determine the empirical formula of an imaginary hydrate.

Question

A student wishes to determine the empirical formula of an imaginary hydrate. The following data is obtained-

1. mass of empty crucible - 23.567g
2. mass of crucible with hydrate before heating- 28.097g
3. mass of crucible with hydrate after heating- 26.463g

Considering the data table, the mass of the hydrate (ionic compound with water) was 4.530 g, and the mass of the anhydrate (dried ionic compound) was 2.896 g. The mass of the water evaporated was 1.634 g.

The molar mass of the imaginary anhydrous salt is 159.6 g/mol.

Your answer should include:

1. all your work
2. mass of components, including water lost
3. moles of components, including water lost
4. hydrate formula (salt* nH2O) including mole ratio

Answer 1

Hydrate → Anhydrous salt + water

28.097-26.463=1.634

1.634 = mass lost due to water boiling off

28.097-23.567=4.53

4.53 = mass of hydrate

4.53-1.634=2.896g anhdyrous salt

Hydrate → Anhydrous salt + water

4.53g → 2.896g+1.634g

moles anydrous salt = 2.896/159.6=0.0181 mol

moles water=1.634/18.02=0.0907mol

anhydrous salt: 0.0181/0.0181=1

water: 0.0907/0.0181=4.99=5

So, if the anhydrous salt is XY, then there are 5 H2O's per XY, so the hydrate formula would be like:

XY⋅5H₂O