Download Chapter 1 - Solutions

CHAPTER 3 Problems: 2, 6 (a,b), 10, 24, 31, 34, 40, 44 (a,b), 50, 52, 54, 56, 64, 70, 76,
80, 84, 88, 92, 94, 100, 106, 116, 128
2) Explain the difference between the terms molecular mass and formula mass. To what
type of compound does each refer?
The term molecular mass refers to the average mass of a single molecular of a
substance, and so applies to substances that exist as molecules. The term formula mass
refers to the average mass of one formula unit of a substance, and so applies to
substances that do not exist as individual molecules, such as ionic compounds.
6) Calculate the molecular mass or formula mass (in amu) for each of the following
substances:
a) Li2CO3
Li
C
O
2 x 6.941 amu
1 x 12.0107 amu
3 x 15.9994 amu
= 13.88 amu
= 12.0107 amu
= 47.9982 amu
M = 73.89 amu
b) C2H6
C
H
2 x 12.0107 amu
6 x 1.00794 amu
= 24.0214 amu
= 6.04764 amu
M = 30.0690 amu
 30.07 amu
Note that we often round off the molecular mass or formula mass to two digits to the
right of the decimal place, as in b, as this is usually sufficient for calculations using M.
10) For many years chloroform (CHCl3) was used as an inhalation anesthetic in spite of
the fact that it is also a toxic substance that may cause severe liver, kidney, and heart
damage. Calculate the percent composition by mass of this compound.
CHCl3
C
H
Cl
1 x 12.0107 amu
1 x 1.00794 amu
3 x 35.453 amu
= 12.0107 amu
= 1.00794 amu
= 106.36 amu
M = 119.38 amu
So
% C = [(12.0107 amu)/(119.38 amu)] · 100 % = 10.06 % C
% H = [(1.00794 amu)/(119.38 amu)] · 100 % = 0.84 % C
% Cl = [(106.36 amu)/(119.38 amu)] · 100 % = 89.10 % C
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24) Balance the following equations using the method outlined in section 3.3
unbalanced
balanced
a) C + O2  CO
b) CO + O2  CO2
c) H2 + Br2  HBr
d) K + H2O  KOH + H2
e) Mg + O2  MgO
f) O3  O2
g) H2O2  H2O + O2
h) N2 + H2  NH3
i) Zn + AgCl  ZnCl2 +Ag
j) S8 + O2  SO2
k) NaOH + H2SO4  Na2SO4 + H2O
l) Cl2 + NaI  NaCl + I2
m) KOH + H3PO4  K3PO4 + H2O
n) CH4 + Br2  CBr4 + HBr
2 C + O2  2 CO
2 CO + O2  2 CO2
H2 + Br2  2 HBr
2K + H2O  2 KOH + H2
2 Mg + O2  2 MgO
2 O3  3 O2
2 H2O2  2 H2O + O2
N2 + 3 H2  2 NH3
Zn + 2 AgCl  ZnCl2 + 2 Ag
S8 + 8 O2  8 SO2
2 NaOH + H2SO4  Na2SO4 + 2 H2O
Cl2 + 2 NaI  2 NaCl + I2
3 KOH + H3PO4  K3PO4 + 3 H2O
CH4 + 4 Br2  CBr4 + 4 HBr
31) If we know the empirical formula for a compound, what additional information do we
need to determine its molecular formula?
You also need to know the molecular mass. Since
(molecular formula) = N (empirical formula)
We can find N by using
N=
N = 1, 2, 3…
M(molecule)
M(empirical formula)
34) How many atoms are there in 5.10 moles of sulfur (S)?
# atoms = 5.10 mole S 6.022 x 1023 atoms = 3.07 x 1024 atoms of S
mole
40) What is the mass in grams of 1.00 x 1012 lead (Pb) atoms?
