Chapter 3:
3.2
Stoichiometry I
See Section 3.1.
1. Assume the equation contains one formula unit of the most complicated substance, and bring the
atoms of this substance into balance by adjusting the coefficients of the substances on the other side
of the equation.
2. Continue by adjusting the coefficients of the reactants and products until the same numbers of each
type of atom appear on both sides of the equation and the coefficients are given in terms of whole
numbers.
3. Check to make sure your final equation contains the same numbers of atoms of each type on both
sides.
3.4
See Section. 3.1
H2
3.6
O2
H2O2
See Section 3.2.
The number of objects in one mole has been determined experimentally to be 6.022 × 1023 (when
expressed to four significant figures) and is known as Avogadro’s number.
3.8
See Section 3.2 and Table 3.2.
The units for atomic, molecular and formula masses are atomic mass units (μ), and the units for molar
masses are grams per mole (g/mol).
3.10
moles
See Section 3.2.
Avogadro’s number
atoms
where the number of moles is multiplied by
3.12
6.022 × 1023 atoms
.
1 mol
See Section 3.3, Figure 3.4, and Question 3.9.
The masses of C and H are obtained as outlined in 3.9. These masses are subtracted from the total mass of
sample burned to obtain the mass of O in the sample. The percentage O by mass is calculated using (mass
of O/mass of sample) × 100%.
3.14
See Section 3.4.
One mol of N2 reacts with two mol H2 to form one mole N2H4.
21
3.16
See Section 3.4.
“The reaction was carried out with the reactants present in stoichiometric amounts,” means the reactants
were present in the exact amounts necessary to react with one another and none was present in excess.
3.18
See Section 3.4
The statement, “C2H4 is the limiting reactant and oxygen is present in excess in a combustion reaction,”
means the amount of carbon dioxide and water that can be produced by the reaction is limited by the
amount of C2H4 present. It also means the C2H4 is completely consumed and an excess of oxygen remains
when the reaction is complete.
3.20
See Section. 3.1
(a) 2SO2 + O2 Æ 2SO3
(b) sulfur trioxide
3.22
See Section 3.1 and Examples 3.1, 3.2.
(a) Unbalanced: Mg3N2 + H2O → NH3 + Mg(OH)2
Step 1: Mg3N2 + H2O → 2NH3 + 3Mg(OH)2
Step 2: Mg3N2 + 6H2O → 2NH3 + 3Mg(OH)2
(b) Unbalanced: Fe + O2 → Fe2O3
Step 1: 2Fe + O2 → Fe2O3
Start with Mg3N2.
Balances Mg, N.
Balances H,O.
Start with Fe.
Balances Fe.
3
Step 2: 2Fe + 2 O2 → Fe2O3
Balances O.
Step 3: 4Fe + 3O2 → 2Fe2O3
Gives whole number coefficients.
−
(c) Unbalanced: Zn + H3PO4 → H2 + Zn3(PO4)2 Start with H3PO4.
Step 1: Zn + 2H3PO4 → H2 + Zn3(PO4)2 Balances PO4 units.
Balances Zn.
Step 2: 3Zn + 2H3PO4 → H2 + Zn3(PO4)2
Step 3: 3Zn + 2H3PO4 → 3H2 + Zn3(PO4)2
Balances H.
3.24
See Section. 3.1
? Al2O3(s)
(a)? Al(s) + ? O2(g)
2 Al(s) + ? O2(g)
1 Al2O3(s)
2 Al(s) +
3
2
O2(g)
4 Al(s) + 3 O2(g)
1 Al2O3(s)
3O
2 Al2O3(s)
? NH3(g)
(b)? N2(g) + ? H2(g)
? N2(g) + 3 H2(g)
2 NH3(g)
1 N2(g) + 3 H2(g)
2 NH3(g)
22
Select order: Al then O
2 Al
Select order: H then N
6H
2N
(c)? C6H6(l) + ? O2(g)
C6H6(l) + ? O2(g)
C6H6(l) + ? O2(g)
? H2O(l) + ? CO2(g)
? H2O(l) + 6 CO2(g)
3 H2O(l) + 6 CO2(g)
C6H6(l) + 152 O2(g)
2 C6H6(l) + 15 O2(g)
3.26
Select order: C, H then O
6C
6H
3 H2O(l) + 6 CO2(g)
15 O
6 H2O(l) + 12 CO2(g)
See Section. 3.1.
H 2O
(a) HF (g) ⎯⎯⎯
→ H+ (aq) + F- (aq)
H 2O
→ 2 Li+ (aq) + SO42- (aq)
(b) Li2SO4 (s) ⎯⎯⎯
3.28
See Section. 3.1.
H 2O
(a) HCl (g) ⎯⎯⎯
→ H+ (aq) + Cl- (aq)
H 2O
(b) Ba(NO3)2 (s) ⎯⎯⎯
→ Ba2+ (aq) + 2NO3- (aq)
3.30
See Section 3.1 and Example 3.3.
(a) Unbalanced: Mg(OH)2 + HF → MgF2 + H2O
Step 1: Mg(OH)2 + 2HF → MgF2 + H2O
Step 2: Mg(OH)2 + 2HF → MgF2 + 2H2O
Start with MgF2.
Balances F.
Balances H, O.
(b) Unbalanced: HCl + NaOH → H2O + NaCl
Step 1: HCl + NaOH → H2O + NaCl
Start with NaCl.
Balances H, O, Na, Cl
(c) Unbalanced: H2SO4 + Sr(OH)2 → H2O + SrSO4
Step 1: H2SO4 + Sr(OH)2 → 2H2O + SrSO4
Start with H2O.
Balances H, O.
3.32
See Section. 3.1.
(a) Unbalanced: ? C4H10(g) + ? O2(g)
Balance C:
C4H10(g) + ? O2(g)
Balance H:
C4H10(g) + ? O2(g)
Balance O:
C4H10(g) + 132 O2(g)
Multiply by 2: 2 C4H10(g) + 13 O2(g)
(b) Unbalanced:
Balance C:
Balance H:
Balance O:
? C6H12O6(s) + ? O2(g)
C6H12O6(s) + ? O2(g)
C6H12O6(s) + ? O2(g)
C6H12O6(s) + 6 O2(g)
? CO2(g) + ? H2O(g)
4 CO2(g) + ? H2O(g)
4 CO2(g) + 5 H2O(g)
4 CO2(g) + 5 H2O(g)
8 CO2(g) + 10 H2O(g)
? CO2(g) + ? H2O(g)
6 CO2(g) + ? H2O(g)6 C’s
6 CO2(g) + 6 H2O(g)
6 CO2(g) + 6 H2O(g)
4 C’s
10 H’s
13 O’s
12 H’s
18 O’s
23
(c) Unbalanced:
Balance C:
Balance H:
? C4H8O(A) + ? O2(g)
C4H8O(A) + ? O2(g)
C4H8O(A) + ? O2(g)
Balance O:
C4H8O(A) + 112 O2(g)
Multiply by 2: 2 C4H8O(A) + 11 O2(g)
3.34
? CO2(g) + ? H2O(g)
4 CO2(g) + ? H2O(g)
4 C’s
4 CO2(g) + 4 H2O(g)
8 H’s
4 CO2(g) + 4 H2O(g)
12 O’s
8 CO2(g) + 8 H2O(g)
See Section 3.1 and Examples 3.3, 3.4.
