CHAPTER 3 | Chemical Reactions and Earth’s Composition
3.12. Collect and Organize
In this exercise, we convert the number of molecules of each gas found in the sample to moles.
Analyze
To convert the number of molecules to moles, we divide by Avogadro’s number.
Solve
1.4  1013 molecules of H 2
 2.3  10 11 mol H 2
(a)
6.022  1023 molecules/mol
1.5  1014 atoms of He
 2.5  1010 mol He
(b)
6.022  1023 atoms/mol
7.7  1012 molecules of N 2 O
 1.3  1011 mol N 2 O
6.022  1023 molecules/mol
3.0  1012 molecules of CO
 5.0  1012 mol CO
(d)
6.022  1023 molecules/mol
(c)
Think about It
The trace gas that has the highest number of atoms or molecules present also has the highest number of moles
present. In this sample of air, the amount of the trace gases decreases in the order He > H 2 > N2O > CO.
3.15. Collect and Organize
In this exercise we convert from the moles of titanium contained in a substance to the number of atoms
present.
Analyze
For each substance we need to take into account the number of moles of titanium atoms present in 1 mol of the
substance. For 0.125 mol of substance, then, a substance that contains two atoms of titanium in its formula
contains 0.125  2 = 0.250 mol of titanium. We can then use Avogadro’s number to convert the moles of
titanium to the number of atoms present in the sample.
Solve
(a) Ilmenite, FeTiO3, contains one atom of Ti per formula unit; so 0.125 mol of ilmenite contains 0.125 mol of Ti.
6.022  1023 Ti atoms
0.125 mol Ti 
 7.53  1022 Ti atoms
1 mol
(b) The formula for titanium(IV) chloride is TiCl4. This formula contains only one Ti atom per formula unit as
well; so the answer is identical to that calculated in (a).
6.022  1023 Ti atoms
0.125 mol Ti 
 7.53  1022 Ti atoms
1 mol
(c) Ti2O3 contains two titanium atoms in its formula; so 0.125 mol of Ti2O3 contains 0.125  2 = 0.250 mol of
titanium.
6.022  1023 Ti atoms
0.250 mol Ti 
 1.51  1023 Ti atoms
1 mol
(d) Ti3O5 contains three titanium atoms in its formula; so 0.125 mol of Ti 2O3 contains 0.125  3 = 0.375 mol
of titanium.
6.022  1023 Ti atoms
0.375 mol Ti 
 2.26  1023 Ti atoms
1 mol
Think about It
The number of atoms of titanium in 0.125 mol of each compound reflects the number of atoms of Ti in the
chemical formula. Ti2O3 has twice the number of Ti atoms, and Ti3O5 has three times the number of Ti atoms
compared to the number of Ti atoms in the same number of moles of FeTiO 3 and TiCl4.
80
Chemical Reactions and Earth’s Composition | 81
3.21. Collect and Organize
We are asked to convert a mass of carbon in grams to moles.
Analyze
We need the mass of 1 mol of carbon to compute the number of moles of carbon in the 500.0 g sample. From
the periodic table, we see that the molar mass of carbon is 12.011 g/mol.
Solve
500.0 g C 
1 mol
 41.63 mol C
12.011 g
Think about It
Because carbon’s molar mass is relatively low at 12 g/mol, 500 g of this substance contain a fairly substantial
number of moles.
3.26. Collect and Organize
From the chemical formulas for various sodium salts, we are asked to determine how many moles of sodium
are in 1 mol of each substance.
Analyze
The chemical formula reflects the molar ratios of the elements in the compound. If there is one atom of sodium
in the compound’s chemical formula, then there is 1 mol of sodium in 1 mol of the compound. Likewise, if
there are two atoms of sodium in the chemical formula, there are 2 mol of sodium present in 1 mol of the
substance.
Solve
(a) There is one atom of sodium in NaCl; therefore, 1 mol of NaCl contains 1 mol of sodium.
(b) There are two atoms of sodium in Na2SO4; therefore, 1 mol of Na2SO4 contains 2 mol of sodium.
(c) There are three atoms of sodium in Na3PO4; therefore, 1 mol of Na3PO4 contains 3 mol of sodium.
(d) There is one atom of sodium in NaNO3; therefore, 1 mol of NaNO3 contains 1 mol of sodium.
Think about It
The number of sodium ions appearing in these chemical formulas balances the charge on the anions in the
salts. For example, three sodium cations are needed to produce the neutral salt of PO 43–.
3.27. Collect and Organize
This exercise has us compute the molar mass of various molecular compounds of oxygen.
