Question

1/ After 0.62 g of CoCl2⋅6H2O is heated, the residue has a mass of 0.321 g . Calculate the % H2O in the hydrate.

2/What was the actual number of moles of water per formula unit CoCl₂?

Answer 1

1)

mass of H2O = mass of hydrated salt - mass of anhydrous salt

mass of H2O = 0.62 g - 0.321 g

mass of H2O = 0.299 g

% H2O = mass of H2O * 100 / mass of sample

= 0.299*100/0.62

= 48 %

Answer: 48 %

2)

Let the formula be CoCl2.XH2O

Molar mass of H2O,

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

= 2*1.008 + 1*16.0

= 18.016 g/mol

mass(H2O)= 0.299 g

use:

number of mol of H2O,

n = mass of H2O/molar mass of H2O

=(0.299 g)/(18.02 g/mol)

= 1.66*10^-2 mol

Molar mass of CoCl2,

MM = 1*MM(Co) + 2*MM(Cl)

= 1*58.93 + 2*35.45

= 129.83 g/mol

mass(CoCl2)= 0.321 g

use:

number of mol of CoCl2,

n = mass of CoCl2/molar mass of CoCl2

=(0.321 g)/(1.298*10^2 g/mol)

= 2.472*10^-3 mol

use:

X = mol (H2O)/mol (CoCl2)

X = 1.66*10^-2 / 2.472*10^-3

X = 6.71

Answer: 6.71