing 200.0 mg of ibuprofen (C13H18O2).
For More Practice 3.12
What is the mass of a drop of water containing 3.55
1022 H2O molecules?
Example 3.13 Mass Percent Composition
Calculate the mass percent of Cl in Freon-112 (C2Cl4F2), a CFC refrigerant.
SOLUTION
SORT You are given the molecular
formula of Freon-112 and asked to
find the mass percent of Cl.
GIVEN C2Cl4F2
FIND mass percent Cl
CONCEPTUAL PLAN
STRATEGIZE The molecular
formula indicates that there are 4 mol
of Cl in each mole of Freon-112. Find
the mass percent composition from
the chemical formula by using the
equation that defines mass percent.
The conceptual plan shows how you
can use the mass of Cl in 1 mol of
C2Cl4F2 and the molar mass of C2Cl4F2
to determine the mass percent of Cl.
RELATIONSHIPS USED
Example 3.13 Mass Percent Composition
Continued
SOLUTION
SOLVE Calculate the necessary parts of
the equation and substitute the values
into the equation to find mass percent Cl.
SOLUTION
CHECK The units of the answer (%) are correct and the magnitude is reasonable because (a) it is between 0
and 100% and (b) chlorine is the heaviest atom in the molecule and there are four of them.
For Practice 3.13
Acetic acid (HC2H3O2) is the active ingredient in vinegar. Calculate the mass percent composition of oxygen in
acetic acid.
For More Practice 3.13
Calculate the mass percent composition of sodium in sodium oxide.
Example 3.14 Chemical Formulas as Conversion Factors
Hydrogen may potentially be used in the future as a fuel to replace gasoline. Most major automobile companies
are developing vehicles that run on hydrogen. These cars are environmentally friendly because their only emission
is water vapor. One way to obtain hydrogen for fuel is to use an emission-free energy source such as wind power
to
split hydrogen from water. What mass of hydrogen (in grams) does 1.00 gallon of water contain? (The density of
water is 1.00 g/mL.)
SOLUTION
SORT You are given a volume of water and
asked to find the mass of hydrogen it contains.
You are also given the density of water.
STRATEGIZE The first part of the
conceptual plan shows how you can convert the
units of volume from gallons to liters and then to
mL. It also shows how you can use the density to
convert mL to g.
The second part of the conceptual plan is the
basic sequence of mass → moles → moles →
mass.
Convert between moles and mass using the
appropriate molar masses, and convert from mol
H2O to mol H using the conversion factor derived
from the molecular formula.
GIVEN 1.00 gal H2O
dH2O = 1.00 g/mL
FIND g H
CONCEPTUAL PLAN
Example 3.14 Chemical Formulas as Conversion Factors
Continued
SOLUTION
RELATIONSHIPS USED
3.785 L = 1 gal (Table 1.3)
1000 mL = 1 L
1.00 g H2O = 1 mL H2O (density of H2O)
Molar mass H2O = 2(1.008) + 16.00 = 18.02 g/mol
2 mol H : 1 mol H2O
1.008 g H = 1 mol H
SOLVE Follow the conceptual plan to SOLUTION
solve the problem.
CHECK The units of the answer (g H) are correct. Since a gallon of water is about 3.8 L, its mass is about 3.8 kg.
H is a light atom, so its mass should be significantly less than 3.8 kg.
Example 3.14 Chemical Formulas as Conversion Factors
Continued
For Practice 3.14
Determine the mass of oxygen in a 7.2 g sample of Al2(SO4)3.
For More Practice 3.14
Butane (C4H10) is a liquid fuel used in lighters. How many grams of carbon are in a lighter containing 7.25 mL of
butane? (The density of liquid butane is 0.601 g/mL.)
Example 3.15 Obtaining an Empirical Formula from Experimental Data
A compound containing nitrogen and oxygen is decomposed in the laboratory and produces 24.5 g nitrogen and
70.0 g oxygen. Calculate the empirical formula of the compound.
SOLUTION
1. Write down (or compute) as given the masses of each element present in
a sample of the compound. If you are given mass percent composition,
assume a 100-g sample and calculate the masses of each element from
the given percentages.
