Empirical formula
Related Topics: Percent Composition, Molecular Formula
This takes the percent composition and suggests the ratio of atoms present.
Example: A compound composed of oxygen and phosphorus is 56.3% oxygen by mass. Find its empirical formula.
Solution: If 56.3% is oxygen, the rest, 43.7% is phosphorus. I set up a table, thus:
|
P |
O |
Elements present. |
|
43.7 31 |
56.3 16 |
Ratio of relative presence by mass. |
|
Divide both sides by respective relative atomic masses to get… |
||
|
1.4 1.4 |
3.5 1.4 |
…ratio of relative number of atoms present |
|
Divide both sides by the least amount (1.4 here) to get… |
||
|
1 |
2.5 |
…easier values to inspect. This suggests whole number ratio of… |
|
2 |
5 |
…or P2O5 for the empirical formula. |
RECALL that you expect to get low, whole numbers out at the end (2 and 5 in the table above). Any small fractions may be attributed to errors in measurement; round accordingly. Any fractions which are close to halves, thirds, quarters, etc. should probably be multiplied out to give whole numbers, none of which need to be 1. Other fractions may well indicate calculation errors on your part.
A compound contains 40.0% carbon and 53.3% oxygen by mass, the balance being hydrogen. Find its empirical formula.
Related Topics: Percent Composition, Molecular Formula