Analyze
The molar mass of each of the compounds can be found by adding the molar mass of each element from the
periodic table, taking into account the number of moles of each atom present in 1 mol of the substance.
Solve
(a) SO2: 32.06 + 2(16.00) = 64.06 g/mol
(b) O3: 3(16.00) = 48.00 g/mol
(c) CO2: 12.01 + 2(16.00) = 44.01 g/mol
(d) N2O5: 2(14.01) + 5(16.00) = 108.02 g/mol
Think about It
Notice that three compounds, SO2, O3, and CO2, have three atoms in their chemical formula, but each has a
different molar mass.
3.34. Collect and Organize
Given a mass of halite, determine the moles of NaCl present.
Analyze
To convert from mass to moles, we divide the mass given by the molar mass of NaCl (22.99 + 35.45 =
58.44 g/mol).
82 | Chapter 3
Solve
6.82 g NaCl
 0.117 mol NaCl
58.44 g/mol
Think about It
The given mass of the halite crystal is about one-tenth that of the formula weight for NaCl; so we expect about
0.1 mol of NaCl to be present in the sample.
3.44. Collect and Organize
To balance the chemical equations we use the three steps described in the textbook.
Analyze
To balance each equation we first write the unbalanced equation using the chemical formulas of the reactants
and products. Next, we balance an element that is present in only one reactant and product. Finally, we balance
the other elements present by placing coefficients in front of the species in the reaction so that the number of
the atoms for each element is equal on both sides of the equation. If there are any fractional coefficients, we
multiply the entire equation through to eliminate all fractions.
Solve
(a) For the reaction of sulfur dioxide with oxygen to form sulfur trioxide:
(1) The unbalanced reaction is
SO2( g) + O2( g) 
Element
Left-Hand Side
S
1
O
2+2=4
SO3( g)
Right-Hand Side
1
3
While the number of sulfur atoms on both sides of the equation is equal, the number of oxygen atoms is not (4
on the reactants side and 3 on the products side). This reaction is not balanced.
(2) We already have the sulfur atoms balanced; therefore the reaction is
SO2( g) + O2( g)  SO3( g)
(3) The number of O atoms on the reactants side is higher by one O atom than on the products side. To balance
the O atoms, therefore, we place 12 as the coefficient before O2 on the left-hand side of the equation.
SO2( g) +
Element
S
O
1
2
O2( g)  SO3( g)
Left-Hand Side
1
2 + 12 (2) = 3
Right-Hand Side
1
3
To eliminate the fractional coefficients we multiply all of the coefficients by 2.
2 SO2( g) + O2( g)  2 SO3( g)
Element
S
O
Left-Hand Side
21=2
(2  2) + (2  ( 12 (2))) = 6
Right-Hand Side
21=2
23=6
The equation is now balanced.
(b) For the reaction of hydrogen sulfide with oxygen to form sulfur dioxide and water:
(1) The unbalanced reaction is
H2S( g)
Element
S
O
H
+ O2( g)  SO2( g) + H2O( g)
Left-Hand Side Right-Hand Side
1
1
2
3
2
2
Chemical Reactions and Earth’s Composition | 83
While the number of hydrogen and sulfur atoms on both sides of the equation is equal, the number of oxygen
atoms is not. This reaction is not balanced.
(2) We already have the sulfur and hydrogen atoms balanced; therefore the reaction is
H2S( g) + O2( g)  SO2( g) + H2O( g)
(3) The number of O atoms on the products side is higher by one O atom than on the reactants side. To balance
the O atoms, therefore, we place 32 as the coefficient before O2 on the right-hand side of the equation.
H2S( g)
Element
S
O
H
+ 32 O2( g)  SO2( g) + H2O( g)
Left-Hand Side Right-Hand Side
1
1
3
2+1=3
(2)
=
3
2
2
2
To eliminate the fractional coefficients we multiply all of the coefficients by 2.
2 H2S( g) + 3 O2( g)  2 SO2( g) + 2 H2O( g)
Element
Left-Hand Side Right-Hand Side
S
21=2
21=2
3
O
2(2 + 1) = 6
2  ( 2 (2)) = 6
H
22=4
22=4
The equation is now balanced.
(c) For the reaction of hydrogen sulfide with sulfur dioxide to form sulfur and water:
(1) The unbalanced reaction is
H2S( g) + SO2( g)  S8( s) + H2O( g)
Element
Left-Hand Side Right-Hand Side
S
1+1=2
8
O
2
1
H
2
2
While the number of hydrogen atoms on both sides of the equation is equal, the number of sulfur and oxygen
atoms is not. This reaction is not balanced.
