An unknown compound contains only C , H , and O . Combustion of 5.90 g of this compound produced 11.8 g CO2 and 4.83 g H2O . What is the empirical formula of the unknown compound

C₂H₄O

Explanation:

In a compound that contains Cabon, hydrogen and oxygen, the combustion produce CO₂ from the carbon, and H₂O from the hydrogens. Using the mass of the products we can solve the moles of Carbon and hydrogen. The empirical formula is the simplest whole-number of atoms present in a molecule.

Moles CO₂ = Moles C:

11.8g CO₂ * (1mol / 44g) = 0.268 moles CO₂ = 0.268 moles C * (12g/mol) =

3.216g C

Moles H₂O = 1/2 moles H:

4.83g H₂O * (1mol / 18g) = 0.268 moles H₂O * (2 mol H / 1 mol H₂O) =

0.537 mol H * (1g/mol) = 0.537g H

Mass O to find moles O:

5.90g Sample - 3.216g C - 0.537g H = 2.147g O * (1mol / 16g) = 0.134 moles O

Ratio of atoms -Dividing in 0.134 moles-:

C = 0.268mol C / 0.134 mol O = 2

H = 0.537mol H / 0.134 mol O = 4

O = 0.134mol O / 0.134 mol O = 1

Empirical formula is:

sorry idk the answer well to you i have a explaination

explanation: a formula which shows the arrangement of atoms in the molecule of a compound