Assessment and Evaluation
ThoughtLab/
ExpressLab/
Investigation
ThoughtLab:
Percent by Mass
and Percent by
Number, page 202
Curriculum
Expectations
Assessment
Tools/Techniques
Developing Skills of Inquiry and Communication
■ [QCR 2.02] determine percentage composition
of a compound through experimentation, as
well as through analysis of the formula and
a table of relative atomic masses
■
Assessment
Checklist 5:
Learning Skills
(see
“Assessment
and Evaluation”
in the front
matter of
Teacher’s
Resource
CD-ROM)
Achievement
Chart Category
■
Knowledge/
Understanding
Learning Skills
■
Works
Independently
Solutions for Practice Problems
Student Textbook page 204
Student Textbook page 204
4 KNO3(s) →
2K2O(s) + 2 N2(g) + 5 O2(g)
See Solutions Manual for solutions to Practice Problems.
This is a decomposition
reaction.
Section Review Answers
Student Textbook pages 205, 206
1. An acetylene molecule contains equal numbers of carbon and hydrogen atoms, but
the mass percent of carbon in acetylene is not 50% because carbon atoms are 12
times as massive as hydrogen atoms.
2. The relative masses are the same, whether expressed as molar masses (g) or entity
masses (u).
3. C16H10N2O2:
M = 16(12.01 g/mol) + 10(1.01 g/mol) + 2(14.01 g/mol) + 2(16.00 g/mol)
= 192.16 g/mol + 10.10 g/mol + 28.02 g/mol + 32.00 g/mol
= 262.28
g/mol 32.00 g/mol
262.28 g/mol
(100)% = 12.2%
(25.0) g = 3.05 g of oxygen.
25.0 g of indigo contains 12.2
100
%O=
M = 39.10
g/mol + 35.45 g/mol + 4(16.00 g/mol) = 138.55 g/mol
64.00 g/mol
% O = 138.55 g/mol (100)% = 46.19%
(24.5) g = 11.3 g of oxygen.
24.5 g of KClO4 contain 46.19
100
M = 2(107.87 g/mol) + 16.00 g/mol
= 215.74 g/mol + 16.00 g/mol
= 231.74
g/mol 215.74 g/mol
% Ag = 231.74 g/mol (100)% = 93.10%
(18.4) g = 17.1 g.
Mass of silver remaining is 93.1
100
6. The claim is 137 mg Na per 500 mg NaHCO3.
Both elements and compounds
have definite proportions. An
element is 100% one kind of
atom. A compound has a
fixed percentage composition,
with the percentages of the
constituent elements adding
up to 100%.
Figure 6.5
4. KClO4:
5. Ag2O:
Student Textbook page 205
Student Textbook page 205
To distinguish between vanillin,
C8H8O3 , and glucose, C6H12O6 ,
chemists would have to calculate
the mass percents of C, H, and O
for each compound. These values
must then be compared to
experimental values of percentage
composition for these compounds.
One method of experimentally
determining the percentage
composition of a compound
containing C, H, and O is to use a
carbon-hydrogen combustion
analyzer (this is explained in
greater detail in section 6.4).
Chapter 6 Chemical Proportions in Compounds • MHR
233
UNIT 2 Chapter 6 CHEMICAL PROPORTIONS IN COMPOUNDS
137 mg
This means a mass percent n of 500 mg (100)% = 27.4%
NaHCO3:
M = 22.99
g/mol + 1.01 g/mol + 12.01 g/mol + 3(16.00 g/mol) = 84.01 g/mol
22.99 g/mol
% Na = 84.01 g/mol (100)% = 27.4%
Thus, the claim is valid.
7. Add the known percents and subtract from 100% to find the percent of the
unknown element. 100% − 70.5% − 11.5% − 10.4% = 7.6%
Consider 100 g of the compound with the following percentage composition:
C (70.5%), H (11.5%), O (10.4%), and x (7.6%).
Find the number of moles of the known elements.
70.5 g
C: n = 12.01 g/mol = 5.870 mol
H: n =
11.5 g
= 11.386 mol
1.01 g/mol
10.4 g
= 0.650 mol
16.00 g/mol
O: n =
Divide each number of moles by the smallest value.
= 9.03
C: 5.870
0.650
H:
11.386
= 17.5
0.650
0.650
= 1.00
0.650
O:
Double all these to get integer values: 18; 35; 2. Since we are told there is one x,
the formula is C18H35O2x.