Start with C8H8.
(a) Unbalanced: C8H8 + O2 → CO2 + H2O
Balances C, H.
Step 1: C8H8 + O2 → 8CO2 + 4H2O
Step 2: C8H8 + 10O2 → 8CO2 + 4H2O Balances O.
(b) Unbalanced: HCl + KOH → H2O + KCl
Step 1: HCl + KOH → H2O + KCl
3.36
Start with KCl.
Balances H, O, K, Cl.
See Section 3.1 and Example 3.4.
Unbalanced: CH3OC(CH3)3 + O2 → CO2 + H2O
Step 1: CH3OC(CH3)3 + O2 → 5CO2 + 6H2O
Start with CH3OC(CH3)3.
Balance C, H.
15
2
O → 5CO2 + 6H2O Balances O.
− 2
Gives whole number coefficients.
Step 3: 2CH3OC(CH3)3 + 15O2 → 10CO2 + 12H2O
Step 2: CH3OC(CH3)3 +
3.38
See Section 3.1 and Example 3.3.
Unbalanced: H3PO3 → H3PO4 + PH3 Start with H3PO3 and H3PO4.
Balances O, P, H.
Step 1: 4H3PO3 → 3H3PO4 + PH3
3.40
See Section 3.1 and Examples 3.1, 3.2.
Unbalanced: UO2 + CCl4 → UCl4 + COCl2 Start with CCl4.
Step 1: UO2 + 2CCl4 → UCl4 + 2COCl2 Balances Cl, C, O.
3.42
See Section. 3.1.
(a) NH4Cl: Assume hydrogen has an an oxidation number of +1 and chlorine -1;
therefore N has an oxidation number of -3.
(b) N2O: Assume oxygen has an oxidation number of -2; therefore each nitrogen has an
oxidation state of +1.
(c) Ag: All elements in their elemental state have an oxidation number of zero
24
(d) AuI3: Assume each iodine has an oxidation number of -1; therefore gold must have an
oxidation number of +3.
3.44
See Section. 3.1.
Na2O2: Assume that each sodium has an oxidation number of +1; therefore each oxygen must
have an oxidation number of -1.
3.46
See Section 3.1 and Example 3.5.
Start with O2.
Unbalanced: Fe + O2 → Fe2O3
Step 1: Fe + 3O2 → 2Fe2O3
Balances O.
Step 2: 4Fe + 3O2 → 2Fe2O3 Balances Fe.
Assigning an ON of -2 to oxygen in Fe2O3 gives 4Fe + 3O2 → 2FeO3
ON:
0
0
+3,-2
The ON of the Fe atoms increase, and the ON of the O atoms decrease.
The Fe metal is oxidized, and the O atoms in O2 are reduced.
3.48
See Section 3.1 and Example 3.5.
Start with O2.
Unbalanced: P4(s) + O2(g) → P4O10(s)
Balances O.
Step 1: P4(s) + 5O2(g) → P4O10(s)
Assigning an ON of −2 to O in P4O10 gives P4(s) + 5O2(g) → P4O10(s)
ON
0
0
+5,−2
The ON of the P atoms increases, and the ON of the O atoms decreases.
The P atoms in P4 are oxidized, and the O atoms in O2 are reduced.
3.50
See Section 3.1 and Example 3.5.
Start with HCl.
Unbalanced: MnO2 + HCl → MnCl2 + Cl2 + H2O
Step 1: MnO2 + 4HCl → MnCl2 + Cl2 + H2O Balances Cl.
Step 2: MnO2 + 4HCl → MnCl2 + Cl2 + 2H2O Balances H, O.
Assigning an ON of −2 to O in MnO2 and H2O, an ON of +1 to H in HCl and H2O and an ON of −1 to Cl
in HCl and MnCl2 gives MnO2 + 4HCl → MnCl2 + Cl2 + 2H2O
ON:
+4,−2 +1,−1 +2,−1 0 +1,−2
The ON of some of the Cl atoms in HCl increases, and the ON of the Mn atoms decreases.
The Cl atoms in HCl that undergo an increase in ON are oxidized, and the Mn atoms in MnO2 are
reduced.
3.52
See Section 3.2 and Example 3.6.
6.022 × 1023 atoms Xe
= 4.69 × 1023 atoms Xe
1 mol Xe
6.022 × 1023 atoms K
× 8.73 × 1023 atoms K
(b) ? atoms K = 1.45 mol K ×
1 mol K
6.022 × 1023 atoms Ti
= 3.36 × 1025 atoms Ti
(c) ? atoms Ti = 55.8 mol Ti ×
1 mol Ti
(a) ? atoms Xe = 0.0778 mol Xe ×
25
3.54
See Section 3.2 and Example 3.6.
(a) ? molecules Cl2 = 0.223 mol Cl2 ×
6.022 × 1023 molecules Cl2
= 1.34 × 1023 molecules Cl2
1 mol Cl2
(b) ? molecules N2H4 = 14.7 mol N2H4 ×
6.022 × 1023 molecules N 2 H 4
= 8.85 × 1024 molecules N2H4
1 mol N 2 H 4
(c) ? molecules C9H18 = 0.334 mol C9H18 ×
6.022 × 1023 molecules C9 H18
= 2.01 × 1023 molecules
1 mol C9 H18
C9H18
(d) ? molecules CO2 = 1.22 mol CO2 ×
3.56
6.022 × 1023 molecules CO2
= 7.35 × 1023 molecules CO2
1 mol CO 2
See Section 3.2 and Example 3.6.
(a) ? mol Br2 = 1.33 × 1026 molecules Br2 ×
1 mol Br2
= 221 mol Br2
6.022 × 1023 molecules Br2
(b) ? mol C5H12 = 7.71 × 1026 molecules C5H12 ×
(c) ? mol B2H6 = 2.34 × 1023 molecules B2H6 ×
(d) ? mol Ne = 7.76 × 1023 atoms Ne ×
3.58
1 mol C5 H12
= 1.28 × 103 mol C5H12
23
6.022 × 10 molecules C5 H12
1 mol B2 H 6
= 0.389 mol B2H6
6.022 × 1023 molecules B2 H 6
1 mol Ne
= 1.29 mol Ne
6.022 × 1023 atoms Ne
See Section 3.2 and Example 3.7.
