Percentage Yield

There will almost certainly be a percentage yield question in your AS or A2 chemistry exam. These can be quite daunting questions, especially to those who aren`t very confident in maths, but once you know the step-by-step process to answering these questions, they are easy marks in your exam.

In practical lab-work, we can never convert all of our reactant material to products. There are several reasons for this: reactants may not fully react, some side-products might be made, we might lose some product through handling, we might drop some of our product on the floor (not the best lab practice, but it happens!). Percentage yield is how we measure the proportion of product that forms from our reactants. In the lab, we are interested in high-yield experiments, because this leads to the least waste, and highest profits.

Let`s look at an example (AQA AS 2021 paper 2, question 5.3)

"In an experiment, 10.3 g of 1-iodopropane (M_r = 169.9) are reacted with an excess of ammonia. 2.3 g of propylamine (M_r = 59.0) are produced."

We are also given the chemical equation:

CH₃CH₂CH₂CH₂I + 2NH₃ __> CH₃CH₂CH₂NH₂ + NH₄I

STEP 1: WORK OUT THE NUMBER OF MOLES OF REACTANT AND PRODUCT

We work this out using "moles = mass ÷ M_r"

Reactant: moles = 10.3 ÷ 169.9 = 0.0606mol

Product: moles = 2.3 ÷ 59.0 = 0.0390 mol

STEP 2: CHECK THE MOLE RATIO FROM THE CHEMICAL EQUATION

Here, there is a 1:1 ratio between 1-iodopropane and propylamine. So, for every mole of 1-iodopropane (reactant), we should produce one mole of propylamine. Here, we started with 0.0606mol of 1-iodopropane (see step 1) so we should produce 0.0606mol of propylamine. But, we only produced 0.0390mol (see step 1), because some product was lost in the experiment.

STEP 3: CALCULATE PERCENTAGE YIELD

We now have everything we need to calculate percentage yield:

% yield = [(moles of product we actually produced) ÷ (moles of product we should have produced - see step 2)] x 100

Here, % yield = [0.0390 ÷ 0.0606] x 100

= 64 %

Let`s now look at a slightly more complicated example (AS 2019, Paper 1 Q23). You should be able to see how the steps required are exactly the same:

"What is the percentage yield when 20 g of aluminium are produced from 50 g of aluminium oxide?"

2Al₂O₃ -> 4Al + 3O₂

STEP 1

Reactant moles: Al₂O₃ has M_r = (2x27) + (3x16) - use the masses from the insert

= 102

So moles = 50 ÷ 102

= 0.490 mol

Product moles: Al has M_r = 27,

So moles = 20 ÷ 27.0

= 0.741 mol

STEP 2

There is a 2:4 ratio (which simplifies to a 1:2 ratio) between Al₂O₃ and Al, so for every mole of Al₂O₃ that we start with, we should produce 2 moles of Al. We started with 0.490 moles of Al₂O₃, so we should produce 0.980 moles of Al (but of course we only produced 0.741 mol).

STEP 3

% yield = (0.741 ÷ 0.980) x 100

= 76%