Question

The deep blue compound Cu(NH₃)₄SO₄ is made by the reaction of copper(II) sulfate and ammonia.

CuSO₄(aq) + 4NH₃(aq) → Cu(NH₃)₄SO₄(aq) If you use 45.0 g of CuSO₄ and excess NH₃, what is the theoretical yield of Cu(NH₃)₄SO₄?

If you isolate 51.1 g of Cu(NH₃)₄SO₄, what is the percent yield of Cu(NH₃)₄SO₄?

________ g Cu(NH₃)₄SO₄

%

Answer 1

1)

Molar mass of CuSO4 = 1*MM(Cu) + 1*MM(S) + 4*MM(O)

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

= 159.62 g/mol

mass of CuSO4 = 45 g

mol of CuSO4 = (mass)/(molar mass)

= 45/159.62

= 0.2819 mol

we have the Balanced chemical equation as:

1CuSO4 ---> 1Cu(NH3)4SO4

From balanced chemical reaction, we see that

when 1 mol of CuSO4 reacts, 1 mol of Cu(NH3)4SO4 is formed

mol of Cu(NH3)4SO4 formed = moles of CuSO4

= 0.2819 mol

Molar mass of Cu(NH3)4SO4 = 1*MM(Cu) + 4*MM(N) + 12*MM(H) + 1*MM(S) + 4*MM(O)

= 1*63.55 + 4*14.01 + 12*1.008 + 1*32.07 + 4*16.0

= 227.756 g/mol

mass of Cu(NH3)4SO4 = number of mol * molar mass

= 0.2819*227.756

= 64.2089 g

Answer: 64.2 g

2)

% yield = actual mass*100/theoretical mass

= 51.1*100/64.2089

= 79.6 %

Answer: 79.6 %