What is the percent by mass of water in $C u S O_{4} \cdot 5 H_{2} O$ $CuSO_4*5H_2O$ ?

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What is the percent by mass of water in $C u S O_{4} \cdot 5 H_{2} O$ $CuSO_4*5H_2O$ ?

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Answer 1 · 2017-04-03T19:44:20

$% water = \frac{Mass of water}{Mass of copper sulfate pentahydrate} ≅ 36 %$ $%"water"="Mass of water"/"Mass of copper sulfate pentahydrate"~=36%$

Explanation:

$% water = \frac{5 \times 18.01 \cdot g \cdot m o l^{- 1}}{249.68 \cdot g \cdot m o l^{- 1}} \times 100 % ≅ 36 %$ $%"water"=(5xx18.01*g*mol^-1)/(249.68*g*mol^-1)xx100%~=36%$

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What is the percent by mass of water in $C u S O_{4} \cdot 5 H_{2} O$ $CuSO_4*5H_2O$ ?

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