Question

A student starts his experiment with an empty crucible and cover having a mass of 10.0450 g. He is working with a hydrated compound of cobalt (II) sulfate. After adding the hydrate to the empty crucible, the mass of the crucible, cover and hydrate is 11.0680 g. After heating the hydrate, he weighs the crucible, cover and dehydrated sample again and the mass is 10.6480 g. Determine the number of moles of water in the original hydrated cobalt (II) sulfate compound. Hint: The hydrate will be cobalt (II) sulphate ● X hydrate. You must determine the number for X. (Note: Use integer numbers in your answer, not decimals).

6

5

2

3

1

Answer 1

mass of cobalt (II) sulphate ● X hydrate = 11.0680 - 10.0450 = 1.0230 g

mass of cobalt (II) sulphate = 10.6480 - 10.0450 = 0.603 g

So,

mass of H2O = 1.0230 g - 0.603 g = 0.420 g

Now find the mol of CoSO4

Molar mass of CoSO4,

MM = 1*MM(Co) + 1*MM(S) + 4*MM(O)

= 1*58.93 + 1*32.07 + 4*16.0

= 155 g/mol

mass(CoSO4)= 0.603 g

number of mol of CoSO4,

n = mass/molar mass

=(0.603 g)/(155 g/mol)

= 0.00389 mol

now find the mol of H2O

Molar mass of H2O,

MM = 2*MM(H) + 1*MM(O)

= 2*1.008 + 1*16.0

= 18.016 g/mol

mass(H2O)= 0.420 g

number of mol of H2O,

n = mass/molar mass

=(0.42 g)/(18.016 g/mol)

= 0.0233 mol

X = mol of H2O / mol of COSO4 = 0.0233 mol /0.00389 mol = 6

Answer: 6