Question

Determination Number	Complex Mass (g)	Consumed HCl volume (mL)
Determination 1	0.260 g	4 mL
Determination 2	0.300 g	5 mL

[Cu(NH3)4]SO4.H2O (aq) +4HCl (aq) → Cu 2+ (aq) + 4NH + 4 (aq) + 4Cl − (aq) +SO 2− 4 (aq) +H2O (l)

Concentration of HCl Solution, mol/L = 0.50M

1. Mass of the complex for both determinations using the stoichiometric ratio

2. Percentage purity of your complex

3. Based on the % purity, what is the correct range of values for the value of ε which you calculated earlier? (Hint: think about percentage error!)

Answer 1

1. Stoichiometric equation for the reaction of the complex with complex with HCl is as follow:

[Cu(NH₃)₄]SO₄.H₂O(aq) + 4HCl(aq) ? Cu²⁺(aq) + 4NH₄⁺(aq) + 4Cl^?(aq) +SO₄^2?(aq) + H₂O(l)

1 mol 4 mol

Here 1 mol of the complex reacts with 4 mol of HCl

To find the mass using the stoichiometric ratio. We have to find the number of moles of HCl used/consumed.

number of moles = concentration x volume of the solution (L)

We know concentration of HCl = 0.50 M

(a)   Determination 1 Volume consumed in first determination = 4 mL

number of moles = 0.50 M x 4 mL = 2 mmol

According to the reaction, 4 mol HCl reacts with 1 mol of the complex.

Hence 2 mmol HCl react with (1/4)x2 mmol = 0.5 mmol of the complex.

Mass of the complex = (number of moles of complex) x (Molecular Mass of complex)

We know the Molecular mass of the complex is 245.79 g/mol

Now,

Mass of the complex = 0.5 mmol x 245.79 g/mol = 122.895 mg = 0.123 g

(b) Determination 2 Volume consumed in first determination = 5 mL

number of moles = 0.50 M x 5 mL = 2.5 mmol

Hence 2.5 mmol HCl react with (1/4)x2.5 mmol = 0.625 mmol of the complex.

Now,

Mass of the complex = 0.625 mmol x 245.79 g/mol = 153.619 mg = 0.154 g

2. % Purity = (theoritical mass or pure complex mass/experimental mass or impure complex mass)x100

Determination 1

  % Purity = (0.123 g/0.260 g)x100 = 47.31%

Determination 2

  % Purity = (0.154 g/0.300 g)x100 = 51.33%

3. % error = [(mass of impure complex - mass of pure complex)/(mass of pure complex)]x100

  Determination 1

  % Error = [(0.260-0.123)/0.123]x100 = 111.38%

  Determination 2

  % Purity = [(0.300-0.154)/0.300]x100 = 94.81%

Hence range of ? %=111.38% to 94.81%

Range of ? = 0.137 g to 0.146 g This is the range in calculated values.