(2) We can best start by balancing the oxygen atoms as they appear in only one reactant and product. We can
do this by placing a 2 in front of H2O on the products side.
H2S( g) + SO2( g)  S8( s) + 2 H2O( g)
Element
Left-Hand Side Right-Hand Side
S
1+1=2
8
O
2
21=2
H
2
22=4
(3) The number of H atoms on the products side is twice the number of that on the reactants side. To balance
the H atoms, therefore, we place 2 as the coefficient before H 2S on the left-hand side of the equation.
2 H2S( g) + SO2( g)  S8( s) + 2 H2O( g)
Element
Left-Hand Side Right-Hand Side
S
8
(2  1) + 1= 3
O
2
21=2
H
22=4
22=4
To balance the sulfur atoms, we place a coefficient of 83 in front of S8 on the products side.
2 H2S( g) + SO2( g)  83 S8( s) + 2 H2O( g)
Element
Left-Hand Side Right-Hand Side
3
S
(8)= 3
(2  1) + 1= 3
8
O
2
21=2
H
22=4
22=4
To eliminate the fractional coefficients we multiply all of the coefficients by 8.
16 H2S( g) + 8 SO2( g)  3 S8( s) + 16 H2O( g)
84 | Chapter 3
Element
S
O
H
Left-Hand Side
8(2  1) + (8  1) = 24
8  2 = 16
8(2  2) = 32
Right-Hand Side
8  ( 83 (8)) = 24
8(2  1) = 16
8(2  2) = 32
The equation is now balanced.
Think about It
Part c has large coefficients due to the formation of S8, the most stable elemental form of sulfur.
3.45. Collect and Organize
To balance the chemical equations we use the three steps described in the textbook.
Analyze
To balance each equation we first write the unbalanced equation using the chemical formulas of the reactants
and products. Next, we balance an element that is present in only one reactant and product. Finally, we balance
the other elements present by placing coefficients in front of the species in the reaction so that the number of
the atoms for each element is equal on both sides of the equation. If there are any fractional coefficients, we
multiply the entire equation through to eliminate all fractions.
Solve
(a) (1) The unbalanced reaction is
FeSiO3(s) + H2O ( )  Fe3Si2O5(OH)4(s) + H4SiO4(aq)
Element
Fe
Si
O
H
Left-Hand Side
1
1
3+1=4
2
Right-Hand Side
3
2+1=3
9 + 4 = 13
4+4=8
None of the atoms in comparing the right-hand and left-hand sides are equal. This reaction is not balanced.
(2) We can best start by balancing the Fe atoms as they appear in only one reactant and product. We can do
this by placing a 3 in front of FeSiO3 on the left-hand side.
3 FeSiO3(s) + H2O ( )  Fe3Si2O5(OH)4(s) + H4SiO4(aq)
Element
Fe
Si
O
H
Left-Hand Side
31=3
31=3
3(3) + 1 = 10
2
Right-Hand Side
3
2+1=3
9 + 4 = 13
4+4=8
(3) This also balances the silicon atoms. The number of H atoms on the right-hand side is four times the
number of that on the left-hand side. To balance the H atoms, therefore, we place 4 as the coefficient before
H2O on the left-hand side of the equation.
3 FeSiO3(s) + 4 H2O ( )  Fe3Si2O5(OH)4(s) + H4SiO4(aq)
Element
Fe
Si
O
H
Left-Hand Side
31=3
31=3
3(3) + 4(1) = 13
42=8
Right-Hand Side
3
2+1=3
9 + 4 = 13
4+4=8
The equation is now balanced.
(b) (1) The unbalanced reaction is
Fe2SiO4(s) + CO2( g) + H2O ( )  FeCO3(s) + H4SiO4(aq)
Element
Fe
Si
Left-Hand Side
2
1
Right-Hand Side
1
1
Chemical Reactions and Earth’s Composition | 85
C
O
H
1
4+2+1=7
2
1
3+4=7
4
While the number of silicon, carbon, and oxygen atoms on both sides of the equation is equal, the number of
iron and hydrogen atoms is not. This reaction is not balanced.
(2) We can best start by balancing the Fe atoms as they appear in only one reactant and product. We can do
this by placing a 2 in front of FeCO3 on the right-hand side.
Fe2SiO4(s) + CO2( g) + H2O ( )  2 FeCO3(s) + H4SiO4(aq)
Element
Fe
Si
C
O
H
Left-Hand Side
2
1
1
4+2+1=7
2
Right-Hand Side
21=2
1
21=2
2(3) + 4 = 10
4
(3) The number of C atoms on the right-hand side is twice the number of that on the left-hand side. To balance
the C atoms, therefore, we place 2 as the coefficient before CO 2 on the left-hand side of the equation.
