Chemistry Reference
In-Depth Information
Steps
Example
Step 1:
Change the percentages to numbers of grams
(by assuming that 100.00 g of sample is pres-
ent). (On exams, state that assumption.)
Because the size of the sample does not matter in deter-
mining an empirical formula, we can assume a 100.00-g
sample. That way, the percentages given are automati-
cally equal numerically to the numbers of grams of the
elements. For example:
39.2 g P
100.00 g compound
100.00 g compound
a
b
39.2 g P
From the percentage
Step 2:
For each element, convert the number of grams
to the number of moles.
39.2 g P and 60.8 g S
a
1 mol P
30.97 g P
39.2 g P
b
1.26
6
mol P
a
1 mol S
32.06 g S
60.8 g S
b
1.89
6
mol S
Step 3:
Try to get an integral ratio by dividing
all
the
numbers of moles by the magnitude of the
smallest number of moles. This will make at
least one number an integer.
1.26
6
mol P
1.26
6
1.00 mol P
1.89
6
mol S
1.26
6
1.50 mol S
Step 4:
If necessary, multiply
all
the numbers of moles
by the same small integer to clear fractions.
Round off the result to an integer only when the
number of moles is within 1% of the integer.
Always use at least three significant digits in
empirical formula calculations; otherwise,
rounding errors may produce an incorrect em-
pirical formula.
1.00 mol P
2
2.00 mol P
1.50 mol S
2
3.00 mol S
The empirical formula is P
2
S
3
.
Never round off more than 1%.
EXAMPLE 7.13
Determine the empirical formula of a compound that has a percent composition
of 43.7% P and 56.3% O.
Solution
Merely change the percentage signs to grams:
43.7 g P
and
56.3 g O
