Answer
0.791 g $CaSO_{4}$ s 0.00581 mol $CaSO_{4}$ or 5.81 x $10^{-3}$ mol $CaSO_{4}$.
0.209 g $H_{2}O$ is 0.01162 mol $H_{2}O$ or 1.162 x $10^{-2}$ mol $H_{2}O$.
The ratio 0.00581 mol $CaSO_{4}$ :0.001162 mol $H_{2}O$ is equivalent to ratio 1 mol $CaSO_{4}$ : 2 mol $H_{2}O$, consistent with the formula of gypsum $CaSO_{4}$$\cdot$$2H_{2}O$.
Work Step by Step
Strategy: We have to convert mass to moles. So the conversion factor is 1 mol/Molar mass.
1- Convert 0.791 g $CaSO_{4}$ into moles $CaSO_{4}$.
Molar mass $CaSO_{4}$ = 1 x 40.07 + 1 x 32.06 + 4 x 15.99
Molar mass $CaSO_{4}$ = 136.09 g/mol
therefore:
0.791 g x $\frac{1 mol (CaSO_{4}}{ 136.09 g (CaSO_{4}}$ = 0.00581 mol $CaSO_{4}$
So 0.791 g $CaSO_{4}$ s 0.00581 mol $CaSO_{4}$ or 5.81 x $10^{-3}$ mol $CaSO_{4}$ .
2- Assuming that the rest of the gypsum sample is water, we find the mass of water 1.000 g - 0.791 g = 0.209 g water.
Convert 0.209 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:
0.209 g x $\frac{1 mol (H_{2}O)}{ 17.99 g (H_{2}O)}$ = 0.01162 mol $H_{2}O$.
So 0.209 g $H_{2}O$ is 0.01162 mol $H_{2}O$ or 1.162 x $10^{-2}$ mol $H_{2}O$ .
3- Based in the formula of gypsum $CaSO_{4}$$\cdot$$2H_{2}O$ the ratio is 1 mol $CaSO_{4}$ : 2 mol $H_{2}O$.
In the sample that is provided, we found that there are 0.00581 mol $CaSO_{4}$ and 0.001162 mol $H_{2}O$.
The ratio 0.00581 mol $CaSO_{4}$ :0.001162 mol $H_{2}O$ is equivalent to ratio 1 mol $CaSO_{4}$ : 2 mol $H_{2}O$.
Therefore the results that we found on the moles of $CaSO_{4}$ and $H_{2}O$ in a sample of 1.000 g gypsum, is consistent with the formula of gypsum $CaSO_{4}$$\cdot$$2H_{2}O$.