Mass = 1.00 x 1012 atoms
1 mol
207.2 g = 3.44 x 10-10 g
6.022 x 1023 atoms mole
2
44) Calculate the molar mass of the following substances:
I prefer to first find the average mass of one molecule or one formula unit (as in
problem 6 above) and then use the concept of moles to rewrite the result as a molar mass.
a) Li2CO3
Li
C
O
2 x 6.941 amu
1 x 12.0107 amu
3 x 15.9994 amu
= 13.88 amu
= 12.0107 amu
= 47.9982 amu
M = 73.89 amu = 73.89 g/mol
b) CS2
C
S
1 x 12.0107 amu
2 x 32.07 amu
= 12.0107 amu
= 64.14 amu
M = 76.15 amu = 76.15 g/mol
50) Calculate the mass in grams of iodine (I2) that will react completely with 20.4 g of
aluminum (Al) to form aluminum iodide (AlI3)
We first need to write a balanced chemical equation for the reaction
2 Al + 3 I2  2 AlI3
.
M(Al) = 26.9815 g/mol
g I2 = 20.4 g Al
1 mol Al
26.9815 g I2
M(I2) = 253.809 g/mol
3 mol I2
2 mol Al
253.809 g I2 = 288. g I2
1 mol I2
52) Determine the empirical formulas for the compounds with the following
compositions:
When you are given the percent composition for a compound it is easiest to
proceed by assuming you have 100.00 g of the compound.
a) 2.1 % H; 65.3 % O; 32.6 % S
mol H = (2.1 g H)/(1.00794 g/mol)
= 2.1 mol H
 1.02 = 2.1
mol O = (65.3 g O)/(15.9994 g/mol) = 4.08 mol O  1.02 = 4.00
mol S = (32.6 g S)/(32.065 g/mol)
= 1.02 mol S  1.02 = 1.00
empirical formula is H2O4S (actual is likely H2SO4)
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b) 20.2 % Al; 79.8 % Cl
mol Al = (20.2 g Al)/(26.9815 g/mol)
= 0.749 mol Al  0.749
= 1.00
mol Cl = (79.8 g Cl)/(35.453 g/mol)
= 2.25 mol Cl  0.749
= 3.00
empirical formula is AlCl3
54) The empirical formula of a compound is CH. If the molar mass of this compound is
about 78 g, what is its molecular formula?
The mass of a CH unit is 13.02 g/mol
So the number of CH units per molecule is
N = (78. g/mol)/(13.02 g/mol) = 6.0
The formula is 6 · (CH) = C6H6
56) Monosodium glutamate (MSG), a food flavoring enhancer, has been blamed for
“Chinese restaurant syndrome”, the symptoms of which are headache and chest pains.
MSG has the following composition by mass: 35.51 percent C; 4.77 percent H; 37.85
percent O; 8.29 percent N; and 13.60 percent Na. What is its molecular formula if its
mass is about 169 g?
We first need to find the empirical formula
mol C = (35.51 g C)/(12.0107 g/mol)
= 2.957 mol C  0.5916 = 4.998
mol H = (4.77 g H)/(1.00794 g/mol)
= 4.73 mol H  0.5916 = 8.00
mol O = (37.85 g O)/(15.9994 g/mol)
= 2.366 mol O  0.5916 = 3.999
mol N = (8.29 g N)/(14.0067 g/mol)
= 0.592 mol N  0.5916 = 1.00
mol Na = (13.60 g Na)/(22.9898 g/mol)
= 0.5916 mol Na  0.5916 = 1.000
empirical formula is C5H8O4NNa
The empirical formula has a mass of 169.11 g/mol
So N = (169. g/mol)/(169.11 g/mol) = 1.00
So the molecular formula is also C5H8O4NNa (likely NaC5H8O4N)
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64) Ascorbic acid (vitamin C) contains C, H, and O. In one combustion analysis, 5.24 g
of ascorbic acid yields 7.86 g CO2 and 2.14 g H2O. Calculate the empirical formula and
molecular formula for ascorbic acid given that its molar mass is about 176 g.
