A chemist heats 100.0 g of $FeSO_4 * 7H_2O$ n a crucible to drive off the water. If all the water is driven off, what is the mass of the remaining salt?

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Answer 1 · 2017-06-15T18:10:10

$54.64$ $"g"$ anhydrate

With $"100.0 g"$ of the heptahydrate, use the molar mass to determine how many mols you have.

Mols are interconvertible, as one mol of anything is the same number of particles as one mol of anything else.

$100.0 cancel("g FeSO"_4cdot7"H"_2"O") xx ("1 mol FeSO"_4cdot7"H"_2"O")/(278.02 cancel("g FeSO"_4cdot7"H"_2"O"))$

$=$ $"0.3597 mols"$

For each $"mol"$ of the heptahydrate, you have $"7 mols"$ of water molecules, so...

$0.3597 cancel("mols FeSO"_4cdot7"H"_2"O") xx "7 mols water"/cancel("1 mol FeSO"_4cdot7"H"_2"O")$

$=$ $"2.518 mols water"$

Consequently, you have this mass of water by using its molar mass:

$2.518 cancel"mols water" xx ("18.015 g water")/cancel"mol water"$

$=$ $"45.36 g water"$

So, your final mass should be $100 - 45.36 = bb("54.64 g FeSO"_4)$ (anhydrate).

A chemist heats 100.0 g of FeSO_4 * 7H_2OFeSO_4 * 7H_2O n a crucible to drive off the water. If all the water is driven off, what is the mass of the remaining salt?