Question

Combustion analysis:

Combustion of a 5.50 g sample of benzene produces 18.59 g of CO2 and 3.81 g of water. Determine the empirical formula and the molecular formula of benzene, given that its molar mass is approximately 78 g/mol.

Answer 1

Solution:-

First , we have to find the amount of C and H in the sample :

Number of moles of CO₂ produced = Mass ÷ molar mass = (18.59 g) / (44.01 g/mol) = 0.4224 moles. Since one mole of CO₂ is produced from one mole of C and two moles of O. Thus, the number of moles of C in sample = 0.4224 mole. Now, the mass of 0.4224 moles of C = (Number of moles of C × molar mass of C) = (0.4224 mol × 12.01 g/mol) = 5.073 g

Number of moles of H₂O = (3.81 g) / (18 g/mol) = 0.2115 moles. Since one mole of H₂O is produced from two moles of H and 1 mole of O. Thus, the number of moles of H in the sample = (2 × 0.2116 moles) = 0.423 moles. Now, the mass of 0.423 moles of H = (0.423 moles × 1.008 g/mol) = 0.426 g

Total mass of C and H = (5.073 + 0.426) g = 5.499 g ~ 5.50 g . Thus, O atoms are not present in the sample.

Therefore, overall we have

0.4224 moles of C

0.423 moles of H

Dividing each by smallest molar amount:

C atoms = (0.4224 / 0.4224) = 1

H atoms = (0.423 / 0.4224) = 1

Hence, empirical formula of benzene = CH

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