Chemistry Reference
In-Depth Information
to hydrogen to oxygen atoms in fructose is 1 to 2 to 1. An empirical formula
always has the smallest integral subscripts that give the correct ratio of atoms
of the elements.
EXAMPLE 7.12
Write the empirical formulas for the compounds containing carbon and hydro-
gen in the following ratios:
(a) 2 mol carbon to 3 mol hydrogen
(b) 1.0 mol carbon to 1.5 mol hydrogen
(c) 0.1712 mol carbon to 0.2568 mol hydrogen
Solution
(a) The mole ratio is 2:3, so the empirical formula is
(b) The mole ratio given is not integral, but we can multiply each value by 2
to get an integral ratio of 2:3. The empirical formula is again
(c) This mole ratio is not integral, and this time it is more difficult to make it
so. If we divide both values by the magnitude of the smaller one, we can
get closer to an integral ratio:
C
2
H
3
.
C
2
H
3
.
0.1712 mol C
0.1712
0.2568 mol H
0.1712
1.000 mol C
1.500 mol H
Now multiply by 2, as in part (b):
1.000 mol C
1.500 mol H
2 mol C
3 mol H
This ratio is also 2:3, and again the empirical formula is
C
2
H
3
.
Practice Problem 7.12
Write the empirical formula for a compound
consisting of elements A and B, for each ratio of A to B:
(a) 1:1
(b) 1:1.5
(c) 1:1.33
(d) 1:1.67
(e) 1:1.25
(f) 1:1.75
(g) 1:1.20
(h) 1:1.40
We can find the empirical formula from percent composition data. The
empirical formula represents a ratio; therefore, it does not depend on the size
of the sample under consideration. Because the empirical formula reflects a
mole
ratio, and percent composition data are given in terms of
mass,
we have to con-
vert the masses to moles. We then convert the mole ratio, which is unlikely to
be an integral ratio, to the smallest possible whole-number ratio, from which
we write the empirical formula.
The steps we take to obtain an empirical formula from percent composi-
tion data are given in the left column (Steps) that follows. In the right column
(Example), the empirical formula of a compound containing 39.2% phosphorus
and 60.8% sulfur is calculated.
