Answer
0.03102 mol $H_{2}O$ or 3.102 x $10^{-2}$ mol $H_{2}O$ were in the original sample .
The ratio 0.0062 mol $CuSO_{4}$ :0.03102 mol $H_{2}O$ is similar to ratio 1 mol $CuSO_{4}$ : 5 mol $H_{2}O$ and it is consistent with its chemical formula $CuSO_{4}$$\cdot$$5H_{2}O$.
Work Step by Step
Strategy: We have to convert mass to moles. So the conversion factor is 1 mol/Molar mass.
1- The moles of water that were in the original sample of 1.547 g of $CuSO_{4}$$\cdot$$5H_{2}O$.
Find first the mass of water that was in the original sample by subtracting the mass of $CuSO_{4}$ from the mass of the sample:
1.547 g - 0.989 g = 0.558 g.
Convert 0.558 g water to moles of water.
Molar mass $H_{2}O$ = 2 x 1.00 1 x 15.99
Molar mass $H_{2}O$ = 17.99 g/mol
therefore:
0.558 g x $\frac{1 mol (H_{2}O)}{ 17.99 g (H_{2}O)}$ = 0.03102 mol $H_{2}O$.
So 0.03102 mol $H_{2}O$ or 3.102 x $10^{-2}$ mol $H_{2}O$ were in the original sample .
2 - Convert 0.989 g $CuSO_{4}$ to moles.
Molar mass $CuSO_{4}$ = 1 x 63.54 + 1 x 32.06 + 4 x 15.99
Molar mass $CuSO_{4}$ = 159.56 g/mol
therefore:
0.989 g x $\frac{1 mol (CuSO_{4}}{ 159.56 g (CuSO_{4}}$ = 0.00620 mol $CuSO_{4}$
So 0.0062 mol $CuSO_{4}$ or 6.2 x $10^{-3}$ mol $CuSO_{4}$ were in the original sample .
3- Based in the formula $CuSO_{4}$$\cdot$$5H_{2}O$ the ratio is 1 mol $CuSO_{4}$ : 5 mol $H_{2}O$.
In the sample that is provided, we found that there are 0.0062 mol $CuSO_{4}$ and 0.03102 mol $H_{2}O$.
The ratio 0.0062 mol $CuSO_{4}$ :0.03102 mol $H_{2}O$ is similar to ratio 1 mol $CuSO_{4}$ : 5 mol $H_{2}O$.
Therefore the results that we found on the moles of $CuSO_{4}$ and $H_{2}O$ in a sample of 1.547 g blu copper (II) hydrate, is consistent with its chemical formula $CuSO_{4}$$\cdot$$5H_{2}O$.
