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9.04: Mass to Mass Calculations

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  • It should be a trivial task now to extend the calculations to mass-mass calculations, in which we start with a mass of some substance and end with the mass of another substance in the chemical reaction. For this type of calculation, the molar masses of two different substances must be used-be sure to keep track of which is which. Again, however, it is important to emphasize that before the balanced chemical reaction is used, the mass quantity must first be converted to moles. Then the coefficients of the balanced chemical reaction can be used to convert to moles of another substance, which can then be converted to a mass.

    For example, let us determine the number of grams of SO3 that can be produced by the reaction of 45.3 g of SO2 and O2:

    \[2\, SO_2 (g) + O_2 (g) → 2\,SO_3 (g) \]

    First, we convert the given amount, 45.3 g of SO2, to moles of SO2 using its molar mass (64.06 g/mol):

    \[45.3\cancel{g\, SO_{2}}\times \frac{1\, mol\, SO_{2}}{64.06\cancel{g\, SO_{2}}}=0.707\, mol\, SO_{2}\]

    Second, we use the balanced chemical reaction to convert from moles of SO2 to moles of SO3:

    \[0.707\cancel{mol\, SO_{2}}\times \frac{2\, mol\, SO_{3}}{2\cancel{mol\, SO_{2}}}=0.707\, mol\, SO_{3}\]

    Finally, we use the molar mass of SO3 (80.06 g/mol) to convert to the mass of SO3:

    \[0.707\cancel{mol\, SO_{3}}\times \frac{80.06\, g\, SO_{3}}{1\cancel{mol\, SO_{3}}}=56.6\, g\, SO_{3}\]

    We can also perform all three steps sequentially, writing them on one line as

    \[45.3\cancel{g\, SO_{2}}\times \frac{1\, mol\, SO_{2}}{64.06\cancel{g\, SO_{2}}}\times \frac{2\, mol\, SO_{3}}{2\cancel{mol\, SO_{2}}}\times \frac{80.06\, g\, SO_{3}}{1\cancel{mol\, SO_{3}}}=56.6\, g\, SO_{3}\]

    We get the same answer. Note how the initial and all the intermediate units cancel, leaving grams of SO3, which is what we are looking for, as our final answer.

    Example \(\PageIndex{3}\):

    What mass of Mg will be produced when 86.4 g of K are reacted?

    MgCl2(s) + 2K(s) → Mg(s) + 2KCl(s)


    We will simply follow the steps

    mass K → mol K → mol Mg → mass Mg

    In addition to the balanced chemical equation, we need the molar masses of K (39.09 g/mol) and Mg (24.31 g/mol). In one line,

    \[86.4\cancel{g\, K}\times \frac{1\, mol\, K}{39.09\cancel{g\, K}}\times \frac{1\, \cancel{mol\, Mg}}{2\cancel{mol\, K}}\times \frac{24.31\, g\, Mg}{1\cancel{mol\, Mg}}=26.87\, g\, Mg\]

    Exercise \(\PageIndex{3}\)

    What mass of H2 will be produced when 122 g of Zn are reacted?

    \[Zn (s) + 2\,HCl (aq) → ZnCl_2 (aq) + H_2 (g) \]


    3.77 g

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