Question

A sample that contains only chromium(III) carbonate (Cr2(CO3)3, 284.03 g/mol) and copper(II) carbonate (CuCO3, 123.56 g/mol) weighs 0.3200 g. When it is dissolved in excess acid, 0.1375 g of carbon dioxide (CO2, 44.01 g/mol) is liberated. What percentage, by mass, of chromium(III) carbonate did the sample contain? Assume all the carbon originally present is converted to carbon dioxide. what is the % by mass chromium(III) carbonate

Answer 1

Sol :-

Given weight of the sample = 0.3200 g

Sample contains Cr₂(CO₃)₃ and CuCO₃.

Let mass of Cr₂(CO₃)₃ = M g

so

Mass of CuCO₃ = (0.3200 - M ) g

We know

Number of moles = Given mass in g / Gram molar mass

so

Number of moles of Cr₂(CO₃)₃ = M g / 284.03 g/mol = M/284.03 mol

and

Number of moles of CuCO₃ = (0.3200 - M ) g / 123.56 g = (0.3200 - M) / 123.56 mol

Now reaction of Cr₂(CO₃)₃ with excess of acid produces 3 moles of CO₂ as :

Cr₂(CO₃)₃ + 6 H⁺ ----------> 2 Cr³⁺ + 3 CO₂ + 3 H₂O

Now 1 moles of Cr₂(CO₃)₃ produces = 3 moles of CO₂

therefore

M/284.03 moles of Cr₂(CO₃)₃ produces = 3M/284.03 moles of CO₂

similarly, reaction of CuCO₃ with excess of acid produces 1 mole of CO₂ as :

reaction of CuCO₃ with excess of acid produces 3 moles of CO₂ as :

CuCO₃ + 2H⁺ ---------> Cu²⁺ + CO₂ + H₂O

Now, 1 mol of CuCO₃ produces = 1 mol of CO₂

therefore

(0.3200 - M ) / 123.56 moles of CuCO₃ produces = (0.3200 - M ) / 123.56 moles of CO₂

Now Total number of moles of CO₂ = 3M/284.03 mol + (0.3200 - M ) / 123.56 mol of CO₂ .....(1)

Also

Given mass of CO₂ formed = 0.1375 g

so

Number of moles of CO₂ = 0.1375 g / 44.01 g/mol = 0.1375 / 44.01 mol ...........(2)

On combining equations (1) and (2), we have

3M/284.03 mol + (0.3200 - M ) / 123.56 mol of CO₂ = 0.1375 / 44.01 mol

370.68 M + 90.8896 - 284.03 M / 35094.7468 = 0.1375 / 44.01  

86.65 M = 109.6462 - 90.8896

M = 18.7566 / 86.65

M = 0.2165 g

Therefore mass of Cr₂(CO₃)₃ in the sample = M = 0.2165 g

and

% by mass chromium(III) carbonate i.e. Cr₂(CO₃)₃ = 0.2165 g x 100 / 0.3200 g = 67.7 %

Hence Mass %age of Cr₂(CO₃)₃ in the sample = 67.7 %