chemistry: unknown carbonate

Question

A 1.1000 gram carbonate sample chosen from Li2CO3, K2CO3, Na2CO3 and CaCO3 was reacted with H2SO4 and was found to lose 0.4831 gram of CO2. (1) Show the calculation of the % CO2 in the unknown carbonate sample. (2) Show the calculation of the % CO2 in each of the carbonate compounds and identify the unknown carbonate from the list.

 

Atomic weights: C = 12.01, O = 16.00. MWs: Li2CO3 = 73.892, K2CO3 = 138.21, Na2CO3 = 105.99 and CaCO3 = 100.09

Answer 1

Actually, there is no CO₂ in each carbonate so you can't calculate %CO₂.  There is however C in each carbonate, so you can calculate the % C in each.

0.4831 g CO₂ x 1 mole CO₂/44.01 g = 0.01098 moles CO₂ = 0.01098 moles C

0.01098 moles C x 12.01 g C/mole C = 0.1318 g C in the unknown

% in unknown = 0.1318 g/1.1000 g (x100%) = 11.98 % C

Li₂CO₃:  %C = 12.01 g/73.892 g (x100%) = 16.25%

K₂CO₃: %C = 12.01 g/138.21 g (x100%) = 8.69% 

Na₂CO₃: %C = 12.01 g/105.99g (x100%) = 11.33%

CaCO₃: %C = 12.01 g/100.09 g (x100%) = 11.99%

UNKNOWN APPEARS TO BE CaCO₃