Chemistry Reference
In-Depth Information
a
16.0 g O
1 mol
O
a
1 mol C
12.01 g C
7.55
(a)
(b)
Same as part (a)
(c)
1.25 mol O
b
20.0 g O
7.91
82.66 g C
b
6.882
6
mol C
a
1 mol H
1.008 g H
17.34 g H
b
17.20
2
mol H
a
32.0 g O
1 mol
O
2
1.25 mol O
2
b
40.0 g O
2
6.882
6
mol C
6.882
6
1.000 mol C
a
1 mol
62.0 g
7.59
12.9 g
b
0.208 mol
17.20
2
mol H
6.882
6
a
16.0 g
1 mol
2.499
3
mol H
1.00 mol
2.66
10
23
g
7.61
(a)
(b)
One molecule has a mass twice that.
1 atom
a
b
b
10
23
atoms
6.02
Multiply each of these by 2 to get 2 mol C to 5 mol H. The
empirical formula is
(b)
C
2
H
5
.
10
23
molecules
1 mol
a
1 mol
64.0 g
a
6.02
Mn
3
O
4
(c)
P
2
O
5
(d)
C
2
H
4
Cl
(e)
C
2
H
4
O
7.62
12.9 g
b
b
7.96 (a)
Sulfuric acid
(b)
Sodium sulfite
7.98 (a)
13.0 amu and 78.0 amu
(b)
14.0 amu and 42.0 amu
(c)
15.0 amu and 30.0 amu
7.100
Each compound consists of 92.26% carbon and 7.74% hy-
drogen. They have the same mass ratio of carbon to hydrogen
because they have the same mole ratio of carbon to hydrogen;
that is, they have the same empirical formula (CH).
7.102
The empirical formula mass is 42.0 amu.
(a)
There is one empirical formula unit in each molecule, so
the molecular formula is
(b)
There are empirical formula
units per molecule, so the molecular formula is
7.105
Empirical formula:
H
2
SO
4
10
23
molecules
1.21
Na
2
SO
3
10
24
molecules CH
4
7.65
3.91
1 mol
6.02
10
23
units
9.25
10
20
units
7.66
(a)
a
b
b
1.54
10
3
mol
a
6.02
10
23
molecules
1 mol
(b)
1.67 mol
1.01
10
24
molecules
a
342 g
1 mol
1 mol
6.02
10
23
molecules
10
24
molecules
7.67
1.91
a
b
b
NO
2
.
108
5
g
1.09 kg
(92.0 amu)/ (46.0 amu)
2
N
2
O
4
.
a
1 mol C
2
H
4
28.05 g C
2
H
4
7.69
(a)
44.7 g C
2
H
4
b
1.59
4
mol C
2
H
4
CH
2
O;
molecular formula:
C
6
H
12
O
6
a
2 mol C atoms
1 mol C
2
H
4
7.107 (a) (b)
(c) (d)
7.108
The mass per mole of empirical formula units is 31.0 g/mol.
Therefore,
C
3
H
8
O
3
H
2
C
2
O
4
(b)
1.59
4
mol C
2
H
4
b
C
6
H
4
N
2
O
6
C
3
H
6
O
2
3.18
7
mol C atoms
10
23
C atoms
1 mol C atoms
a
6.02
(c)
3.18
7
mol C atoms
b
124 g/mol
31.0 g/mol empirical formula units
1.92
10
24
C atoms
a
1.008 g H
1 mol H
3 mol H
1 mol NH
3
4 mol empirical formula units
1 mol
7.72
4.11 mol NH
3
a
b
b
12.4 g H
The formula is
P
4
.
1 mol O atoms
6.02
10
23
O atoms
10
24
O atoms
7.76
6.78
a
b
7.110
7.114 (a)
The law of conservation of mass enables us to calculate
the mass of oxygen used:
C
2
H
6
a
1 mol H
3
PO
4
4 mol O atoms
b
2.82 mol H
3
PO
4
17.9 mg
11.0 mg
6.09 mg
22.8
1
mg
a
1 mol C
6
H
6
78.0 g C
6
H
6
6 mol H
1 mol C
6
H
6
7.80
125 g C
6
H
6
b
a
b
(b)
The number of millimoles of carbon in the reactant is the
same as that in the
CO
2
:
10
23
H atoms
1 mol H atoms
a
6.02
5.79
10
24
H atoms
b
a
1 mmol CO
2
44.0 mg CO
2
1 mmol C
1 mmol CO
2
17.9 mg CO
2
b
a
b
a
1 molecule C
5
H
12
5 C atoms
0.406
8
mmol C
1.44
10
25
C atoms
7.82
b
a
1 mmol H
2
O
18.0 mg H
2
O
2 mmol H
1 mmol H
2
O
2.88
10
24
molecules C
5
H
12
(c)
11.0 mg H
2
O
b
a
b
7.85
The only empirical formula:
(c)
K
2
Cr
2
O
7
1.22
2
mmol H
7.87
(a)
85.63 g of carbon and 14.37 g of hydrogen
(b)
7.130 mol C and 14.26 mol H
(c)
1:2
(d)
The millimole ratio is 0.406
8
mmol C : 1.22
2
mmol H
or 1.00 mmol C : 3.00 mmol H
The empirical formula is
7.116
We treat this problem as an empirical formula problem with
as one of the “elements.”
(d)
CH
2
CH
3
.
a
1 mol C
12.01 g C
7.88
79.12 g C
b
6.588 mol C
H
2
O
a
1 mol H
1.008 g H
a
1 mol C
12.0 g C
9.79 g H
b
9.71 mol H
40.0 g C
b
3.33 mol C
a
1 mol O
16.00 g O
a
1 mol H
2
O
18.0 g H
2
O
11.10 g O
b
0.6938 mol O
60.0 g H
2
O
b
3.33 mol H
2
O
The ratio of moles is
9.496 mol C : 14.0 mol H : 1.000 mol O
In integral numbers 19 mol C, 28 mol H, 2 mol O: C
19
H
28
O
2
The ratio is 1:1 and the empirical formula is
CH
2
O.
a
1 mol H
1.008 g H
a
2 mol C
6 mol H
7.118
4.14 g H
b
b
1.37 mol C
