Calculating Percent Yield

IntroductionThe percent yield of a reaction is a measure of how efficient we were at running the reaction.

High percent yields (>95%) are excellent.

Low percent yields (>50%) are usually poor.

Most reactions have percent yields between 50% and 95%.

IntroductionThe percent yield of a reaction is ...

... the actual yield ...

... divided by the theoretical yield ...

... times 100%.

% yield= actual yieldtheoretical yield

×100%

IntroductionIn order to calculate % yield, we need to know ...

... the actual yield (from experiment)

... the theoretical yield (from calculation)

which is a stoichiometric calculation.

% yield= actual yieldtheoretical yield

×100%

Chlorobenzene, C6H5Cl, is used in the production of many important chemicals, such as aspirin, dyes, and disinfectants. One industrial method of preparing chlorobenzene is to react benzene, C6H6, with chlorine, as represented by the following equation.

C6H6(l) + Cl2(g) → C6H5Cl(s) + HCl(g)

When 36.8 g of C6H6 react with an excess of Cl2, the actual yield of C6H5Cl is 38.8 g. What is the percent yield of C6H5Cl?

C6H6(l) + Cl2(g) → C6H5Cl(s) + HCl(g)

When 36.8 g of C6H6 react with an excess of Cl2, the actual yield of C6H5Cl is 38.8 g. What is the percent yield of C6H5Cl?

M (g/mol) 78.1 70.9112.

636.4

5

The first thing we need to do is to find the molar masses, M, of the reactants and products.Next, we put in the masses we know.

m (g) 36.8 38.8

We need to find the moles of C6H6 we started with and the theoretical yield of C6H5Cl.

n (mol)

nC6H6 = mC6H6

MC6H6

= 36.8 g78.1 g/mol

= 0.471 mol

0.471

0.471

nC6H5Cl = nC6H6 coeff

C6H5Clcoeff C6H6

= 0.471 mol

1

1= 0.471

mol

C6H6(l) + Cl2(g) → C6H5Cl(s) + HCl(g)

When 36.8 g of C6H6 react with an excess of Cl2, the actual yield of C6H5Cl is 38.8 g. What is the percent yield of C6H5Cl?

M (g/mol) 78.1 70.9112.

636.4

5

Now we can find the mass of C6H5Cl that would have been formed if all of the C6H6 reacted with all of the Cl2.

m (g) 36.8 38.8

n (mol) 0.471

0.471

mtheoretical 53.0

mtheoretical = ntheoretical MC6H5Cl(0.471 mol)

(112.6 g/mol)

= = 53.0 g

C6H6(l) + Cl2(g) → C6H5Cl(s) + HCl(g)

When 36.8 g of C6H6 react with an excess of Cl2, the actual yield of C6H5Cl is 38.8 g. What is the percent yield of C6H5Cl?

M (g/mol) 78.1 70.9112.

636.4

5

Finally, we have enough information to calculate the percent yield.

m (g) 36.8 38.8

n (mol) 0.471

0.471

mtheoretical 53.0

percent yield =

mactual

mtheoretical

38.8 g

53.0 g=100%× 100%× = 73.2

%