The total mass without x is 283.53 g, and, adding the mass percent of carbon, hydrogen, and oxygen, this is92.4% of the mass of the compound. Thus, the total mass of
100
(283.53) g = 306.85 g.
the compound is 92.4
Mass of x = 306.86 g − 283.53 g = 23.32 g. The molar mass of sodium is 23 g/mol.
Therefore, the alkali metal cation is a sodium ion.
8. These steps would be required:
(a) Measure the initial mass of sucrose.
(b) Carry out the reaction.
(c) Clean and dry the resulting carbon.
(d) Find the mass of the carbon.
(e) Do the mass percent calculation.
The problems lie in purifying the carbon, removing excess acid and water, avoiding
loss of carbon during purification, and ensuring a complete reaction.
Student Textbook page 207
In chemistry, the word
“formula” implies a ratio
between atoms within a
compound. As the student
textbook states, the word
“empirical” comes from the
Greek word empeirikos,
meaning “by experiment.”
Since the simplest ratio of
atoms within a compound is
determined through
experimentation, it makes
sense that the simplest
formula of a compound be
called its empirical formula.
234
6.2 The Empirical Formula of a Compound
Student Textbook pages 207–214
Section 6.2 introduces the empirical and molecular formulas for compounds and then
focusses on empirical formulas. It shows how percentage composition data can be used to
find the empirical formula of a compound. It concludes with an investigation in which
students find the percentage composition of a compound and use their data to obtain the
empirical formula.
MHR • Unit 2 Chemical Quantities
UNIT 2 Chapter 6 CHEMICAL PROPORTIONS IN COMPOUNDS
(100)%
product was 0.90 g. The student might predict the percent Mg = 0.60
0.90
= 66.6%. However, in reality, the product contains less than 0.60 g of
magnesium, because some was lost. The actual percent Mg is lower than the
calculated value.
(b) If some magnesium is unreacted, the student will calculate a percent of magnesium
that is too high. Suppose that the mass of the original magnesium were 0.70 g and
0.10 g remained unreacted. The 0.60 g that reacted would give 1.0 g (approximately)
of product. The mass of solid afterward
would be (0.10 + 1.0) g or 1.1 g, and the
0.70 g
calculated percent Mg would be 1.1 g (100)% = 63.6%.
(c) If the student’s data produced a percent Mg that was too high, then either scenario
(a) or (b) would be possible.
Assessment and Evaluation
ThoughtLab/
ExpressLab/
Investigation
Curriculum
Expectations
Assessment
Tools/Techniques
Investigation 6-A:
Determining the
Empirical Formula of
Magnesium Oxide,
pages 212–213
Developing Skills of Inquiry and Communication
■ [QCR 2.02] determine percentage composition of
a compound through experimentation, as well
as through analysis of the formula and a table
of relative atomic masses
■
Achievement
Chart Category
Rubric for
Investigation 6-A
(see
“Assessment
and Evaluation”
in the front
matter of
Teacher’s
Resource
CD-ROM)
■
■
Learning Skills
Inquiry
Communication
■
■
Teamwork
Organization
Section Review Answers
Student Textbook page 214
1. (a) This formula provides the simplest ratio of the atoms of the elements found in
the compound.
(b) The subscripts of a molecular formula are whole number multiples of the corre-
sponding subscripts of the empirical formula.
2. Assume a 100 g sample of compound and use n =
m
M
.
Element
m (g)
M (g/mol)
n (mol)
Ratio to
Smallest n
Revised
Ratio
C
63.1
12.01
5.254
2.66
7.98
1.01
5.257
2.66
7.98
16.00
1.975
1.000
3.00
H
5.31
O
31.6
The empirical formula is C8H8O3.
3. (a)
Element
MHR • Unit 2 Chemical Quantities
Revised Ratio
Cl
0.315
0.315
= 1.00
2.00
O
1.1
0.315
= 3.49
7.00
The empirical formula is Cl2O7.
238
Ratio to Smaller n
(b)
m (g)
Element
Si
n (mol)
4.90
Cl
24.8
Ratio to Smaller n
4.90
28.09
= 0.174
1.00
24.8
35.45
= 0.699
4.00
The empirical formula is SiCl4.
4. Consider 100 g of the compound:
m (g)
Element
C
n (mol)
40.0
H
40.0
12.01
= 3.330
1.00
6.71
1.01
= 6.643
1.99
53.3
16.00
= 3.331
1.00
6.71
0
53.3
Ratio to Smallest n
The empirical formula is CH2O.
