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    This is probably easy to work out, but I'm having trouble... can anyone help?

    1.60 g of a metal sulphate were dissolved in water. Addition of an excess of barium chloride solution results in the precipitation of 2.33 g of barium sulphate.

    Given that the relative atomic mass of barium is 137, the original substance was which of the following?

    magnesium sulphate

    copper(II) sulphate

    sodium sulphate

    calcium sulphate

    Since you've added an excess of barium chloride, you can assume that all the sulphate has ended up as barium sulphate. That means that you need to do the following:

    1) Work out the molecular mass of barium sulphate (look up the mass of sulphate's constituent elements on a periodic table if you have to or look it up on Wiki)

    2) Calculate the number of moles of barium sulphate (and hence also the number of moles of sulphate ions, since there's 1 mole of barium sulphate formed for every mole of sulphate)

    3) This gives you the number of moles of the original sulphate, so you can find the molecular mass of that sulphate (by dividing the 1.60g by the number of moles you calculated in step 2).

    4) Subtract the molecular mass of sulphate and then compare the remaining mass with the atomic mass of the metals (also note the valence of the metals, since sodium is Na+, not Na2+).
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