CHAPTER THREE STOICHIOMETRY Questions

CHAPTER THREE
STOICHIOMETRY
Questions
18.
The two major isotopes of boron are lOB and liB. The listed mass of 10.81 is the average mass of
a very large number of boron atoms.
19.
The molecular formula tells us the actual number of atoms of each element in a molecule (or
formula unit) of a compound. The empirical formula tells only the simplest whole number ratio
of atoms of each element in a molecule. The molecular formula is a whole number multiple of
the empirical formula. Ifthat multiplier is one, the molecular and empirical formulas are the
same. For example, both the molecular and empirical formulas of water are H20. They are the
same. For hydrogen peroxide, the empirical formula is OH; the molecular formula is H2 0 2 •
20.
Side reactions may occur. For example, in the combustion ofCH4 (methane) to CO2 and H20,
some CO is also formed. Also, some reactions only go part way to completion and reach a state
of equilibrium where both reactants and products are present (see Ch. 13).
Exercises
Atomic Masses and the Mass Spectrometer
21.
A = atomic mass = 0.7899(23.9850 amu) + 0.1000(24.9858 amu) + 0.1101(25.9826 arnu)
A = 18.95 amu + 2.499 amu + 2.861 arnu = 24.31 amu
22.
0.3460(283.4 amu) + 0.2120(284.7 arnu) + 0.4420(287.8 amu) = 285.6 amu
23.
A = atomic mass = 0.7553(34.96885 arnu) + 0.2447(36.96590 amu)
A = 26.41 amu + 9.046 = 35.46 amu; From atomic masses in the period table, this is chlorine.
24.
A = 0.0140(203.973) + 0.2410(205.9745) + 0.2210(206.9759) + 0.5240(207.9766)
A = 2.86 + 49.64 + 45.74 + 109.0 = 207.2 arnu; From the periodic table, the element is Pb.
25.
Letx =% of IS lEu andy=%of IS3 Eu,thenx+y= 100andy= 100-x.
151.96= x(150.9196) +(100 -x)(152.9209)
100
28
CHAPTER 3
STOICHIOMETRY
29
15196 = 150.9196 x + 15292.09 - 152.9209 x, -96
= -2.0013 x
x = 48%; 48% 151Eu and 100 - 48 = 52% 153Eu
26.
Let A = mass of l85 Re: 186.207 = 0.6260(186.956) + 0.3740(A), 186.207 - 117.0 = 0.3740(A)
A=
27.
69.2
0.3740
= 185 amu (A = 184.95 amu without rounding to proper significant figures.)
There are three peaks in the mass spectrum, each 2 mass units apart. This is consistent with two
isotopes, differing in mass by two mass units. The peak at 157.84 corresponds to a Br2 molecule
composed of two atoms of the lighter isotope. This isotope has mass equal to 157.84/2 or 78.92.
This corresponds to 79Br. The second isotope is 81Br with mass equal to 161.84/2 = 80.92. The
peaks in the mass spectrum correspond to 79Br2, 79Br81Br and 81Br2 in order of increasing mass.
The intensities of the highest and lowest mass tell us the two isotopes are present at about equal
abundance. The actual abundance is 50.69% 79Br and 49.31 % 81Br. The calculation of the
abundance from the mass spectrum is beyond the scope of this text..
28.
Scaled Intensity
Largest Peak = 100
Intensity
Compound
Mass
H212°Te
121.92
0.09
0.3
H 2122Te
123.92
2.46
7.1
H/ 23 Te
124.92
0.87
2.5
H/ 24Te
125.92
4.61
13.4
H 2125Te
126.92
6.99
20.3
H/ 26Te
127.92
18.71
54.3
H 2128Te
129.92
31.79
92.2
H 2130Te
131.93
34.48
100.0
100
50
O---l------1~-----1r_....I...-_+-....L---+--_+--__+--____r
122
124
126
128
130
132
134
CHAPTER 3
30
STOICIllOMETRY
Moles and Molar Masses
29.
When more than one conversion factor is necessary to determine the answer, we will apply the
conversion factors into one calculation instead of determining intermediate answers. This
method reduces round-off error and is a time saver.
500. atoms Fe x
30.
500.0 g Fe x
464 x 10.20 g Fe
I mol Fe
x 55.85 g Fe =.
23
mol Fe
6.022 x 10 atoms Fe
I mol Fe
= 8.953 mol Fe
55.85 g Fe
23
8.953 mol Fe x 6.022 x 10 atoms Fe = 5.391 x 1024 atoms Fe
mol Fe
31 .
23
100 x 1022 atoms C
. carat x 0.200gC x ImolC x 6.022x10 atomsC =.
100
12.01 g C
mol C
carat
32.
21
5.0 x 10 atoms C x
I mol C
_)
= 8.3 x 10 mol C
23
6.022 x 10 atoms C
8.3x 1O-3 mo lCx 12.01gC =O.IOgC
molC
33.
A1 20 3 : 2(26.98) + 3(16.00) = 101.96 g/mol
Na3AIF6: 3(22.99) + 1(26.98) + 6(19.00) = 209.95 glmol
34.
HFC - 134a, CH2FCF3 : 2(12.01) + 2(1.008) + 4(19.00) = 102.04 glmol
HCFC-124, CHCIFCF): 2(12.01) + 1(1.008) + 1(35.45) + 4(19.00) = 136.48 glmol
35.
a.
NH3 : 14.01 glmol + 3(1.008 g/mol) = 17.03 glmol
b. N2H4: 2(14.01) + 4(1.008) = 32.05 glmol
36.
37.
c.
(NH4)2Cr207: 2(14.01) + 8(1.008) + 2(52.00) + 7(16.00) = 252.08 glmol
a.
PP6: 4(30.97 glmol) + 6(16.00 glmol) = 219.88 glmol
b.
CalP04)2: 3(40.08) + 2(30.97) + 8(16.00) = 310.18 glmol
c.
N~HP04:
a.
1.00 g NH) x
2(22.99) + 1(1.008) + 1(30.97) + 4(16.00) = 141.96 glmol
I molNH3
17.03 gNH3
= 0.0587 mol NH)
CHAPTER 3
38.
39.
STOICHIOMETRY
1 mol N 2H4
b.
1.00 g N 2H4 X
c.
1.00 g (NH4)2CrP7 x
a.
1.00 g P40 6 x
b.
1.00 g CaiP04)2 x
c.
1.00 g Na2HP0 4 x
a.
5.00 mol NH3 x
32.05 gN 2H4
= 0.0312 mol N 2H4
-3
I mol (NH4)2 Cr20 7 _
- 3.97 x 10 mol (NH4)2CrP7
252.08 g (NH4)2Cr207
1 mol PO
4_6
219.88 g
b. 5.00 mol N 2H4 x
c.
31
= 4.55
x 10-3 mol P40 6
1 mol Ca3(PO )
42 = 3.22
310.18g
I
X
moIN~HPO
4 = 7.04
141.96 g
17.03 g NH4
molNH3
=
X
10-3 mol Ca3(P04)2
10-3 mol N~HP04
85.2 g NH3
32.05 gN 2H4
= 160. g N 2H4
molN 2H4
5.00 mol (NH4)2CrP7 x
252.08 g (NH4)2Cr207
mol (NH4)2Cr207
40.
41.
a.
5.00 mol P40 6 x 219.88 g = 1.10
mol P40 6
b.
5.00 mol Ca3(P0 4)2 x
c.
5.00moIN~HP04x
X
103 g PP6
310.18 g
= 1.55
mol Ca3(P04)2
141.96g
moIN~HP04
= 1260 g (NH4)2CrP7
X
103 g CaiP04)2
=7.IOxI02gN~HP04
Chemical formulas give atom ratios as well as mol ratios.
a.
5.00 mol NH 3 x 1 mol N x 14.01 gN = 70.1 g N
molNH3
molN
b. 5.00 mol N 2H4 x 2 mol N x 14.01 g N = 140. g N
mol N2H4
mol N
42.
c.
5.00 mol (NH4)2Cr20 7 x
a.
5.00mo1PP6x
2 mol N
x 14.01 g N = 140. g N
mol (NH4)2Cr207
mol N
4molP x 30.97gP=619gP
mol P40 6
mol P
CHAPTER 3
32
43.
c.
5.00 mol N~HP04 x
a.
1.00 g NH3
1 mol P
x 30.97 g P mol N~HP04
mol P - 155 g P
1 mol NH
X
3 x
17.03 gNH3
1 mol N zH4
b.
1.00 g N 2H4 X
c.
