Question

A mineral consists of an equimolar mixture of the carbonates of two bivalent metals. One metal is present to the extent of 13.2% by weight. 2.58 g. of the mineral on heating lost 1.233 g. Of C02. Calculate the % by weight of the other metal.

Answer 1

Moles of XCO₃= x
Moles of YCO₃=y

We know that:
x = y

So, XCO₃→ XO + CO₂
YCO₃ → YO + CO₂

Moles of CO₂ = x+y = 1.233/molar mass of CO₂ = 1.233 / 44 = 2.80*10^-2
then x = y = 2.80*10^-2 / 2 = 1.40*10^-2

x*molar mass (X) = 2.58 * (13.2/100) = 0.33354g
⇒ molar mass X= 0.33354 / 1.40*10^-2 = 23.94
⇒Molar mass of (XCO₃) = 23.94 + 12 + 3*16 = 83.94
⇒ x*molar mass of (XCO₃) = mass of XCO₃ = 1.40*10^-2 * 83.94 = 1.175g

% of XCO₃ = 1.175g / 2.58g = 0.455 → 45.5%
% of YCO₃=100 - 45.5 = 54.5 %