GIVEN 24.5 g N, 70.0 g O
FIND empirical formula
2. Convert each of the masses in step 1 to moles by using the appropriate
molar mass for each element as a conversion factor.
3. Write down a pseudoformula for the compound using the number of
moles of each element (from step 2) as subscripts.
N1.75O4.38
4. Divide all the subscripts in the formula by the smallest subscript.
5. If the subscripts are not whole numbers, multiply all the subscripts by a
small whole number (see table) to get whole-number subscripts.
N1O2.5 2 → N2O5
The correct empirical formula is N2O5.
Example 3.15 Obtaining an Empirical Formula from Experimental Data
Continued
Fractional
Subscript
0.20
0.25
0.33
0.40
0.50
0.66
0.75
0.80
Multiply by
This
5
4
3
5
2
3
4
5
For Practice 3.15
A sample of a compound is
decomposed in the laboratory
and produces 165 g carbon, 27.8
g hydrogen, and 220.2 g oxygen.
Calculate the empirical formula of
the compound
Example 3.16 Obtaining an Empirical Formula from Experimental Data
A laboratory analysis of aspirin determined the following mass percent composition:
C 60.00%
H 4.48%
O 35.52%
Find the empirical formula.
SOLUTION
1. Write down (or compute) as given the masses of each element present in GIVEN In a 100-g sample: 60.00 g C,
a sample of the compound. If you are given mass percent composition,
4.48 g H, 35.52 g O
assume a 100-g sample and calculate the masses of each element from
FIND empirical formula
the given percentages.
2. Convert each of the masses in step 1 to moles by using the appropriate
molar mass for each element as a conversion factor.
3. Write down a pseudoformula for the compound using the number of
moles of each element (from step 2) as subscripts.
4. Divide all the subscripts in the formula by the smallest subscript.
C4.996H4.44O2.220
Example 3.16 Obtaining an Empirical Formula from Experimental Data
Continued
5.
If the subscripts are not whole numbers, multiply all the subscripts
by
a small whole number (see table) to get whole-number subscripts.
Fractional
Multiply by
Subscript
This
0.20
5
0.25
4
0.33
3
0.40
5
0.50
2
0.66
3
0.75
4
0.80
5
C2.25H2O1 2 → C9H8O4
The correct empirical formula is C9H8O4.
For More Practice 3.16
Ibuprofen, an aspirin substitute, has the
following mass percent composition:
C 75.69%, H 8.80%, O 15.51%.
What is the empirical formula of
ibuprofen?
Example 3.17 Calculating a Molecular Formula from an Empirical Formula
and Molar Mass
Butanedione—a main component in the smell and taste of butter and cheese—contains the elements carbon,
hydrogen, and oxygen. The empirical formula of butanedione is C2H3O, and its molar mass is 86.09 g/mol. Find its
molecular formula.
SOLUTION
SORT You are given the empirical formula and molar
mass of butanedione and asked to find the molecular
formula.
STRATEGIZE A molecular formula is always a wholenumber multiple of the empirical formula. Divide the
molar mass by the empirical formula mass to get the
whole number.
SOLVE Calculate the empirical formula mass.
Divide the molar mass by the empirical formula mass to
find n.
Multiply the empirical formula by n to obtain the
molecular formula.
GIVEN Empirical formula = C2H3O
molar mass = 86.09 g/mol
FIND molecular formula
Example 3.17 Calculating a Molecular Formula from an Empirical Formula
and Molar Mass
Continued
SOLUTION
CHECK Check the answer by calculating the molar mass of the formula as follows:
4(12.01 g/mol) + 6(1.008 g/mol) + 2(16.00 g/mol) = 86.09 g/mol
The calculated molar mass is in agreement with the given molar mass. The answer is correct.
For Practice 3.17
A compound has the empirical formula CH and a molar mass of 78.11 g/mol. Find its molecular formula.
For More Practice 3.17
A compound with the percent composition shown below has a molar mass of 60.10 g/mol. Find its molecular
formula.