1.00 g (NH4)2Cr207 x
32.05 g N 2H 4
6.022 x 10z3 molecules NH
3 = 3.54
X
1 mol (NH4)2Cr207
6.022 x 1023 formula units (NH4)2CrP7
x --------------
mol (NH4)2Cr207
2.39 x 1021 formula units (NH4)2Cr207
23
I mol P40 6 x 6.022 x 10 molecules
219.88g
molP 0
4 6
=
2.74 x 1021 molecules P 0
4 6
a.
1.00gPP6
b.
1 mol Ca3(P04)2 6.022 x 1023 formula units
1.00 g CaiP04)2 x
310.18 g
x
mol Ca (P0 )2
3
4
=
c.
1.00 g N~HP04 x
1.94
X
1021 formula units C~(P04)2
mol N~HP04 6.022 x ]023 formula units
141.96 g
x
mol Na HP0
2
4
= 4.24
45.
46.
]022 molecules NH3
22
6.022 x 1023 molecules N zH4
= 1.88 x 10 molecules N 2H 4
mol N 2H4
=
x
X
molNH3
252.08 g (NH4)2CrP7
44.
STOICHIOMETRY
x
1021 formula units Na2HP0 4
Using answers from Exercise 43:
a.
22
3.54 x 10 molecules NH3 x
1 atom N
22
= 3.54 x 10 atoms N
molecule NH3
b.
22
1.88 x 10 molecules N 2H4 x
c.
1Units
' (NH4)2Cr207 x
2.39 x 102
formula
2 atoms N
22
= 3.76 x 10 atoms N
molecule N 2H 4
2 atomsN
21
= 4.78 x 10 atoms N
formula unit (NH4)2Cr207
Using answers from Exercise 44:
a.
21
2.74 x 10 molecules P40 6 x
4 atoms P
22
= 1.10 x 10 atoms P
molecule P40 6
b.
1umts
' Ca (P0 )2 x
1.94 x 102
formula
3
4
2 atoms P
21
= 3.88 x 10 atoms P
formula unit Ca3(P04)2
CHAPTER 3
c.
47.
STOICHIOMETRY
4.24 x 1021 formula units N~HP04 x
2.839
50.
1g
x 1 mol = 2.839 x 10.3 mol
1000mg 176.12g
23
10-3 mol x 6.022 x 10 molecules = 1.710 x 1021 molecules
mol
X
a.
9(12.01) + 8(1.008) + 4(16.00) = 180.15 glmol
b.
500. mg x
2.78
49.
X
1g
x 1 mol = 2.78 x 10-3 mol
1000 mg 180.15 g
23
10-3 mol x 6.022 x 10 molecules = 1.67 x 1021 molecules
mol
a.
100 molecules H20 x
b.
100.0 g H20
c.
150 molecules O2 x
a.
150.0 g FeP3 x
c.
a.
1.5
X
°
1 molH
2
= 1.661
6.022 x 1023 molecules H20
1 molH
X
°
2
18.02 g~O
b.10.OmgN02 x
51.
1 atom P
= 4.24 x 1021 atoms P
formula unitN~HP04
Molar mass ofC 6Hs0 6 = 6(12.01) + 8(1.008) + 6(16.00) = 176.12 glmol
500.0 mg x
48.
33
10-22 mol Hp
= 5.549 mol H20
1 molO
2
= 2.491
6.022 x 1023 molecules 02
X
10-22 mol O 2
1 mol
= 0.9393 mol Fe20 3
159.70 g
Ig
x 1 mol =2.17 x I0-4 mo IN0 2
1000 mg 46.01 g
10 16 molecules BF3 x
3.00 x 1020 molecules N 2 x
6.02
x
1 mol
= 2.5 x lO's mol BF3
23
10 molecules
1 mol N 2
28.02 g N
x
2 = 1.40 x 10-2 g N 2
23
6.02 x 10 molecules
mol N 2 .
28.02 gN
b. 3.00 x 10-3 mol N 2 x
2 = 8.41
molN 2
c.
X
X
10-2 g N 2
28.02 gN
1.5 x 102 mol N 2 x
2 = 4.2 x 103 g N 2
molN 2
d. 1 molecule N 2 x
6.022
X
1 mol N 2
28.02 g N
x
2
1023 molecules N 2
mol N 2
= 4.653
X
10-23 g N 2
34
CHAPTER 3
1O- 1s mol N 2 x
e.
2.00
f.
18.0 pmol N 2 x
x
g.5.0nmoIN 2 x
52.
a.
28.02 gN
molN2
2 = 5.60
1 mol N
1
x 10 12
2 x
pmol
1 mol N
2 x
1 x 109 nmol
10- 14 g N 2 = 56.0 fg N 2
X
28.02 gN
2 = 5.04
X
mol N2
28.02 gN
mol N 2
STOICHIOMETRY
10- 10 g N 2 = 504 pg N 2
2=I.4 x lO- 7 gN 2 =140ng
A chemical formula gives atom ratios as well as mole ratios. We will use both ideas to show
how these conversion factors can be used.
Molar mass ofC 2Hs0 2N = 2(12.01) + 5(1.008) + 2(16.00) + 14.01 = 75.07 g/mol
1 mol C2Hs0 2N
6.022 x 1023 molecules C2H s0 2N
x ----------5.00 g C2Hs0 2N x
75.07 g C2Hs0 2N
mol C2Hs0 2N
I atom N
molecule C H
x
2
°
N
S 2
=4.01
22
10 atoms N
x
b. Molar mass of Mg3N 2 = 3(24.31) + 2(14.01) = 100.95 g/mol
5.00 g Mg3N 2 X
1 mol Mg3N 2
X
6.022 x 1023 formula units Mg3N 2
100.95 g Mg3N 2
2 atoms N
X ----
mol Mg3N 2
mol Mg3N 2
= 5.97
c.
Molar mass ofCa(N03)2 = 40.08 + 2(14.01) + 6(16.00) = 164.10 g/mol
5.00 g Ca(N03)2 x
1 mol Ca(N03)2
2 mol N
x
164.10 g Ca(N03)2
6.022 x 1023 atoms N
x -------
mol Ca(N03)2
mol N
= 3.67
d.
°
a.
X
1022 atoms N
Molar mass ofN20 4 = 2(14.01) + 4(16.00) = 92.02 g/mol
5.00 g N 2
53.
1022 atoms N
X
4 X
1 mol N20 4
92.02 g N 20 4
X
2 mol N
mol N 20 4
X
6.022 x 1023 atoms N =.
654 x 1022 at oms N
mol N
14 mol C ( 12.01 g) + 18 mol H ( 1.008 g) + 2 mol N ( 14.01 g) + 5 mol
molC
molH
molN
°(
16.00 g)
molO
= 294.30 g/mol
b.
10.0 g aspartame x
1 mol = 3.40 x 10-2 mol
294.30 g
CHAPTER 3
c.
STOICHIOMETRY
1.56 mol x _29_4_.3_0-=g
mol
d . 5 .0 mg x
e.
35
= 459 g
Ig
I mol x 6.02
x
1000 mg 294.30 g
x
23
10 molecules = I .0 x 10 19 mo Iecu I es
mol
The chemical formula tells us that I molecule of aspartame contains two atoms ofN. The
chemical formula also says that I mol of aspartame contains two mol ofN.
I mol aspartame
2 mol N
6.02 x 1023 atoms N
x
x ------I .2 g aspartame x
294.30 g aspartame mol aspartame
mol N
= 4.9
I mol
x 294.30 g
6.02 x 1023 molecules
mol
f.
1.0 x 109 molecules x
g.
I molecule aspartame x
6.022
54.
a.
I mol
x 294.30 g
1023 molecules
mol
x
10- 13 g or 490 fg
= 4.887
x
10-22 g
Molar mass = 21(12.01) + 30(1.008) + 5(16.00) = 362.45 g/mol
b. 275 mg C21H300S
c.
x
= 4.9
1021 atoms of nitrogen
x
0.600 mol
x
Ig
x
x
1000mg
I mol C21H300S
362.45 gC21H300S
_
-4
-7.59 x 10 mol C21H300S
362.4~ g = 217 g C21H300S
mo
= 4.98
e.
1.00 x 109 molecules x
6.022
f.
x
10 10 atoms H
I mol
x 362.45 g = 6.02 x 10- 13 g C21H300S
23
10 molecules
mol
I mol
x 362.45 g = 6.019
1023 molecules
mol
I molecule x
6.022
x
x
X
10-22 g C21H300S
Percent Composition
55 .
mass OJ'
/0 Cd = mass of Cd in I mol compound
molar mass of compound
CdS: %Cd = 112.4 g Cd
144.5 gCdS
CdSe: %Cd
=
112.4 g
19\.4 g
x
x
x
100
100 = 77.79% Cd
100 = 58.73% Cd; CdTe: %Cd = 112.4 g
240.0 g
x
100 = 46.83% Cd
CHAPTER 3
36
56.
