Question

Combustion of a 34.8mg sample of benzaldehyde, which contains only carbon, hydrogen and oxygen,produced 101mg of CO2 and 17.7mg of H2O.

a, what was the mass of carbon and hydrogen in the sample.

b, Assuming that the original sample contained only carbon,hydrogen and oxygen. What was the mass of oxygen in the sample.

c,what was the mass percentage of oxygen in the sample

d, what is the emperical fomula of benzaldehyde.

Answer 1

Amount of carbon present in 101 mg of CO₂ =$\dfrac{12}{44}*101=27.5mg$

amount of hydrogen present in 17.7 mg of H₂O =$\dfrac{2}{18}*17.7=1.96mg$

amount of oxygen present in 34.8 mg of the compound = (34.8-27.5-1.96)mg = 5.34 mg

$27.5C*\dfrac{1}{12}=2.29$

$1.96H*1=1.96$

$5.34O*\dfrac{1}{16}=0.33$

Converting this into simple ratio=C:H:O=7:6:1

Hence the empirical formula $C_7H_6O$