Question

A sample that contains only rubidium carbonate (Rb2CO3, 230.95 g/mol) and strontium carbonate (SrCO3, 147.63 g/mol) weighs 0.3293 g. When it is dissolved in excess acid, 0.0778 g of carbon dioxide (CO2, 44.01 g/mol) is liberated. What percentage, by mass, of rubidium carbonate did the sample contain? Assume all the carbon originally present is converted to carbon dioxide.

Answer 1

no of mol of CO2 liberated = 0.0778/44 = 0.00177 mol

no of mol of CO2 = no of mol of carbonate(CO3^-2)

We know that, 1 mol Rb2CO3 = 1 mol carbonate(CO3^-2)

              1 mol SrCO3 = 1 mol carbonate(CO3^-2)

mass of sample = mass of Rb2CO3 + mass of SrCO3

         0.3293   = x*(230.95) + (0.00177-x)*(147.63)

   x = 0.000816

no of mol of Rb2CO3 in sample = x = 0.000816 mol

amount of Rb2CO3 in sample = 0.000816*230.95

                           = 0.188 g

percentage,by mass, of rubidium carbonate in the sample = 0.188/0.3293*100

               = 57.1%