STOICHIOMETRY
a. C3 HPz: Molar mass = 3(12.01) + 4(1.008) + 2(16.00) = 36.03 + 4.032 + 32.00 = 72.06
g/mo1
%C=
36.03gC
x 100=50.00%C; %H=
4.032gH
x 100= 5.595% H
72.06 g compound
72.06 g compound
%0 = 100.00 - (50.00 + 5.595) = 44.41%
°or %0 = 32.00
g x 100 = 44.41% °
72.06 g
b. C4H 60 Z: Molar mass = 4(12.01) + 6(1.008) + 2(16.00) = 48.04 + 6.048 + 32.00 = 86.09 g/mol
%C = 48.04 g x 100 = 55.80% C; %H = 6.048 g x 100 = 7.025% H
86.09 g
86.09 g
%0 = 100.00 - (55.80 + 7.025) = 37.18%
°
c. C3H3N: Molar mass = 3(12.01) + 3(1.008) + 1(14.01) = 36.03 + 3.024 + 14.01 = 53.06 g/mol
%C = 36.03 g x 100 = 67.90% C;
53.06 g
%N = 14.01 g
53.06 g
57.
x
%H = 3.024 g x 100 = 5.699% H
53.06 g
100 = 26.40% N or %N = 100.00 - (67.90 + 5.699) = 26.40% N
In 1 mole ofYBazCU307, there are 1 mole ofY, 2 moles ofBa, 3 moles ofCu and 7 moles ofO.
Molar mass = 1 mol Y ( 88.91 g Y) + 2 mol Ba ( 137.3 g Ba)
molY
molBa
+ 3 mol Cu ( 63.55 g cu) + 7 mol
molCu
Molar mass = 88.91 + 274.6 + 190.65 + 112.00 = 666.2 g/mol
%Y = 88.91 g x 100 = 13.35% Y; %Ba = 274.6 g x 100 = 41.22% Ba
666.2 g
666.2 g
%CU = 190.65 g
666.2 g
58.
x
100 = 28.62% Cu; %0 = 112.00 g
666.2 g
a. NO: %N = 14.01 gN
30.01 gNO
100 = 46.68% N
x
b. N02 : %N =
14.01 gN
46.01 gN0 2
c. N Z0 4 : %N =
28.02 g N
92.02 gN 20 4
d. NzO: %N =
28.02 gN
44.02 gN 20
x
x
100 = 30.45% N
x
100 = 30.45% N
100 = 63.65% N
x
100 = 16.81%
°
°(16.00
g 0)
molO
CHAPTER 3
59.
STOICHIOMETRY
37
C SH ION4 0 2 : molar mass = 8(12.01) + 10(1.008) + 4(14.01) + 2(16.00) = 194.20 g/mol
%C =
8(12.01) g C
194.20 g C gH ION40 2
x
100 = 96.08
194.20
x
100 = 49.47% C
C 12 H2PIl: molar mass = 12(12.01) + 22(1.008) + 11(16.00) = 342.30 g/mol
%C = __
12---,(_12_.0_1-,-)-=g_C_
342.30 g CI2H220t J
x
100 = 42.10% C
C 2HsOH: molar mass = 2(12.01) + 6(1.008) + 1(16.00) = 46.07 g/mol
%C=
2(12.01)gC
46.07 g C2H sOH
x
100=52.14%C
The order from lowest to highest mass percentage of carbon is: sucrose (C 12 H2P11) < caffeine
(CSH ION 40 2) < ethanol (C 2HsOH)
60.
From results in Exercise 58: N0 2 = N 2 0 4 < NO < Np
61.
There are many valid methods to solve this problem. We will assume 100.00 g of compound,
then determine from the information in the problem how many mol of compound equals 100.00 g
of compound. From this information, we can determine the mass of one mol of compound (the
molar mass) by setting up a ratio. Assuming 100.00 g cyanocobalamin:
.
mol cyanocobalamm = 4.34 g Co
x
1 mol Co
58.93 g Co
1 mol cyanocobalamin
mol Co
x --~-----
= 7.36
x g cyanocobalamin =
--=---=--------1 mol cyanocobalamin
62.
x
10-2 mol cyanocobalamin
100.00 g
x = mo Iar mass = 1360 g/mo I
7.36 x 10-2 mol'
There are 0.390 g Cu for every 100.00 g offungallaccase. Assuming 100.00 g fungallaccase:
mo I fiungaIIaccase =.
0 390 g Cu
x
1 mol Cu
63.55 g Cu
x
1mol fungallaccase - .
1 53
4 mol Cu
x g fungallaccase =__1_0_0_.0_0....::g:::.....-_, x = molar mass = 6.54
1 mol fungallaccase
1.53 x 10-3 mol
x
x
10-3 mo I
104 g/mol
Empirical and Molecular Formulas
63.
a.
Molar mass of CH20 = 1 mol C
12.01 g) + 2 mol H ( 1.008 g H)
( molC
molH
+ 1 mol
°( 16.00
g)
molO
= 30.03 g/mol
CHAPTER 3
38
STOICHIOMETRY
%C =
12.01 g C x 100 = 39.99% C; %H = 2.016 g H x 100 = 6.713% H
30.03 g CHp
30.03 g CHzO
%0 =
16.00 gO x 100 = 53.28%
30.03 gCH20
°
or %0 = 100.00 - (39.99 + 6.713) = 53.30%
b. Molar Mass of C6H I20 6 = 6(12.01) + 12( 1.008) + 6(16.00) = 180.16 g/mol
%C=
n.06gC
x 100=40.00%; %H= 12(1.008)g x 100=6.714%
180.16 g C6H 120 6
180.16 g
%0 = 100.00 - (40.00 + 6.714) = 53.29%
c.
Molar mass of HCzHJO z = 2(12.01) + 4(1.008) + 2(16.00) = 60.05 g/mol
%C = 24.02 g x 100 = 40.00%;
60.05 g
%H = 4.032 g x 100 = 6.714%
60.05 g
%0 = 100.00 - (40.00 + 6.714) = 53.29%
64.
All three compounds have the same empirical formula, CHzO, and different molecular formulas.
The composition of all three in mass percent is also the same (within rounding differences).
Therefore, elemental analysis will give us only the empirical formula.
65.
a.
CJ H 40 J
66.
a.
SNH: Empirical formula mass = 32.07 + 14.01 + 1.008 = 47.09 g
188.35
47.09
b.
CH
= 4.000;
c.
CH
d.
P zOs
e. CHzO
f. CHzO
So the molecular formula is (SNH)4 or S4N4H4'
b. NPCl z: Empirical formula mass = 14.01 + 30.97 + 2(35.45) = 115.88 g/mol
347.64
115.88
c.
=
3.0000; Molecular formula is (NPClz)J or NJPJCI6 •
COC40 4: 58.93 + 4(12.01) + 4(16.00) = 170.97 g/mol
341.94
170.97
- - = 2.0000; Molecular formula: COzCsOs
d.
SN: 32.07 + 14.01 = 46.08 g/mol;
184.32
= 4.000; Molecular formula: S4N4
46.08
CHAPTER 3
67.
STOICHIOMETRY
39
Out of 100.0 g of the pigment, there are:
59.9gTi x
ImolTi = 1.25 mol Ti; 40.lgOx ImolO =2.5ImoI0
47.88 g Ti
16.00 gO
Empirical formula = Ti02 since mol 0 to mol Ti is in a 2: I mol ratio (2.51/1.25 = 2.0 I).
68.
Out of 100.00 g of adrenaline, there are:
56.79 g C x
28.37 gO x
I mol C = 4.729 mol C; 6.56 g H x Imol H = 6.51 mol H
12.01 g C
1.008 g H
I molO
= 1.773 mol 0; 8.28 g N x I mol N = 0.591 mol N
14.01gN
16.00 gO
Dividing each mol value by the smallest number:
4.729 = 8.00' ~ = 11.0' 1.773 = 3.00' 0.591 = 1.00
0.591
' 0.591
' 0.591
' 0.591
This gives adrenaline an empirical formula ofCsH 1l 0 3N.
69.
Compound I: mass 0 = 0.6498 g HgxOy - 0.6018 g Hg = 0.0480 g 0
0.6018 g Hg x
I mol Hg = 3.000
200.6 gHg
0.0480 g 0 x 1 mol 0 = 3.00
16.00 gO
X
X
10-3 mol Hg
10.3 mol 0
The mol ratio between Hg and 0 is 1: 1, so the empirical formula of compound I is HgO.
Compound II: mass Hg = 0.4172 g HgxOy - 0.016 gO = 0.401 g Hg
-3
1 mol 0
-3
1 mol Hg
0.401 g Hg x
= 2.00 x 10 mol Hg; 0.016 g 0 x
= 1.0 x 10 mol 0
.
