Question

. A 100 mg sample of the blue hydrate CuSO4·5H2O is heated in a crucible. The white, anhydrous salt (CuSO4) is obtained, and weighs 64.0 mg after being cooled to room temperature. Calculate the percent water of hydration for the hydrated salt. Confirm the empirical formula as shown above (penta hydrate).

Answer 1

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

mass of H2O = 0.100 g - 0.064 g

mass of H2O = 0.036 g

1)

% H2O = mass of H2O *100 / mass of hydrated compound

= 0.036 * 100 / 0.100

= 36 %

Answer: 36 %

2)

Molar mass of H2O,

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

= 2*1.008 + 1*16.0

= 18.016 g/mol

mass(H2O)= 0.036 g

use:

number of mol of H2O,

n = mass of H2O/molar mass of H2O

=(3.6*10^-2 g)/(18.02 g/mol)

= 1.998*10^-3 mol

Molar mass of CuSO4,

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

= 1*63.55 + 1*32.07 + 4*16.0

= 159.62 g/mol

mass(CuSO4)= 0.064 g

use:

number of mol of CuSO4,

n = mass of CuSO4/molar mass of CuSO4

=(6.4*10^-2 g)/(1.596*10^2 g/mol)

= 4.01*10^-4 mol

use:

X = mol (H2O)/mol (CuSO4)

X = 1.998*10^-3 / 4.01*10^-4

X = 5

So, the formula is confirmed