Question

A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate:

1. empirical formula, 
2. molar mass of the gas, and 
3. molecular formula

Answer 1

Given:

Mass of CO₂ = 3.38 g
Mass of H₂O = 0.690 g
Volume of gas at STP = 10.0 L
Mass of gas = 11.6 g

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(i) Empirical formula

Step 1: Calculate mass of carbon

From CO₂:

44 g CO₂ contains 12 g C

Carbon = (12 / 44) × 3.38 = 0.922 g

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Step 2: Calculate mass of hydrogen

From H₂O:

18 g H₂O contains 2 g H

Hydrogen = (2 / 18) × 0.690 = 0.0767 g

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Step 3: Convert masses to moles

Moles of C = 0.922 / 12 = 0.0768 mol

Moles of H = 0.0767 / 1 = 0.0767 mol

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Step 4: Find simplest whole-number ratio

C : H = 0.0768 : 0.0767 ≈ 1 : 1

Empirical formula = CH

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(ii) Molar mass of the gas

At STP, 22.4 L of gas = 1 mol

Moles of gas in 10.0 L = 10.0 / 22.4 = 0.446 mol

Molar mass = Mass / Moles

Molar mass = 11.6 / 0.446 = 26 g mol⁻¹

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(iii) Molecular formula

Empirical formula mass of CH = 12 + 1 = 13 g mol⁻¹

n = Molecular mass / Empirical formula mass

n = 26 / 13 = 2

Molecular formula = (CH)₂ = C₂H₂

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Final Answers:

(i) Empirical formula = CH
(ii) Molar mass = 26 g mol⁻¹
(iii) Molecular formula = C₂H₂