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