1. The big deal about low yield
Chemistry is a wasteful science. Reactions are not 100% efficient. When we mix reactants, not 100% will react, giving us less product than expected. In other words, the actual mass of product we get is a percentage of the theoretical mass of product calculated using stoichiometry. We call this percentage the percentage yield.
A low percentage yield may be because the reaction is incomplete, like the partial dissociation of weak acids in water. There may also be side reactions to form unwanted side products. Other times, the problem lies with the laboratory technique: we lose some products during the separation and purification steps.
2. How to calculate percentage yield
When you look up cheem research papers, the percentage yield of each reaction is written above the arrow. The example above shows a three-step method to synthesise Lacosamide, an important medication for epilepsy. Step 2 is a stumbling block, with the lowest percentage yield of 37%. This means that you only get 37 g of product in the lab, even though you have put in enough reactants to make 100 g of it. To say this in a cheem manner appropriate for the O Level:
- The actual yield of 37 g is the amount of product that is weighed at the end of a reaction
- The theoretical yield of 100 g is the calculated amount of product, assuming that the reaction is 100% efficient
- The percentage yield is the ratio of actual yield to theoretical yield expressed as a percentage: (37 g/100 g) × 100% = 37%
3. Formula for percentage yield
percentage yield = actual yield ÷ theoretical yield × 100%
Here, the actual and theoretical yields could be expressed in:
- Number of moles
- Mass (usually for solid products)
- Gaseous volume (usually for gas products)
4. Worked example: calculating the percentage yield of salt preparation
In this example, we will be finding the percentage yield of silver iodide formed by precipitation.
10 cm3 of 1 mol/dm3 silver nitrate solution was added to an excess of potassium iodide solution. The silver iodide precipitated was filtered, dried, and weighed. The measured mass was 2 g.
STEP 1: Write the chemical equation
AgNO3 + KI ⟶ AgI + KNO3
STEP 2: Use the amount of limiting reactant to calculate the theoretical yield of product
No. of moles of AgNO3 = concentration × volume in dm3 = 1 × (10/1000) = 0.01 mol
Mole ratio of AgNO3 to AgI = 1:1
No. of moles of AgI = 1/1 × 0.01 = 0.01 mol
Molar mass of AgI = 108 + 127 = 235 g/mol
Theoretical yield of AgI (in mass) = 0.01 × 235 = 2.35 g
STEP 3: Find the percentage mass
Actual yield of AgI = 2 g (given in question)
% yield = (2/2.35) × 100% = 85.1% (3 s.f.)
5. Complexity from impurity
Another reason why we get less product than expected is due to impurities mixed with reactants. When a sample is impure, its mass we measure includes the mass of impurities. This means that the actual mass of the reactants is lower than expected, translating to less product.
We can quantify how pure a sample is by its percentage purity. It is the ratio of the mass of the pure substance to the mass of the impure sample, but expressed in percentage. In the picture above, a 10% percentage purity would translate to 0.2 g of reactant in 2 g of the impure sample.
6. Formula for percentage purity
percentage purity = mass of pure substance ÷ mass of impure sample × 100%
To find the mass of pure substance, we can react a sample and leave the impurities behind. From the amount of product formed, we can back calculate to find out the amount of the pure substance that actually reacted.
7. Worked example: calculating the percentage purity of an impure sample of calcium carbonate
In this example, we will calculate the percentage purity of an impure sample containing calcium carbonate. 10 g of the impure sample is heated to decompose any calcium carbonate into calcium oxide and carbon dioxide. Only 2 dm3 of carbon dioxide was formed.
STEP 1: Write the chemical equation
CaCO3 ⟶ CaO + CO2
STEP 2: Work backwards to calculate the mass of pure calcium carbonate
No. of moles of CO2 = gaseous volume ÷ molar volume = 2 ÷ 24 = 0.0833 mol
Mole ratio of CO2 to CaCO3 = 1:1
No. of moles of CaCO3 = (1/1) × 0.0833 = 0.0833 mol
Molar mass of CaCO3 = 40 + 12 + 16×3 = 100 g/mol
Mass of pure CaCO3 = 0.0833 × 100 = 8.3 g
STEP 3: Find the percentage purity
Mass of impure sample = 10 g (given in question)
Percentage purity = (8.33/10) × 100% = 83.3%