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3 February, 02:54

If 8.65 g of a tin - fluorine compound contains 5.28 g of tin, what is its empirical formula?

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  1. 3 February, 03:08
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    First, we need to know the amounts of the elements in the compound.

    Tin (Sn) = 5.28 g

    Fluorine (F) = 8.65 - 5.28 = 3.37 g

    Convert these to units of moles by dividing the molar masses.

    Tin (Sn) = 5.28 g / 118.71 g/mol = 0.044 mol

    Fluorine (F) = 3.37 g / 19.00 g/mol = 0.177 mol

    Divide both by the least number of moles of the two.

    Tin (Sn) = 0.044 mol / 0.044 mol = 1

    Fluorine (F) = 0.177 mol / 0.044 mol = 4

    Therefore, the empirical formula would be:

    SnF4
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