0.250 g of a hydrocarbon known to contain carbon, hydrogen and oxygen was subject to complete combustion and produced 0.3664 g of CO2 and 0.1500 g of H2O. What is the empirical formula of this hydrocarbon?

The first step here is to determine the mass of C in CO2 and the mass of H in H2O. This is done by dividing the relative atomic mass, Mr, by the relative molecular mass of the compound, MCO2 or MH2O, and then multiplying by the mass of compound produced.

For C:

(12.011 g / 44.009 g) x 0.3664 g = 0.1000 g

For H:

(1.0079 x 2 / 18.0148) x 0.1500 g = 0.0168 g

From here we can determine the mass of oxygen in the hydrocarbon by subtraction of the two above masses:

O: 0.2500 g - 0.1000 g - 0.0168 g = 0.1332 g

From here we determine the number of moles of each element, this is done by dividing by the mass of each element by its atomic mass:

C: 0.1000 g / 12.011 g mol-1 = 0.0083 mol

H: 0.01678 g / 1.0079 g mol-1 = 0.0166 mol

O: 0.1332 g / 15.999 g mol-1 = 0.0083 mol

The final step is to divide by the smallest number that obtained, in this case that is the number of moles of C and O:

C: 0.0083 mol / 0.0083 mol = 1

H: 0.0166 mol /0.0083 mol = 2

O: 0.0083 mol / 0.0083 mol = 1

We have now arrived at a ratio of 1:2:1 for C:H:O, thus we have an empirical formula of CH2O.

JH
Answered by Joshua H. Chemistry tutor

11992 Views

See similar Chemistry A Level tutors

Related Chemistry A Level answers

All answers ▸

Explain why transition metal compounds are often coloured in solution.


Calculate the pH of a 0.025 mol dm-​3​ solution of methanoic acid. For HCOOH, Ka = 1.58 x 10-​4​ mol dm-​3


X, a gas, has a mass of 0.270g and is present in a gas syringe with a volume of 105.0cm^3 at 97C and 100kPa. Calculate the Mr of X. (5 marks)


State and explain the trend in ionisation energies and its effect on the reactivity of groups containing metals.


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences