Calculating Yields and Atom Economy

Two different ways to judge a reaction

Chemists need to know both how much product was actually obtained and how efficiently a reaction equation uses its starting materials. Percentage yield and atom economy answer these different questions.

Percentage yield compares an experimental result with the maximum calculated result. Atom economy examines the balanced equation itself and asks what proportion of the reactants becomes the desired product. A reaction can therefore have a high percentage yield but a low atom economy, or the reverse.

Theoretical and actual yield

The theoretical yield is the maximum amount of product predicted from the balanced equation. To calculate it from a reactant mass, convert the mass to moles, use the mole ratio in the balanced equation, and convert the product moles back to mass.

The actual yield is measured after carrying out the reaction and separating the product. It is usually below the theoretical yield because the reaction may be incomplete, some product may be lost during separation or transfer, or competing side reactions may form other products.

Calculating percentage yield

The calculation is

$$\text{percentage yield}=\frac{\text{actual yield}}{\text{theoretical yield}}\times 100$$

The two yields must refer to the same product and be expressed in the same unit. The units then cancel, leaving a percentage.

For example, if an experiment produces $3.25\text{ g}$ of magnesium oxide when the theoretical yield is $4.00\text{ g}$,

$$\text{percentage yield}=\frac{3.25}{4.00}\times100=81.25\%=81.3\%\text{ (3 s.f.)}$$

This means that the experiment recovered 81.3% of the maximum mass predicted by the balanced equation. It does not mean that 81.3% of the reactant atoms entered the desired product—that is what atom economy considers.

Calculating atom economy

Atom economy is calculated from the balanced equation rather than from an experimental measurement:

$$\text{atom economy}=\frac{\text{total relative mass of desired product}}{\text{total relative mass of all reactants}}\times100$$

“Total relative mass” means that balancing numbers must be included. For a substance with coefficient 2, count $2\times M_r$, not just one $M_r$. Because mass is conserved, the denominator is also equal to the total relative mass of all products.

Consider the balanced equation

$$2\mathrm{H_2O}\rightarrow \mathrm{O_2}+2\mathrm{H_2}$$

If hydrogen is the desired product, the reactants have a total relative mass of $2\times18=36$. The desired hydrogen has a total relative mass of $2\times2=4$. Therefore,

$$\text{atom economy}=\frac{4}{36}\times100=11.1\%\text{ (3 s.f.)}$$

Only 11.1% of the reactant mass becomes the desired hydrogen; the remaining mass becomes oxygen. Atom economy is fixed by the chosen balanced reaction and desired product. Losing some hydrogen while collecting it would lower the percentage yield, but it would not change the atom economy.

Why both values matter

A high percentage yield means that the process obtains an amount close to its calculated maximum. A high atom economy means that little reactant mass is directed into unwanted products. Industrial chemists prefer pathways that perform well in both respects, although they may also consider reaction rate, equilibrium position and whether any by-products are useful.