Chemistry 121
Mines, Spring 2011
Answer Key, Problem Set 4—Complete, with explanations
1. 2.80 [2.82 in Mas]; 2. 2.82(a&c) [2.84 in Mast]; 3. 2.84(in grams) [2.86 in Mast]; 4. 2.86(b&d) [2.88 in Mast];
5. 2.128 [2.134 in Mast]; 6. NT1; 7. 3.58(a&c) [3.64 in Mast]; 8. NT2; 9. 3.64 [3.70 in Mast]; 10. NT3; 11. NT4; 12. NT5;
13. 3.68 [3.74 in Mast]; 14. 3.72 [3.78 in Mast]; 15. 3.82(a) [3.88 in Mast]; 16. 3.86 [3.92 in Mast]; 17. 3.88 [3.94 in Mast]; 18. NT6;
19. 3.118 [3.126 in Mast]; 20. 3.120 [3.128 in Mast]; 21. 3.124 [3.132 in Mast]
-------------------------------1. 2.80.
24
How many moles of aluminum do 5.8 x 10 aluminum atoms represent?
Answer: 9.6 mol Al
24
(reasonable, because 6 x 10 atoms is more (almost 10 times more!) than one mole)
Work / Reasoning:
A mole of any “item” is, by definition, Avogadro’s number (NA) of items (which, to 4 SF, is 6.022 x 1023 items).
Thus:
5.8 x 1024 Al atoms x
2. 2.82(a&c).
1 mol Al atoms
 9.63… 9.6 mol Al
6.022 x 10 23 Al atoms
(You could also just divide
the # of atoms by NA)
What is the mass, in grams, of each elemental sample?
Strategy: Use the periodic table to determine how much mass one mole of that element’s atoms will
contain (“molar mass”). Then you can use that value to figure out how much mass is in any number of
moles of it (by multiplying). This is like calculating “cost” from “price per lb” and “lb” values.
(a) 2.3 x 10-3 mol Sb Answer: 2.3 x 10 -3 mol Sb x
(c) 43.9 mol Xe
3. 2.84.
Answer: 43.9 mol Xe x
121.76 g Sb
 0.280...  0.28 g Sb
1 mol Sb
131.29 g Xe
 5763.6...  5760 or 5.76 x 103 g Xe
1 mol Xe
21
What is the mass of 4.91 x 10 platinum atoms (in grams)? [Assume a “natural” sample from Earth]
Answer: 1.59 g Pt
21
(reasonable, because 5 x 10 atoms is less than one mole, and 2 g is also less than one mole of Pt)
Work: 4.91 x 1021 Pt atoms x
1 mol Pt atoms
195.08 g Pt
x
 1.5905...  1.59 g Pt
23
1 mol Pt
6.022 x 10 Pt atoms
21
23
Note: If you wanted the mass in amus, it would simply be: 4.91 x 10 Pt atoms x 195.08 amu/atom = 9.578…x 10 amu! That means
23
23
that 1.59 g must equal 9.58 x 10 amu. Recall, a g is 6.022 x 10 times as much mass as an amu. Do you see this better now?
4. 2.86(b&d).
Calculate the number of atoms in each sample. [Again, assume “natural” samples from Earth]
(b) 39.733 g S
Answer: 39.733 g S x
1 mol S 6.022 x 10 23 atoms
x
 7.4609...  7.461 x 10 23 atoms *
32.07 g
1 mol atoms
(d) 97.552 g Sn Answer: 97.552 g Sn x
1 mol Sn 6.0221 x 10 23 atoms
x
 4.94876...  4.9488 x 10 23 atoms **
118.71 g
1 mol atoms
*This result only has 4 SF because of the molar mass of S found in your text. It’s typically better to use more SF’s in a molar mass
than what is “needed” in a problem to maintain precision, but in this case, your text only provided 4 so I used that value.
**This result can only have 5 SF if you use NA from the back cover of your text , rounded to 5 SF (NA is not an exact quantity!). If
23
23
you used 6.022 x 10 , you technically should have gotten 4.949 x 10 (rounded to 4 SF) (which would still have been “correct”).
5. 2.128.
Without doing any calculations, determine which of the samples contains the greatest amount of the element in
moles. Which contains the greatest mass of the element?
