Use the masses of carbon dioxide, CO2, and water, H2O, to determine the empirica

Question

Use the masses of carbon dioxide, CO2, and water, H2O, to determine the empirical formula of the alkane component

A 1.687g sample of a component of the light petroleum distillate called naphtha is found to yield 5.171g CO2(g) and 2.469g H2O(l) on complete combustion. This particular compound is also found to be an alkane with one methyl group attached to a longer carbon chain and to have a molecular formula twice its empirical formula. The compound also has the following properties: Use the masses of carbon dioxide, CO2, and water, H2O, to determine the empirical formula of the alkane component.

Explanation / Answer

5.171 g CO2 x (12.01g C/44.01g CO2) = 1.41 g C

1.41 g C / 12.01 = .0.117 mol C atoms

2.469 g H2O x (2.02g H/18.01g H2O) = 0.277 g H

0.277 g H / 1.008 =.0.275 mol H atoms

for H: 0.275 /0.117 = 2.35 mole

for C: 0.117/0.117 = 1

C1H2.35

C3H7=empirical formula

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