To find the empirical formula we need to know how many grams of C, H, and O
there are in the sample. Since the sample only contains C, H, and O
grams C + grams H + grams O = total mass
We may use the results from the combustion analysis to find the number of grams
of C and H in the sample. We will use the following information in doing this
M(C) = 12.011 g/mol
M(H) = 1.0079 g/mol
M(CO2) = 44.010 g/mol
M(H2O) = 18.015 g/mol
grams C = 7.86 g CO2 12.011 g/mol C = 2.145 g C
44.010 g/mol CO2
grams H = 2.14 g H2O 2·(1.0079 g/mol) H = 0.239 g H
18.015 g H2O
And so g O = 5.24 g - (2.145 g + 0.239 g) = 2.86 g O
We can now find the empirical formula
mol C = (2.145 g C)/(12.0107 g/mol)
= 0.1786 mol C  0.1786 = 1.000
mol H = (0.239 g H)/(1.00794 g/mol)
= 0.237 mol H  0.1786 = 1.33
mol O = (2.86 g O)/(15.9994 g/mol)
= 0.179 mol O  0.1786 = 1.00
Notice that the relative moles of H is not close to an integer value. However, if we
multiply through by 3 we get values that are all close to integers.
So the empirical formula is C3H4O3 (M = 88.07 g/mol)
Since the molecular mass is approximately 176. g/mol, then N (the number of
empirical formula units per molecule) is
N = (176. g/mol)/(88.07 g/mol) = 2.00  2
And so the molecular formula is C6H8O6
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70) Consider the combustion of carbon monoxide (CO) in oxygen gas:
2 CO(g) + O2(g)  2 CO2(g)
Starting with 3.60 moles of CO, calculate the number of moles of CO2 produced if there
is enough oxygen to react with all of the CO.
Moles CO2 formed = 3.60 mol CO 2 mol CO2 = 3.60 mol CO2
2 mol CO
76) When potassium cyanide (KCN) reacts with acids, the deadly poisonous gas
hydrogen cyanide (HCN) is given off. Here is the equation:
KCN(aq) + HCl(aq)  KCl(aq) + HCN(g)
If a sample of 0.140 g of KCN is treated with an excess of HCl, calculate the amount of
HCN formed, in grams.
M(KCN) = 65.116 g/mol
grams HCN = 0.140 g KCN
M(HCN) = 27.025 g/mol
1 mol KCN
65.116 g KCN
1 mol HCN
1 mol KCN
27.025 g HCN
1 mol HCN
= 0.0581 g HCN
80) Limestone (CaCO3) is decomposed by heating to quicklime (CaO) and carbon
dioxide. Calculate how many grams of quicklime can be produced from 1.00 kg of
limestone.
The reaction is
CaCO3(s)  CaO(s) + CO2(g)
M(CaCO3) = 100.09 g/mol
M(CaO) = 56.077 g/mol
grams quickline
= 1.00 kg limestone 1000 g 1 mol limestone 1 mol quickline
56.077 g
1 kg
100.09 g
1 mol limestone 1 mol quickline
= 560. g
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84) Define limiting reactant and excess reactant. What is the significance of the limiting
reactant in predicting the amount of product obtained in a reaction? Can there be a
limiting reactant if only one reactant is present?
The limiting reactant is the reactant that first runs out in a chemical reaction,
therefore limiting the amount of products that can be formed. Excess reactants refer to
any reactants that would still be present in the system when the limiting reactant has been
completely consumed.
The theoretical maximum amount of products formed in a chemical reaction is
determined by the number of moles of the limiting reactant, along with the stoichiometry
of the reaction.
In a reaction with a single reactant (such as a decomposition reaction, like that in
problem 80) the reaction stops when that reactant is completely consumed. I would argue
that this meets the definition of a limiting reactant, and so would answer "yes" to this
question.