(a) 55.0 g Cr
(b) 45.0 g Ti
(c) 60.0 g Zn
Answers: 55.0 g has the greatest amount of moles (of atoms); 60.0 g of Zn has the greatest amount of mass
PS4-1
Answer Key, Problem Set 4
Reasoning: From the Periodic Table, you can see that (left two columns):
One mole of ___
has a mass of:
Sample in Problem is
Cr
52.00 g

55.0 g > one mole
Ti
47.87 g

45.0 g < one mole
Zn
65.38 g

60.0 g < one mole
Since Cr is the only sample that has a mass greater than its molar mass, it is the only one that has
greater than one moles’ worth of atoms in it. Thus it has the “most moles”. Since the values given
are in grams (a unit of mass), clearly the 60.0 g sample has the “most mass”.
6. NT1 2.116. The U.S. Environmental Protection Agency (EPA) sets limits on healthful levels of air pollutants.
The
3
maximum level that the EPA considers safe for lead air pollution is 1.5 g/m . If your lungs were filled with air
containing this level of lead, how many lead atoms would be in your lungs? (Assume a total lung volume of 5.50 L)
(Also stop in the middle of the calculation to report/determine the mass, in grams, of all the lead atoms).
5.50 L x
1000 mL 1 cm 3 10 -2 m 10 -2 m 10 -2 m 1.5 g 10 -6 g
x
 8.25 x 10 -9 g  8.3 x 10 -9 g Pb
x
x
x
x
x
1 g
1L
1 mL
1 cm
1 cm
1 cm
1m 3
8.25 x 10 -9 g Pb x
1 mol Pb 6.022 x 10 23 atoms
x
 2.397...  2.4 x 10 13 atoms (of Pb)
207.2 g
1 mol atoms
→ Note that although the number of atoms seems enormous (24 trillion!), it is an extremely tiny
amount of mass (about 10,000 times smaller than 0.1 mg, which is the detection limit of the
extremely sensitive balances in our lab). Wow!
7. 3.58(a&c).
How many moles (of molecules or formula units) are in each sample?
(a) 55.98 g CF2Cl2
Answer: 0.4630 mol (of CF2Cl2 molecules)
(Reasonable: 121 g of CF2Cl2 is one mole,
so 56 g should be about half a mole)
MM of CF2Cl2: 1(12.01) + 2(19.00) + 2(35.45) = 120.91 g/mol
55.98 g CF2 Cl 2 x
1 mol CF2 Cl 2
 0.462989  0.4630 mol CF2 Cl 2
120.91 g
Answer: 0.001039 (or 1.039 x 10-3) mol (of C8H18 molecules)
(c) 0.1187 g C8H18
MM of C8H18: 8(12.01) + 18(1.008) = 114.22 g/mol
0.1187 g C 8H18 x
1 mol C 8H18
 0.0010392...  0.001039 mol C 8H18
114.22 g
8. NT2 3.62(a&c)* Calculate the mass (in g) of each sample. Add: (c’) calculate the (average) mass
of a water molecule in amus; and (d) Calculate the mass, in grams, of 0.00000010 mol of water;
(a) 4.5 x 1025 O3 molecules Answer: 3.6 x 103 g
4.5 x 1025 O3 molecules x
1 mol O 3 molecules
6.022 x 10
23
O 3 molecules
x
48.00 g O 3
 3.586...  3.6 x 103 g
1 mol O 3
(c) 1 water molecule (average) Answer: 2.992 x 10-23 g
MM H2O: 2(1.008) + 1(16.00) = 18.016 g/mol
1 H2O molecule x
1 mol H 2 O molecules
6.022 x 10
23
H 2 O molecules
x
18.016 g H 2 O
 2.9916...  2.992x 10-23 g
1 mol H 2 O
(c’) 1 water molecule (in amu) Answer: 18.016  18.02 amu (on average)
PS4-2
Answer Key, Problem Set 4
Note again (as in #3 above) that the masses in (c) and (c’) are for the same sample and are thus the same amount of mass. (c) is
23
a way smaller “number” because a g is such an enormously bigger amount of mass (6.022 x 10 times more) than an amu. Do
you see this better now?
(d) 0.00000010 mol water
Answer: 0.0000018 g (or 1.8 x 10-6 g or 1.8 g)
0.00000010 mol H 2 O x
9. 3.64.
18.02 g
 0.000001802...  0.0000018 g
1 mol H 2 O
A salt crystal has a mass of 0.12 mg. How many NaCl formula units does it contain?
