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Complete combustion of 3.90 g of a hydrocarbon produced 12.5 g of CO2 and 4.28 g of H2O. What is the empirical formula for the hydrocarbon?

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\(44g \ CO_{2} \ \ \ \ \rightarrow \ \ \ 12g \ C\\ 12,5g \ CO_{2} \ \ \rightarrow \ \ \ x\\\\ x=\frac{12,5g*12g}{44g}=3,41g \ \ \ \Rightarrow \ \ \ n=\frac{3,41g}{12\frac{g}{mol}}=0,28mol\\\\\\ 18g \ H_{2}O \ \ \ \ \ \ \rightarrow \ \ \ 2g \ \ H\\ 4,28g \ H_{2}O \ \ \ \rightarrow \ \ \ x\\\\ x=\frac{4,28g*2g}{18g}=0,48g \ \ \ \Rightarrow \ \ \ n=\frac{0,48g}{1\frac{g}{mol}}=0,48mol\\\\\\\\ n_{c}:n_{x}=0,28:0,48\approx1:2\\\\\\ empirical \ formula: \ CH_{2}\)

Answer:

Empirical formula of the hydrocarbon is \(C_1H_2\).

Explanation:

Mass of carbon dioxide produced = 12.5 g

Moles of carbon dioxide = \(\frac{12.5 g}{44 g/mol}=0.2841 mol\)

Moles of carbon atom in 0.2841 moles of carbon dioxide:

1 × 0.2841 mol = 0.2841 mol

Mass of water produced = 4.28 g

Moles of water = \(\frac{4.28 g}{18 g/mol}= 0.2378 mol\)

Moles of hydrogen in water = 2 × 0.2378 mol = 0.4756 mol

For empirical formula divide the moles of element which are is less amount with all the moles of every element.

Carbon = \(\frac{0.2841 mol}{0.2841 mol}=1\)

Hydrogen = \(\frac{0.4756 mol}{0.2841 mol}=1.67\approx 2\)

Empirical formula of the hydrocarbon is \(C_1H_2\).

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