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What is the empirical formula of the hydrocarbon?

Combustion analysis of a hydrocarbon produced 33.01gCO2 and 10.13 gH2O what is the empirical formula of the hydrocarbon?

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    (33.01 g CO2) / (44.00964 g CO2/mol) x (1 mol C / 1 mol CO2) = 0.750063 mol C (10.13 g H2O) / (18.01532 g H2O/mol) x (2 mol H / 1 mol H2O) = 1.12460 mol H Divide by the smaller number of moles: (0.750063 mol C) / 0.750063 mol = 1.000 (1.12460 mol H) / 0.750063 mol = 1.499 In order to achieve integer coefficients, multiply by 2 then round to the nearest whole numbers to find the empirical formula: C2H3

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Combustion analysis of a hydrocarbon produced 33.01gCO2 and 10.13 gH2O what is the empirical formula of the hydrocarbon? ---------------------------------------... Assuming because it is a hydrocarbon) that the only elements present are carbon and hydrogen, we have 12/44 x 33.01 grams C = 8.00 grams carbon and 2/18 x 10.13 grams H = 1.125 grams hydrogen which gives 8/(8+1.125) x 100 = 87.13 % C / 12 = 7.261 and 1.125/9.125 x 100 = 12.162 % H = 12.162/1.007 = 12.077 Letting the value for carbon = 1, the empirical formula is CH(1.66), or cH(5/3), which means that the intermediate formula is CH(5/3) x 3 = C3H5. Since the number of hydrogens is odd, double it to have the molecular formula. C6H10, which could be cyclohexene.

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