5. Students should point out that the empirical formula can only show that the substance could be the alleged substance. The molecular formula is needed to confirm the
identity of the substance.
6.
n (mol)
Element
Ratio to Smallest n
C
76.54
12.01
= 6.373
9.00
H
12.13
1.01
= 12.009
16.96
0
11.33
16.00
= 0.708
1.00
The empirical formula is C9H17O.
7.
n (mol)
Element
Ratio to Smallest n
Revised Ratio
C
74.13
12.01
= 6.172
5.50
11
H
7.92
1.01
= 7.841
6.99
14
0
17.95
16.00
= 1.121
1.00
2
The empirical formula is C11H14O2.
8. (a) Consider a 100 g sample:
n (mol)
Element
Ratio to Smallest n
C
64.56
12.01
= 5.375
10.00
H
5.42
1.01
= 5.366
9.98
Fe
30.02
55.85
= 0.5375
1.00
The empirical formula is C10H10Fe.
(b) The molecular formula is the same. We know this because the question says there
is one iron atom between two hydrocarbon rings.
6.3 The Molecular Formula of a Compound
Student Textbook pages 215–218
This section of the student textbook provides the necessary information to allow students
to derive the molecular formula of a compound, given the empirical formula and some
Chapter 6 Chemical Proportions in Compounds • MHR
239
Solutions for Practice Problems
Unit Investigation Prep
Student Textbook page 218
Student Textbook page 218
See Solutions Manual for solutions to Practice Problems.
There are two possible ways
to answer this question:
Section Review Answers
Student Textbook page 218
Use Mm and M e as defined above.
1. The mass spectrometer allows determination of the molar mass of the molecular
formula. This can be divided by the molar mass of the empirical formula to get the
factor needed to “multiply up” the empirical formula to yield the molecular formula.
2. Given that there are (3.61)(1024) oxygen atoms in a mole of tartaric acid, and given
the empirical formula C2H3O3, 1 mole of the empirical formula contains 3 moles
of O atoms or 3(6.02)(1023) = (1.81)(1024) O atoms.
24
)
= 21 . The molecular formula is C4H6O6.
The ratio we want to find is (3.61)(10
(1.81)(1024)
3. Several compounds with very different properties can have the same empirical formula.
The molecular formula allows the identification of a specific compound, or in the
case of an organic compound, a set of isomeric compounds.
4. (a) To get the empirical formula, divide all subscripts in the molecular formula by the
common factor, in this case 2, to get C2H3O.
(b) The molecular formula is twice the empirical formula, so its molar mass is also
twice the molar mass of the empirical formula.
5. M e (C6x H5x Ox ) = 93.11 g/mol
Mm = 186 g/mol
186 g/mol
n = 93.11 g/mol
n = 2.00
Therefore, x is equal to 2, and the molecular formula of the compound is C12H10O2.
6.4
Finding Empirical and Molecular
Formulas by Experiment
Method 1:
Molar mass of
CuCl2 = 134.45 g/mol
percentage composition of
CuCl2 → Cu = 47.3%,
Cl = 52.7%
If 5.00 g of Cu accounts for
47.3% of the mass in the CuCl2
sample, then what is the mass
of 52.7% of the sample?
(5.00 g × 52.7%)
mass =
47.3%
= 5.57 g
Total Mass = 5.00 g + 5.57 g
= 10.6 g of CuCl2
Method 2:
Mass of Cu in sample of
CuCl2 = 5.00 g
Number mol of Cu in
5.00 g = 0.079 mol
If there is 0.079 mol of Cu in
the CuCl2 sample, then there
must
be 2 × 0.079 mol = 0.158 mol
of Cl in the sample.
Mass of Cl in the sample
of CuCl2
= 0.158 mol × 35.45 g/mol
= 5.60 g of Cl
Total mass of sample
= 5.00 g + 5.60 g = 10.6 g
Both methods yield the same
answer to three significant
digits.
Student Textbook pages 219–228
The focus of this section is on obtaining experimental data that allows determination of
empirical formulas. The student textbook describes the carbon-hydrogen combustion
analyzer and how to process data obtained from it. The students textbook also shows how
molar mass data from mass spectrometer can be combined with data from a
carbon-hydrogen combustion analyzer to determine molecular formulas. The section ends
with an investigation into the amount of water trapped within a hydrate.