Fe2SiO4(s) + 2 CO2( g) + H2O ( )  2 FeCO3(s) + H4SiO4(aq)
Element
Fe
Si
C
O
H
Left-Hand Side
2
1
21=2
4 + 2(2) + 1 = 9
2
Right-Hand Side
21=2
1
21=2
2(3) + 4 = 10
4
To balance the H atoms we place a coefficient of 2 in front of H 2O on the left-hand side. This also balances the
O atoms on each side.
Fe2SiO4(s) + 2 CO2( g) + 2 H2O ( )  2 FeCO3(s) + H4SiO4(aq)
Element
Fe
Si
C
O
H
Left-Hand Side
2
1
21=2
4 + 2(2) + 2(1) = 10
22=4
Right-Hand Side
21=2
1
21=2
2(3) + 4 = 10
4
The equation is now balanced.
(c) (1) The unbalanced reaction is
Fe3Si2O5(OH)4(s) + CO2( g) + H2O ( )  FeCO3(s) + H4SiO4(aq)
Element
Fe
Si
C
O
H
Left-Hand Side
3
2
1
9 + 2 + 1 = 12
4+2=6
Right-Hand Side
1
1
1
3+4=7
4
While the number of C atoms on both sides of the equation is equal, the number of Fe, Si, O, and H atoms are
not. This reaction is not balanced.
(2) We can best start by balancing the Fe and Si atoms as they appear in only one reactant and product. We can
do this by placing a 3 in front of FeCO3 and a 2 in front of H4SiO4 on the right-hand side.
Fe3Si2O5(OH)4(s) + CO2( g) + H2O ( )  3 FeCO3(s) + 2 H4SiO4(aq)
Element
Fe
Si
Left-Hand Side
3
2
Right-Hand Side
31=3
21=2
86 | Chapter 3
C
O
H
1
9 + 2 + 1 = 12
4+2=6
31=3
3(3) + 2(4) = 17
24=8
(3) The number of C atoms on the right-hand side is three times the number of that on the left-hand side. To
balance the C atoms, therefore, we place 3 as the coefficient before CO 2 on the left-hand side of the equation.
Fe3Si2O5(OH)4(s) + 3 CO2( g) + H2O ( )  3 FeCO3(s) + 2 H4SiO4(aq)
Element
Fe
Si
C
O
H
Left-Hand Side
3
2
31=3
9 + 3(2) + 1 = 16
4+2=6
Right-Hand Side
31=3
21=2
31=3
3(3) + 2(4) = 17
24=8
To balance the H atoms we place a coefficient of 2 in front of H2O on the left-hand side.
Fe3Si2O5(OH)4(s) + 3 CO2( g) + 2 H2O ( )  3 FeCO3(s) + 2 H4SiO4(aq)
Element
Fe
Si
C
O
H
Left-Hand Side
3
2
31=3
9 + 3(2) + 2(1)
= 17
4 + 2(2) = 8
Right-Hand Side
31=3
21=2
31=3
3(3) + 2(4) = 17
24=8
The equation is now balanced.
Think about It
These complex formulas looked difficult at first glance to balance. A good strategy here is to balance all atoms
except O and H atoms first, then finish by balancing the O and H with a coefficient in front of H 2O.
3.60. Collect and Organize
Here we have two methods that produce chromium metal, one that uses silicon and another that uses
aluminum. We are asked to compute the amounts of each reactant required for each process to yield 400.0 g of
Cr metal.
Analyze
In the first reaction the ratio of the moles of Cr produced to Al required is 2:2 (or 1:1). Therefore, when we
compute the moles of Cr in 400.0 g, that amount must equal the number of moles of aluminum required. To
find the mass of aluminum, we multiply that result by the molar mass of aluminum. In the second reaction the
ratio of moles of Cr to Si is 4:3. These are not equal, and we have to take that into account. We require fewer
(three-quarters) moles of Si than the moles of Cr produced. Once we have the moles of Si required, we can use
the molar mass of Si to compute the mass of Si required.
Solve
1 mol Cr 2 mol Al 26.98 g Al


 207.5 g Al
52.00 g 2 mol Cr 1 mol Al
1 mol Cr 3 mol Si 28.09 g Si
(b) 400.0 g Cr 


 162.1 g Si
52.00 g 4 mol Cr 1 mol Si
(a) 400.0 g Cr 
Think about It
The reaction to produce Cr using silicon requires about 20% less of the reactant (in terms of mass).
