Edexcel IGCSE Chemistry · Spec 1.27-1.30
The Mole, Reacting Masses & % Yield
The mole, reacting-mass calculations and percentage yield.
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The Mole, Reacting Masses & % Yield
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Explained
Moles, reacting masses and yield
A balanced equation tells you the ratio in which particles react. A balance tells you a mass. The mole is what converts between the two, and every calculation in this topic is that conversion done in one direction or the other.
Moles equals mass divided by relative formula mass. Rearranged, mass equals moles multiplied by relative formula mass. Those are the only two forms you need.
The three step route
Almost every reacting mass question follows the same path, and writing the three steps down before starting makes the arithmetic obvious.
First, convert the mass you were given into moles, by dividing by its relative formula mass.
Second, use the ratio in the balanced equation to find the moles of the substance you want. If the equation says two of one for one of the other, halve it or double it accordingly.
Third, convert those moles back into a mass, by multiplying by the relative formula mass of the new substance.
Moles in, ratio across, mass out. The step in the middle is the chemistry, and it is the one most often skipped. Dividing by one relative formula mass and multiplying by another, without ever looking at the equation, gives an answer that is wrong by exactly the ratio.
A worked example
What mass of carbon dioxide is produced when 5 g of calcium carbonate decomposes, given CaCO3 goes to CaO plus CO2?
The relative formula mass of CaCO3 is 100, so 5 divided by 100 gives 0.05 moles.
The ratio is one to one, so 0.05 moles of carbon dioxide are produced.
The relative formula mass of CO2 is 44, so the mass is 0.05 multiplied by 44, which is 2.2 g.
Limiting reactants
Where a question gives you two masses, one of them will run out first, and that one decides how much product you can get. The other is in excess and some of it is left over.
Convert both masses to moles, then compare those moles against the ratio in the equation. The reactant that provides fewer moles than the ratio requires is the limiting one, and it is the one to base the rest of the calculation on.
Using the excess reactant instead gives an answer that is too large, and it is a mistake that produces a plausible looking number, so it is worth checking which is which before going further.
Percentage yield
The theoretical yield is the mass the calculation says you should get. The actual yield is the mass you collected. Percentage yield is the actual divided by the theoretical, multiplied by 100.
It is almost never 100 per cent, and questions ask why. Some product is lost when transferring between containers or when filtering. Some reactants may react in a different way, giving a different product. And if the reaction is reversible it never goes fully to completion, so some reactant is always still present.
What the mark scheme accepts and rejects
An Edexcel International GCSE Chemistry mark scheme runs this calculation backwards. Given a yield of 113 per cent, it asks for the mass actually recorded, and awards three marks: finding the moles of aluminium sulfate, calculating the theoretical mass as 13.68 g, and then multiplying by 1.13 to reach 15.46 g.
Its notes are generous in three ways worth knowing about. A correct answer without any working scores all three. An error carried forward is allowed if the relative formula mass used was wrong. And any number of significant figures is accepted except one.
There is also a specific concession recorded: an answer of 46.38 scores two of the three marks. That is the value you get from a wrong step late in the calculation, and the mark scheme has anticipated it rather than treating it as simply wrong.
The next part asks why the yield could exceed 100 per cent, and credits three explanations: that the crystals have not fully dried and still contain water, that the acid used was more concentrated than stated, or that a greater volume of acid was used. It rejects adding more aluminium and ignores calculation errors.
A yield above 100 per cent always means the product weighed more than the chemistry allows, so something extra is in the sample. Water is the usual culprit, and it is why crystals are dried before weighing rather than after.
Habits that save marks
Write the balanced equation first, every time, even when the question seems to be about masses only. The ratio is in it and nowhere else.
Set the work out in labelled lines rather than as one long calculation. Mark schemes for these questions award separate marks for the moles, for the ratio and for the final mass, so a visible middle step is worth credit even when the last number is wrong.
And check the answer is the right size. If 5 g of a reactant has produced 60 g of product, something has been multiplied that should have been divided.
Spec 1.27-1.30
What you need to know
- Use the relationship moles equals mass divided by M r
- Work out reacting masses from a balanced equation
- Calculate a percentage yield
Active recall
Quick check
Answer each question before opening the answer.
How do you calculate the number of moles from a mass?
Moles = mass ÷ relative formula mass (Mr).
How is percentage yield calculated?
Percentage yield = (actual yield ÷ theoretical yield) × 100.
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