Answer: 1.2 x 1018 FU NaCl
0.12 mg NaCl x
10 3 g 1 mol NaCl 6.022 x 10 23 FU
x
x
 1.236...  1.2 x 1018 FU (of NaCl)
1 mg
58.44 g
1 mol FUs
10. NT3.
In each case below, which sample has the greater mass? NOTE: If the two samples have essentially the same
mass (within experimental uncertainty), say "SAME". Assume that all the "1 g" or "1 mol" quantities in this problem
represent "exact" amounts (not 1 SF). The focus of this problem is NOT on uncertainty--it is about understanding and
distinguishing terms related to amounts of chemical substances and entities (mol, g, atoms, molecules, amus, etc.)
Answers in bold, with very brief reasoning (detailed reasoning follows after):
(a) 1 mol of iron or 1 mol of aluminum
Each Fe atom weighs more than each Al atom [~56 amu > ~27 amu]; OR
55.85 g (1 mol Fe) > ~26.98 g (1 mol Al)
(b) 6.022 x 1023 lead atoms or 1 mol of lead ~SAME
6.022 x 1023 “things” is 1 mol of “things”
(c) 1 mol of copper atoms or 1 copper atom
6.022 x 1023 Cu atoms >> 1 Cu atom !!! (and so it will have way more mass!)
(d) 1 mol of Cl2 or 1 mol of Cl
There are twice as many (total) Cl atoms in the sample of Cl2 (b/c each FU has 2 vs. 1); OR
70.9 g > 35.45 g
(e) 1 g of oxygen atoms or 1 g of oxygen molecules SAME
1 g = 1 g ! (“Which weighs more, a pound of feathers or a pound of bricks?” The same! But
there will be lot more feathers than bricks, because each feather weighs less.)
(f) 24.3 g of Mg or 1 mol of Mg ~SAME
The molar mass of Mg is 24.3 g (to 3 SF)
(g) 1 mol of NH3 or 1 g of NH3
17 g > 1 g
(h) 1 mol of I2 or 1 molecule of I2
6.022 x 1023 molecules >> 1 molecule !!! (a mol is not a thing; it’s a “moleion” things!)
(i) 1 oxygen molecule or 1 oxygen atom
two O atoms have more mass than just one
(j) 24.00 amus of C-12 OR 24.00 g of C-12
a gram >> an amu !!! (the “thing” here (C-12) is irrelevant b/c “gram” and “amu” are both
mass units)
PS4-3
Answer Key, Problem Set 4
More Detailed Reasoning:
IMPORTANT NOTES:
1) a mole is a general unit, like a “dozen”; it should ALWAYS be followed by some “thing” in order to give it
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some meaning. A mole of eggs means 6.022 x 10 eggs. **You must always get a picture of what the
“thing” is that you have x moles “of”; that will clarify what you actually have and so then you can confidently
answer questions like these.** For example:
2) a “molecule” of something, by definition, contains MORE THAN ONE ATOM. Therefore, a “mole of
oxygen molecules” MUST contain more total atoms than a “mole of oxygen atoms”. Since each atom of
oxygen has the same (average) mass, a mole of oxygen molecules will have a greater mass than a mole
of oxygen atoms.
3) a mole of atoms contains 6.022 x 1023 times more atoms than a single atom; therefore a mole of atoms
23
has a mass that is 6.022 x 10 times the mass of a single atom, and is therefore MUCH, MUCH, MUCH
more massive than a single atom!
(a) 1 mol of iron or 1 mol of aluminum. Since a single atom of iron has a greater mass than a
single atom of aluminum (55.847 amu vs. 26.9815 amu, on average), a mole of iron will have a
greater mass than a mole of aluminum. Why? There are the same number of items in a mole of
any type of “item”. Clearly if you have the same number of bowling balls as ping pong balls, the
sample of bowling balls will have a greater mass since each bowling ball weighs more than each
ping pong ball.
(b) 6.022 x 1023 lead atoms or 1 mol of lead. These two will have essentially the SAME MASS since
they represent essentially the same quantity of lead.
(c) 1 copper atom or 1 mol of copper atoms. See Note (3) above! A mole is a macroscopic amount
of a substance; an atom is a nanoscopic amount of a substance. THESE ARE HUGELY
DIFFERENT!!!!
(d) 1 mol of Cl or 1 mol of Cl2. See note 2 above! There are twice as many atoms of Cl in a mol of
Cl2 as in a mol of Cl because there are two atoms of Cl in each molecule of Cl2. So clearly the
sample of Cl2 will have a greater mass.
