How to Calculate the Efficiency of a Chemical Reaction

In chemistry, the theoretical amount is the maximum amount of product a chemical reaction can produce based on the chemical equation. In fact, most reactions are incomplete. If you do the experiment, you will get less product called the actual quantity . You can calculate the reaction yield using the following formula: %efficiency = (actual quantity/theoretical quantity) x 100 . A 90% reaction efficiency means that the reaction is 90% productive, and 10% of the materials have been wasted (they either don’t react, or the product isn’t recovered at all).

Table of Contents

Find all reactants

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Start with balancing chemical equations . The chemical equation describes the reaction between reactants (on the left side) to form products (on the right side). Some exercises provide this equation, others require you to write it yourself. Since atoms are neither created nor destroyed in the reaction, each element will have the same number of atoms on both sides. ^{[1] X Research Source}

For example, oxygen and glucose react with each other to form carbon dioxide and water: $6$ $O$ $2$ $+$ $OLD$ $6$ $H$ $twelfth$ $O$ $6$ ${displaystyle 6O_{2}+C_{6}H_{12}O_{6}}$ $6O_{2}+C_{6}H_{{12}}O_{6}$ → $6$ $OLD$ $O$ $2$ $+$ $6$ $H$ $2$ $O$ ${displaystyle 6CO_{2}+6H_{2}O}$ $6CO_{2}+6H_{2}O$
Each side has 6 carbon atoms (C), 12 hydrogen atoms (H) and 18 oxygen atoms (O). The equation is balanced.
Read this guide if the problem requires balancing an equation.

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Calculate the mass mp of each reactant . Find the mass mp of each atom in the compound, then add them together to get the mp mass of the compound. Find the mass mp for one molecule of the compound.

For example, an oxygen molecule ( $O$ $2$ ${displaystyle O_{2}}$ $O_{2}$ ) has two oxygen atoms.
The mp mass of oxygen is 16 g/mp. (You can find a more precise value on the periodic table.)
2 oxygen atoms x 16 g/mp = 32 g/mp, molecular mass mp $O$ $2$ ${displaystyle O_{2}}$ $O_{2}$ .
For example another reactant, glucose ( $OLD$ $6$ $H$ $twelfth$ $O$ $6$ ${displaystyle C_{6}H_{12}O_{6}}$ $C_{6}H_{{12}}O_{6}$ ) has a mass mp of (6 C atoms x 12 g C/mp) + (12 H atoms x 1 g H/mp) + (6 O atoms x 16 g O/mp) = 180 g/mp.

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Convert the mass of each reactant from grams to mp. Now it’s time to look at the experiment we’re doing. Record the mass of each reactant in grams, dividing this value by the mass mp of the compound to get the number of mp. ^{[2] X Research Source}

For example, let’s say you have 40 grams of oxygen and 25 grams of glucose.
40 g $O$ $2$ ${displaystyle O_{2}}$ $O_{2}$ / (32 g/mp) = 1.25 mp of oxygen.
25g $OLD$ $6$ $H$ $twelfth$ $O$ $6$ ${displaystyle C_{6}H_{12}O_{6}}$ $C_{6}H_{{12}}O_{6}$ / (180 g/mp) = about 0.139 mp of glucose.

Calculate theoretical quantity

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Get to know the desired product. The right hand side of a chemical equation is the products formed from the reaction. Each product has a theoretical amount, that is, the amount of product you would expect to get if the reaction was complete.

Continuing with the example above, you are analyzing the reaction $6$ $O$ $2$ $+$ $OLD$ $6$ $H$ $twelfth$ $O$ $6$ ${displaystyle 6O_{2}+C_{6}H_{12}O_{6}}$ $6O_{2}+C_{6}H_{{12}}O_{6}$ → $6$ $OLD$ $O$ $2$ $+$ $6$ $H$ $2$ $O$ ${displaystyle 6CO_{2}+6H_{2}O}$ $6CO_{2}+6H_{2}O$ . The right side has two products, carbon dioxide and water. Calculate the amount of carbon dioxide, $OLD$ $O$ $2$ ${displaystyle CO_{2}}$ $CO_{2}$ .

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Record the mp number of the reactant exhausted. Theoretical quantity is the amount of product produced under perfect conditions. To calculate this value, we start with the number of mp of reactant. (This mp number has been found in the section on finding all reactants).

In the example above, you know glucose is the end reactant so you would start with 0.139 mp of glucose.

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Find the ratio of molecules in products and reactants. Back in the balanced equation, you divide the number of molecules of reactant by the number of molecules of the desired product.

