The mass of 0.00392 mol of gas is 0.345 g, so the mass of one mole is
88.0 g/mol
You should now work Exercises 51 and 54.
Let’s carry the calculation one step further in the next example.
EXAMPLE 12-14 Molecular Formula
Additional analysis of the gaseous compound in Example 12-13 showed that it contained 54.5%
carbon, 9.10% hydrogen, and 36.4% oxygen by mass. What is its molecular formula?
Plan
We first find the simplest formula for the compound as we did in Section 2-9 (Examples 2-12
and 2-13). Then we use the molecular weight that we determined in Example 12-13 to find
the molecular formula. To find the molecular formula, we reason as in Example 2-16. We use
the experimentally known molecular weight to find the ratio
n
The molecular weight is ntimes the simplest-formula weight, so the molecular formula is n
times the simplest formula.
Solution
Relative Number Smallest Whole-
Relative Mass of Atoms Divide by Number Ratio
Element of Element (divide mass by AW) Smallest Number of Atoms
C 54.5
5
1
4
2
.
.
5
0
4.54
4
2
.
.
5
2
4
8
1.99 2
H 9.10
9
1
.
.
1
0
0
1
9.01
9
2
.
.
0
2
1
8
3.95 4 C 2 H 4 O
O 36.4
3
1
6
6
.
.
4
0
2.28
2
2
.
.
2
2
8
8
1.00 1
The simplest formula is C 2 H 4 O and the simplest-formula weight is 44.0 amu.
Division of the molecular weight by the simplest-formula weight gives
2
The molecular formula is therefore 2(C 2 H 4 O)C 4 H 8 O 2.
The gas could be either ethyl acetate or butyric acid. Both have the formula C 4 H 8 O 2. They
have very different odors, however. Ethyl acetate has the odor of nail polish remover. Butyric
acid has the foul odor of rancid butter.
You should now work Exercise 52.
88.0 amu
44.0 amu
molecular weight
simplest-formula weight
molecular weight
simplest-formula weight
0.345 g
0.00392 mol
_?_g
mol
12-10 Determination of Molecular Weights and Molecular Formulas of Gaseous Substances 455
H
H
H
C
CCH
C
O
O
O
H H
HH
H
H
H
C
H
H
C
H
H
CC
OH
ethyl acetate
butyric acid