(e) 1 g of oxygen atoms or 1 g of oxygen molecules. VERY TRICKY! You must look carefully here to
see that they say 1 gram of each. This is not the same as (d) where they say 1 mol of each! If
both samples have a mass of 1 g, then by definition they have the SAME MASS!! (What this
means is that there must be half as many oxygen molecules in the molecular sample as the
number of atoms in the first sample.)
(f) 24.3 g of Mg or 1 mol of Mg. SAME MASS. This is essentially the same as in (b). One mole of
Mg should have a mass of 24.3050 g, which is essentially the same as 24.3 g.
(g) 1 mol of NH3 or 1 g of NH3. A mole of NH3 (molecules) has a mass of about 14.0 + 3(1.0) = 17.0
grams (because it contains one mole of N atoms, and 3 moles of H atoms), so clearly it is more
massive than 1 g of NH3.
(h) 1 molecule of I2 or 1 mol of I2. See answer to (c)! A mole of I2 means “a mole of I2 molecules”
which means 6.022 x 1023 I2 molecules; a macroscopic amount vs. a nanoscopic amount.
(i) 1 oxygen molecule or 1 oxygen atom. An oxygen molecule has two atoms of oxygen in it, so it
clearly will have twice the mass of 1 oxygen atom.
(j) 24.00 amus of C-12 OR 24.00 g of C-12. The mass in a gram is enormously greater than the
mass in an amu (1 g = 6.022 x 1023 amu) so 24 g is enormously greater in mass than 24 amus.
A gram is a macroscopic amount of mass and an amu is a nanoscopic amount of mass. A gram
is about the mass of half a penny, and an amu is roughly the mass of a single proton or neutron!
PS4-4
Answer Key, Problem Set 4
11. NT4 3.76(a&c) Determine the number of moles of oxygen atoms in each sample.
Add: For b&d, calculate the
number of O atoms in each sample)
NOTES: Going from “moles of AxBy” to “moles of B” is like going from “dozens of bicycles” to “dozens of wheels”. [which
you do NOT need any MASSES for, right?]. You must use the subscript of “B” to create the “interconversion
factor”.
Going from “moles of B” to “number of B atoms” is like going from “dozens of wheels” to “wheels”. Instead of
multiplying by “12” (things per dozen things), you multiply by NA (things per mol things)
(a) 4.88 mol H2O2
Answer: 4.88 mol H2O2 x
2 mol O
 9.76 mol O (atoms)
1 mol H 2 O 2
(c) 0.0237 mol H2CO3 Answer: 0.0237 mol H2CO3 x
3 mol O
 0.0711 mol O (atoms)
1 mol H 2 CO 3
Be careful not to mix up “moles of O” with “number of O atoms”! The following
two parts (as I amended them) need the “extra step” to get to number of atoms
(b’) 2.15 mol N2O
Answer: 2.15 mol N2O x
1 mol O
6.022 x 10 23 atoms
x
 1.294… 1.29 x 1024 O atoms
1 mol N 2 O
1 mol (of atoms)
(d’) 24.1 mol CO2
Answer: 24.1 mol CO2 x
12. NT5.
2 mol O 6.022 x 10 23 atoms
x
 2.902… 2.90 x 1025 O atoms
1 mol CO 2
1 mol (of atoms)
th
(From Zumdahl, 7 ed.) A given sample of a xenon fluoride compound contains molecules of the type XeFn
20
where n is some whole number. If a sample containing 9.03 x 10 molecules of XeFn weighs 0.368 g, determine the
value for n in the formula.