The equilibrium equation is $6$ $O$ $2$ $+$ $OLD$ $6$ $H$ $twelfth$ $O$ $6$ ${displaystyle 6O_{2}+C_{6}H_{12}O_{6}}$ $6O_{2}+C_{6}H_{{12}}O_{6}$ → $6$ $OLD$ $O$ $2$ $+$ $6$ $H$ $2$ $O$ ${displaystyle 6CO_{2}+6H_{2}O}$ $6CO_{2}+6H_{2}O$ . There are six desired product molecules, carbon dioxide ( $OLD$ $O$ $2$ ${displaystyle CO_{2}}$ $CO_{2}$ ). One molecule of reactant is all glucose ( $OLD$ $6$ $H$ $twelfth$ $O$ $6$ ${displaystyle C_{6}H_{12}O_{6}}$ $C_{6}H_{{12}}O_{6}$ ).
The ratio of carbon dioxide to glucose is 6/1 = 6. In other words, the reaction produces 6 carbon dioxide molecules from one glucose molecule.

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Multiply this ratio by the number of mp of reactant. The result obtained is the theoretical amount of the desired product in mp.

With 0.139 mp glucose and the ratio of carbon dioxide to glucose is 6. So the theoretical amount of carbon dioxide is (0.139 mp glucose) x (6 mp carbon dioxide / mp glucose) = 0.834 mp carbon dioxide.

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Convert the result to grams. Multiply the number of mp by the mass mp of the compound to find the theoretical amount in grams. This is a convenient unit to use in experiments.

For example, the mp mass of CO ₂ is 44 g/mp. (The mp mass of carbon is ~12 g/mp and of oxygen is ~16 g/mp, so the total is 12 + 16 + 16 = 44.)
Take 0.834 mp CO ₂ x 44 g/mp CO ₂ = ~36.7 grams. So the theoretical amount of this example is 36.7 grams of CO ₂ .

Calculate percentage efficiency

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Understand what percentage performance is. The theoretical amount you just calculated is the amount of substance obtained when everything goes perfectly. In the actual experiment, this never happened: impurities and unforeseen problems would prevent the reactants from converting completely into products. That’s why chemists use three different concepts to refer to yield:

Theoretical amount is the maximum amount of product obtained by the reaction.
Actual quantity is the actual amount of product obtained, and is measured directly on the balance.
Percent efficiency = $Actual Amount$ $Amount of Theory$ $*$ $100$ $%$ ${displaystyle {frac {text{Actual Amount}}{text{Theory Amount}}}*100%}$ ${frac {{text{Actual Amount}}}{{text{Theory Amount}}}}*100%$ . For example, the percentage efficiency is 50%, which means that the amount of product obtained is only 50% of the theoretical amount.

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Record the actual amount of the experiment. If you do the experiment yourself, collect the pure product and weigh it to find its mass. If you do your homework, the test will give you a realistic amount.

Assume the actual amount of _CO2 product is 29 grams.

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Divide the actual amount by the theoretical amount. Remember to use the same units for both values (usually grams). The result will be a unitless value.

The actual amount is 29 grams while the theoretical amount is 36.7 grams. $29$ $g$ $36$ $,$ $7$ $g$ $=$ $0$ $,$ $79$ ${displaystyle {frac {29g}{36.7g}}=0.79}$ ${frac {29g}{36.7g}}=0.79$ .

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Convert to a percentage by multiplying by 100. This is the percentage yield.

0.79 x 100 = 79, so the percentage efficiency of this experiment is 79%. You generate 79% of the theoretical maximum CO ₂ .

Advice

Some people confuse percentage efficiency (actually obtained in total that can be obtained) with percentage error (deviation of experimental results from desired results). The exact percentage yield formula is $Actual Amount$ $Amount of Theory$ $*$ $100$ ${displaystyle {frac {text{Actual Amount}}{text{Theory Amount}}}*100}$ ${frac {{text{Actual Amount}}}{{text{Theory Amount}}}}*100$ . If you subtract these two amounts, that’s the formula for percentage error.
If you get abnormal results then unit test. If the actual amount is greater than the theoretical amount, then you have used the wrong units in the calculations. Redo the calculations and remember to double check the units of each calculation.
If the percentage efficiency is more than 100% (and you are sure the calculations are correct) then your product is probably not pure. You need to clean the product (like drying or filtering) and reweigh.

Steps

Find all reactants

Calculate theoretical quantity

Calculate percentage efficiency

Advice

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