Question

With 40.01% C (carbon), 6.72% H (hydrogen), and 53.27% O (oxygen), what is its empirical formula?

a) CH
b) C2H4O
c) C4H8O2
d) C6H12O6

Answer 1

The empirical formula of the compound is C2H4O (option b).

Step-by-step explanation:

The empirical formula of the compound with 40.01% C, 6.72% H, and 53.27% O is C2H4O (option b).

To determine the empirical formula, we first convert the percentages to moles. From the given percentages, we have 40.01 g C, 6.72 g H, and 53.27 g O. We then divide the moles by their respective atomic masses to find the ratios: C: 40.01 g / 12.01 g/mol = 3.332 mol; H: 6.72 g / 1.008 g/mol = 6.667 mol; O: 53.27 g / 16.00 g/mol = 3.33 mol. We simplify these ratios to the nearest whole numbers, giving us a ratio of 1:2:1. Therefore, the empirical formula is C2H4O.