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Percent yield calculations differ from limiting stoichiometry problems by only one extra step.

The question includes a mass recovered. Here is the formula you will use at the end of the problem.

When you read the question the mass of the product must not be used until the very end.

So.....put it in your pocket for later.

You are allowed to use moles, grams, liters in this equation, as long as actual and theoretical are both in these units.

EXAMPLE QUESTION #1 (Plain Stoichiometry not limiting)

2CO_(g) + O_2(g) --> 2CO_2(g)

Calculate the % yield if 69.1g of CO combines with excess O₂ to form an experimental yield of 48.3L of CO₂ @STP.

See how it says yields 48.3L of CO₂ save that for later (in your pocket).

Now determine the volume of of CO₂ from 69.1g of CO using stoichiometry

Find the moles, the mol ratio, then find the volume of CO₂

69.1g CO(1 mol CO/28.0g)= 2.57mol CO

2.57mol CO(2mol CO₂/2mol CO)= 2.57mol CO₂

2.57mol CO₂(22.4L/1 mol)= 55.3L of CO₂

Now take out that volume recovered and plug into the % yield equation.

(48.3L/55.3L) x 100%= 77.3% yield

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