Chemical Reactions and Earth’s Composition | 87
3.74. Collect and Organize
We are not able to tell simply by looking at the chemical formula how to order the compound according to
their percentage of lead by mass. We have to find the percent composition of the elements present in each
compound and then compare the results. Because we are only interested in the percentage of Pb in each, we
need not compute the percent composition of the other elements.
Analyze
For each compound, assume we have 1 mol of the substance. To determine the percentage of Pb in each
compound, we divide the mass of lead present in that 1 mol by the molar mass. Multiplying that result by 100
will give the percentage of Pb in the compounds.
Solve
PbS:
PbCO3:
207.2 g Pb
207.2 g Pb
 100  86.59% Pb
% Pb =
 100  77.54% Pb
239.3 g
267.2 g
PbCl(OH):
Pb2Cl2CO3:
207.2 g Pb
(2  207.2) g Pb
% Pb =
 100  79.78% Pb
% Pb =
 100  75.99% Pb
259.7 g
545.3 g
In order of increasing percentage by mass of lead: Pb2Cl2CO3 < PbCO3 < PbCl(OH) < PbS.
% Pb =
Think about It
It may have been tempting to look at this question and guess that Pb 2Cl2CO3 had the most lead in terms of
percent composition because its molecular formula contains two atoms of Pb. However, this compound has the
lowest percentage of Pb. This shows that we must also take into account the masses of the other atoms present.
3.81. Collect and Organize
We are given the masses of the reactants and products and are asked to determine the empirical formula for the
product of the reaction of magnesium with oxygen. Then we write a balanced chemical reaction.
Analyze
The masses of the reactants add up to give the mass of the products; so no other products are formed in this
reaction. Therefore, the product contains 2.43 g Mg and 1.60 g O. The mass composition of the product, then,
is the mass of each element divided by the total mass of the product, multiplied by 100. We can then assume
that there are 100 g of product. In that case, the percent composition gives the mass of each element in the
product, which we can then convert to moles using the molar mass of the elements. We can get the mole ratio
for the two elements in the products by determining the lowest whole-number ratio of the moles. From that we
can write the empirical formula and the chemical equation for balancing.
Solve
(a) The mass percentage of each element in the product is
2.43 g
% mass of Mg =
 100  60.3%
4.03 g
1.60 g
% mass of O =
 100  39.7%
4.03 g
Assuming 100 g of product,
1 mol
60.3 g Mg 
 2.48 mol Mg
24.31 g
1 mol
39.7 g O 
 2.48 mol O
16.00 g
This is a 1:1 molar ratio of Mg to O; so the empirical formula is MgO.
(b) The balanced equation is 2 Mg(s) + O2( g)  2 MgO(s).
88 | Chapter 3
Think about It
This question reflects how we would experimentally determine the formula for a new compound.
3.82. Collect and Organize
From the mass percentage of phosphorus and sulfur, we have to find the number of moles of phosphorus and
sulfur in the compound and then determine the lowest whole-number ratio of P to S to determine the empirical
formula. The molecular formula will be a whole-number multiple of the empirical formula. Once we have the
molecular formula for the other product, we can write the balanced chemical equation.
Analyze
First, we assume 100 g of the substance, and we then know that the substance has 27.87 g of P and 72.13 g of
S. Using the molar mass of each of these from the periodic table, we convert these masses into moles. Dividing
each result by the lowest number of moles (which may be for either P or S) and finding the lowest wholenumber ratio of P and S gives us the empirical formula. To obtain the molecular formula, we compare the
molar mass of the empirical formula to that given for the molecular formula (444.56 g/mol). This should be a
whole-number ratio.
Chemical Reactions and Earth’s Composition | 89
Solve
(a) The moles of P and S in the compound are found by
1 mol
27.87 g P 
 0.8999 mol P
30.97
1 mol
72.13 g S 
 2.249 mol S
32.07
Dividing each by 0.8999 gives a ratio of 1 P to 2.5 S. Multiplying this ratio by 2 to give whole numbers gives
the empirical formula of P2S5. The molar mass of the empirical formula is (2  30.97) + (5  32.07) =
222.29 g/mol. This is half that of the given molecular molar mass, so the molecular formula is P4S10.
(b) Knowing that the formula for iron(II) sulfide is FeS and that the phosphorus–sulfur product is P4S10, we can
write the balanced equation:
4 Fe2P(s) + 18 FeS2(s)  26 FeS(s) + P4S10(s)
Think about It
This kind of problem reflects the multilayered problem that you would encounter in determining the products
of a reaction.