200.6 g Hg
16.00 gO
The mol ratio between Hg and 0 is 2: 1, so the empirical formula is Hg20.
70.
1.121 g N x 1 mol N = 8.001
14.01 g N
0.480 g C x
I mol C = 4.00
12.01 g C
X
X
10-2 mol N; 0.161 g H x 1 mol H = 1.60
1.008 g H
10-2 mol C; 0.640 g 0 x 1 molO = 4.00
16.00 g 0
Dividing all mol values by the smallest number:
2
8.001 x 10- = 2.00;
4.00 x 10-2
1
2
1.60 x 10- = 4.00; 4.00 x 10- = 1.00
4.00 x 10-2
4.00 x 10-2
X
X
10. 1 mol H
10-2 mol 0
CHAPTER 3
40
71.
Out of 100.0 g compound: 30.4 g N
x
1 molN
= 2.17 mol N
14.01 gN
%0 = 100.0 - 30.4 = 69.6% 0; 69.6 gO
2.17 = 1.00;
2.17
STOICHIOMETRY
x
1 mol 0
= 4.35 mol 0
16.00 gO
4.35 = 2.00; Empirical fonnula is N02 •
2.17
The empirical formula mass ofN02
'"
14 + 2(16) = 46 g/mol.
92 g = 2.0; Therefore, the molecular fonnula is NP4'
46g
72.
Out of 100.0 g, there are:
69.6gS x
1 mol S
1 molN
=2.17moIS; 30.4gNx
=2.17moIN
32.07 g S
14.01 gN
Empirical fonnula is SN since mol values are in a I: I mol ratio.
The empirical fonnula mass of SN is - 46 g. Since 184 = 4.0, then the molecular fonnula is S4N4'
46
73.
Assuming 100.00 g of compound:
7.74 g H x
1 mol H
I molC
= 7.68 mol H; 92.26 g C x
= 7.682 mol C
1.008gH
12.01 gC
The mole ratio between C and H is 1: 1 so the empirical fonnula is CH.
empirical fonnula mass = 12.01 + 1.008 = 13.02 g/mol
= ~ = 6.00
molar mass
empirical fonnula mass 13.02
molecular fonnula = (CH)6 = C6H6
74.
Assuming 100.00 g of compound (mass hydrogen = 100.00 g - 49.31 g C - 43.79 gO = 6.90 g H):
49.31 g C x
43.79 g 0 x
1 mol C
I mol H
= 4.106 mol C; 6.90 g H x
= 6.85 mol H
12.01 g C
1.008 g H
1 molO = 2.737 mol 0
16.00 g 0
Dividing alI mole values by 2.737 gives:
4.106 = 1.500' 6.85 = 2.50' 2.73 7 = 1.000
2.737
' 2.737
' 2.737
Since a whole number ratio is required, then the empirical fonnula is C3Hs0 2•
CHAPTER 3
41
STOICHIOMETRY
The empirical formula mass is: 3(12.01) + 5(1.008) +2(16.00) = 73.07 g/mol
molar mass
empirical formula mass
75.
=
°
146.1 = 1.999', mo Iecu Iar fiormu Ia = (C3 H50)
2 2 = C 6 H 10
73.07
4
When combustion data is given, it is assumed that all the carbon in the compound ends up as
carbon in CO2 and all the hydrogen in the compound ends up a hydrogen in H20. In the sample
of propane combusted, the moles of C and Hare:
mol C = 2.641 g CO2
1 mol CO
X
mol H = 1.442 g H20 x
mol H
mol C
=
2
44.01 g CO2
x
°
1 mol C
mol CO2
= 0.06001 mol C
I molH
2
IH
2
x mo
= 0.1600 mol H
18.02 g H20
mol H20
0.1600 = 2.666
0.06001
Multiplying this ratio by three gives the empirical formula of C3Hs.
76.
This compound contains nitrogen and one way to determine the amount of nitrogen in the
compound is to calculate composition by mass percent. We assume that all of the carbon in 33.5
mg CO2 came from the 35.0 mg of compound and all of the hydrogen in 41.1 mg Hp came from
the 35.0 mg of compound.
3.35 x 10-2 g CO
X
2
1 mol CO2 x 1 mol C x 12.01 g C = 9.14
44.01 g CO2 mol CO2
mol C
9.14x 1O- gC
x 100=26.1%C
3.50 x 10-2 g compound
4.11
10-2 g H20
%H =
10-3 g C
3
%C=
X
X
X
°
1 mol H2
x 2 mo I H x1' 008 g H = 4.60
18.02 g H20
mol ~O
mol H
X
10-3 g H
3
x 100 = 13.1% H
4.60 x 10- gH
3.50 x 10-2 g compound
The mass percent of nitrogen is obtained by difference:
%N = 100.0 - (26.1 + 13.1) = 60.8% N
Now perform the empirical formula determination by first assuming 100.0 g of compound. Out
of 100.0 g of compound, there are:
26.1gCx
ImolC =2.17 molC; 13.1gHx ImolH =13.0moIH
12.01 g C
1.008 g H
60.8 g N x
1 mol N = 4.34 mol N
14.01 gN
CHAPTER 3
42
Dividing all mol values by 2.17 gives: 2.17 = 1.00; 13 .0
2.17
2.17
=
5.99; 4.34
2.17
=
STOICHIOMETRY
2.00
The empirical formula is CH6N 2 •
77.
The combustion data allows determination of the amount of hydrogen in cumene. One way to
determine the amount of carbon in cumene is to determine the mass percent of hydrogen in the
compound from the data in the problem, then determine the mass percent of carbon by difference
(100.0 - mass %H = mass %C).
42.8 mg H2
%H =
°
x
1g
1000 mg
4.79 mgH
47.6 mg cumene
X
2.016 g H
18.02 g H20
x
1000 mg =.
4 79 mg H
g
100 = 10.1% H; %C
x
=
100.0 - 10.1
=
89.9% C
Now solve this empirical formula problem. Out of 100.0 g cumene, we have:
89.9 g C
10.0
7.49
=
1 mol C
12.01gC
x
.±,
3
1.34 '"
=
7.49 mol C; 10.1 g H
x
1 molH
1.008gH
=
10.0 mol H
i.e., mol H to mol C are in a 4:3 ratio. Empirical formula = C3H4
Empirical formula mass'" 3(12) + 4(1) = 40 g/mol
The molecular formula is (C3H4)3 or C9H 12 since the molar mass will be between 115 and 125
g/mol (molar mass'" 3 x 40 g/mol = 120 g/mol).
78.
First, we will determine composition by mass percent:
16.01 mg CO2
%C =
x
1g
1000 mg
x
4.369 mgC
10.68 mg compound
. mg HO
2
437
x
Ig
1000 mg
x
12.01 gC
44.01 g CO2
x
x
1000 mg = 4.369 mg C
g
100 = 40.91% C
2.016gH
18.02 g H20
X
1000mg =.
0489 mg H
g
%H = 0.489 mg x 100 = 4.58% H; %0 = 100.00 - (40.91 + 4.58) = 54.51%
10.68 mg
So, in 100.00 g of the compound, we have:
40.91 g C
x
54.51 gO x
1 mol C
12.01 g C
= 3.406 mol C;
°
°
1 mol
= 3.407 mol
16.00 gO
4.58 g H
x
1 molH
1.008 g H
= 4.54 mol H
°
CHAPTER 3
43
STOICHIOMETRY
Dividing by smallest number:
4.54 = 1.33 = .i. Therefore, empirical formula is C 3H40 3.
3.406
3'
The empirical formula mass ofC]HP3 is'" 3(12) + 4(1) + 3(16) = 88 g.
Since 176.1
88
= 2.0, then the molecular formula is C6 Ha0 6 •
Balancing Chemical Equations
79.
When balancing reactions, start with elements that appear in only one of the reactants and one of
the products, then go on to balance the remaining elements.
a.
°
Fe + Oz ---+ Fez0 3 • Balancing Fe first then gives: 2 Fe + 3/2 Oz ---+ FezO]. The best
balanced equation contains smallest whole numbers. To convert to whole numbers, multiply
each coefficient by two which gives: 4 Fe(s) + 3 Oz (g) ---+ 2 FezOls)
b. Ca + HzO ---+ Ca(OH)z + Hz; Calcium is already balanced, so concentrate on oxygen next.
Balancing gives: Ca(s) + 2 HzO(l) ---+ Ca(OH)z(aq) + Hz (g). The equation is balanced.
Note: Hydrogen is the most difficult element to balance since it appears in both products. It
is generally easiest to save these atoms for last when balancing an equation.
°
c.
