Empirical Formulas From Mass Percent Composition
The empirical formula of a compound is the simplest formula we can write for that compound. The subscripts in an empirical formula consist of the smallest integers possible to show the ratio of atoms (or moles) of each element in the compound. In the mass percent composition of a compound, we specify the ratios of each element in terms of mass, in the empirical formula the ratio must be on a mole basis.
We can state an algorithm for obtaining an empirical formula from mass percent composition in terms of four steps as follows.
| Step 1. Convert percentages to masses. |
| Step 2. Convert masses to moles. |
| Step 3. Obtain tentative subscripts by dividing each number of moles by the smallest. |
| Step 4. If the tentative subscripts from step 3 are not integers(or very close), multiply each by a common |
| factor to convert them all to integers. |
Since mass percentages are ratios, they are the same regardless of how much sample we have. The most convenient amount of sample to work with is 100 g, this way the masses of the elements are numerically equal to their mass percentages. We will illustrate this with a couple of examples.
Find the empirical formula of a compound with the mass percent composition; 36.48% Na, 25.44% S, 38.08% O
Solution
| Step 1. A 100.00 g sample of the compound contains 36.48 g Na, 25.44 g S, and 38.08 g O. |
| Step 2. Convert the masses of Na,S, and O to moles. |
1 mol Na |
||||
36.48 g Na |
x |
------------------ |
= |
1.587 mol Na |
22.99 g Na |
1 mol S |
||||
25.44 g S |
x |
------------------ |
= |
0.793 mol S |
32.07 g S |
1 mol O |
||||
38.08 g O |
x |
------------------ |
= |
2.380 mol O |
16.00 g O |
| Step 3. Divide each of the mole quantities by the smallest and if the results are integers use as subscripts. |
| 1.587 / 0.793 = 2.00 0.793 / 0.793 = 1.00 2.380 / 0.793 =3.00 |
| Since the results of Step 3. are integers Step 4. is not necessary and the empirical formula is Na2SO3. |
Find the empirical formula of a compound with the mass percent composition; 45.27% C, 9.50% H, 45.23% O
Solution
| Step 1. A 100.00 g sample of the compound contains 45.27 g C, 9.50 g H, and 45.23 g O. |
| Step 2. Convert the masses of Na,S, and O to moles. |
1 mol C |
||||
45.27 g C |
x |
------------------ |
= |
3.769 mol C |
12.01 g C |
1 mol H |
||||
9.50 g H |
x |
------------------ |
= |
9.425 mol H |
1.008 g H |
1 mol O |
||||
45.23 g O |
x |
------------------ |
= |
2.827 mol O |
16.00 g O |
| Step 3. Divide each of the mole quantities by the smallest and if the results are integers use as subscripts. |
| 3.769 / 2.827 = 1.33 9.425 / 2.827 = 3.33 2.827 / 2.827 =1.00 |
| Since the results of Step 3. are not integers Step 4. necessary. |
| Step 4. Multiply each of the mole quantities from Step 3. by the factor 3. The empirical formula is C4H10O3. |