A 2.446-g sample thought to contain only one volatile substance, MgCO3, was heated in a nitrogen atmosphere for 30 min. After cooling, the residue weighed 2.216 g. Calculate the percentage of magnesium carbonate in the sample.

Answer 1

MgCO3 ==> MgO + CO2

Wt loss on heating = mass CO2 = 2.446 - 2.216 = 0.230g CO2
Convert to moles. moles = grams/molar mass
moles CO2 = moles MgCO3 in the sample.
g MgCO3 in the sample = moles MgCO3 x molar mass MgCO3

%MgCO3 = (mass MgCO3/mass sample)*100 = ?

Answer 2

Thanks DrBob222 :)

Answer 3

The answer is 18.01% :))