°
Ba(OH)z + H ZS04 ---+ BaS04 + HzO; Ba and S are already balanced. There are 6 atoms on
the reactant side and in order to get 6 atoms on the product side, we will need 2 HzO
molecules. The balanced equation is: Ba(OH)z(aq) + H ZS04(aq)---+ BaS04(s) + 2 HzO(l).
°
Equation is balanced.
Balance Fe atoms: FeZS3 + HCI ---+ 2 FeCI] + HzS
Balance S atoms: FeZS] + HCI ---+ 2 FeCI] + 3 HzS
There are 6 Hand 6 Cion right, so balance with 6 HCI on left:
Fe 2S](s) + 6 HCI(g) ---+ 2 FeCI)(s) + 3 H2S(g). Equation is balanced.
CHAPTER 3
44
c.
CS 2(1) + NH3(g)
---+
STOICHIOMETRY
H 2S(g) + NH 4SCN(s)
C and S balanced; balance N:
CS2 + 2 NH3 ---+ H 2S + NH4SCN
H is also balanced. So: CSil) + 2 NH3(g)
81.
82.
83.
a.
Cu(s) + 2 AgN0 3(aq)
---+
---+
H 2S(g) + NH4SCN(s)
2 Ag(s) + Cu(N03)iaq)
b. Zn(s) + 2 HCl(aq)
---+
ZnCliaq) + Hig)
c.
Au 2SJCs) + 3 H 2(g)
---+
2 Au(s) + 3 H 2S(g)
a.
3 Ca(OH)iaq) + 2 H3POiaq)
b.
Al(OH)ls) + 3 HCl(aq)
c.
2 AgNOJCaq) + H 2SOiaq)
---+
a.
CI2H22011(S) + 12 02(g)
12 CO2(g) + 11 HP(g)
b.
C6H 6(1) +
---+
---+
---+
6 Hp(I) + CaJCP04)is)
AlCl 3(aq) + 3 Hp(l)
Ag 2S04(S) + 2 HNOJCaq)
J2 0ig) ---+ 6 COig) + 3 HP(g);
2
2 C6H 6(1) + 15 0ig)
c.
3
2 Fe + - 02
2
d.
C4H IO +
---+
---+
Multiply by two to give whole numbers.
12 COig) + 6 HP(g)
FeP3; For whole numbers: 4 Fe(s) + 3 0ig)
1.2 02 ---+ 4 CO 2+ 5 HP;
2
---+
2 FepJCs)
Multiply by two to give whole numbers.
2 C 4H IO(g) + 13 0ig) ---+ 8 CO 2(g) + 10 HP(g)
e.
2 FeO(s) +
1. 02(g) ---+ FepJCs);
2
4 FeO(s) + 02(g)
84.
a.
---+
For whole numbers, multiply by two.
2 Fe20 3(s)
16 Cr(s) + 3 S8(S) ---+ 8 Cr2S3(s)
Na2COb) + COig) + HP(g)
b. 2 NaHCOb)
---+
c.
2 KClOb)
2 KCl(s) + 3 02(g)
d.
2 Eu(s) + 6 HF(g)
---+
---+
2 EuFb) + 3 Hig)
CHAPTER 3
85.
a.
45
STOICHIOMETRY
Si02(s) + C(s)
Si(s) + CO(g)
~
Balance oxygen atoms: Si02 + C
~
Si + 2 CO
Balance carbon atoms: SiOis) + 2 C(s) ~ Si(s) + 2 CO(g)
b. SiCIll) + Mg(s)
~
Si(s) + MgCI 2(s)
Balance CI atoms: SiCl4 + Mg
~
Si + 2 MgCl 2
Balance Mg atoms: SiCI 4(l) + 2 Mg(s)
c.
N~SiF6(s)
~
+ Na(s) ~ Si(s) + NaF(s)
Na2SiF6 + Na ~ Si + 6 NaF
Balance F atoms:
Balance Na atoms: Na2 SiFis) + 4 Na(s)
86.
Si(s) + 2 MgClis)
~
Si(s) + 6 NaF(s)
Unbalanced equation:
Balancing Ca2+, F-, and
pot:
On the right hand side there are 20 extra hydrogen atoms, 10 extra sulfates, and 20 extra water
molecules. We can balance the hydrogen and sulfate with 10 sulfuric acid molecules. The extra
waters came from the water in the sulfuric acid solution. The balanced equation is:
Note: The insecticide used is PbHAs0 4 and is commonly called lead arsenate. This is not the
correct name, however. Correctly, lead arsenate would be Pb3(As04)2 and PbHAs0 4 should be
named lead hydrogen arsenate.
88.
2 NaCI(s) + 2 H20(l)
~
CI2(g) + H2(g) + 2 NaOH(aq)
Reaction Stoichiometry
89.
The stepwise method to solve stoichiometry problems is outlined in the text. Instead of
calculating intermediate answers for each step, we will combine conversion factors into one
calculation. This practice reduces round-off error and saves time.
CHAPTER 3
46
I mol (NH4)zCrZ0 7 _
10.8 g (NH4)2CrP7 x
4.28 x 10-2 mol (NH4)2Cr20 7 x
4.28 x 10-2 mol (NH4)2CrP7 x
4.28
X
-2
4.28 x 10 mol (NH4)2CrP7
-
252.08 g
I mol Cr
°
152.00 g Cr
3 x
mol (NH4)zCrZ0 7
Z
I mol N
28.02 g N
mol N z
mol (NH4)zCrZ0 7
4 mol HzO
FepJCs) + 2 AI(s)
15.0 g Fe x
---+
0.269 mol Fe x
91.
1.000 kg Al x
92.
a.
I mol Fez0 3
2 mol Fe
I mol Al z0 3
x
mol Fez0 3
101.96 g Al z0 3
1000 g Al
x
kg Al
mol AIP3
I mol Al
x
=:
x
°
3.09 g H20
2 mol Al
2 mol Fe
x
26.98 gAl = 7.26 g Al
mol Al
= 13.7 g AIP3
117.49 g NH4CI04
x
3 mol Al
---+
x
= 21.5 g FeP3
mol NH4CI04
3 I 5.4 g
= 0.0206 mol = 0.02 I mol
2 mol NH4SCN
I mol Ba(OH)/8HzO
x
76.13 g NH4SCN
mol NH4 SCN
= 3.2 g NH4SCN
907 kg x 1000g x I molCuO x I molC x 12.01 gC x 100. gcoke
ton
kg
79.55 g CuO 2 mol CuO
mol C
95 gC
= 7.2
94.
=4355g
Ba(SCNMs) + 10 Hp(l) + 2 NH3(g)
I mol Ba(OH)ze8HP
0.021 mol Ba(OH)2e8H20 x
x
mol HzO
3 mol NH4CI04
26.98 gAl
Ba(OH)2e8HP(s) + 2 NH4SCN(s)
I .Oton C u
= 1.20 g N2
18.02 g HzO
0.269 mol Fe
159.70 g Fez0 3
x
2 mol Fe
b. 6.5 g Ba(OH)2e8H20
93.
3 = 6.51 g Cr20 3
2 Fe(l) + AI 20JCs)
I mol Fe
= 0.269 mol Fe;
55.85 g Fe
0.269 mol Fe x
z
x
mol (NH4)zCrp7
90.
°
Z
mol CrZ0 3
x
Z
10-2 mol (NH4)2CrP7 x
STOICHIOMETRY
1.0 x 104 kg waste x
3.0 kg NH4+
100 kg waste
x
x
1000 g
kg
x
I mol NH4+
18.04 g NH4+
X
104 g coke
I mol C H70 N
z
S
x ---55 mol NH4+
3 2
I 1 .I g C SH70 zN = 3.4 x 104 g tissue if all NH4+ converted
mol C SH70 zN
CHAPTER 3
47
STOICHIOMETRY
Since only 95% of the NH4+ ions react:
mass of tissue = (0.95) (3.4
95.
a.
x
104 g) = 3.2
X
104 g or 32 kg bacterial tissue
Molar mass = 195.1 + 2(14.01) + 6(1.008) + 2(35.45) = 300.1 g/mol
%Pt= 195.1g x 100=65.01%Pt; %N= 28.02g x 100=9.337%N;
300.1 g
300.1 g
% H = 6.048 g
300.1 g
x
100 = 2.015% H; % Cl = 70.90 g
300.1 g
x
100 = 23.63% Cl
65.01% Pt; 9.337% N; 2.015% H; 23.63% Cl
= 72.3 g Pt(NH3)2CI2
100. g K PtCl
4
2
x
1 mol ~PtCI4
415.1 g ~PtCI4
x
_2_m_o_lK_C_l_
mol ~PtCI4
x
74.55 g KCl = 35.9 g KCl
mol KCl
96.
= 1.30
x
102 g aspirin
Limiting Reactants and Percent Yield
97.
a.