The Carbon-Hydrogen Combustion Analyzer
Student Textbook page 219
Chemistry Background
Two sets of information must be obtained by experimentation in order to determine the
molecular formula of an unknown compound. The molar mass can be obtained by mass
Chapter 6 Chemical Proportions in Compounds • MHR
241
UNIT 2 Chapter 6 CHEMICAL PROPORTIONS IN COMPOUNDS
Section Review Answers
Student Textbook page 228
1. If the nitrogen does not form any product with the carbon or hydrogen (e.g., NH3),
2.
3.
4.
5.
6.
but just oxides of nitrogen that escape from the analyzer, then the nitrogen content
can be found by applying conservation of mass.
Note: This is similar to question 23 in the Practice Problems on page 225 of the
student textbook.
Molar mass of MgSO4 • 7H2O = 246.52 g/mol
Mass of 7H2O = 126.14 g/mol
126.14 g/mol
% water in MgSO4 • 7H2O = 246.52 g/mol (100)% = 51.17%.
In 1000 g (1.0 kg) of MgSO4 • 7H2O, there are 0.5117 × 1000 g = 511.7 g of water.
The remaining mass comes from the MgSO4. Therefore, the bag of anhydrous
magnesium sulfate would have a mass of 1000 g − 511.7 g = 488.3 g.
No, we cannot find the amounts of oxygen and chlorine. Assuming that the chlorine
does not react with either “trap,” then the masses of carbon and hydrogen can be
found. Using conservation of mass, total mass of the oxygen and chlorine can be
found but not the individual masses.
There should be two branches to the flowchart. One should begin with the question:
“What elements are present in this compound?” (This is needed to decide how to
obtain the percentage composition.) To answer this question, techniques of qualitative
analysis such as flame tests and precipitation formation must be used. Once this is
answered, the second question in this branch is: “What is the percentage composition?”
(This is needed to determine the empirical formula.) To answer this question, a
quantitative analysis technique such as the combustion analyzer must be used. The
technique chosen will depend on the elements present in the compound. The third
question in this branch is: “What is the empirical formula?” (This is needed to obtain
M e, the molar mass of the empirical formula.)
The second branch of the flowchart is much shorter. The only question is: “What
m
is Mm, the molar mass of the compound?” (This is needed so the ratio M
may be
Me
obtained.) To answer this question, mass spectrometry must be used. Connect the
m
two branches by computing the ratio M
and multiplying the subscripts of the
Me
empirical formula accordingly to obtain the molecular formula.
It suggests that the sodium hydroxide would absorb water as well as carbon dioxide,
making it impossible to find the masses of carbon and hydrogen in the original sample.
Consider 100 g of the hydrate. There are 63.67g of Zn(NO3)2 and 36.33 g of H2O.
For the zinc nitrate
m = 63.67 g
m = 36.33 g
M = 189.41 g/mol
M = 18.02 g/mol
n=
x =
2.016
0.336
For the water
m
M
= 0.336 mol
n=
m
M
= 2.016 mol
= 6, so the hydrate is Zn(NO3)2 • 6H2O.
7. (a) Consider the carbon dioxide:
m
= 0.0841 mol
m = 3.703 g and M = 44.01 g/mol so n = M
Thus, there were 0.0841 mol of C in the original compound.
That is, m = nM = (0.0841)(12.01) g = 1.0105 g of C.
Consider the water:
m
m = 1.514 g and M = 18.02 g/mol so n = M
= 0.0840 mol
246
MHR • Unit 2 Chemical Quantities
Thus, there were 2(0.0840) mol = 0.1680 mol of H in the original compound.
That is, m = nM = (0.1680)(1.01) g = 0.1697 g of H.
The mass of O is found using conservation of mass.
Mass (g)
n (mol)
ratio to smallest n
compound
2.524
carbon
1.0105
0.0841
1
hydrogen
0.1697
0.1680
2
oxygen
1.3438
1.3438
16.00
= 0.0839
1
Thus, the empirical formula is CH2O.
(b) Since it contains 12 hydrogen atoms, the molecular formula must be six times the
empirical formula or C6H12O6.
Chapter 6 Review Answers
Student Textbook pages 229 –231
Answers to Knowledge/Understanding Questions
1. It is for convenience. Using a one-mole sample allows use of the molar mass values
from the periodic table.