(a) Molecular mass for N2O4:
2[N] × 14.01
= 28.02
4[O] × 16.00
= 64.00
92.02; Hence, molar mass for N2O4 is 92.02 g/mol.
(b) Formula mass for Na2SO4:
1[Na] × 22.99 = 45.98
2[S] × 32.07
= 32.07
4[O] × 16.00
= 64.00
142.05; Hence, molar mass for Na2SO4 is 142.05 g/mol.
(c) Molecular mass for C6H10O2:
6[C] × 12.01
= 72.06
10[H] × 1.01
= 10.10
2[O] × 16.00
= 32.0
114.16; Hence, molar mass for C6H10O2 is 114.16 g/mol.
3.60
See Section 3.2 and Example 3.7.
(a) Molecular mass N2O2:
26
2[N] × 14.01
= 28.02
2[O] × 16.00 = 32.00
60.02 Hence, molar mass for N2O2 is 60.02 g/mol.
2[N] × 14.01 = 28.02
(b) Formula mass for (NH4)CO3:
8[H] × 1.01
= 8.08
1[C] × 12.01 = 12.01
3[O] × 16.00 = 48.00
96.11 Hence, molar mass for (NH4)2CO3 is 96.11 g/mol.
(c) Molecular mass for C8H15N: 8[C] × 12.01 = 96.08
15[H] × 1.01 = 15.15
1[N] × 14.01 = 14.01
125.24 Hence, molar mass for C8H15N is 125.24 g/mol.
3.62
See Section. 3.2.
(a) Molar mass H2SO4 = 2 × (1.0079 g) + (32.065 g) + 4 × (15.9994 g) = 98.078 g/mol
39.2 g H 2SO4 ×
1 mol H 2 SO4
= 0.400 mol H 2 SO4
98.078 g H 2 SO4
(b) Molar mass O2 = 2 × (15.9994 g) = 31.9988 g/mol
8.00 g O 2 ×
1 mol O2
= 0.250 mol O2
31.9988 g O2
(c) Molar mass NH3 = 14.0067 g + 3 × (1.0079 g) = 17.0304 g/mol
10.7 g NH 3 ×
3.64
1 mol NH 3
= 0.628 mol NH 3
17.0304 g NH 3
See Section 3.2 and Examples 3.7, 3.8.
44.01 g CO 2
= 3.45 × 103 g CO2
1 mol CO 2
1 mol AgNO3
(b) ? mol AgNO3 = 192 g AgNO3 ×
= 1.13 mol AgNO3
169.9 g AgNO3
(a) ? g CO2 = 78.4 mol CO2 ×
(c) ? molecules CH4 = 9.22 g CH4 ×
1 mol CH 4 6.022 × 1023 molecules CH 4
×
= 3.46 × 1023
16.05 g CH 4
1 mol CH 4
molecules CH4
3.66
See Section 3.2 and Example 3.7.
92.02 g N 2 O 4
= 695 g N2O4
1 mol N 2 O 4
111.0 g CaCl2
= 1.0 × 103 g CaCl2
(b) ? g CaCl2 = 9.2 mol CaCl2 ×
1 mol CaCl2
(a) ? g N2O4 = 7.55 mol N2O4 ×
27
(c) ? CO = 0.44 mol CO ×
3.68
28.0 g CO
= 12 g CO
1 mol CO
See Section 3.2 and Example 3.8.
1 mol H 2 6.022 × 1023 molecules H 2
×
= 1.0 × 1024 molecules H2
2.02 g H 2
1 mol H 2
2 atoms H
= 2.0 × 1024 atoms H
(b) ? atoms H = 1.0 × 1024 molecules H2 ×
1 molecule H 2
(a) ? molecules H2 = 3.4 g H2 ×
3.70
See Section 3.2 and Examples 3.6, 3.7, 3.8.
(a) ? mol SO3 = 3.31 g SO3 ×
1 mol SO3
= 0.413 mol SO3
80.07 g SO3
6.022 × 1023 molecules SO3
= 2.49 × 1023 molecules SO3
1 mol SO3
1 atoms S
(c) ? atoms S = 2.49 × 1023 molecules SO3 ×
= 2.49 × 1023 atoms S
1 molecule SO3
3 atoms O
? atoms O = 2.49 × 1023 molecules OS3
= 7.47 × 1023 atoms O
1 molecule SO3
(b) ? molecules SO3 = 0.413 mol SO3 ×
3.72
See Sections. 1.4 and 3.2.
? moles C2H5OH = 0.9 fl. oz. ×
0.7894 g C 2 H 5OH
1 mol C 2 H 5 OH
29.56 mL
×
×
= 0.4559 mol
1 fl. oz.
1 mL C2 H5 OH
46.07 g C 2 H 5OH
C2H5OH
3.74
See Section 3.2 and Examples 3.6, 3.7, 3.8.
(a) Molecular mass for Ni(CO)4: 1[Ni] × 58.69 = 58.69
4[C] × 12.01 = 48.04
4[O] × 16.00 = 64.00
170.73
Hence, molar mass for Ni(CO)4 is 170.73 g/mol.
(b) ? mol Ni(CO)4 = 3.22 g Ni(CO)4 ×
1 mol Ni(CO) 4
= 1.89 × 10−2 mol Ni(CO)4
170.73 g Ni(CO) 4
(c) ? molecules Ni(CO)4 = 5.67 g Ni(CO)4
= 2.00 × 1022 molecules Ni(CO)4
28
1 mol Ni(CO)4
6.022 × 10 23 molecules Ni(CO) 4
×
170.73 g Ni(CO) 4
1 mol Ni(CO) 4
(d) ? atoms C = 34 g Ni(CO)4 ×
1 mol Ni(CO) 4
6.022 × 10 23 molecules Ni(CO) 4
×
×
1 mol Ni(CO) 4
170.73 g Ni(CO) 4
4 atom C
1 molecule Ni(CO) 4
= 4.8 × 1023 atoms C
Note: Ni(CO)4 is composed of a metal and nonmetals and would be predicted to be an ionic compound
(2.6). However, the observation that it is a volatile compound rather than a crystalline solid suggests it is a
molecular compound (2.7).
3.76
See Section. 3.2.
Formula for hydrogen peroxide: H2O2
? moles H atoms = 0.011 mol H2O2 ×
3.78
2 mol H
= 0.022 mol H atoms
1 mol H 2 O 2
See Section 3.3 and Example 3.9.
(a) Molecular mass for C6H12:
6[C] × 12.01 = 72.06
12[H] × 1.01 = 12.12
72.06 g C
× 100% = 85.60% C
84.18 g C6 H12
12.12 g H
%H =
× 100% = 14.40% H
84.18 g C6 H12
%C =
84.18
Molar mass for C6H12 is 84.18 g/mol.