Mg(s) + Iz(s)
~
Mglz(s)
From the balanced equation, 100 molecules of Iz reacts completely with 100 atoms of Mg.
We have a stoichiometric mixture. Neither is limiting.
b.
c.
150 atoms Mg
x
1 molecule 12
1 atom Mg
= 150 molecules Iz needed
We need 150 molecules Iz to react completely with 150 atoms Mg; we only have 100
molecules 12, Therefore, 12 is limiting.
1 molecule 12
200 atoms Mg x
= 200 molecules 12.; Mg is limiting since 300 molecules 12
1 atom Mg
are present.
CHAPTER 3
48
d. 0.16 mol Mg x
1 mol 12
1 molMg
1 mol I
= 0.16 mol 12; Mg is limiting since 0.25 mol 12 are present.
e.
0.14 mol Mg x
f.
0.12 mol Mg x
g.
6.078 g Mg x 1 mol Mg x 1 mol 12
253.8 g 12
24.31 g Mg 1 mol Mg x mol 1 = 63.46 g 12
2
2
1 molMg
STOICHIOMETRY
= 0.14 mol 12 needed; Stoichiometric mixture. Neither is limiting.
1 mol I
2 = 0.12 mol 12 needed; 12 is limiting since only 0.08 mol 12 are
1 mol Mg
present.
Stoichiometric mixture. Neither is limiting.
h.
1.00gMgx
1 mol Mg
24.31 g Mg
x
1 mol 12
1 mol Mg
x
253.8 g 12
mol 12
=10.4g1 2
10.4 g 12 needed, but we only have 2.00 g. 12 is limiting.
I.
98.
From h above, we calculated that 10.4 g 12 will react completely with 1.00 g Mg. We have
20.00 g 12. 12 is in excess. Mg is limiting.
2 H2(g) + 02(g) -+ 2 HP(g)
a.
50 molecules H2 x
1 molecule
°2 = 25 molecules O2
2 molecules H2
Stoichiometric mixture. Neither is limiting.
b.
c.
100 molecules H2 x
1 molecule 02
2 molecules H2
= 50 molecules 02; O2 is limiting since only 40 molecules
O2 are present.
From b, 50 molecules of02 will react completely with 100 molecules ofH2. We have 100
molecules (an excess) of02. So, H2 is limiting.
d. 0.50 mol H2 x
e.
0.80 mol H2 x
f.
1.0 g H2 X
1 mol 02
2molH2
1 mol 02
2 molH 2
= 0.25 mol 02; H2 is limiting since 0.75 mol O2 are present.
= 0.40 mol 02; H2 is limiting since 0.75 mol O2 are present.
1 molH
1 molO
2 x
2 = 0.25 mol O2
2.016 g H2 2 mol H2
Stoichiometric mixture, neither is limiting.
CHAPTER 3
g.
5.00 g Hz x
1 mol Hz
x
2.016 g Hz
99.
a.
49
STOICHIOMETRY
mol Ag = 2.0 gAg
moIS s =2.0gS s x
x
1 molO z
2 mol Hz
32.00 g 0z
x
molOz
1 mol Ag = 1.9
107.9 gAg
1 mol S
g =7.8
256.56 g Sg
X
x
= 39.7 g 0z; Hz is limiting since
56.00 g Oz are present.
10-z mol Ag
l0-3 moIS s
From the balanced equation the required mol Ag to mol Ss ratio is 16: 1. The actual mol
ratio is:
z
1.9 x lO- mol Ag = 2.4
7.8 x 10-3 mol Sg
This is well below the required ratio so Ag is the limiting reagent.
1.9
10- z mol Ag
X
8 mol AgzS
247.9 g AgzS
x
x
16 mol Ag
b.
1.9
10-z mol Ag
X
1 mol Sg
256.56 g Sg
x
= 2.4 g AgzS
mol AgzS
x
16 mol Ag
= 0.30 g Ss
mol Sg
0.30 g Ss are required to react with all of the Ag present.
Ss in excess = 2.0 g Ss - 0.30 g Ss = 1.7 g Ss in excess
100.
a.
10.0 g Hg
x
9.00 g Brz x
1 mol Hg = 4.99
200.6 gHg
1 mol Br
X
lO- z mol Hg
'
z = 5.63
159.80 g Brz
X
10-z mol Brz
The required mol ratio from the balanced equation is 1 mol Brz to 1 mol Hg. The actual mol
ratio is:
5.63 x lO- z mol Br = 1.13
4.99 x 10- z mol Hg
This is higher than the required ratio so Hg is the limiting reagent.
4.99 x 10-z mol Hg x
1 mol HgBrz
mol Hg
4.99
x
IO- z mol Hg
x
1 mol Brz
1 mol Hg
x
360.4 g HgBrz
mol HgBrz
x
159.80 g Brz
mol Br2
excess Brz = 9.00 g Brz - 7.97 g Brz = 1.03 g Brz
= 18.0 g HgBr z produced
= 7.97 g Brz reacted
CHAPTER 3
50
b.
STOICHIOMETRY
5.00 mL Hg x 13.6 g Hg x 1 mol Hg = 0.339 mol Hg
mL Hg
200.6 g Hg
5.00 mL Br2
3.10gBr2
ImolBr2
x
= 0.0970 mol Br2
mLBr2
159.80gBr2
x
Br2 is limiting since the actual moles of Br2 present is well below the required 1: 1 mol ratio.
0.0970 mol Br2 x
101.
1 mol HgBr2
---+
103 g Ca3(P04 )2
x
x
360.4 g HgBr2
mol Br2
Ca3(P04 )2 + 3 H2 S04
1.0
x
mol HgBr2
= 35.0 g HgBr2 produced
3 CaS04 + 2 H3P0 4
1 mol Ca (PO)
3
4 2
= 3.2 mol Ca3(P04)2
310.18 g Ca3(P04)2
1.0 x 103 g cone. H2 S04 x
98 g H2S04
100 g cone.
~S04
x
1 mol H2S04
98.09 g H2S04
=
10. mol H2 S04
The required mol ratio from the balanced equation is 3 mol H2 S04 to 1 mol CaJCP04 )2' The
10. mol H2S04
actual ratio is:
= 3.1
3.2 mol Ca3(P04)2
This is higher than the required mol ratio so CaJCP04)2 is the limiting reagent.
3.2 mol Ca3(P04 )2
x
x
mol Ca3(P04)2
3.2 mol Ca3(P0 4 )2 x
102.
3moiCaS04
2 mol H3P04
136.15gCaS04
mol CaS04
x
mol Ca3(P04)2
97.99 g H3P04
mol H 3P04
= 1300 g CaS04 produced
= 630 g H3P04 produced
An alternative method to solve limiting reagent problems is to assume each reactant is limiting
then calculate how much product could be produced from each reactant. The reactant that
produces the smallest amount of product will run out first and is the limiting reagent.
5.00 x 106 g NH 3
5.00 x 106 g O2
X
5.00 x 106 g CH 4
1 molNH
X
3
17.03 g NH 3
x
2
lHCN
mo
= 2.94 x 105 mol HCN
2 mol NH 3
1 mol °2 x 2 mol HCN = 1.04 x 105 mol HCN
32.00 g 02
3 mol 02
1 mol CH
X
2
I HCN
4 x mo
= 3.12 x 105 mol HCN
16.04 g CH4 2 mol CH4
O2 is limiting since it produces the smallest amount ofHCN. Although more product could be
produced from NH 3 and CH4 , only enough O2 is present to produce 1.04 x 105 mol HCN. The
mass of HCN produced is:
CHAPTER 3
1.04
5.00
51
STOICHIOMETRY
X
X
lOs mol HCN x 27.03 g HCN = 2.81 x 106 g HCN
molHCN
106 g O 2
1 mol 02
X
X
6 mol H 20
32.00 g 02
x
3 mol 02
18.02 g H
2
°
=
5.63
X
1 mol H 20
106 g HzO
The balanced equation requires a 1: 1 mol ratio between reactants. 9.17 mol of C 2H 6 will react
with all of the CI 2 present (9.17 mol). Since 9.98 mol C 2H 6 is present, then Cl 2 is the limiting
reagent.
The theoretical yield of C 2H sCI is:
1 mol C2H sCI
9.17 mol CI 2 x
Percent yield
=
64.51 g C 2H sCI
mol C 2H sCI
mol Cl 2
= 592 g C 2H sCI
actual x 100 = 490. g x 100 = 82.8%
theoretical
592 g
1.50 g C 7H6 0 3 x
2.00 g C4 H 6 0 3
x
1 mol C 7H 60 3
138.12 gC 7H6 0 3
1 molC H
4
X
°
6
3
= 1.09 X 10-2 mol C 7HP3
.=
1.96
X
102.09 g C4H 6 0 3
10- 2 mol C 4H6 0 3
C 7H 60 3 is the limiting reagent since the actual moles of C 7H 60 3 is below the required 1: 1 mol
ratio. The theoretical yield of aspirin is:
1.09 x 10-2 mol C 7 H6 0 3 x
% yield = 1.50 g
1.96 g
105.