2. It is exactly the same according to the law of definite proportions. Each molecule of
water has two hydrogen atoms and one oxygen atom.
3. (a) The required measurements are: the initial mass of the compound, the water trap,
and the carbon dioxide trap; and the final mass of the water trap and the carbon
dioxide trap.
(b) Having the same empirical formula means they have the same percentage composition. Thus, the combustion analysis results would be the same.
4. No, there is insufficient information. The relative proportions of the constituent
elements must be known—that is, either percentage composition or empirical formula.
Answers to Inquiry Questions
5. Find the molar mass of the hydrate.
M = 2(22.99 g/mol) + 4(10.81 g/mol) + 7(16.00 g/mol)
+ 10[2(1.01 g/mol) + 16.00 g/mol]
= 45.98 g/mol + 43.24 g/mol + 112.00 g/mol + 180.20 g/mol
= 201.22 g/mol + 180.20 g/mol
= 381.42 g/mol
201.22 g/mol
The percent of anhydrous compound is 381.42 g/mol (100)% = 52.8%
The mass
of anhydrous compound from 5.00 g of hydrate would be
52.8
(5.00) g = 2.64 g.
100
6. (a) Find the molar mass for CCl2F2.
M = 12.01 g/mol + 2(35.45 g/mol) + 2(19.00 g/mol)
= 12.01 g/mol + 70.90 g/mol + 38.00 g/mol
= 120.91
g/mol 12.01 g/mol
(100)% = 9.93%
120.91 g/mol 70.90 g/mol
% Cl = 120.91 g/mol (100)% = 58.6%
38.00 g/mol
% F = 120.91 g/mol (100)% = 31.4%
%C=
Chapter 6 Chemical Proportions in Compounds • MHR
247
UNIT 2 Chapter 6 CHEMICAL PROPORTIONS IN COMPOUNDS
Freon-12 is 9.93% C; 58.6% Cl; 31.4% F
(b) Find the molar mass for white lead, Pb3(OH)2(CO3)2.
M = 3(207.20 g/mol) + 2(16.00 g/mol + 1.01 g/mol)
+ 2[12.01 g/mol + 3(16.00 g/mol)]
= 775.64 g/mol
List the masses of the constituent elements for 1 mol of the compound.
Pb: 621.60
g; H: 2.02g; C: 24.02 g
g; O: 128.00
621.60 g/mol
% Pb = 775.64 g/mol (100)% = 80.1%
128.00 g/mol
% O = 775.64 g/mol (100)% = 16.5%
2.02 g/mol
% H = 775.64 g/mol (100)% = 0.260%
24.02 g/mol
% C = 775.64 g/mol (100)% = 3.10%
White lead is 80.1% Pb; 16.5% O; 0.260% H; 3.10% C.
7. (a) Find the molar mass of MgCl2 • 2H2O.
M = 24.31 g/mol + 2(35.45 g/mol)
+ 2[2(1.01 g/mol) + 16.00 g/mol]
= 95.21 g/mol + 36.04 g/mol
= 131.25
g/mol
36.04 g/mol
% H2O = 131.25 g/mol (100)% = 27.46%
(25.0) g = 6.86 g
The mass of water in 25.0 g of hydrate is 27.46
100
(b) Find the molar mass of KMnO4.
M = 39.10 g/mol + 54.94 g/mol + 4(16.00 g/mol)
= 158.04
g/mol 54.94 g/mol
% Mn = 158.04 g/mol (100)% = 34.8%
(5.00) g = 1.74 g
Mass of Mn in 5.00 g of KMnO4 is 34.8
100
8. (a) Find the molar mass of AgNO3.
M = 107.87
g/mol +14.01 g/mol + 3(16.00 g/mol) = 169.88 g/mol
107.87 g/mol
% Ag = 169.88 g/mol (100)% = 63.5% 63.5
(2.00)(102) kg or
(b) Mass of Ag in (2.00)(102) kg of AgNO3 is
100
2
(1.27)(10 ) kg
9. Find the molar mass of BaSO4.
M = 137.33
g/mol +32.07 g/mol + 4(16.00 g/mol) = 233.40 g/mol
137.33 g/mol
% Ba = 233.40 g/mol (100)% = 58.8%
(45.8) g = 26.9 g.
