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 495.3-mg sample produced 610.5 mg of CO₂ and 180.6 mg of H₂O.

Use atomic masses: C 12.011; H 1.00794; O 15.9994. 1) Calculate the % hydrogen, by mass, in the sample. 2) Calculate the % oxygen, by mass, in the sample 3)Calculate the % citric acid, by mass, in the sample.

Answer 1

Solution :-

Lets calculate the moles of the mass of of the C and H using the percent of the C and H in CO2 and H2O

Mass of C = (610.5 mg CO2 *27.29 % C / 100 %)*(1000 mg/1 g) = 166.6 mg C

Mass of H = (180.6 mg H2O *11.19 % H / 100 %) = 20.21 mg H

Now lets calculate the mass of oxygen

Mass of O = total mass – (mass of C+ mass of H)

                    = 495.3 mg – (166.6 mg + 20.21 mg)

                    = 308.49 mg O

Now lets calculate the percent of the H and O

% H= (mass of H / total mass )*100%

      = (20.21 mg/495.3 mg)*100%

     = 4.08 %

% O = (mass of O / total mass) *100%

        = (308.49 mg/ 495.3mg)*100%

     = 62.28 %

Both citric acid and tartaric acid contains total 6+8 = 14 mol H

lets calculate the percent of the H that comes from citric acid

(6/14)*100% = 42.85 %

So the citric acid is 42.85 %