Answer: n = 6
Strategies: There are several strategies one could pursue here! Most chemists (including myself)
would do this at the "macroscopic" level and deal with grams and moles. However, you should realize
that a perhaps more "visualizable" means of doing this would be to do it at the molecular/nanoscopic
level. I'll show two macro approaches and one nano approach:
#1: Use molecules to get moles of compound; use moles and grams to get molar mass; recognize
that although the subscript of F is unknown, the subscript for Xe is “one”, and thus there is one mole of
Xe in one mole of XeFn; thus, assume one mole of compound and subtract away the mass of one
mole of Xe to get grams of F in one mole of compound; divide by 19.0 g/mol F to get the number of
moles of F per mole of compound (n). An abbreviated flow diagram showing this is:
molecules XeFn → moles XeFn → g/mol XeFn → g F (in a mole of XeFn) → mol F (in a mole of XeFn)
Execution of #1:
9.03 x 1020 molecules
 0.00150 mol XeFn ;
6.022 x 1023 molecules/mol
0.368 g XeFn
 245.3 g/mol
0.00150 mol XeFn
245.3 g XeFn – 131.3 g Xe = 114 g F (in one mole of XeFn);
114 g F
 6.00 mol F  n = 6
19.00 g/mol F
-------------------------------------#2: Use molecules to get moles of compound; this equals the moles of Xe in the 0.368 g sample; use
moles of Xe with 131.3 g/mol Xe to get the mass of Xe in the sample; subtract away the mass of Xe
from 0.368 g to get grams of F in the sample; divide by 19.0 g/mol F to get the number of moles of F
in the sample; divide moles of F by moles of Xe to get n. In abbreviated form:
PS4-5
Answer Key, Problem Set 4
molecules XeFn → moles XeFn → mol Xe → g Xe → g F → mol F → mol F / mol Xe ( n)
Execution of #2:
9.03 x 1020 molecules
1 mol Xe
131.3 g
x
 0.00150 mol XeFn  x
 0.197 g Xe ;
1 mol XeFn
1 mol Xe
6.022 x 1023 molecules/mol
0.368 g XeFn – 0.197 g Xe = 0.171 g F;
0.171 g F
0.00900 mol F
 0.00900 mol F ;
6  n=6
19.00 g/mol F
0.00150 mol Xe
-------------------------------------#3: Divide mass by #molecules to get the mass per molecule; convert to amu; subtract away the
(average) mass of one Xe atom (131.3 amu) to get the mass of F atoms in one molecule; divide by
the (average) atomic mass to get n. In abbreviated form:
g / molecule XeFn → amu / molecule XeFn → amu F (in a molecule of XeFn) → atoms of F (in a
molecule of XeFn)
Execution of #3
0.368 g
6.022 x 1023 amu
= 245.4 amu (per molecule)
 4.075 x 10-22 g (per molecule)  x
20
1g
9.03 x 10 molecules
245.3 amu XeFn – 131.3 amu Xe = 114 amu F (in 1 molecule);
114 amu F
 6 atoms F  n = 6
19.00 amu/atom F
13. 3.68.
Iron from the earth is in the form of iron ore. Common ores include Fe2O3 (hematite), Fe3O4 (magnetite), and
FeCO3 (siderite). Calculate the mass percent composition of iron for each of these iron ores. Which ore has the
highest iron content?
Answers: 69.94%, 72.36% and 48.20% respectively. Fe3O4 has the highest iron content.
Strategy: Pretend you had one mole of each compound. Find the mass of the sample (molar
mass) and the mass of (just) the iron in the sample. % is “partial over total” x 100, and
here “partial” means “mass of Fe” and “total” means “mass of whole sample [Fe + O]”
mass Fe
hence: %Fe 
x 100
mass sample
︵
Fe3O4: MM = 3(55.85) + 4(16.00)  231.55 g  %Fe 
FeCO3:
MM = 1(55.85) + 1(12.01) + 3(16.00)  115.86 g


3(55.85) g
x 100  72.360...  72.36% Fe
261.55 g
 %Fe 
14. 3.72.
︶
e
F
%
4
9
.
9
6
.
.
.
3
4
9
.
9
6
Fe2O3: MM = 2(55.85) + 3(16.00)  159.7 g  %Fe 
0
0
1
x
g
5 g
7
8 .
.
9
5
5 5
1
2
Execution of Strategy:
1(55.85) g
x 100  48.204...  48.20% Fe
115.86 g
-
The American Dental Association recommends that an adult femaile should consume 3.0 mg of fluoride (F ) per
day to prevent tooth decay. If the fluoride is consumed in the form of sodium fluoride (45.24% F), what amount of
sodium fluoride contains the recommended amount of fluoride?
Answer:
0.0066 g (or 6.6 mg) NaF
PS4-6
Answer Key, Problem Set 4
Reasoning: 45.24%F means
45.24 g F
. Use this like a “conversion factor” to calculate “g NaF”
100.0 g NaF
from “g F”. Most chemists would convert mg to g first and report the answer in grams, although you
45.24 mg F
to go directly to mg (since the Q does not specify “grams”!)
could also use the factor
100.0 mg NaF
Setup: 3.0 mg F x
15. 3.82(a).