6.0 g Al x
x
=
mol C 7H6 0 3
180.15 g C9 H S0 4
x ------
mol C 9 H S0 4
= 1.96 g C9Hg04
100 = 76.5%
1 mol AI
26.98 mol AI
% yield
1 mol C 9 H S0 4
x
2 mol AIBr3
2 mol Al
50.3 g x 100 = 85%
59 g
x
266.68 g AIBr3
mol AIBr3
=
59 g AIBr3
52
106.
CHAPTER 3
P4(s) + 6 F2(g)
~
STOICHIOMETRY
4 PF3(g); The theoretical yield ofPF 3 is:
100.0 gPF 3 (theoretical)
120. g PF3 (actual) x
78.1 g PF3 (actual)
=
154 g PF 3 (theoretical)
1 mol PF3
6 mol F2
38.00 g F2
= 99.8 g F2
154 g PF 3 x 87.97 g PF x 4 mol PF x
mol F2
3
3
99.8 g F2 are needed to produce an actual PF 3 yield of78.1%.
Additional Exercises
107.
9.123 X 10-23 gx 6.022 x 1023 atom
atom
mol
='
54.94g
mol
The atomic mass is 54.94 amu. From the periodic table, the element is manganese (Mn).
108.
a.
2(12.01) + 3(1.008) + 3(35.45) + 2(16.00) = 165.39 g/mol
b.
500.0 g x
d.
5.0 g C H ClP2
2 3
1 mol
= 3.023 mol;
165.39 g
c.
. x 10-2 rna I x 165.39g = 33
. g
20
mol
23
1 mol x 6.02 x 10 molecules x 3 atoms CI
165.39 g
mol
molecule
X
=
I mol CI
e.
1.0 g CI x
f.
500 molecules x
35.45 g
x
1 mol C2H3Cl30 2
3 mol Cl
I mol
x
23
6.022 x 10 molecules
109.
165.39 g C2H3CI 30 2
mol C2H3Cl 30 2
165.39 g
=
=
X
1022 atoms of chlorine
1.6 g chloral hydrate
19
1.373 x 10- g
mol
Empirical formula mass = 12.01 + 1.008 = 13.02 glmol; Since 104.14/13.02 = 7.998 : : : 8, then the
molecular formula for styrene is (CH)s = CsHs.
2 .00 g CsHs x
110.
x
5.5
925 x 1022 at oms H
I mol CgHg x 8 mol H x 6.022 x 1023 atoms H =.
mol H
104.14 g CgHg mol CgHg
41.98 mg CO2 x
6.45 mg Hp x
12.01 mg C = 11.46 mg C; %C = 11.46 mg x 100 = 57.85% C
44.01 mg CO2
19.81 mg
2.016 mg H = 0.722 mg H; %H = 0.722 mg x 100 = 3.64% H
18.02 mg H20
19.81 mg
%0 = 100.00 - (57.85 + 3.64) = 38.51 %
°
CHAPTER 3
53
STOICmOMETRY
Out of 100.00 g terephthalic acid, there are:
57.85gCx
ImolC =4.817moIC; 3.64gHx ImolH =3.6ImoIH
12.01 g C
1.008 g H
°
= 2.407 mol
38.51 gO x I mol
16.00 gO
°
4.817 = 2.001' ~ = 1.50' 2.407 = 1.000
2.407
' 2.407
' 2.407
C:H:O mol ratio is 2: 1.5: I or 4:3 :2. Empirical formula: C4H30 2
Mass ofC4H30 2
III.
4(12) + 3.(1) + 2(16) = 83;
::::
166
= 2; Molecular formula: CSH6 0 4
83
Assuming 100.00 g E3Hs:
mol E = 8.73 g H x I mol H x 3 mol E = 3.25 mol E
1.008 g H 8 mol H
x gE
I mol E
=
91.27 g E , x = molar mass ofE = 28.1 g/mol; atomic mass ofE = 28.1 amu
3.25 mol E
0.483 g CuS04 x
0.272 g H20 x
0.0151
I molCuS0 4
159.62 g CuS04
I mol H2 0
18.02 g H20
mol~O
= 0.00303 mol CuS04
= 0.0151 mol H20
4.98 mol H20
----; Compound formula = CuS0 4-5H20, x = 5
I mol CuS04
0.00303 mol CuS04
lB.
2.00 xl 06 g CaCO3x
114.
a.
I mol CaC03
100.09 g CaC03
x
I mol CaO x 56.08 g CaO -_ 1.12 x 106 g CaO
mol CaC03
mol CaO
Only acrylonitrile contains nitrogen. Ifwe have 100.00 g of polymer:
8.80 g N x
% C3 H3N =
I mol C3H3N
----=-------'-----
14.01gN
x
53.06 g C3H3N
ImolC 3H3N
33.3 gC H N
3_3_
100.00 g polymer
= 33.3 g C3H3N
= 33.3% C3H3 N
54
CHAPTER 3
STOICHIOMETRY
Only butadiene in the polymer reacts with Br2:
1 mol Br2
0.605 g Br2 x
x
1 mol C4 H6
159.80 g Br2
x
54.09 g C4 H6
mol Br2
mol C4 H6
= 0.205 g C 4H6
%C 4H6 = 0.205g x 100= 17.1%C4 H6
1.20 g
b. Ifwe have 100.0 g of polymer:
33.3 g C3H3N x
17.1 g C4 H6 x
1 molC 3H3N
53.06 g
1 mol C4H6
54.09 gC 4H6
49.6 g CsHs x
Dividin b 0.316:
g y
= 0.628 mol C3H3N
= 0.316 mol C 4H6
1 molCgH g
104.14 g CgH g
= 0.476 mol CsHs
0.628 = 1.99' 0.316 = 1.00' 0.476 = 1.51
0.316
' 0.316
' 0.316
This is close to a mol ratio of 4:2:3. Thus, there are 4 acrylonitrile to 2 butadiene to 3
styrene molecules in the polymer or (A4B2S3)n'
115.
4 Al(s)+ 3 02(g)
a.
---+
1.0 mol Al x
b. 2.0 mol Al x
c.
0.50 mol Al x
d. 64.75 g Al
x
e.
75.89 g Al
x
f.
51.28 g Al
x
2 AI 20 3(s)
°
3 mol
2 = 0.75 mol 02; AI is limiting since 1.0 mol O2 is present.
4 mol Al
3 molO
2
4 molAl
= 1.5 mol 02; Al is limiting since 4.0 mol O2 is present.
3 molO
2 = 0.38 mol 02; Al is limiting since 0.75 mol O2 is present.
4 mol Al
1 mol Al
x
26.98 gAl
1 mol Al
x
4 mol Al
x
26.98 gAl
1 mol Al
26.98 gAl
3 mol 02
x
°2
4 mol Al
3 mol °
2
3 mol
4 mol Al
32.00 g 02
mol 02
x
32.00 gO
mol 02
x
= 57.60 g 02; Al is limiting.
2 = 67.51 g 02; Al is limiting.
°
32.00 g
2 = 45.62 g 02; Al is limiting.
mol 02
CHAPTER 3
116.
55
STOICHIOMETRY
a.
CH4(g) + 4 S(s)
b.
120. g CH4 x
~
CSll) + 2 HzS(g) or 2 CHig) + Sg(s) ~ 2 CSz(l) + 4 HzS(g)
1 mol CH4
16.04 g CH4
= 7.48 mol CH4 ;
1 mol S
120. g S x
= 3.74 mol S
32.07 g S
The required S to CH4 mol ratio is 4: 1. The actual S to CH4 mol ratio is:
3.74 mol S
7.48 mol CH4
=
0.500
This is well below the required ratio so sulfur is the limiting reagent.
The theoretical yield ofCSz is: 3.74 mol S x
I mol CS
4 mol S
76.15 g CS
2 =
2 x
mol CS 2
71.2 g CS z
The same amount of CS z would be produced using the balanced equation with Sg.
117.
453 g Fe x
1 mol Fe
55.85 g Fe
mass % Fez0 3 =
118.
a.
x
1 mol Fe20 3
2 mol Fe
648 g Fe 03
2
752 gore
x
x
159.70 g Fe20 3
=
648 g Fep3
100 = 86.2%
Mass of Zn in alloy = 0.0985 g ZnCl z x
%Zn =
mol Fep3
65.38 g Zn
136.28 g ZnCI 2
=
0.0473 g Zn
0.0473 gZn
x 100 = 9.34% Zn; %Cu = 100.00 - 9.34 = 90.66% Cu
0.5065 g brass
b. The Cu remains unreacted. After filtering, washing, and drying, the mass of the unreacted
copper could be measured.
