College Chem

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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.

  • College Chem -

    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 = ?

  • College Chem -

    Thanks DrBob222 :)

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