A 4.626-grain sample of a hydrocarbon, upon combustion in a combustion analysis

Question

A 4.626-grain sample of a hydrocarbon, upon combustion in a combustion analysis apparatus, yielded 6.527 grams of water. The percent, by weight, of hydrogen in the hydrocarbon is, therefore: 14.11% 15.79% 41.09% 66.22% 85.89% in a quantitative analysis study, 4.624 grams of a hydrocarbon (which contains carbon and hydrogen only) sample yielded CO_2 and 7.556 g of H_2 O in a combustion analysis apparatus. Determine the empirical formula or the hydrocarbon. CH_3 C_2 H_5 C_3 H_8 C_10 H_26 C_10 H_27

Explanation / Answer

10)
mol of H2O formed = mass of H2O / molar mass of H2O
= 6.527/18
=0.3626 mol

mol of H = 2*mol of H2O
= 2*0.3626 mol
= 0.7252 mol

mass of H = mol of H * molar mass of H
= 0.7252 mol * 1 g/mol
= 0.7252 g

mass % of H = mass of H * 100 / total mass
= 0.7252*100/4.626
= 15.68 %

Answer: b

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