119.
Assuming one mol of vitamin A (286.4 g Vitamin A):
mol C = 286.4 g Vitamin A x 0.8386 g C x 1 mol C = 20.00 mol C
g Vitamin A 12.01 g C
mol H = 286.4 g Vitamin A x 0.1056 g H x 1 mol H
g Vitamin A 1.008 g H
=
30.00 mol H
Since one mol of Vitamin A contains 20 mol C and 30 mol H, then the molecular formula of
Vitamin A is Czo H30E. To determine E, lets calculate the molar mass ofE.
286.4 g = 20(12.01) + 30(1.008) + molar mass E, molar mass E = 16.0 g/mol
From the periodic table, E = oxygen and the molecular formula of Vitamin A is CZOH 300.
CHAPTER 3
56
STOICHIOMETRY
Challenge Problems
120.
8SRb
- - = 2.591; Assuming 100 atoms, let x = number of 8sRb atoms and 100 - x = number of 87Rb
87Rb
atoms.
~ = 2.591 x = 259.1 - 2.591 x x = 259.1 = 72.15% 8sRb
100 - x '
,
3.591
0.7215 (84.9117) + 0.2785 (A) = 85.4678, A = 85.4678 - 61.26 = 86.92 amu = atomic mass of 87Rb
0.2785
121.
First, we will determine composition in mass percent. We assume all of the carbon in 0.213 g
CO2 came from 0.157 g of the compound and that all ofthe hydrogen in the 0.0310 g H20 came
from the 0.157 g of the compound.
0.213 g CO2
X
12.01 g C
44.01 g CO2
0.0310 g H20 x
= 0.0581 g C; %C =
0.0581 g C
x 100 = 37.0% C
0.157 g compound
3
2.016 g H = 3.47
18.02 g H20
X
10-3 g H; %H = 3.47 x 10- g = 2.21 % H
0.157 g
We get %N from the second experiment:
0.0230
g
NH x
3
14.01 gN = 1.89 x 10-2 N
17.03 g NH
g
3
2
%N = 1.89 x 10- g x 100 = 18.3% N
0.103 g
The mass percent of oxygen is obtained by difference:
%0 = 100.00 - (37.0 + 2.21 + 18.3) = 42.5%
So out of 100.00 g of compound, there are:
37.0 g C x
1 mol C
1 mo1H
= 3.08 mol C; 2.21 g H x
= 2.19 mol H
12.01 g C
1.008 g H
18.3gNx
1 molN
1 molO
=1.31moIN; 42.5g0 x
=2.66moI0
14.01gN
16.00g0
Lastly, and often the hardest part, we need to find simple whole number ratios. Divide all mole
values by the smallest number:
3.08
1.31
=
2.35;
2.19 = 1.67;
1.31
1.31 = 1.00;
1.31
2.66 = 2.03
1.31
Multiplying all these ratios by 3 gives an empirical formula of C7HsNP6'
CHAPTER 3
122.
1.0
x
57
STOICHIOMETRY
106 kg HN0 3
X
1000 g HN03
kg HN0 3
X
1 mol HN03
63.02 g HN0 3
7
= 1.6 x 10 mol HN0 3
We need to get the relationship between moles ofHN03 and moles ofNH3 • We have to use all 3
equations.
2 mol
3 x
_
_ HN0
__
2 mol N02
3 mol N02
x
2 mol NO
4 mol NO
=
16molHN03
24 molNH3
4 mol NH3
Thus, we can produce 16 mol HN0 3 for every 24 mol NH3 that we begin with:
This is an oversimplified answer. In practice the NO produced in the 3rd step is recycled back
into the process in the second step.
123.
Total mass of copper used:
10,000 boards x (8.0 cm x 16.0 cm x 0.060 cm) x 8.96 g
board
cm 3
=
6.9 x lOs g Cu
Amount ofCu removed = 0.80 x 6.9 x lOs g = 5.5 x lOs g Cu
5.5 x lOs g Cu x
1 mol Cu
x
4 mol NH 3
63.55 g Cu
124.
a.
x
mol Cu
17.03 g NH
mol NH3
3
=
5.9 x lOs g NH3
From the reaction stoichiometry we would expect to produce 4 mol of acetaminophen for
every 4 mol ofC6H s0 3N reacted. The actual yield is 3 moles of acetaminophen compared to
a theoretical yield of 4 moles of acetaminophen. Solving for percent yield by mass (where M
= molar mass acetaminophen):
% yield = 3 mol
4mol
x
x
M
M
x
100 = 75%
b. The product ofthe percent yields of the individual steps must equal the overall yield, 75%.
(0.87) (0.98) (x) = 0.75, x
=
0.88; Step III has a % yield = 88%.
CHAPTER 3
58
125.
10.00 g XCI 2 + excess CI2 ---+ 12.55 g XCI4 ; 2.55 g CI reacted with XCI 2 to form XCI 4 . XCI 4
contains 2.55 g CI and 10.00 g XCI 2 • From mol ratios, 10.00 g XCI2 must also contain 2.55 g CI;
mass X in XCI 2 = 10.00 - 2.55 = 7.45 g X.
2.55 g CI x
So, 3.60
X
I mol CI x 1 mol XCI 2 x 1 mol X = 3.60
35.45 g CI
2 mol Cl
mol XCI 2
3.60 x
X
10-2 mol X
10-2 mol X must equal 7.45 g X. The molar mass of X is:
7.45 g X
------==::.......---
126.
STOICHIOMETRY
10-2
207
=-g;
mol X
A
' mass = 207 amu so X'IS Pb •
tomlC
mol X
4.000 g M 2 S3 ---+ 3.723 g M0 2
There must be twice as many mol ofM02 as mol ofM2 S3 in order to balance M in the reaction.
Setting up an equation for 2 mol M02 = mol M2 S3 where A = molar mass M:
2(
4.000 g
) _
3.723 g
8.000
_ 3.723
A + 2(16.00)' 2 A + 96.21 A + 32.00
2 A + 3(32.07)
8.000 A + 256.0 = 7.446 A + 358.2, 0.554 A = 102.2, A = 184 glmol; atomic mass = 184 amu
127.
Consider the case of aluminum plus oxygen. Aluminum forms AI3+ ions; oxygen forms 0 2- anions.
The simplest compound of the two elements is AIP3' Similarly we would expect the formula of any
group 6A element with Al to be A1 2 X3. Assuming this, out of 100.00 g of compound there are 18.56
g Al and 81.44 g of the unknown element, X. Let's use this information to determine the molar mass
of X which will allow us to identify X from the periodic table.
18.56 g Al
x
1 mol Al
26.98 gAl
x
3 mol X = 1.032 mol X
2 mol Al
81.44 g of X must contain 1.032 mol of X.
The molar mass of X =
81.44 g X
= 78.91 glmol X.
1.032 mol X
From the periodic table, the unknown element is selenium and the formula is A1 2 Se3.
128.
NaCI(aq) + Ag+(aq)
8.5904 g AgCI x
---+
AgCl(s); KCl(aq) + Ag\aq)
---+
AgCI(s)
1 mol AgCI x 1 mol CI- = 5.991 x 10-2 mol CI143.4 g AgCI 1 mol AgCI
The molar masses of NaCI and KCI are 58.44 and 74.55 glmol, respectively. Let x = g NaCI
andy= g KCI:
x + Y = 4.000 g and _x_
58.44
+
- y - = 5.991
74.55
X
10-2 total mol Cl- or 1.276 x +y = 4.466
CHAPTER 3
59
STOICHIOMETRY
Solving using simultaneous equations:
1.276 x + y = 4.466
- x - y = -4.000
0.276 x
= 0.466,
x = 1.69 g NaCI and y
=
2.31 g KCl
% NaCI = 1.69 g x 100 = 42.3% NaCl; % KCl = 57.7%
4.000 g
129.
1.252 g Cu x
1 mol Cu = 1.970
63.55 g Cu
X
10-2 mol Cu
The molar mass of Cup is 143.10 g/mol and the molar mass ofCuO is 79.55 g/mol and note that
Cu 20 contains twice the mol Cu as compared to CuO. Let x = g Cu20 and y = g CuO, then x + y
= 1.500 and:
2(
x ) + _y_ = 1.970
143.10
79.55
X
10-2 total mol Cu or 1.112 x + y = 1.567
Solving by the method of simultaneous equations:
1.112 x + y = 1.567
-x - y = -1.500
0.112 x = 0.067
x = 0.067/0.112 = 0.60 g = mass Cup
%CuP =
0.60 g = 40.%; %CuO = 100. - 40. = 60.%
1.500 g