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On complete combustion in oxygen of 0.42grams of a gaseous hydrocarbon x,it gave 1.32grams of carbon dioxide and 0.54grams of water .Determine the empirial formula

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Molar mass of CO2 = 44.01 g/mol

Molar mass of H2O = 18.02 g/mol

Molar mass of C = 12.01 g/mol

Molar mass of H = 1.01 g/mol

Moles of CO2 = 1.32 g / 44.01 g/mol = 0.03 mol

Moles of H2O = 0.54 g / 18.02 g/mol = 0.03 mol

Moles of C = 0.03 mol

Moles of H = 0.5 * 0.03 mol = 0.015 mol

Mass of C = 0.03 mol * 12.01 g/mol = 0.36 g

Mass of H = 0.015 mol * 1.01 g/mol = 0.015 g

To simplify the ratio, we divide both masses by the smaller mass (0.015 g):

Mass of C / 0.015 g = 0.36 g / 0.015 g = 24

Mass of H / 0.015 g = 0.015 g / 0.015 g = 1

The empirical formula of the hydrocarbon X is CH4.

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C2H2
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Answer 


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CH

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