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5 February, 11:46

A sample of raw mining ore contains a hydrated salt called copper sulfate tetrahydrate, CuSO4.4H2O, along with other impurities. If 10.854g of the ore loses 2.994g of water when heated strongly, what is the mass percentage of the salt CuSO4.4H2O in the raw ore sample?

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  1. 5 February, 15:12
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    88.5 % of the raw ore is CuSO4*4H2O

    Explanation:

    Step 1: Data given

    Mass of the ore = 10.854 grams

    Mass of water = 2.994 grams

    Step 2: Calculate moles H2O

    Moles H2O = mass H2O / molar mass H2O

    Moles H2O = 2.994 grams / 18.02 g/mol

    Moles H2O = 0.166 moles

    Step 3: Calculate moles CuSO4*4H2O

    For 1 mol CuSO4*4H2O we have 4 moles H2O

    For 0.166 moles H2O we have 0.166/4 = 0.0415 moles CuSO4*4H2O

    Step 4: Calculate mass CuSO4*4H2O

    Mass CuSO4*4H2O = moles * molar mass

    Mass CusO4*4H2O = 0.0415 moles * 231.67 g/mol

    Mass CuSO4*4H2O = 9.61 grams

    Step 5: Calculate mass % of CuSO4*4H2O

    Mass % = (9.61 grams / 10.854 grams) * 100 %

    Mass % = 88.5 %

    88.5 % of the raw ore is CuSO4*4H2O
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