(b) Molecular mass for C5H12:
5[C] × 12.01 = 60.05
12[H] × 1.01 = 12.12
1[O] × 1600 = 16.00
60.05 g C
× 100% = 68.11% C
88.17 g C5 H12 O
12.12 g H
× 100% = 13.75% H
%H =
88.17 g C5 H12 O
16.00 g O
%O =
= 100% = 18.15% O
88.17 g C5 H12 O
%C =
88.17
Molar mass for C5H12O is 88.17 g/mol.
(c) Formula mass for NiCl2:
58.69 g Ni
× 100% = 45.29% Ni
129.59 g NiCl2
70.90 g Cl
2[Cl] × 35.45 = 70.90 %Cl =
× 100% = 54.71% Cl
129.59 g NiCl2
129.59
Molar mass for NiCl2 is 129.59 g/mol.
1[Ni] × 58.69 = 58.69
%Ni =
29
3.80
See Section. 3.3.
Mass of CuS to provide 10.0g of Cu:
To calculate the weight percent of Cu in CuS, we need the respective atomic weights:
Cu = 63.546
S = 32.066
adding CuS = 95.612
63.546 g Cu
× 100 = 66.46% Cu
The % of Cu in CuS is then:
95.612 g CuS
3.82
See Sections 2.8, 3.3, and Example 3.9.
Formula mass for MgCO3:1[Mg] × 24.30 %Mg =
24.30 g Mg
× 100% = 28.83% Mg
84.31 g MgCO3
12.01 g C
× 100% = 14.25% C
84.31 g MgCO3
48.00 g O
3[O] × 16.00 = 48.00 %O =
× 100% = 56.93% O
84.31 g MgCO3
84.31
Molar mass for MgCO3 is 84.31 g/mol.
1[C] × 12.01 = 12.01 %C =
3.84
See Section 2.8, 3.3, and Example 3.9.
(a)
Formula mass for CaCO3:
(b)
1[Ca] × 40.08 = 40.08
%Ca =
1[C] × 12.01 = 12.01
3[O] × 16.00 = 48.00
40.08 g Ca
× 100% = 40.04% Ca
100.09 g CaCO3
12.01 g C
× 100% = 12.00% C
%Ca =
100.09 g CaCO3
48.00 g O
%O =
× 100% = 47.96% O
100.09 g CaCO3
100.09
Molar mass for CaCO3 is 100.09 g/mol.
3.86
See Section 3.3 and Example 3.9.
Formula mass for C6H4(OH)Cl: 6[C] × 12.01 = 72.06 %C =
72.06 g C
× 100% 56.05%C
128.56 g C6 H 4 (OH) Cl
5.05 g H
× 100% = 3.93%H
128.56 g C6 H 4 (OH) Cl
16.00 g O
× 100% = 12.45%O
1[O] × 16.00 %O =
128.56 g C6 H 4 (OH) Cl
35.45 g Cl
1[Cl] × 35.45 = 35.45 %Cl =
× 100% = 27.57%Cl
128.56 g C6 H 4 (OH) Cl
5[H] × 1.01 = 5.05 %H =
30
128.56
Molar mass for C6H4(OH)Cl is 128.56 g/mol.
Analytical percent composition of 56.05% C, 3.93% H, and 27.57% Cl agrees with percent composition
of C6H4(OH)Cl. Compound is C6H4(OH)Cl.
3.88
See Section 3.3.
(a) Molecular mass for C4H10O:
4[C] × 12.01 =
48.04
10[H] × 1.01 = 10.10
1[O] × 16.00 = 16.00
74.14 Hence, molar mass for C4H10O is 74.14 g/mol.
48.04 g C
? g C = 1.80 g C4H10O ×
= 1.17 g C
74.14 g C 4 H10 O
(b) Formula mass for Na2CO3:
2[Na] × 22.99 = 45.98
1[C] × 12.01 = 12.01
3[O] × 16.00 = 48.00
105.99 Hence, molar mass for Na2CO3 is 105.99 g/mol
12.01 g C
? g C = 0.00223 g Na2CO3 ×
= 2.53 × 10−4 g C
105.99 g Na 2 CO3
(c) Molecular mass for C5H11N:
5[C] × 12.01 = 60.05
11[H] × 1.01 = 11.11
1[N] × 14.01 = 14.01
85.17 Hence, molar mass for C5H11N is 85.17 g/mol.
60.05 g C
? g C = 22.1 g C5H11N ×
= 15.6 g C
85.17 g C5 H11 N
Note: See direct method used in working 3.73.
3.90
See Section 3.3.
2.02 g H
= 0.485 g H
18.02 g H 2 O
2.02 g H
(b) ? g H = 1.22 g C2H2 ×
= 0.0946 g H
26.04 g C2 H 2
4.04 g H
(c) ? g H = 4.44 g N2H4 ×
× 0.560 g H
32.06 g N 2 H 4
(a) ? g H = 4.33 g H2O ×
3.92
See Section 3.3 and Example 3.10.
? g C = 4.06 g CO2 ×
12.01 g C
1.11 g C
= 1.11 g C %C =
× 100% = 40.1% C
44.01 g CO 2
2.770 g sample
31
2.02 g H
0.186 g H
= 0.186 g H %H =
× 100% = 6.7% H
18.02 g H 2 O
2.770 g sample
1.47 g O
? g O = g sample - g C - g H
%O =
× 100% = 53.1% O
2.770 g sample
= 2.770 g − 1.11 g − 0.186 g = 1.47 g O
? g H = 1.66 g H2O ×
3.94
See Section 3.3.
12.01 g C
0.284 g C
= 0.284 g C
%C =
× 100% = 55.4% C
44.01 g CO 2
0.513 g sample
2.02 g H
0.0789 g H
? g H = 0.704 g H2O ×
= 0.0789 g H %H =
× 100% = 15.4% H
18.02 g H 2 O
0.513 g sample
0.150 g N
? g N = g sample − g C − gH
%N =
× 100% = 29.2% N
0.513 g sample
? g C = 1.04 g CO2 ×
= 0.513 g − 0.284 g − 0.0789 g = 0.150 g N
3.96
See Section 3.3 and Example 3.11.
1 mol C
0.067 mol C
= 1.00 mol C
= 0.067 mol C relative mol C =
0.067
12.011 g C
1 mol H
0.20 mol H
= 3.0 mol H
? mol H = 0.20 g H ×
= 0.20 mol H relative mol H =
0.067
1.008 g H
The empirical formula is CH3.
? mol C = 0.80 g C ×
3.98
See Section 3.3 and Example 3.11.
1 mol N
0.0109 mol N
= 0.0109 mol Nrelative mol N =
= 1.00 mol N
0.0109
14.007 g N
1 mol O
0.0218 mol O
? mol O = 0.348 g O ×
= 0.0218 mol O
relative mol O =
2.00 mol O
0.0109
15.9994 g O
The empirical formula is NO2.
? mol N = 0.152 g N ×
3.100
See Section 3.3 and Example 3.11.