88) Nitric oxide (NO) reacts with oxygen gas to form nitrogen dioxide (NO2), a dark
brown gas:
2 NO(g) + O2(g)  2 NO2(g)
In one experiment, 0.886 mol of NO is mixed with 0.503 mol of O2. Determine which of
the two reactants is the limiting reactant. Calculate also the number of moles of NO2
produced.
I am always going to do these limiting reactant problems by finding the yield of
product for each reactant. The smallest yield of product will correspond to the limiting
reactant. Note that there are other ways to do these sorts of problems.
If NO is limiting
mol NO2 = 0.886 mol NO 2 mol NO2 = 0.886 mol NO2
2 mol NO
If O2 is limiting
mol NO2 = 0.503 mol O2 2 mol NO2 = 1.006 mol NO2
1 mol O2
The NO has the smaller yield, and so NO is the limiting reactant. The number of moles
of NO2 that can be produced is 0.886 mol NO2.
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92) Titanium (IV) oxide (TiO2) is a white substance produced by the action of sulfuric
acid on the mineral ilmenite:
FeTiO3 + H2SO4  TiO2 + FeSO4 + H2O
Its opaque and nontoxic properties make it suitable as a pigment in plastics and paints. In
one process, 8.00 x 103 kg of FeTiO3 yielded 3.67 x 103 kg of TiO2. What is the percent
yield of the reaction? (Assume FeTiO3 is the limiting reactant.)
We first need to find the theoretical yield. We are told that FeTiO3 is the limiting
reactant.
M(FeTiO3) = 151.71 g/mol
M(TiO2) = 79.866 g/mol
Theoretical yield TiO2
= 8.00 x 106 g FeTiO3 1 mol FeTiO3 1 mol TiO2 79.866 g TiO2
151.71 g
1 mol FeTiO3
1 mol
= 4.21 x 106 g TiO2 = 4.21 x 103 kg TiO2
The percent yield of the reaction is then
% yield = actual yield · 100 % = 3.67 x 103 kg · 100 % =87.2 % yield
theo. yield
4.21 x 103 kg
94) When heated, lithium reacts with nitrogen to form lithium nitride:
6 Li(s) + N2(g)  2 Li3N(s)
What is the theoretical yield of Li3N in grams when 12.3 g of Li is heated with 33.6 g of
N2? If the actual yield of Li3N is 5.89 g, what is the percent yield of the reaction?
We first need to find the limiting reactant
M(Li) = 6.941 g/mol
M(N2) = 28.013 g/mol
M(Li3N) = 34.830 g/mol
If Li is limiting
mol Li3N = 12.3 g Li 1 mol Li 2 mol Li3N = 0.591 mol
6.941 g
6 mol Li
If N2 is limiting
mol Li3N = 33.6 g N2 1 mol N2 2 mol Li3N = 2.40 mol
28.013 g 1 mol N2
The Li has the smaller yield and so is the limiting reactant.
8
We can use the above calculations to find the theoretical yield of Li3N.
theoretical yield Li3N = 0.591 mol Li3N 34.830 g Li3N = 20.6 g
1 mol
The percent yield is then
% yield = 5.89 g · 100 % = 28.6 % yield
20.6 g
100) Determine whether each of the following equations represents a combination
reaction, a decomposition reaction, or a combustion reaction:
a) 2 NaHCO3  Na2CO3 + CO2 + H2O
Decomposition. A single reactant is broken apart into several products.
b) NH3 + HCl  NH4Cl
Combination. Two reactants form a single product.
c) 2 CH3OH + 3 O2  2 CO2 + 4 H2O
Combustion. A single substance reacts with oxygen to form combustion products.
106) The aluminum sulfate hydrate [ Al2(SO4)3 · xH2O ] contains 8.10 % Al by mass.
Calculate x, that is, the number of water molecules associated with each Al2(SO4)3 unit.
