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AP Chemistry

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A 28.4858 g sample of impure magnesium carbonate was heated to complete decomposition according to the equation
MgCO3(s)→MgO(s) + CO2(g).After the reaction was complete, the solid
residue (consisting of MgO and the original impurities) had a mass of 17.3014 g. Assuming that only the magnesium carbonate had decomposed, how much magnesium carbonate was present in the original sample?
Answer in units of g

  • AP Chemistry -

    mass CO2 = 28.4858 - 17.3014 = ?
    mass MgCO3 = mass CO2 x (molar mass MgCO3/molar mass CO2) = ?

  • AP Chemistry -

    Known:
    Mass of sample: s = 28.4858 g
    Mass of residue: r = 17.3014 g

    Unknown:
    Mass of Impurities: x
    Mass of MgCO3: m0 = s - x
    Mass of MgO: m1 = r - x
    Therefore: m1 = m0 + r - s

    Lookup:
    Molecular mass of MgCO3: w0 = 84.31413 ± 0.00005 g/mol
    Molecular mass of MgO: w1 = 40.30449 ± 0.00002 g/mol
    Stoichiometry: m0/w0 = m1/w1

    Find m0:
    Substituting: m1 = m0+r-s
    m0/w0 = (m0 + r - s)/w1

    Rearranging:
    m0 = (s-r)w0/(w0-w1)

    Evaluate:
    m0 = (28.4858g-17.3014g)*(84.31413)/(84.31413-40.30449)
    = 21.4272g

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