Assume the sample has a mass of 100.00 g and therefore contains 52.7 g Se and 47.3 g Cl.
1 mol Se
0.667 mol Se
= 1.00 mol Se
? mol Se = 52.7 g Se ×
= 0.667 mol Se
relative mol Se =
0.667
78.96 g Se
1 mol Cl
1.33 mol Cl
? mol Cl = 47.3 g Cl ×
= 1.99 mol Cl
= 1.33 mol Cl
relative mol Cl =
0.667
35.453 g Cl
The empirical formula is SeCl2.
3.102
32
See Section 3.3 and Example 3.11.
? mol Cr = 0.173 g Cr ×
1 mol Cr
0.00333 mol Cr
= 1.00 mol
= 0.00333 mol Cr relative mol Cr =
0.00333
51.996 g Cr
Cr
1 mol O
= 0.0100 mol O
15.9994 g O
The empirical formula is CrO3.
? mol O = 0.160 g O ×
3.104
relative mol O =
0.0100 mol O
= 3.00 mol O
0.00333
See Section. 3.3.
1 mol CO 2
1 mol C
×
= 0.06635 mol C
44.01 g CO2
1 mol CO 2
1 mol H 2O
2 mol H
? mol H = 1.22 g H 2O ×
×
= 0.1254 mol H
18.02 g H 2 O
1 mol H 2 O
12.01 g C
? g C = 0.06635 mol C ×
= 0.7969 g C
1 mol C
1.01 g H
? g H = 0.1354 mol H ×
= 0.1365 g H
1 mol H
? g O = 1.20 g total - [ 0.7969 g C + 0.1365 g H ] = 0.2666 g O
? g C = 0.104 g CO 2 ×
? mol O = 0.2666 g O ×
1 mol O
= 0.01666 mol O
15.99 g O
Element
C
Moles of Element
0.06635
H
0.1354
O
0.01666
Divide by Smallest
0.06635
= 4.00
0.01666
0.1354
= 8.00
0.01666
0.01666
= 1.00
0.01666
The simplest formula is C4H8O
3.106
See Section 3.3 and Example 3.11.
Assume the sample has a mass of 100.00 g and therefore contains 79.95 g C, 9.40 g H and 10.65 g O.
1 mol C
6.656 mol C
= 10.00 mol C
? mol C = 79.95 g C ×
= 6.656 mol C
relative mol C =
0.6656
12.011 g C
1 mol H
9.33 mol H
? mol H = 9.40 g H ×
= 9.33 mol H relative mol H =
= 14.0 mol H
1.008 g H
0.6656
1 mol O
0.6656 mol O
= 1.000 mol O
? mol O = 10.65 g O ×
= 0.6656 mol O
relative mol O =
0.6656
15.9994 g O
The empirical formula is C10H14O.
33
3.108
See Section 3.3.
1 mol CO 2
1 mol C
×
= 0.003855 mol C
44.01 g CO 2
1 mol CO 2
1 mol H 2 O
2 mol H
= 0.003863 mol H
×
? mol H = 0.0348 g H2O ×
18.02 g H 2 O
1 mol H 2 O
12.01 g C
? g C = 0.003855 mol C ×
= 0.0464 g C
1 mol C
1.01 g H
? g H = 0.003863 mol H ×
= 0.003894 g H
1 mol H
0.0464 g C
× 100 = 10.1% C
%C=
0.459 g sample
0.003894 g H
%H=
× 100 = 0.84% H
0.459 g sample
? mol C = 0.170 g ×
3.110
See Section 3.3 and Example 3.11.
? g C = 3.94 g CO2 ×
12.01 g C
2.02 g H
= 1.08 g C ?g H = 1.89 g H2O ×
= 0.212 g H
44.01 g CO 2
18.02 g H 2O
0.235 g N
= 0.417 g N
1.23 g sample
? g = g sample − g C − g H − g N = 2.18 g − 1.08 g − 0.212 g − 0.417 g = 0.47 g O
1 mol C
0.0899 mol C
? mol C = 1.08 g C ×
= 0.0899 mol C
relative mol C =
= 3.1 mol C
0.029
12.011 g C
? g N in 2.18 g sample = 2.18 g sample ×
? mol H = 0.212 g H ×
1 mol H
= 0.210 mol H
1.008 g H
relative mol H =
0.210 mol H
= 7.2 mol H
0.029
? mol N = 0.417 g N ×
1 mol N
= 0.0298 mol N
14.01 g N
relative mol N =
0.0298 mol N
= 1.0 mol N
0.029
1 mol O
0.029 mol O
= 0.029 mol O
relative mol O =
= 1.0 mol O
0.029
15.999 g O
The empirical formula is C3H7NO.
Note: An alternative method of solving this problem involves calculating % C from g C in 2.18 g
sample, %H from g H in 2.18 g sample, % N from g N in 1.23 g sample, % O from 100.00% − % C − %
H − % N and working with a 100.00 sample.
? mol O = 0.47 g O ×
3.112
See Section 3.3 and Example 3.12.
1[H] × 1.0
= 1.0
1[O] × 16.0
= 16.0
17.0
Hence, molar mass for HO is 17.0 g/mol.
molar mass compound
34 g / mol
n=
=
=2
molar mass HO
17.0 g/ mol
The molecular formula is H2O2.
(a) Formula mass for HO:
34
3.114
See Section 3.3 and Example 3.12.
5[C] × 12.0
= 60.0
10[H] × 1.0
= 10.0
1[O] × 16.0
= 16.0
86.0
Hence, molar mass for C5H10O is 86.0 g/mol.
molar mass compound 258 g / mol
=
=3
n=
86.0 g / mol
molar mass C5 H10O
(a) Formula mass for C5H10O:
The molecular formula is C15H30O3.
1[P] × 31.0
= 31.0
3[Cl] × 35.5
= 106.5
137.5 Hence, molar mass for PCl3 is 137.5 g/mol.
molar mass compound 137.3 g / mol
n
=
=1
molar mass PCl3
137.5 g / mol
(b) Formula mass for PCl3:
The molecular formula is PCl3
3.116
See Section. 3.3.
Empirical and Molecular formula for Mandelic Acid:
1 mol C
= 5.258 mol C
12.0115 g C
1 mol H
5.30 g H •
= 5.28 mol H
1.0079 g H
1 mol O
31.55 g O •
= 1.972 mol O
15.9994 g O
63.15 g C •
Using the smallest number of atoms, we calculate the ratio of atoms:
5.258 mol C
2.666 mol C
22/3 mol C
8/3 mol C
=
or
or
1.972 mol O
1 mol O
1 mol O
1 mol O
So 3 mol O combine with 8 mol C and 8 mol H so the empirical formula is C8H8O3. The
formula mass of C8H8O3 is 152.15. Given the data that the molar mass is 152.15 g/mL, the
molecular formula for mandelic acid is C8H8O3.
