Question

A dry-cleaning solvent that contains only C, H, and Cl is suspected to be a cancercausing agent. When a 0.25-g sample was analyzed using combustion analysis, 0.451 g CO2 and 0.0617 g of H2O formed. Determine the empirical formula of this compound.

Answer 1

Answer:

The empirical formula is = $C_3H_2Cl$

Explanation:

Mass of water obtained = 0.0617 g

Molar mass of water = 18 g/mol

Moles of $H_2O$ = 0.0617 g /18 g/mol = 0.003428 moles

2 moles of hydrogen atoms are present in 1 mole of water. So,

Moles of H = 2 x 0.003428 = 0.006856 moles

Molar mass of H atom = 1.008 g/mol

Mass of H in molecule = 0.006856 x 1.008 = 0.006911 g

Mass of carbon dioxide obtained = 0.451 g

Molar mass of carbon dioxide = 44.01 g/mol

Moles of $CO_2$ = 0.451 g  /44.01 g/mol = 0.01025 moles

1 mole of carbon atoms are present in 1 mole of carbon dioxide. So,

Moles of C = 0.01025 moles

Molar mass of C atom = 12.0107 g/mol

Mass of C in molecule = 0.01025 x 12.0107 = 0.12311 g

Given that the dry-cleaning solvent only contains hydrogen, chlorine and carbon. So,

Mass of Cl in the sample = Total mass - Mass of C  - Mass of H

Mass of the sample = 0.25 g

Mass of Cl in sample = 0.25 - 0.12311 - 0.006911 = 0.119979 g  

Molar mass of Cl = 35.453 g/mol

Moles of Cl  = 0.119979  / 35.453  = 0.003384 moles

Taking the simplest ratio for H, Cl and C as:

0.006856 : 0.003384 : 0.01025

= 2 : 1 : 3

The empirical formula is = $C_3H_2Cl$