Question

Complete combustion of 7.70 g of a hydrocarbon produced 23.6 g of CO2 and 11.3 g of H2O. What is the empirical formula for the hydrocarbon? Insert the subscripts:

CH

Answer 1

There will be the same number of moles of C and H on both sides of the equation, so all the Carbon in your original

hydrocarbon is now present in CO2, and all H is in the H2O.

Hydrogen
Calculate moles of H in H2O
molar mass H2O = 16.00 + (2 x 1.008) = 18.016 g/mol
molar mass H = 1.008 g/mol

moles = mass / molar mass
moles H2O = 11.3 g / 18.016 g/mol
= 0.6272 moles of H2O

In ever H2O there are 2 moles of H
therefore moles of H in H2O, and thus in your original hydrocarbon = 2 x 1.421 x 10^-4
= 1.2544 moles of H

Carbon
Calculate moles C in CO2
molar mass CO2 = (16.00 x 2) + 12.01 = 44.01 g/mol
moles CO2 = mass / molar mass = 23.6 g / 44.01 g/mol
= 0.536 moles of CO2
Every CO2 molecule has 1 Carbon, therefore moles of C in CO2 and thus in the original hydrocarbon
=0.536 moles of C

Now work out the ratio of C : H
C : H
= 0.536 : 1.2544
To get it into whole number divide both numbers in the ratio by the lowest number (both here are the same, but this will

not always be the case
C : H
= 0.536/0.536 : 1.2544/0.536
= 1:2.34
= 3:7
answer= C3H7

So empirical formula is CH

The molar mass of CH = 12.01 + 1.008 = 13.018 g/mol

To work out the molecular mass divide the molar mass of the actual compound by the molar mass of the empirical

formula, this will tell you how may time the empirical formula "repeats in the molecular formula.

40 / 13.018
= 3.07

molecular formula = C3H3