When heated a sample consisting of only CaCO3 and MgCO3 yields a mixture of CaO and Mgo. If the weight of the combined oxides is equal to 51.00% of the initial sample weight, calculate the per cent of CaCO3 and MgCO3 in the sample. (12 M.)

38.83 % of CaCO3

61.17 % of MgCO3

Explanation:

where Moles of CaCO3 is equals to x and MgCO3 is y we have that ...

CaCO3 molar mass = 100.09 g / mol = 100.09 x

MgCO3 molar mass = 84.31 g / mol = 84.31 y

decomposition reactions:

CaCO3 - --> CaO + CO2

MgCO3 - --> MgO + CO2

So we have that, Moles of CaO = Moles of CaCO3 = x

and Moles of MgO = Moles of MgCO3 = y

CaO molar mass = 56.08 g / mol

MgO molar mass = 40.30 g / mol

CaO = 56.08 x

MgO = 40.30 y

"If the weight of the combined oxides is equal to 51.00% of the initial sample weight,"

total mass of MgO and CaO = 51.00 % of Total Mass of MgCO3 and CaCO3

thus

56.08 x + 40.30 y = 0.51 (100.09 x + 84.31 y)

56.08 x + 40.30 y = 51.04 x + 42.99y

5.04 x = 2.7 y

y = 1.87 x

CaCO3 % in the sample

= 100.09 x * 100 / (100.09 x + 84.31 y)

= 10009 x / (100.09 x + 84.31 * 1.87 x)

= 10009 x / (x (100.09 + 157.66))

= 10009 / 257.75

= 38.83 %

MgCO3 % in the sample

= 100 - 38.83

= 61.17 %