3.118
See Section 3.3 and Example 3.12.
Assume the sample has a mass of 100.0 g and therefore contains 40.0 g C, 6.71 g H and 53.3 g O.
3.33 mol C
1 mol C
= 3.33 mol C relative mol C =
= 1.00 mol C
? mol C = 40.0 g C ×
3.33
12.011 g C
35
? mol H = 6.71 g H ×
1 mol H
6.66 mol H
= 6.66 mol H relative mol H =
= 2.00 mol H
3.33
1.008 g H
1 mol O
3.33 mol O
× 3.33 mol O relative mol O =
= 1.00 mol O
3.33
15.999 g O
The empirical formula is CH2O.
Formula mass for CH2O:
1[C] × 12.0
= 12.0
2[H] × 1.0
= 2.0
1[O] × 16.0
= 16.0
= 30.0 Hence, molar mass for CH2O is 30.0 g/mol.
molar mass compound 180 g / mol
=
=6
n=
30.0 g / mol
molar mass CH 2O
? mol O = 53.3 g O ×
The molecular formula for fructose is C6H12O6.
3.120
See Sections 3.2, 3.4, and Examples 3.4, 3.13, 3.14.
(a) Unbalanced: C4H8O + O2 → CO2 + H2O
Step 1: C4H8O + O2 → 4CO2 + 4H2O
Start with C4H8O.
Balances C, H.
11
Step 2: C4H8O+ 2 O2 → 4CO2 + 4H4O Balances O.
−
Step 3: 2C4H8O + 11O2 → 8CO2 + 8H2O
Gives whole number coefficients.
(b) Strategy: g C4H8O → mol C4H8O → mol O2 → g O2
11 mol O 2
32.0 g O 2
1 mol C4 H 8O
? g O2 = 5.33 g C4H8O ×
= 13.0 g O2
×
×
72.0 g C4 H 8O 2 mol C4 H 8O 1 mol O 2
3.122
See Sections 3.1, 3.4, and Examples 3.1, 3.2, 3.13.
Unbalanced: N2 + H2 → NH3 Start with NH3.
Step 1: N2 + H2 → 2NH3
Balances N.
Step 2: N2 + 3H2 → 2NH3
Balances H.
Strategy: g N2 → mol N2 → mol NH3 → g NH3
1 mol N 2 2 mol NH 3 17.0 g NH 3
? g NH3 = 5.33 g N2 ×
= 6.47 g NH3
×
×
28.0 g N 2
1 mol N 2
1 mol NH 3
3.124
See Section. 3.4.
Use the stoichiometry of the balanced equation as a conversion factor to convert the moles of
product to moles of reactant. The balanced equation says: 4 mol HCl are needed to make 1 mol
Cl2.
12.5 mol Cl2 ×
3.126
36
4 mol HCl
= 50.0 mol HCl
1 mol Cl2
See Sections 3.1, 3.4, and Examples 3.1, 3.2, 3.13, 3.14.
Balanced:
HCl + NaOH → NaCl + H2O
Strategy: g H2O → mol H2O → mol NaCl → g NaCl
1 mol H 2O 1 mol NaCI 58.5 g NaCI
×
×
= 254 g NaCl
? g NaCl = 78.2 g H2O ×
18.0 g H 2O 1 mol H 2O 1 mol NaCI
3.128
See Section. 3.5.
(a) 2Sb + 3Cl2 Æ 2SbCl3
(b) According to the figure, there appears to be one Sb atom left over after the reaction is complete;
therefore Cl2 is the limiting reactant.
3.130
See Section 3.5 and Example 3.16.
Balanced:
Zn(s) + 2AgNO3(aq) → 2Ag(s) + Zn(NO3)2(aq)
Strategy: g Zn → mol Zn → mol Ag → g Ag
1 mol Zn 2 mol Zn 107.9 g Zg
? g Ag based on Zn = 3.22 g Zn ×
×
×
= 10.6 g Ag
65.4 g Zn 1 mol Zn 1 mol Ag
Strategy: g AgNO3 → mol AgNO3 → mol Ag → g Ag
1 mol AgNO3
2 mol Ag
107.9 g Ag
×
×
= 2.76 g Ag
? g Ag based on AgNO3 = 4.35 g AgNO3 ×
169.9 g AgNO3 2 mol AgNO3 1 mol Ag
AgNO3 is the limiting reactant because it produces less Ag. The maximum amount of Ag which can be
produced from 3.22 g Zn and 4.35 g AgNO3 is 2.76 g.
3.132
See Sections 3.1, 3.5, and Examples 3.1, 3.2, 3.16.
Unbalanced: P4 + O2 → P4O10
Start with P4O10.
Step 1: P4 + 5O2 → P4O10
Balances O.
Strategy: g P4 → mol P4 → mol P4O10 → g P4O10
1 mol P4 1 mol P4O10 284.0 g P4O10
? g P4O10 based on P4 = 2.2 g P4 ×
×
×
= 5.0 g P4O10
124.0 g P4
1 mol P4
1 mol P4O10
Strategy: g O2 → mol O2 → mol P4O10 → g P4O10
1 mol O 2 1 mol P4O10 284.0 g P4O10
? g P4O10 based on O2 = 4.2 g O2 ×
×
×
= 7.5 g P4O10
32.0 g O2
5 mol O 2
1 mol P4O10
P4 is the limiting reactant because it produces less P4O10. The maximum amount of P4O10 that can be
produced from 2.2 g P4 and 4.2 g O2 is 5.0 g. Hence, the theoretical yield is 5.0 g P4O10.
3.134
See Sections 3.1, 3.4, and Examples 3.1, 3.2, 3.13, 3.14.
Unbalanced: CS2 + Cl2 → CCl4 + S2Cl2
Start with Cl2.
Step 1: CS2 + 3Cl2 → CCl4 + S2Cl2
Balances Cl2.
Strategy: g CS2 → mol CS2 → mol CCl4 → g CCl4
1 mol CS2 1 mol CCI 4 153.8 g CCI 4
? g CCl4 = 43.1 g CS2 ×
×
×
= 87.0 g CCl4
76.2 g CS2 1 mol CS2
1 mol CCI 4
37
percent yield CCl4 =
actual yield CCI 4
theoretical yield CCI 4
× 100% =
45.2 g CCI 4
87.0 g CCI 4
× 100% = 52.0%
Strategy: g CS2 → mol CS2 → mol S2Cl2 → g S2Cl2
1 mol CS2 1 mol S2CI 2 135.1 g S2CI 2
? g S2Cl2 = 43.1 g CS2 ×
= 76.4 g S2Cl2
×
×
76.2 g CS2 1 mol CS2
1 mol S2CI 2
percent yield S2Cl2 =
3.136
actual yield S2CI 2
theoretical yield S2CI 2
× 100% =
41.3 g S2CI 2
76.4 g S2CI 2
× 100% = 54.1%
See Sections 3.1, 3.5, and Examples 3.1, 3.2, 3.17.