1 0 -3 g 100.0 g NaF
 0.00663...  0.0066 g (or 6.6 mg)
x
45.24 g F
1 mg
Calculate the empirical formula for each natural flavor based on its elemental mass percent composition.
(a) methyl butyrate (component of apple taste and smell): C 58.80%, H 9.87%, O 31.33%
Answer: C5H10O2
masses
Strategy:
mass %'s of each element

of each element in any specific
mass of compound
[100 g is most convenient
unless molar mass is provided]

moles
of each
reduced
mole ratio

(and formula)
Execution: Assuming a 100-gram sample of the compound, you’d have 58.80 g of C (atoms), 9.87 g
of H (atoms), and 31.33 g O (atoms).
58.80 g C
12.01 g / mol C
9.87 g H
1.008 g / mol H
31.33 g O
16.00 g / mol O
16. 3.86.
 4.8959 mol C
 9.7916 mol H
 1.9581 mol O
C 4.89 59H9.7916O1.95 81
Divide all subscripts by the smallest to get:
C 4.89 59 H 9.7916 O 1.95 81  C2.50 0H5.0 01O1
1.95 81
1.95 81
1.95 81
Multiply all subscripts by 2
(because 0.5 is ½ )
C5H10O2
A 45.2-mg sample of phosphorus reacts with selenium to form 131.6 mg of the selenide. Determine the
empirical formula of phosphorus selenide.
Answer: P4Se3
Strategy:
1) (Optional: Recognize that since the mass ratios and mass % of elements are fixed for a
compound, you can use gram amounts rather than mg amounts [saves a conversion step])
2) Recognize that in order to determine EF, you need masses (or % masses) of all elements (see
prior problem’s “Strategy”), and that you can obtain mSe by: mSe  msample  mP
(NOTE: the “selenide” here is the compound formed when P and Se react)
3) Calculate moles of each element(‘s atoms), reduce the ratio, and express as a formula.
Execution:
mSe= 131.6 g of PxSey (the “selenide”) – 45.2 g P = 86.4 g Se
45.2 g P
30.97 g / mol P
 1.459 mol P
P1.4 59 Se1.0 9 4
86.4 g Se
78.96 g / mol Se
 1.09 4 mol Se
Divide all subscripts by the smaller to get:
P1.459 Se 1.09 4  P1.333Se1
1.0 9 4
1.0 9 4
PS4-7
Multiply all subscripts by 3
(because 0.333  1/3)
P4.001Se3  P4Se3
Answer Key, Problem Set 4
17. 3.88.
The molar mass and empirical formula of several compounds are listed below. Find the molecular formula of
each compound.
(a) C4H9, 114.22 g/mol
Answers: (a) C8H18
(b) C6Cl6
(c) C18H12N6
(b) CCl, 284.77 g/mol
(c) C3H2N, 312.29 g/mol
Strategy:
Recognize that the molecular formula is always a multiple of the empirical formula (because the
empirical formula is a reduced version of the molecular formula). It follows then that the molar mass
must be a multiple of the empirical (molar) mass (your text calls this the “empirical formula molar mass”)
Thus, to determine the molecular formula from the empirical formula and molar mass:
1) Calculate the molar mass of the empirical formula (think of this as the molar mass of a hypothetical substance
whose molecular formula IS the empirical formula given).
2) Divide the molar mass of the compound by the empirical (molar) mass to see how many times
larger the formula units are. Round this to a whole number (call it "n").
3) Multiply all the subscripts in the empirical formula by "n". (This makes the formula units n times bigger than
those given by the empirical formula.)
Execution:
(a) C4H9, 114.22 g/mol
Empirical (molar) mass: 4(12.01) + 9(1.008)  57.112 g/mol
114.22 g/mol (MM)
57.112 g/mol (EM)
 1.9999...  2  Formula units are 2 x “larger”  (C4H9)2 
C8H18
(b) CCl, 284.77 g/mol
Empirical (molar) mass: 1(12.01) + 1(35.45)  47.46 g/mol
284.77 g/mol (MM)
47.46 g/mol (EM)
 6.0002...  6  Formula units are 6 x “larger”  (CCl)6 
C6Cl6
(c) C3H2N, 312.29 g/mol
Empirical (molar) mass: 3(12.01) + 2(1.008) + 1(14.01)  52.056 g
312.29 g/mol (MM)
52.056 g/mol (EM)
 5.9999...  6  Formula units are 6 x “larger”  (C3H2N)6 
C18H12N6
18. NT6 3.92.
Tartaric acid is the white, powdery substance that coats tart candies such as Sour Patch Kids.
