Question

A compound is 53.31% C, 11.18% H, and 35.51% O by mass. What is its empirical formula?

Answer 1

`C_2H_5O_1` - ethanol

Explanation:

The atomic weights:

• C - carbon - 12.0107
• H - hydrogen - 1.00794
• O - oxygen - 15.9994

Therefore, 100 g would contain:

Carbon: 53.31 g, which is 53.31 / 12.0107 = 4.4385 moles
Hydrogen: 11.18 g, which is 11.18 / 1.00794 = 11.0919 moles
Oxygen: 35.51 g , which is 35.51 / 15.9994 = 2.2194 moles

Working the ratios:
Carbon: 4.4385 / 2.2194 = 1.9999
Hydrogen: 11.0919 / 2.2194 = 4.9977
Oxygen: 2.2194 / 2.2194 = 1

So, the answer is `C_2H_5O_1`, which is ethanol