Unbalanced: P4 + Cl2 → PCl5 Start with PCl5
Step 1: P4 + Cl2 → 4PCl5
Balances P.
Step 2: P4 + 10Cl2 → 4PCl5
Balances Cl.
Strategy: g P4 → mol P4 → mol PCl5 → g PCl5
4 mol PCI5 208.2 g PCI5
1 mol P4
×
×
= 15 g PCl5
? g PCl5 based on P4 = 2.3 g P4 ×
124.0 g P4 1 mol PCI5
1 mol PCI5
Strategy: g Cl2 → mol Cl2 → mol PCl5 → g PCl5
1 mol CI 2 4 mol PCI5 208.2 g PCI5
= 8.2 g PCl5
? g PCl5 based on Cl2 = 7.0 g Cl2 ×
×
×
70.9 g CI 2 10 mol CI 2
1 mol PCI5
Cl2 is the limiting reactant because it produces less PCl5. The maximum amount of PCl5 that can be
produced from 2.3 g P4 and 7.0 g Cl2 is 8.2 g. Hence, the theoretical yield is 8.2 g PCl5.
7.1 g PCI5
actual yield
percent yield =
× 100% =
× 100% = 87%.
theoretical yield
8.2 g PCI5
3.138
See Section. 3.5 and Example 3.20.
The mole ratio comes from the balanced equation.
5.0 × 103 g H 2 ×
1 mol H2
1 mol CH3OH 32.0417 g CH 3OH
×
×
= 4.0 × 10 4 g CH 3OH
2.0158 g H 2
2 mol H2
1 mol CH3OH
The given mass of CH3OH is the actual yield. Use these two masses to calculate the percent
yield.
3.5 × 103 g CH 3OH actual
× 100% = 8.8% yield
4.0 × 104 g CH 3OH theoretical
3.140
See Section. 3.5 and Example 3.20.
Balanced equation: 2KClO3
2KCl + 3O2
(a) Calculate the theoretical yield of O2.
(1)
(2)
(3)
Plan: g KClO3 Þ mol KClO3 Þ mol O2 Þ g O2 (theoretical)
38
g KClO3
3.75 g
(1) ? mol KClO3 = FW KClO = 122.55 g/mol = 0.0306 mol KClO3
3
3 mol O2
(2) ? mol O2 = 0.0306 mol KClO3 x 2 mol KClO = 0.0459 mol O2
3
(3) ? g O2 = mol O2 x AW O2 = 0.0459 mol x 32.0 g/mol = 1.47 g O2 (theoretical yield)
Alternatively by dimensional analysis:
1 mol KClO3
3 mol O2
32.0 g O2
? g O2 = 3.75 g KClO3 x 122.55 g KClO x 2 mol KClO x 1 mol O = 1.47 g O2
2
3
3
(b) Calculate the percent yield of O2.
1.05 g
actual yield
% yield = theoretical yield x 100% = 1.47 g x 100% = 71.4%
3.142
See Section. 3.5 and Problem 3.141.
(a) Unbalanced:
NO + O2 → NO2
Start with NO
Balances N and O.
Step 1:
2NO + O2 → 2NO2
Strategy: g NO → mol NO → mol NO2
2 mol NO 2
1 mol NO
? g NO2 based on NO = 75.0 g NO ×
×
= 2.50 mol NO2
30.01 g NO
2 mol NO
Strategy: g O2 → mol O2 → mol NO2
1 mol O 2
2 mol NO 2
×
= 2.81 mol NO2
32.00 g O 2
1 mol O 2
NO is the limiting reactant because it produces less NO2.
? g NO2 based on O2 = 45.0 g O2 ×
(b) ? mol O2 reacted = 2.50 mol NO ×
? mol O2 initial = 45.0 g O2 ×
1 mol O2
= 1.25 mol O2
2 mol NO
1 mol O 2
32.00 g O 2
= 1.41 mol O2
Number of moles of O2 remaining = # mol initial - # mol reacted:
1.41 mol O2 – 1.25 mol O2 = 0.16 mol O2
Strategy: mol O2 → g O2
? g O2 = 0.16 mol O2 ×
3.144
32.00 g O 2
= 5.12 g O2 remaining
1 mol O 2
See Sections 3.1, 3.4, and Examples 3.1, 3.2, 3.13, 3.14.
Balanced:
2 NaOH + X2 → NaX + NaXO + H2O
Strategy: Let y equal molar mass of X2 in g X2 → mol X2 → mol NaX → g NaX set up.
39
? g NaX = 3.11 g X2 ×
1 mol X 2
y g X2
×
1 mol NaX (23.0 + 0.5 y) g NaX
×
= 2.00 g NaX
1 mol X 2
1 mole NaX
Thus, (3.11)(23.0 + 0.5 y) = 2.00 y, 71.5 + 1.56 y = 2.00 y, and y = 162.
Since Br2 has a molar mass of 159.8 g/mol, the halogen is Br.
3.146
See Sections 2.6, 2.8, 3.3, and Example 3.9.
Iron(III) sulfate is Fe2(SO4)3.
Formula mass for Fe2(SO4)3:
2[Fe] × 55.8 = 111.6
%F =
111.6 g Fe
× 100% = 27.9% Fe
399.9 g Fe 2 (SO 4 )3
3[S] × 32.1 = 96.3
%S =
96.3 g S
× 100% = 24.1% S
399.9 g Fe 2 (SO 4 )3
12[O] × 16.0 =
192.0
399.9
%O =
192.0 g O
× 100% = 48.0% O
399.9 g Fe 2 (SO 4 )3
Molar mass for Fe2(SO4)3 is 399.9 g/mol.
3.148
See Sections 3.1, 3.2, 3.3, and Examples 3.1, 3.2, 3.13.
(a) Unbalanced: In2S3 + O2 → In2O3 + SO2
Step 1: In2S3 + O2 → In2O3 + 3SO2
Step 2: In2S3 + 92 O2 → In2O3 + 3SO2
Start with In2S3.
Balances S.
Balances O.
Unbalanced: In2O3 + CO → In + CO2 Start with In2O3.
Balances In.
Step 1: In2O3 + CO → 2In + CO2
Step 2: In2O3 + 3CO → 2In + 3CO2
Balances O.
