4.3.3.1,4.3.3.2 - Percentage yield and atom economy

Actual yield and theoretical yield

The amount of product collected can be smaller than the amount predicted from the intended reaction.

The actual yield is the amount of product obtained. The maximum theoretical yield is the greatest amount the equation predicts from the available reactants, assuming the intended reaction goes to completion and none of its product is lost. A low actual yield does not contradict conservation of mass: atoms can remain in unreacted material, form other substances, or be present in product that is not collected.

Three important causes of a yield below the theoretical maximum are:

• A reversible reaction may not go to completion. Products can react to reform reactants, leaving some starting material present.
• Product may be lost while it is separated from the mixture. Some crystals can remain dissolved in the liquid, or product may be left on a filter or in a transfer vessel.
• Some reactants may undergo different reactions from the intended one, forming unwanted products instead of the desired product.

These mechanisms require different improvements. A better transfer and separation method can reduce collection losses but cannot by itself change how much reactant is converted in a reversible reaction. Altering conditions may reduce an unwanted reaction, but indiscriminately heating more strongly could instead cause decomposition.

In an Illustrative preparation with a theoretical yield of 10.0 g, a student collects 8.0 g of dry crystals. The missing 2.0 g need not all be in one place: some could remain dissolved and some might never have formed. We need observations or other evidence to identify the cause. Wet crystals can give an apparently high mass because water is included; purity and drying matter when measuring actual yield.

Calculating percentage yield

Percentage yield compares the actual amount of a particular product with the maximum theoretical amount of that same product.

Use matching mass units in numerator and denominator. For a theoretical yield of 20.0 g and actual yield of 15.0 g:

The denominator is the theoretical product mass, not the initial reactant mass. The collected product is three quarters of its theoretical maximum.

Rearranging finds actual yield when the percentage is known. For theoretical yield 40.0 g and percentage yield 80.0%:

The recovered mass is below 40.0 g, consistent with a yield below 100%. A correctly identified dry pure product cannot exceed its theoretical maximum. An apparent yield over 100% suggests water or impurities in the weighed sample, incorrect reactant information, or a measurement error.

Theoretical yield from equations (Higher)

A balanced equation connects the available reactant mass with the maximum mass of the desired product, before collection losses are considered.

Use the mole route first: reactant mass → reactant amount → product amount → theoretical product mass. Only then compare the measured actual mass with that theoretical maximum. In $\ce{CaCO3 -> CaO + CO2}$, 25.0 g pure calcium carbonate decomposes completely. With $M_r(\ce{CaCO3})=100$ and $M_r(\ce{CaO})=56$:

Dividing 11.2 g by the starting 25.0 g would wrongly count the carbon dioxide's mass as uncollected calcium oxide. Carbon dioxide is another intended product.

For $\ce{CuO + H2 -> Cu + H2O}$, 7.95 g copper(II) oxide reacts with excess hydrogen. Use $M_r(\ce{CuO})=79.5$ and $A_r(\ce{Cu})=63.5$, with 5.08 g dry copper actually collected:

The two reactions have the same percentage yield even though their reactants, product masses and other products differ. Yield compares actual with theoretical mass of the same desired product. When neither reactant is explicitly in excess, identify the limiting reactant before finding the maximum. Keep unrounded values through the calculation and round only the requested final answer.