Question

When a chemist treated 28 g of C4H8 with 48 grams of Cl2, she isolated 51 g of C4H8Cl2 in pure form. No other products were formed in this reaction. What was the % yield?

Answer 1

Molar mass of C4H8 = 4*MM(C) + 8*MM(H)

= 4*12.01 + 8*1.008

= 56.104 g/mol

mass of C4H8 = 28.0 g

we have below equation to be used:

number of mol of C4H8,

n = mass of C4H8/molar mass of C4H8

=(28.0 g)/(56.104 g/mol)

= 0.4991 mol

Molar mass of Cl2 = 70.9 g/mol

mass of Cl2 = 48.0 g

we have below equation to be used:

number of mol of Cl2,

n = mass of Cl2/molar mass of Cl2

=(48.0 g)/(70.9 g/mol)

= 0.677 mol

we have the Balanced chemical equation as:

C4H8 + Cl2 ---> C4H8Cl2

1 mol of C4H8 reacts with 1 mol of Cl2

for 0.4991 mol of C4H8, 0.4991 mol of Cl2 is required

But we have 0.677 mol of Cl2

so, C4H8 is limiting reagent

we will use C4H8 in further calculation

Molar mass of C4H8Cl2 = 4*MM(C) + 8*MM(H) + 2*MM(Cl)

= 4*12.01 + 8*1.008 + 2*35.45

= 127.004 g/mol

From balanced chemical reaction, we see that

when 1 mol of C4H8 reacts, 1 mol of C4H8Cl2 is formed

mol of C4H8Cl2 formed = (1/1)* moles of C4H8

= (1/1)*0.4991

= 0.4991 mol

we have below equation to be used:

mass of C4H8Cl2 = number of mol * molar mass

= 0.4991*1.27*10^2

= 63.38 g

% yield = actual mass*100/theoretical mass

= 51*100/63.38

= 80.5 %

Answer: 80.5 %