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upon cumbustion of a 0.144g sample of Mg, 0.090g of O2 is used. Though the sample likly contains impurites (substance other than Mg). What is the percent of imputities by mass?,

please show the steps I'm really confused with this one.
Thank you

  • Chem -

    help balancing S+HNO3=H2SO4=N02=H2O

  • Chem -

    I suppose we are assuming that the impurities were unchanged as the Mg was oxidized.
    2Mg + O2 ==> 2MgO
    mols O2 = 0.090/32 = about 0.0028 but you should redo and get a more accurate answer.
    Convert mols O2 to mols Mg, the to g Mg.
    mols O2 x (2 mols Mg/1 mol O2) = aboaut 0.0056 mols Mg.
    0.0056 mols Mg x atomic mass Mg = about 0.137 g Mg.
    So impurities = 0.144 - 0.137 = about 0.007g.
    Then % impurities = (0.007/0.144)*100 = ?

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