Mass of Ba in 45.8g of BaSO4 is 58.8
100
10. Find the molar mass of Bi(NO3)2.
M = 208.98
g/mol +2[14.01 g/mol + 3(16.00 g/mol)] = 333.00 g
208.98 g/mol
% Bi = 333.00 g/mol (100)% = 62.8%
(268) g = 168 g.
Mass of Bi in 268 g of Bi(NO3)2 is 62.8
100
11. Let M e and Mm be the molar masses of the empirical and molecular formulas of the
compound. Given Mm ≅ 121 g/mol:
from the empirical formula CH2O,
M e = 12.01 g/mol + 2(1.01 g/mol) + 16.00 g/mol = 30.03 g/mol
Mm
121
= 30.03
= 41
Me
The molecular formula is C4H8O4.
12. Given Mm = 322 g/mol, empirical formula C6H2OCl2:
248
MHR • Unit 2 Chemical Quantities
Me = 6(12.01 g/mol) + 2(1.01 g/mol) + 16.00 g/mol + 2(35.45 g/mol)
= 160.98 g/mol
Ratio Mm : M e = 160.98 : 322 = 2 : 1
The molecular formula is C12H4O2Cl4.
13. (a) The subscripts represent numbers of atoms, so they must be whole numbers.
(b) Since 2.67 is the decimal equivalent of 8:3, multiply the subscripts by 3 to
obtain C3H8.
14. Consider a 100 g sample.
Element
n=
C
80.2
12.01
O
H
m
M
(mol)
Ratio to Smallest n
Revised Ratio
= 6.677
10.498
21
10.18
16.00
= 0.636
1.00
2
9.62
1.01
= 9.524
14.975
30
Ratio to Smallest n
Revised Ratio
The empirical formula is C21O2H30.
15. Consider a 100 g sample.
m
M
Element
n=
Na
17.6
22.99
= 0.766
1.00
2
Cr
39.7
52.00
= 0.763
1.00
2
O
42.8
16.00
= 2.675
3.50
7
(mol)
The empirical formula is Na2Cr2O7.
16. Consider a 100 g sample.
m
M
Element
n=
Hg
67.6
200.59
= 0.337
1.00
S
10.8
32.07
= 0.337
1.00
O
21.6
16.00
= 1.35
4.00
(mol)
Ratio to Smallest n
The empirical formula is HgSO4.
17. (a) Consider a 100 g sample.
Element
n=
Ca
38.8
40.08
P
O
m
M
(mol)
Ratio to Smallest n
Revised Ratio
= 0.968
1.50
3
20.0
30.97
= 0.646
1.00
2
41.2
16.00
= 2.575
3.99
8
The empirical formula is Ca3P2O8.
(b) Since each formula unit contains 2 PO43− ions, the molecular formula
is Ca3(PO4)2.
Chapter 6 Chemical Proportions in Compounds • MHR
249
UNIT 2 Chapter 6 CHEMICAL PROPORTIONS IN COMPOUNDS
18. (a) Consider a 100 g sample.
m
M
Element
n=
C
71.0
12.01
= 5.91
18.02
H
8.60
1.01
= 8.51
25.95
O
15.8
16.00
= 0.988
3.01
N
4.60
14.01
= 0.328
1.00
(mol)
Ratio to Smallest n
The empirical formula is C18H26O3N.
(b) Since each molecule contains just one nitrogen atom, the molecular formula is the
same as the empirical formula.
19. Let the molar mass of x be x g.
Then, for the compound, the molar mass is
m = 2x + 5(16.00
g/mol) = 2x + 80.00 g/mol.
80.00
+ 80.00 (100)% = 44.0% (given)
So, % O =
2x
8000
(2x + 80)
= 44 and then 8000 = 44(2x + 80) or
8000 = 88x + 3520 which gives 88x = 4480 or x = 50.9
The molar mass of x is 50.9 g/mol. From the periodic table, x is vanadium.
20. First, find the mass of H in the HCl produced.
HCl
x
1.01
H
Mass in sample
4.730 g
xg
Mass in one mole
36.46 g
1.01 g
4.730
36.46
=
or x = 0.1310 g
Now, find mass of C in the CCl4 produced.
