Could someone show me a quick technique to calculate this question? It’s on the BMAT 2009 past paper (section 2)

Mass of CO2 produced = 4.77g

Molar Mass of CO2 = 12 + (16 x 2) = 44

Moles of CO2 = Mass/Molar Mass = 4.77/44 ≈ 0.1

The molar ratio of (C: CO2) = 1: 1

Moles of C = 0.1

Mass of C = Moles x Molar Mass = 0.1 x 12 = 1.2 g

Percentage = (1.2 / 2) x 100 = 60%

This is closest to 65% which is option E.

1 Like

thank you! I was wondering how to calculate the moles without a calculator but it’s okay to round off haha

1 Like