12.48 g of sample of MgO and MgCO3 on strong heating lost 4.4g weight. What is composition of original mixture?

Molar mass of MgCO₃ = (24.3 + 12.0 + 16.0×3) g/mol = 84.3 g/mol
Molar mass of CO₂ = (12.0 + 16.0×2) g/mol = 44.0 g/mol

MgO is stable on heating, but MgCO₃ decomposes on heating.
MgCO₃ → MgO + CO₂
Mole ratio MgCO₃ : CO₂ = 1 : 1

Mass of CO₂ produced = lost in mass = 4.4 g
No. of moles of CO₂ produced = (4.4 g) / (44.0 g/mol) = 0.1 mol
No. of moles of MgCO₃ decomposed = 0.1 mol
Mass of MgCO₃ in the sample = (0.1 mol) × (84.3 g/mol) = 8.43 g
Mass of MgO in the sample = (12.48 - 8.43) g = 4.05 g

Percent by mass of MgCO₃ in the sample = (8.43/12.48) × 100% = 67.5%
Percent by mass of MgO in the sample = 1 - 67.5% = 32.5%

On heating the carbonate will lose CO2 forming the oxide according to the equation:
MgCO3 ------------> CO2 + MgO
You simply need to use a table of atomic masses to calculate quantities of reactants and products. What quantity of MgCO3 will produce 4.4g of CO2 on heating?