M(Al) = 26.982 g/mol
M(Al2(SO4)3) = 342.15 g/mol
M(H2O) = 18.015 g/mol
We can find the formula mass of the compound as follows
mass Al
· 100 % = 8.10 %
formula mass
formula mass = mass Al 100 % = 2 (26.982 g/mol) 100 % = 666. g/mol
8.10 %
8.10 %
formula mass hydrate = formula mass (Al2(SO4)3) + x molecular mass H2O
x = [ formula mass of hydrate - formula mass (Al2(SO4)3) ]
molecular mass H2O
= [ 666. g/mol - 342.15 g/mol ] = 17.97  18
18.015 g/mol
Formula is Al2(SO4)3 · 18 H2O
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116) A mixture of NaBr and Na2SO4 contains 29.96 % Na by mass. Calculate the percent
by mass of each compound in the mixture.
Assume 100.00 g of mixture. Let x = g NaBr, then (100.0 - x) = g Na2SO4
Since the percent Na in the mixture is 29.96 %, that means there are 29.96 g Na in the
100.00 g of mixture.
So 29.96 g Na = (mass NaBr) (% Na in NaBr) + (mass Na2SO4) (% Na in Na2SO4)
100%
100 %
M(Na) = 22.9898 g/mol
M(NaBr) = 102.89 g/mol
M(Na2SO4) = 142.04 g/mol
% Na in NaBr = 1 · (22.9898 g/mol) · 100 % = 22.34 %
102.89 g/mol
% Na in Na2SO4 = 2 · (22.9898 g/mol) · 100 % = 32.37 %
142.04 g/mol
So
29.96 = x (0.2234) + (100.00 - x) (0.3237)
29.96 = 0.2234 x + 32.37 - 0.3237 x = 32.37 - 0.1003 x
x = (29.96 - 32.37) = 24.0
( - 0.1003)
So the mixture is 24.0% NaBr (and 76.0 % Na2SO4)
128) A mixture of methane (CH4) and ethane (C2H6) of mass 13.43 g is completely
burned in oxygen. If the total mass of CO2 + H2O produced is 64.84 g, calculate the
fraction of CH4 in the mixture.
This one is tricky. There are several ways to do this problem. I've decided to
proceed as follows. I am first going to find out the mass of products formed when 1.00 g
of CH4 is burned, and when 1.00 g of C2H6 is burned. The relevant combustion reactions
are
methane
CH4 + 2 O2  CO2 + 2 H2O
ethane
C2H6 + 5/2 O2  2 CO2 + 3 H2O
M(CH4) = 16.042 g/mol
M(CO2) = 44.010 g/mol
M(C2H6) = 30.069 g/mol
M(H2O) = 18.015 g/mol
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For the combustion of 1.000 g methane
g CO2 = 1.000 g CH4 1 mol CH4
16.042 g
1 mol CO2 44.010 g CO2 = 2.743 g CO2
1 mol CH4
1 mol
g H2O = 1.000 g CH4 1 mol CH4
16.042 g
2 mol H2O 18.015 g H2O = 2.246 g H2O
1 mol CH4
1 mol
Total mass of products = 2.743 g + 2.246 g = 4.989 g
For the combustion of 1.000 g ethane
g CO2 = 1.000 g C2H6 1 mol C2H6
30.069 g
2 mol CO2 44.010 g CO2 = 2.927 g CO2
1 mol CH4
1 mol
g H2O = 1.000 g C2H6 1 mol C2H6
30.069 g
3 mol H2O 18.015 g H2O = 1.797 g H2O
1 mol CH4
1 mol
Total mass of products = 2.927 g + 1.797 g = 4.724 g
Now, let x = mass of CH4, then ( 13.43 g - x) = mass of C2H6
x 4.989 g product
1.000g methane
+ (13.43 - x) 4.724 g product = 64.84 g
1.000 g ethane
4.989 x + 63.443 - 4.724 x = 64.84
0.265 x = 1.397
x = (1.397/0.265) = 5.27 g
So the fraction of methane in the mixture is (5.27 g methane)/(13.43 g mixture) = 0.393
(so 39.3 % by mass methane, 60.7 % by mass ethane).
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