C, 39.97%
H, 13.41%
N, 46.62%
Example 3.18 Determining an Empirical Formula from Combustion Analysis
Upon combustion, a compound containing only carbon and hydrogen produced 1.83 g CO2 and 0.901 g H2O. Find
the empirical formula of the compound.
SOLUTION
1. Write down as given the masses of each combustion product and the
mass of the sample (if given).
2. Convert the masses of CO2 and H2O from step 1 to moles by using the
appropriate molar mass for each compound as a conversion factor.
3. Convert the moles of CO2 and moles of H2O from step 2 to moles of C
and moles of H using the conversion factors inherent in the chemical
formulas of CO2 and H2O.
GIVEN 1.83 g CO2, 0.901 g H2O
FIND empirical formula
Example 3.18 Determining an Empirical Formula from Combustion Analysis
Continued
SOLUTION
4. If the compound contains an element other than C and H, find the mass
of
the other element by subtracting the sum of the masses of C and H
(obtained in step 3) from the mass of the sample. Finally, convert the
mass of the other element to moles.
This sample contains no
elements besides C and H, so
proceed to next step.
C0.0416H0.100
5. Write down a pseudoformula for the compound using the number of
moles of each element (from steps 3 and 4) as subscripts.
6. Divide all the subscripts in the formula by the smallest subscript. (Round
all subscripts that are within 0.1 of a whole number.)
7. If the subscripts are not whole numbers, multiply all the subscripts by a
small whole number to get whole-number subscripts.
C1H2.4 5 → C5H12
The correct empirical formula is
C5H12.
For Practice 3.18
Upon combustion, a compound containing only carbon and hydrogen produced 1.60 g CO2 and
0.819 g H2O. Find the empirical formula of the compound.
Example 3.19 Determining an Empirical Formula from Combustion Analysis
Upon combustion, a 0.8233 g sample of a compound containing only carbon, hydrogen, and oxygen produced
2.445 g CO2 and 0.6003 g H2O. Find the empirical formula of the compound.
SOLUTION
1. Write down as given the masses of each combustion product and the
mass of the sample (if given).
2. Convert the masses of CO2 and H2O from step 1 to moles by using the
appropriate molar mass for each compound as a conversion factor.
3. Convert the moles of CO2 and moles of H2O from step 2 to moles of C
and moles of H using the conversion factors inherent in the chemical
formulas of CO2 and H2O.
GIVEN 0.8233 g sample, 2.445 g CO2,
0.6003 g H2O
FIND empirical formula
Example 3.19 Determining an Empirical Formula from Combustion Analysis
Continued
SOLUTION
4. If the compound contains an element other than C and H, find the mass
of the other element by subtracting the sum of the masses of C and H
(obtained in step 3) from the mass of the sample. Finally, convert the
mass of the other element to moles.
5. Write down a pseudoformula for the compound using the number of
moles of each element (from steps 3 and 4) as subscripts.
6. Divide all the subscripts in the formula by the smallest subscript. (Round
all subscripts that are within 0.1 of a whole number.)
C0.05556H0.06662O0.00556
Example 3.19 Determining an Empirical Formula from Combustion Analysis
Continued
SOLUTION
7. If the subscripts are not whole numbers, multiply all the subscripts by a
small whole number to get whole number subscripts.
The subscripts are whole
numbers; no additional
multiplication is needed.
The correct empirical formula
is C10H12O.
For Practice 3.19
Upon combustion, a 0.8009 g sample of a compound containing only carbon, hydrogen, and oxygen produced
1.6004 g CO2 and 0.6551 g H2O. Find the empirical formula of the compound.
Example 3.20 Balancing Chemical Equations
Write a balanced equation for the reaction between solid cobalt(III) oxide and solid carbon to produce solid
cobalt and carbon dioxide gas.
SOLUTION
1. Write an unbalanced equation by writing chemical formulas for each
of the reactants and products. Review Sections 3.5 and 3.6 for
nomenclature rules. (If an unbalanced equation is provided, go to step 2.)
