Percent Yield Problem

Question

Lithium nitride is prepared by the reaction of lithium metal and nitrogen gas. What is the percent yield of

     litium nitride if 76.8 grams of lithium nitride is recovered from reacting 56.0 grams of nitrogen gas with          

     56.0 grams of lithium metal?

                                                                       6 Li + N₂ →   2 Li₃N

 

Answer 1

In any complex problem it is always a good idea to start with the questions:

1. What am I looking for?
2. What does this mean?

So what are we looking for? 

We are looking for the percent yield of lithium nitride (Li₃N).

What does this mean?

Yield in this case means the amount of product you make. The percent yield is a ratio of what you actually made to what you could have made (theoretically), represented as a percent. 

So in this problem we are looking for:

Percent yield = [(Actual yield of Li₃N)/(Theoretical yield of Li₃N)] x 100%

Notice that we have been given the actual yield of 76.8 g Li₃N. So now we have:

Percent yield = [(76.8 g)/(Theoretical yield of Li₃N)] x 100%

So what we really need to find is the theoretical yield of Li₃N. Once we have this, we can put the number in the equation above and solve it. 

Notice we are asked for one thing (percent yield) and it turns out that we first need to find something else (theoretical yield). This happens a lot in chemistry problems, which is why I like this problem-solving approach.

Our next question is: What are we given?

We know the amounts of each reactant we are starting with. From these, using stoichiometry, we can determine the amount of product we could theoretically produce. Because we are given masses of reactants, and the balanced equation has a mole ratio, we do need to do some conversions. 

We are given amounts of both of the reactants, meaning one of them could be limiting, There are many ways of solving problems with a limiting reactant. We'll just look at one.

For this problem, let’s determine the number of moles we have of each reactant and compare them to what the balanced equation tells us. Time for the math!

Atomic mass of Li = 6.94 g/mol

Atomic mass of N = 14.01 g/mol, so molar mass of N₂ = 28.02 g/mol

56.0 g Li x (1mole Li)/(6.94 g Li) = 8.07 moles Li

56.0 g N₂ x (1 mole N₂)/(28.02 g N₂) = 2.00 moles N₂

According to the balanced equation, 6 Li + N₂ →   2 Li₃N, we need six times as much Li as N₂ in terms of moles. But we only have about four times the number of moles of Li (8.07/2 = 4.035), meaning that Li is limiting. We can see this mathematically as:

2.00 moles N₂ x (6 moles Li/1 mole N₂) = 12 moles Li needed to react all of the N₂ and we only have 8.07 moles of Li, so not enough Li.

or

8.07 moles Li x (1 mole N₂/6 moles Li) = 1.345 moles N₂ needed to react with all of the Li, and we have 2.00 moles, so we have plenty of N₂.

In either case, we can see we N₂ is in excess and Li is the limiting reactant.

The limiting reactant limits the amount of product we make, so we use the amount of lithium to determine the theoretical yield of Li₃N:

Molar mass of Li₃N = (3)(6.94g/mol) + 14.01 g/mol = 34.83 g/mol

8.07 moles Li x (2 moles Li₃N/6 moles Li) x (34.83 g Li₃N/1 mole Li₃N) = 93.7 g Li₃N.

The percent yield = (76.8g/93.7g) x 100% = 82.0% yield

There are other ways we could have determined the theoretical yield in this problem — as I said, there are always more than one way to solve a complex problem. As your tutor, I would be happy to work with you so that you understand how to solve problems in general, and not just memorize separate solutions for separate problems.