# Stoichiometry hydrated NiSO4 question

## Homework Statement

given that 40.14g sample of hydrated NiSO4 is reduced in mass to 22.14g upon heating.
Show that the formula of the hydrate is NiSO4.7H20

n / M

## The Attempt at a Solution

22.14g/ 40.14g = 0.5516 (mol of NiSO4)

40.14g - 22.14 = (18 g H20 lost in heating)

18.0 / 18.02 = 0.9989 (mol of H20)

∴ mole ratio = 0.5516 : 0.9989

but this does not give the result in the ratios of 1:7 (NiSO4.7H20)

stoichiometry (or chem for that matter) isn't really my thing and I have about 10 questions similar to this, can someone please point me in the right direction please?

22.14/ 150.76 = 0.1469 (mols of NiSO4)

0.1469:0.9989
1.469:9.989 (x10)

which is a ratio of 1:6.79

but is there a more accurate way of calculating this?

Your second calculation is much better than the first. Have another try at calculating the molar mass of NiSO4. 150.76 g/mol is wrong.

thanks for the tip johnRC, I needed to check my copy of the periodic table better!

does this calculation look more correct? the ratios are almost perfect, which my chem teacher has said is to be expected.

22.14 / 154.76 = 0.14306 (mols of NiSO4)
40.14g - 22.14 = 18.00 (H20 lost in heating)
18.0 / 18.02 = 0.9989 (mols of H20)
0.14306:0.9989
0.14306/0.9989 = 0.1432
1/0.1432 = 6.982

1:6.982
6.98 ≈ 7
∴ NiSO4.7H2O

Right on, symmetrica1!

thanks john!

could anyone take a look at this for me please?

Given that a 139.4g sample of hydrated MnSO4 is reduced in mass to 94.38g upon heating, find the empirical formula of the Hydrate. Also, write the name of the hydrate of manganese sulphate.

94.38g / 151g = 0.6250 (mols of MnSO4)
139.4g – 94.38g = 45.02g (H2O lost in heating)
45.02 / 18.02 = 2.4983 (mols of H2O)

0.6250:2.4983
0.6250/2.4983 = 0.250166
1 / 0.250166 = 3.997

1:3.997
3.997 ≈ 4
∴ MnSO4.4H2O

would the name of the hydrate be Manganese (II) sulfate monohydrate?

Last edited:
Borek
Mentor
MnSO4.4H2O

Yes.

would the name of the hydrate be Manganese (II) sulfate monohydrate?

No.

Mono means one, not four.