Please help with empirical formula.

Question

An unknown compound contains only C,H, and O. Combustion of 5.10g of this compound produced 11.6g CO2 and 4.75g H2O. What is the empirical formula of the unknown compound? Insert subscripts as needed.

Answer 1

The goal is to determine moles of C, H and O present in the unknown. We are given mass of CO2 and H2O, so from these we can obtain moles of C and H. But to find moles of O, we need to deduct grams of C and H from the original grams of sample (5.10 g) and then convert that to moles of O,

Moles C = 11.6 g CO2 x 1 mol CO2 / 44 g x 1 mol C/mol CO2 = 0.2636 moles C

Moles H = 4.75 g H2O x 1 mol H2O / 18 g x 2 mol H/mol H2O = 0.5278 moles H

Mass of C = 0.2636 g C x 1 mol / 12 g = 3.16 g

Mass of H = 0.5278 g H x 1 mol / 1 g = 0.53 g

Mass of O = 5.10 g - 3.16 g - 0.53 g = 1.41 g O

Moles O = 1.41 g O x 1 mol O / 16 g = 0.0881 moles O

Divide all moles by 0.0881 to try to get whole numbers...

Moles C = 0.2636/0.0881 = 3

Moles H = 0.5278/0.0881 = 6

Moles O = 0.0881/0.0881 = 1

Empirical formula = C₃H₆O