Combustion analysis of a 12.01-g sample of tartaric acid—which contains only carbon, hydrogen, and oxygen—
produced 14.08 g CO2 and 4.32 g H2O. Determine the empirical formula for tartaric acid.
Answer: C2H3O3
Strategy: As in Q’s 15 and 16, we need ratios of “moles of atoms” of each element (here, it’s C, H,
and O atoms. In a combustion analysis, all the carbon ends up in the CO2, so the number of moles of
C atoms simply equals the number of moles of CO2 (because there’s one mole of C atoms per mole of
CO2). All of the H ends up in the H2O, so the number of moles of H atoms in the sample is simply
equal to the number of moles of H atoms in the H2O. This will be TWO times the number of moles of
H2O (because there are two moles of H atoms per mole of H2O. What about the oxygen? As in
Experiment 4 (CuxCly·z H2O), you must realize that once you know the mass of the C and H in the
sample (which you can calculate from the moles of atoms of each, by multiplying by the molar masses
of C and H, respectively), you can figure out the mass of O in it by difference (i.e., all the “remaining”
mass that isn’t C or H must be O!). Once you have that mass, calculate moles of O by dividing by its
molar mass. Then you can reduce the mole ratio of atoms and express it as a formula.
PS4-8
Answer Key, Problem Set 4
Execution:
From CO2:
From H2O:
14.08 g CO 2
44.01 g / mol CO 2
4.32 g H 2 O
18.02 g / mol H 2 O
 0.31992 mol CO 2  x
 0.2397 mol H2 O  x
1 mol C
1 mol CO 2
2 mol H
1 mol H 2 O
 0.31992 mol C atoms
 0.47946 mol H atoms
The masses of the C atoms and the H atoms are thus:
12.01 g C
1.008 g H

 

0.31992 mol C x 1 mol C  3.8422 g C  0.47946 mol H x 1 mol H  0.48329 g H  4.325 49 g C & H

 

So, mass of O atoms in the sample = 12.01 g (total) – 4.32549 g (C + H) = 7.684... g O
which means there are
7.68 4 g O
16.00 g / mol O
 0.4802 mol O atoms
So the empirical formula must be:
C 0.319 9 2H0.47 95 O 0.48 0 2  C 0.319 9 2 H 0.47 95 O 0.48 0 2  C1H1.4 98 O1.5 01
0.319 9 2
19. 3.118.
0.319 9 2
0.319 9 2
Multiply all subscripts by 2
C2H3O3
(because 0.498  ½ )
A metal (M) forms an oxide with the formula M2O. If the oxide contains 16.99% O by mass, what is the identity
of the metal?
Answer: M is K (potassium)
Strategy(ies) and Solutions:
In this problem, you ultimately have to realize that the only way to determine M’s identity is to
figure out the molar mass of M. Then you can use the Periodic Table to match the molar mass to
a single element. There are many correct approaches to determining the molar mass from the
info given. Most will involve assuming a 100-g (exact) sample so that you “have” 16.99 g of O
and (100-16.99) = 83.01 g of M. Let’s assume that. Also, you must realize that you can calculate
molar mass of a substance “experimentally” by knowing the grams and moles of a single sample
of the substance: MM =
# of grams in sample
# of moles (of FUs) in sample
I’ll now give two approaches for moving forward:
(A) “My” approach:
1) Calculate the moles of O in the sample (by dividing g O by MM of O).
2) Use the subscript ratio in the formula to calculate moles of M in the sample.
3) Divide mass of M by moles of M to get MM.
Execution of (A):
16.99 g O
2 mol M
x
 2.12375 mol M (atoms) ;
16.00 g / mol O 1 mol O
MM(of M) =
83.01 g M
2.12375 mol M
 39.09 g/mol
The element on the Periodic Table with MM closest to 39.09 g/mol is K (MM = 39.10 g/mol)
(B) A former student’s approach that I found very clever:
1) Let MM = the MM of M
2) Because of the subscript ratio, the mass ratio of M:O must be:
PS4-9
2 x MM
gM

g O 1(16.00)
Answer Key, Problem Set 4
3) Set the mass ratio in (2) equal to 83.01 g / 16.99 g (mass values from problem), and solve
for MM!
Execution of (B):
83.01 g M
16.99 g O
20. 3.120.

