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    5.00 g of a mixture of MgSO4.7H2O and CuSO4.5H2O was heated at 120C until a mixture of the anhydrous salts was formed, which weighed 3.00 g. Calculate the percentage by mass of MgSO4.7H2O in the mixture.
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    Have you tried anything?

    Posted from TSR Mobile
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    I'll give you a hint
    Calculate the amount of mass losses and that's the mass of all the water in the mixture

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    so would u do 5-2 = 3 and thats 13H20
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    5-3=2

    The water molecules add up to 12 (5+7)
    Can you find the relative molecular mass of water?

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    (Original post by GabbytheGreek_48)
    5-3=2

    The water molecules add up to 12 (5+7)
    Can you find the relative molecular mass of water?

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    jjjj
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    Right so there are 12 molecules of water so the overall mass is 216 (12*18)

    So you have a mass and a relative formula mass Wat do you think that would help you to find out?

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    Sorry for the multiple steps it's a pretty long method but I don't want to give too much away I guess

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    (Original post by GabbytheGreek_48)
    Sorry for the multiple steps it's a pretty long method but I don't want to give too much away I guess

    Posted from TSR Mobile
    you work out moles
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    Yeah correct so work out the moles and see what you get

    I'll leave the rest of the answer here for anyone else who wants to solve the problem.

    Sorry I don't know how to do a spoiler

    Then from the moles you get multiply them by 7/12 to find out the moles of the MgSO4.7H2O (because this has 7 out of the 12 water molecules)

    Then multiply your answer by the relative molecular mass of MgSO4.7H2O to get the mass of MgSO4.7H2O that was present.

    The divide you answer by the original mass of the overall mixture (5) and multiply by 100

    Posted from TSR Mobile
 
 
 
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