Question

A white powder, consisting of a simple mixture of tartaric acid (C4H6O6) and citric acid (C6H8O7) was analysed to determine the elemental composition. Combustion of a 676.4-mg sample produced 900.7 mg of CO2 and 251.5 mg of H2O. Use atomic masses: C 12.011; H 1.00794; O 15.9994.

Calculate the % carbon, by mass, in the sample.

Calculate the % hydrogen, by mass, in the sample.

Calculate the % oxygen, by mass, in the sample.

Answer 1

1)

Combustion reaction=

C4H6O6 + C6H8O7 + 7O2→10 CO2 + 7H2O

The carbon in the sample is converted into CO2,so we need to calculate Carbon in the amount of CO2 produced on combustion of the sample.

Mass of CO2=900.7 mg

Molar mass of CO2=12.011+2*15.9994=44.0098g/mol

Mass % of C per mole of CO2=(12.011 g/mol) / (44.0098 g/mol) *100=27.29 %

So mass of C in the sample=27.29% of 900.7 mg=27.29/100*900.7 mg=245.8mg

% C by mass in the sample=245.8mg/676.4mg*100=36.34% (answer)

2) Total hydrogen content of the sample gets converted into H2O on combustion

Molar mass of H2O=1.00794*2+15.9994=18.015 g/mol

Mass % of H per mol of H2O=(1.00794g/mol)/( 18.015 g/mol)*100=5.595%

So mass of H in the sample =5.595% of 251.5 mg=14.071mg

Mass % of H in the sample=14.071 mg/676.4 mg*100=2.080 %

3)% of O=100-(% of C+% of H in the sample)=100-(36.34%+2.080%)=100-38.42=61.58 %(answer)