Gypsum is a hydrate of calcium sulphate 1.0 g of the sample contains 0.791 g of `CaSO_(4)`. How many of `CaSO_(4)` are there in the sample ? Assuming that the rest of sample is water, how many moles of water are there in the sample ? Show that the result is in consistent with the formula `CaSO_(4).2H_(2)O`

Explanation

Mass of `CaSO_(4)=40 +32 + 64 = 136 "g mol"^(-1)`
Molar mass of `CaSO_(4)=40+32+64=136 "g mol"^(-1)`
Mass of water present in 1.0 g of the sample `= 1.0 - 0.791 = 0.209 g`
Moles of `CaSO_(4)=("Mass of "CaSO_(4))/("Molar mass")=((0.791g))/((136"g mol"^(-1)))=5.816xx10^(-3)mol`
Moles of water `= ("Mass of water")/("Molar mass")=((0.209g))/((18 gmol^(-1)))=11.6xx10^(-3)mol`
Ratio of `CaSO_(4) : H_(2)O = 5.816 xx 10^(-3) : 11.6 xx 10^(-3) = 1:2`
`:.` Molecular formula of hydrated salt `= CaSO_(4).2H_(2)O`.