2 x MM
1(16.00)
 MM 
83.01 x 16.00
2 x 16.99
 39.09 g/mol  M is K (potassium)
Fructose is a common sugar found in fruit. Elemental analysis of fructose gives the following mass percent
composition: C 40.00%, H 6.72%, O 53.28%. The molar mass of fructose is 180.16 g/mol. Find the molecular
formula of fructose.
Answer: C6H12O6
Strategy:
Although you could solve this problem by assuming 100 g of sample and finding the empirical
formula (as in #14 above), and then dividing MM by EM (as in #16 above), the more
straightforward approach here is to just
1) assume a sample of 180.16 grams, since that must be one mole of the compound.
2) Use the %’s to calculate masses of each element in the sample (as “usual”, although here you
actually have to do the multiplication since the answer is not equal to the %!).
3) Calculate moles of each element in the sample (as “usual”).
4) Recognize the following: Since there must be a whole number of moles of each atom-type in
a mole of a substance (equal to the subscript in the formula), when you calculate the number
of moles of each element from the masses now (i.e., #3 above), you will get whole numbers
that are the subscripts in the molecular formula (i.e., you don’t ever calculate an empirical
formula here—you go “directly” to the molecular formula)!
Execution:
Mass of C (in a one-mole [i.e., 180.16 g] sample) = 180.16 g x 0.4000 [i.e., 40.00%] = 72.064 g C
Mass of H (in a one-mole [i.e., 180.16 g] sample) = 180.16 g x 0.0672 [i.e., 6.72%] = 12.106 g H
Mass of O (in a one-mole [i.e., 180.16 g] sample) = 180.16 g x 0.5328 [i.e., 53.28%] = 95.989 g O
72.06 4 g C
12.01 g / mol C
12.106 g H
1.008 g / mol H
95.989 g O
16.00 g / mol O
 6.0003 mol C
 12.0099 mol H
C6H12O6 (is the molecular formula)
 5.9993 mol O
NOTE: If you had gone the “traditional” route of assuming 100 g of sample, you would have ended up with ~3.33 moles of
C, 6.666 mol H, and 3.33 mol O, which would have given an empirical formula of CH2O. The EM of this is 30.03
g/mol. 180.16 / 30.03 = 6.00, so the molecular formula would be C6H12O6 (same as result shown above). Either
way is fine!
21. 3.124.
A hydrate of copper(II) chloride has the following formula: CuCl2 • xH2O. The water in a 3.41-g sample of the
hydrate was driven off by heating. The remaining sample had a mass of 2.69 g. Find the number of waters of
hydration (x) in the hydrate.
Answer: x = 2; formula of hydrate is CuCl2• 2H2O
NOTE: Please try to recognize the similarity of this problem to what you did in Experiments 4 and 6! (Although the way I
made you work up Experiment 6, you found the ratio of the actual number of FU’s of each in the sample rather than
the moles (since we had not learned about moles yet and I wanted to stress that you could do the problem without
“moles”)
PS4-10
Answer Key, Problem Set 4
Strategy:
1) Recognize that x here represents the ratio of “waters” : “FU’s of CuCl2”, and thus it is also the
ratio of moles of H2O (molecules) : moles of CuCl2 (FU’s).
2) This means you need to figure out the two mole values in #1 to calculate x.
3) Recognize that 2.69 g represents the mass of the “anhydrous compound”, which is CuCl2!
Thus, you can calculate the molar mass of CuCl2 [= 1(63.55) + 2(35.45) = 134.45 g/mol) and
use it with the mass of CuCl2 (2.69 g) to calculate “moles of CuCl2”. [You are half done with
the problem! Now you need to get “moles of H2O”]
4) Recognize that mwater = mhydrate – manhydrous. Once you have this mass, use it, along with the
molar mass of H2O (= 2(1.008) + 16.00 = 18.02 g/mol) to calculate “moles of H2O”.
5) Divide the moles of H2O by moles of CuCl2 (and round to nearest whole number) to get x.
Execution:
2.69 g CuCl 2
134.45 g / mol CuCl 2
 0.0200 mol CuCl 2 (in sample of hydrate)
Mass of water in original hydrate = 3.41 g – 2.69 g = 0.72 g
0.72 g H2 O
18.02 g / mol H2 O
x=
 0.03996 mol H2 O (in sample of hydrate)
0.03996 mol H2 O
 1.9977  2  There are 2 molecules of H2O per FU of hydrate compound
0.0200 mol CuCl 2
PS4-11