CCl4
Mass in sample
Mass in one mole
y
12.01
C
9.977 g
yg
153.81 g
12.01 g
9.977
153.81
=
or y = 0.7790 g
The original compound contains 1.254 g − (0.1310 g + 0.7790 g) = 0.3440 g of O
(using conservation of mass). Thus, the masses are:
Compound
Carbon
Hydrogen
Oxygen
1.254 g
0.7790 g
0.1310 g
0.3440 g
Element
n=
C
m
M
(mol)
Ratio to Smallest n
0.7790
12.01
= 0.06486
3
H
0.1310
1.01
= 0.1297
6
O
0.3440
16.00
= 0.0215
1
The empirical formula is C3H6O.
21. Mass of FeSO4 • xH2O = 2.78 g
Mass of FeSO4 = 1.52 g
Mass of H2O in sample = 2.78 g − 1.52 g = 1.26 g
1.52 g
Moles of FeSO4 = 151.92 g/mol
= 0.01 mol FeSO4
250
MHR • Unit 2 Chemical Quantities
Moles of H2O =
1.26 g
18.02 g/mol
= 0.07 mol H2O
Divide moles of iron sulfate and water by 0.01 mol, this gives
the ratio 1 mol FeSO4 : 7 mol H2O
Therefore, x = 7 and the formula for the compound is FeSO4 • 7H2O.
22. (a) First, find the mass of C in the CO2 produced.
CO2
Mass in sample
xg
0.6871 g
Mass in one mole
x
12.01 g
C
44.01
g
12.01 g
0.6871 g
44.01 g
=
or x = 0.1875 g
Now find the mass of H in the H2O produced.
H2O
Mass in sample
yg
0.1875 g
Mass in one mole
y
2.02
H
18.02
g
2.02 g
0.1875
18.02
=
or y = 0.0210 g
In the original compound using conservation of mass:
Mass (g)
Mass (%)
compound
0.5000
100
C
0.1875
0.1875
0.500
(100) = 37.50
H
0.0210
0.0210
0.500
(100) = 4.200
O
0.2915
0.2915
0.500
(100) = 58.30
Citric acid is 37.50% C; 4.20% H; 58.30% O.
(b) Consider 100 g of citric acid.
Element
n=
C
37.50
12.01
H
O
m
M
(mol)
Ratio to Smallest n
Revised Ratio
= 3.122
1.00
6
4.20
1.01
= 4.158
1.33
8
58.30
16.00
= 3.643
1.17, which is 7
6
7
The empirical formula is C6H8O7.
(c) From the empirical formula, M e = 6(12.01) + 8(1.01) + 7(16.00) = 192.14 g/mol.
m
= 11 , the molecular formula is the same as the
Given Mm = 192 g/mol: Since M
Me
empirical formula.
23. Find the molar mass of CH3OH.
m = 12.01 + 3(1.01) + 16.00 + 1.01 = 32.05 g/mol
m
1.00
n= M
= 32.05
= 0.03120 mol
1 mole of CH3OH contains 1 mole of C and 4 moles of H. So the 0.03120 mole
sample contains 0.03120 moles of C and 4(0.03120) or 0.1248 moles of H.
When the products are formed, 0.03120 moles of C would yield 0.03120 moles of
CO2 and the 0.1248 moles of H would yield 0.1248
or 0.0624 moles of H2O.
2
Now use m = nM .
Mass of CO2 produced is (0.03120 mol)(44.01) g/mol = 1.373 g.
Mass of H2O produced is (0.0624 mol)(18.02) g/mol = 1.124 g.
Chapter 6 Chemical Proportions in Compounds • MHR
251
UNIT 2 Chapter 6 CHEMICAL PROPORTIONS IN COMPOUNDS
24. (a) Use small quantities to minimize the amount of heat released. Have sand on hand
to extinguish any fire that gets out of control. Use safety goggles and gloves.
(b) Materials: C (powdered charcoal) and Cux O. Apparatus: Evaporating dish, retort
stand, ring clamp, wire gauze, Bunsen burner, stirring rod, and balance.
(c) Choose a small sample of Cux O (1.00 g perhaps) and calculate how much C
(d)
(e)
25. (a)
(b)
(c)
would be needed to react with this much Cu2O (which has a higher percent
Cu than the other possibility). Then, choose an amount of C that is in excess to
ensure complete reaction of the copper oxide, but not a huge excess that would
generate large amounts of heat. Find the mass of an empty evaporating dish. Add
the sample of Cux O and find the total mass. Add the charcoal and stir. Heat
strongly for several minutes until no further reaction occurs. Let dish and contents
cool, then find total mass again.
The data needed for the calculations are:
mass of dish, mass of dish plus Cux O, and mass of dish plus Cu.
