1) When 6.580 grams of a hydrocarbon, C x H y , were burned in a combustion anal

Question

1) When 6.580 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 20.16 grams of CO2 and 9.631 grams of H2O were produced.

In a separate experiment, the molar mass of the compound was found to be 86.18 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon.

2)When 2.823 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 8.859 grams of CO2 and 3.627 grams of H2O were produced.

In a separate experiment, the molar mass of the compound was found to be 70.13 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon.

Explanation / Answer

1)

moles CO2 = moles C = 20.16 g/44 g/mol = 0.46 mol

moles H = 2 x moles H2O = 2 x 9.631 g/18 g/mol = 1.070 mol

divide by smallest factor

C = 0.46/0.46 = 1

H = 1.070/0.46 = 2.33

empirical formula = CH2.33 = C3H7

empirical formula eight = 12 x 3 + 7 x 1 = 43

ratio (molar mass/empirical formula weight) = 86.18/43 = 2

So,

molecular formula = C6H14

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2)

moles CO2 = moles C = 8.859 g/44 g/mol = 0.201 mol

moles H = 2 x moles H2O = 2 x 3.627 g/18 g/mol = 0.403 mol

divide by smallest factor

C = 0.201/0.201 = 1

H = 0.403/0.201 = 2

empirical formula = CH2

empirical formula eight = 12 + 2 x 1 = 14

ratio (molar mass/empirical formula weight) = 70.13/14 = 5

So,

molecular formula = C5H10

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