2. Balance atoms that occur in more complex substances first. Always
balance atoms in compounds before atoms in pure elements.
Co2O3(s) + C(s) → Co(s) + CO2(g)
Example 3.20 Balancing Chemical Equations
Continued
SOLUTION
3. Balance atoms that occur as free elements on either side of the
equation last. Always balance free elements by adjusting their
coefficients.
Example 3.20 Balancing Chemical Equations
Continued
SOLUTION
4. If the balanced equation contains coefficient fractions, clear these by
multiplying the entire equation by the denominator of the fraction.
This step is not necessary in this
example. Proceed to step 5.
5. Check to make certain the equation is balanced by summing the total
number of each type of atom on both sides of the equation.
2 Co2O3(s) + 3 C(s) →
4 Co(s) + 3 CO2(g)
Left
Right
4 Co atoms
4 Co atoms
6 O atoms
6 O atoms
3 C atoms
3 C atoms
The equation is balanced.
For Practice 3.20
Write a balanced equation for the reaction between solid silicon dioxide and solid carbon that produces solid
silicon carbide and carbon monoxide gas.
Example 3.21 Balancing Chemical Equations
Write a balanced equation for the combustion of gaseous butane (C4H10), a fuel used in portable stoves and grills,
in which it combines with gaseous oxygen to form gaseous carbon dioxide and gaseous water.
SOLUTION
1. Write an unbalanced equation by writing chemical formulas for each
of the reactants and products. Review Sections 3.5 and 3.6 for
nomenclature rules. (If an unbalanced equation is provided, go to step 2.)
2. Balance atoms that occur in more complex substances first. Always
balance atoms in compounds before atoms in pure elements.
C4H10(g) + O2(g) → CO2(g) + H2O(g)
Example 3.21 Balancing Chemical Equations
Continued
SOLUTION
3. Balance atoms that occur as free elements on either side of the
equation last. Always balance free elements by adjusting their
coefficients.
4. If the balanced equation contains coefficient fractions, clear
these by multiplying the entire equation by the denominator
of the fraction.
Example 3.21 Balancing Chemical Equations
Continued
SOLUTION
5. Check to make certain the equation is balanced by summing the total
number of each type of atom on both sides of the equation.
2 C4H10(g) + 13 O2(g) →
8 CO2(g) + 10 H2O(g)
Left
Right
8 C atoms
8 C atoms
20 H atoms
20 H atoms
26 O atoms
26 O atoms
The equation is balanced.
For Practice 3.21
Write a balanced equation for the combustion of gaseous ethane (C2H6), a minority component of natural gas, in
which it combines with gaseous oxygen to form gaseous carbon dioxide and gaseous water.
Example 3.22 Balancing Chemical Equations Containing Ionic Compounds
with Polyatomic Ions
Write a balanced equation for the reaction between aqueous strontium chloride and aqueous lithium phosphate
to form solid strontium phosphate and aqueous lithium chloride.
SOLUTION
1. Write an unbalanced equation by writing chemical
formulas for each of the reactants and products.
Review Sections 3.5 and 3.6 for naming rules. (If an
unbalanced equation is provided, go to step 2.)
2. Balance metal ions (cations) first. If a polyatomic
cation exists on both sides of the equation, balance it
as a unit.
SrCl2(aq) + Li3PO4(aq) → Sr3(PO4)2(s) + LiCl(aq)
Example 3.22 Balancing Chemical Equations Containing Ionic Compounds
with Polyatomic Ions
Continued
SOLUTION
3. Balance nonmetal ions (anions) second. If a
Polyatomic anion exists on both sides of the
equation, balance it as a unit.
4. Check to make certain the equation is balanced
by summing the total number of each type of ion
on both sides of the equation.
3 SrCl2(aq) + 2 Li3PO4(aq) → Sr3(PO4)2(s) + 6 LiCl(aq)
Left
Right
2+
3 Sr ions
3 Sr2+ ions
6 Li+ ions
6 Li+ ions
2 PO43– ions
2 PO43– ions
6 Cl– ions
6 Cl– ions
The equation is balanced.