This procedure assumes that no Cu was lost from the dish, all the C was removed
as CO2, and all the Cux O reacted.
Find the mass of a sample of the substance. Heat the sample. Find the mass again.
If the sample is a hydrate, the mass should decrease.
For anhydrous crystals, the mass would be unchanged.
Answers to Communication Questions
26.
Unknown Compound
Percentage Composition
× molar mass of elements
Determination of
Empirical Formula
molar mass of compound
molar mass of empirical formula
Molecular Formula
27. If a one-mol sample of dimethyl ether (C2H6O) were passed through a combustion
analyzer, the following would result. Once complete combustion of the dimethyl
ether has occurred, the resulting carbon dioxide and water vapour will pass into the
water-absorbing chamber. At this point, three moles of water would be absorbed
by the desiccant, and only the carbon dioxide would pass on to the carbon dioxide
absorber. Here, two moles of CO2 would be absorbed, and any impurities would
be passed on to the final filter. From the mass data collected at each absorber, the
percentage composition of dimethyl ether could be calculated.
CO2 and H2O
Furnace
H2O Absorber
−H2O
Filter
CO2 Absorber
−CO2
252
MHR • Unit 2 Chemical Quantities
Answers to Making Connections Questions
28. (a) Consider 3.8 L of gasoline. The volume of tetraethyl lead is 2.0 mL.
to find mass of the tetraethyl lead.
Use density, D = M
V
M = DV = (1.653 g/mL)(2.0 mL) = 3.306 g
Now use ratio and proportion.
3.306 g of tetraethyl lead are found in 3.8 L of gasoline.
x g of tetraethyl lead are found in 1.0 L of gasoline.
x
= 1.0
or x = 0.87 g
3.306
3.8
Thus there are 0.87 g of Pb(C2H5)4 in 1.0 L of leaded gasoline.
(b) Find the molar mass of Pb(C2H5)4.
M = 207.20
+ 5(1.01)] = 207.20 + 116.28 = 323.48 g/mol.
+ 4[2(12.01)
207.20
% Pb = 323.48 (100)% = 64.1%
(0.87) g = 0.56 g.
The mass of lead in 1.0 L of leaded gasoline is 64.1
100
29. (a) Sodium carbonate heptahydrate
126.14 g
= 54%
(b) Mass percent of H2O =
232.13 g
(c) Mass of H2O in the body = 78% × 80 000 g = 62 400 g H2O
62 400 g
Moles of H2O in the body = 18.02 g/mol = 3462.8 mol
Ratio of Na2CO3 to H2O in hydrated sodium carbonate
= 1 mol Na2CO3 : 7 mol H2O
mol
= 495 mol Na2CO3
Number of moles of Na2CO3 required = 3462.8
7
Mass of anhydrous Na2CO3 required = 495 mol × 105.99 g/mol = 52 432 g
Therefore, 52.432 kg of anhydrous Na2CO3 is required to absorb all the water in
an 80 kg human.
30. (a) If a contaminated sample contains the same elements from two different
compounds, then the experimentally determined percentage composition of the
sample will differ from that of a pure compound. The experimentally determined
empirical formula will be different from the true empirical formula of ASA, as the
ratio of elements will differ.
(b) Molar mass of SA (MSA) = 138.06 g/mol
Molar mass of ASA (MASA) = 180 g/mol
Mass of SA in sample = 0.35 g
Mol of SA in sample = 0.0025 mol
Mass of ASA in sample = 5.73 g − 0.35 g = 5.38 g
Mol of ASA in sample = 0.030 mol
Mass of Element
SA (g)
ASA (g)
Total mass (g)
Carbon
0.21
3.24
3.45
Hydrogen
0.015
0.24
0.26
Oxygen
0.12
1.92
2.01
Grams per 100 g
Sample (g)
Molar Mass
(g/mol)
Element
Mass (%)
C
60.0
H
4.5
4.5
1.01
O
35.5
35.5
16.00
60
12.01
Number of
Moles (mol)
Molar Amount
÷ Lowest
Molar Amount
2.25 × 4 = 9
5
4.5
2×4=8
2.225
1×4=4
Therefore, the empirical formula you will obtain for the contaminated mixture is
C9H8O4.
Note: Students should realize that the sample contains a mixture of compounds,
neither of which has the empirical formula C9H8O4.
Chapter 6 Chemical Proportions in Compounds • MHR
253