Question

If a researcher isolates 58.33 g of copper(II) phosphate after reacting 143.47 g of copper(II) sulfate with 172.33 g of sodium phosphate, what is the percent yield of copper(II) phosphate. Report your answer rounded to the tenths of a percent without units.

Answer 1

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)= 143.47 g

use:

number of mol of CuSO4,

n = mass of CuSO4/molar mass of CuSO4

=(1.435*10^2 g)/(1.596*10^2 g/mol)

= 0.8988 mol

Molar mass of Na3PO4,

MM = 3*MM(Na) + 1*MM(P) + 4*MM(O)

= 3*22.99 + 1*30.97 + 4*16.0

= 163.94 g/mol

mass(Na3PO4)= 172.33 g

use:

number of mol of Na3PO4,

n = mass of Na3PO4/molar mass of Na3PO4

=(1.723*10^2 g)/(1.639*10^2 g/mol)

= 1.051 mol

Balanced chemical equation is:

3 CuSO4 + 2 Na3PO4 ---> Cu3(PO4)2 + 3 Na2SO4

3 mol of CuSO4 reacts with 2 mol of Na3PO4

for 0.8988 mol of CuSO4, 0.5992 mol of Na3PO4 is required

But we have 1.051 mol of Na3PO4

so, CuSO4 is limiting reagent

we will use CuSO4 in further calculation

Molar mass of Cu3(PO4)2,

MM = 3*MM(Cu) + 2*MM(P) + 8*MM(O)

= 3*63.55 + 2*30.97 + 8*16.0

= 380.59 g/mol

According to balanced equation

mol of Cu3(PO4)2 formed = (1/3)* moles of CuSO4

= (1/3)*0.8988

= 0.2996 mol

use:

mass of Cu3(PO4)2 = number of mol * molar mass

= 0.2996*3.806*10^2

= 1.14*10^2 g

% yield = actual mass*100/theoretical mass

= 58.33*100/1.14*10^2

= 51.15%

Answer: 51.1