(b) Strategy: kg In2S3 → g In2S3 → mol In2S3 → mol In2S3 → mol In → g In → kg In
2 mol In 2O3
103 g In 2S3 1 mol In 2S3
2 mol In
? kg In = 35.7 kg In2S3 ×
×
×
×
1 kg In 2S3 325.9 g In 2S3 2 mol In 2S3 1 mol In 2S3
114.8 g In 1 kg In
×
= 25.2 kg In
1 mol In 103 g In
Alternatively, using the composition of In2S3 gives
229.6 g kg In
? kg In = 35.7 kg In2S3 ×
= 25.2 kg In
325.9 kg In 2S3
×
3.150 See Section. 3.4.
? mass of H2O = 3.650 g – 1.782 g = 1.868 g H2O
Strategy: g H2O → mol H2 O
1.868 g H2O ×
40
1 mol H 2 O
= 0.104 mol H2O
18.02 g H 2 O
Strategy: g MgSO4 → mol MgSO4
1.782 g MgSO4 ×
1 mol MgSO 4
= 0.0148 mol MgSO4
120.37 g MgSO 4
Note: By heating the sample all of the water was driven off; therefore the 1.782 g contains only MgSO4.
0.104 mol H 2 O
7 mol H2O
0.0148 mol MgSO 4
There are 7 moles of H2O for every one mole of MgSO4.
3.152
See Sections 3.2 and 3.3.
Assume the sample contains 100.0 g and therefore contains 25.5 g Cu, 12.8 g S, 57.7 g O and 4.0 g H.
Strategy: g Cu → mol Cu → mol CuSO4
? mol CuSO4 = 25.5 g Cu ×
1 mol CuSO 4
1 mol Cu
×
= 0.402 mol CuSO4
1 mol Cu
63.5 g Cu
Strategy: g H → mol H → mol H2O
1 mol H 2O
1 mol H
×
= 2.0 mol H2O
? mol H2O = 4.0 g H ×
2 mol H
1.008 g H
mol H 2O
2.0 mol
=
= 5.0
0.402 mol
mol CuSO 4
The value of × in CuSO4 ⋅× H2O is 5.0
3.154
See Section 3.3 and Examples 3.12.
Assume the sample contains 100.00 g and therefore contains 71.56 g C, 6.71 g H, 4.91 g N and 16.82 g O.
1 mol C
5.96 mol C
? mol C = 71.56 g C ×
= 5.96 mol C relative mol C =
= 17.0 mol C
0.350
12.011 g C
? mol H = 6.71 g H ×
1 mol H
= 6.66 mol H
1.008 g H
? mol N = 4.91 g N ×
1 mol N
0.350 mol N
= 0.350 mol N relative mol N =
= 1.00 mol N
0.350
14.007 g N
relative mol H =
6.66 mol H
= 19.0 mol H
0.350
1 mol O
1.05 mol O
= 1.05 mol O relative mol O =
= 3.00 mol O
0.350
15.999 g O
The empirical formula is C17H19NO3.
Empirical formula mass for C17H19NO3:
204.0 The molar mass of the empirical formula for
17[C] × 12.0 =
19[H] × 1.0
=
19.0 morphine is numerically equal to the molar mass
1[N] × 14.0
=
14.0 of morphine. Hence, the molecular formula for
? mol O = 16.82 g O ×
41
3[O] × 16.0
3.156
=
=
48.0
285.0
morphine is C17H19NO3.
See Sections 3.1, 3.4, and Examples 3.1, 3.2, 3.15..
NaWCl6
→
Step 1: 2NaWCl6
Unbalanced:
Na2WCl6 + WCl6
Start with Na2 WCl6.
→
Na2WCl6 + WCl6
Balances Na, W, Cl.
Strategy: g NaWCl6 → mol NaWCl6 → mol WCl6 → g WCl6
1 mol NaWCI 6
1 mol WCI 6
396.6 g WCI 6
? g WCl6 = 5.64 g NaWCl6 ×
×
×
= 2.67 g WCl6
419.6 g NaWCI 6
2 mol NaWCI 6
1 mol WCI 6
percent yield =
3.158
1.52 g WCI 6
actual yield
× 100% =
× 100% = 56.9%
2.67 g WCI 6
theorectical yield
See Section 3.4.
g Cu
0.306 g Cu
× 100% =
× 100% = 0.255%
g sample
1.20 g sample
mol H 2O
% Cu =
×=
mol CuSO4
1 mol CuSO 4
1 mol Cu
×
= 0.00482 mol Cu
1 mol Cu
63.55 g Cu
159.62 g CuSO 4
? g CuSO4 in sample = 0.00482 mol CuSO4 ×
= 0.769 g CuSO4
1 mol CuSO 4
? mol CuSO4 in sample = 0.306 g Cu ×
? g H2O in sample = g sample − g CuSO4 = 1.20 g − 0.769 g = 0.43 g H2O
1 mol H 2O
? mol H2O in sample = 0.43 g H2O ×
= 0.024 mol H2O
18.02 g H 2O
mol H 2O
mol CuSO4
3.160
=
0.024 mol
= 5.0, so the value of × in CuSO4 ⋅ ×H2O is 5.
0.00482 mol
See Sections 3.1, 3.5, and Examples 3.1, 3.2, 3.16.
Unbalanced:
Step 1:
Step 2:
Step 3:
SCl2 + NaF
SCl2 + 4NaF
SCl2 + 4NaF
3SCl2 + 4NaF
→
→
→
→
SF4 + S2Cl2 + NaCl
SF4 + S2Cl2 + NaCl
SF4 + S2Cl2 + 4NaCl
SF4 + S2Cl2 + 4NaCl
Start with NaF.
Balances F.
Balances Na.
Balances S, Cl.
Strategy: g SCl2 → mol SCl2 → mol SF4 → g SF4
1 mol SF4
108.1 g SF4
1 mol SCI 2
×
×
= 4.352 g SF4
? g SF4 based on SCl2 = 12.44 g SCl2 ×
103.0 g SCI 2
3 mol SCI 2
1 mol SF4
Strategy: g NaF → mol NaF → mol SF4 → g SF4
42
? g SF4 based on NaF = 10.11 g NaF ×
1 mol SF4
108.1 g SF4
1 mol NaF
×
×
= 6.505 g SF4
42.00 g NaF
4 mol NaF
1 mol SF4
SCl2 is the limiting reactant because it produces less SF4. The maximum amount of SF4 that can be
produced from 12.44 g SCl2 and 10.11 g NaF is 4.352 g SF4.
3.162
See Sections. 3.1 and 2.8.
(a) Reactants: solid copper; concentrated nitric acid
Products: aqueous copper(II) nitrate; gaseous nitrogen dioxide; liquid water
(b) Unbalanced: Cu + HNO3
Step 1:
Cu + 4HNO3
Step 2:
Cu + HNO3
→
→
→
Cu(NO3)2 + NO2 + H2O
Cu(NO3)2 + NO2 + 2H2O
Cu(NO3)2 + 2NO2 + H2O
Start with HNO3
Balances H
Balances N and O
(c) Reactants: Cu = 0; HNO3 (H = +1, N = +5, O = -2)
Products: Cu(NO3)2 (Cu = +2, N = +5, O = -2); NO2 (N = +4, O = -2); H2O (H = +1, O = -2)
Because the oxidation numbers of Cu and N change this is a redox reaction.
43