Complete combustion of 8.70 g of a hydrocarbon produced 27.8 g of CO2 and 9.96 g of H2O. What is the empirical formula for the hydrocarbon? Insert subscripts as necessary.

Question

Complete combustion of  g of a hydrocarbon produced  g of CO2 and  g of H2O. What is the empirical formula for the hydrocarbon? Insert subscripts as necessary.



Answer 1

Combustion of a hydrocarbon (CxHy) results in production of CO2 and H2O.

(Note: it states the combustion is of a hydrocarbon which means it contains C and H only, and no oxygen)

moles of C in the compound: 27.8 g CO2 x 1 mol CO2/44 g x 1 mol C/mole CO2 = 0.632 moles of C

moles of H in the compound: 9.96 g H2O x 1 mol H2O/18 g x 2 mol H/mol H2O = 1.11 moles of H

To be sure no oxygen is in the original compound, we can find the mass and be sure it adds to 8.70 g

mass C = 0.632 mol C x 12 g/mol = 7.58 g C

mass H = 1.11 mol H x 1 g/mol = 1.11 g H

Sum = 8.69 g total mass (close enough to 8.70 so our assumption of no oxygen is correct)

To find the lowest whole number of moles, we can divide both by the lowest value (0.632) to obtain...

moles C = 0.632/0.632 = 1.0

moles H = 1.11/0.632 = 1.75

Now to get a whole number for H we can multiply both by 4 to obtain...

moles C = 4

moles H = 7

Empirical formula = C₄H₇