Yield, Atom Economy, Molar Concentrations and Gas Volumes
Percentage Yield
In any chemical reaction, the amount of product actually made is often less than the theoretical maximum. The percentage yield compares what you actually get with what you theoretically could get:
$$\text{Percentage yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100$$
Reasons why the yield is less than 100%:
- The reaction may be reversible — not all reactants convert to products because the reaction reaches equilibrium.
- Product may be lost during transfer between containers, or during filtration, evaporation, or other separation steps.
- Side reactions may produce unwanted by-products instead of the desired product.
A high percentage yield is desirable in industrial chemistry because it means less waste, lower costs, and more efficient use of raw materials. A yield of 100% is rarely achievable in practice.
Atom Economy
Atom economy measures how efficiently a reaction uses its starting materials. It compares the mass of the desired product with the total mass of all products:
$$\text{Atom economy} = \frac{M_r \text{ of desired product}}{\text{Total } M_r \text{ of all products}} \times 100$$
Consider the reaction: \(\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2\)
If the desired product is calcium oxide (CaO), \(M_r\) of CaO = 56, \(M_r\) of CO2 = 44.
Total \(M_r\) of all products = 56 + 44 = 100.
$$\text{Atom economy} = \frac{56}{100} \times 100 = 56\%$$
This means 44% of the mass of the reactants ends up as the unwanted by-product (carbon dioxide).
High atom economy is important for sustainability. Reactions with low atom economy produce more waste by-products. Addition reactions have 100% atom economy because all atoms end up in the single product. Substitution and elimination reactions have lower atom economy because they produce by-products.
Molar Concentrations
The concentration of a solution can be expressed in mol/dm³ (moles per cubic decimetre). This tells you how many moles of solute are dissolved in each dm³ of solution.
The key formula is:
$$\text{concentration (mol/dm}^3\text{)} = \frac{\text{moles of solute}}{\text{volume of solution (dm}^3\text{)}}$$
Remember that 1 dm³ = 1000 cm³ = 1 litre. To convert cm³ to dm³, divide by 1000.
Example: 0.5 mol of NaCl is dissolved in 250 cm³ of water. Calculate the concentration.
Volume = 250 ÷ 1000 = 0.25 dm³
$$\text{Concentration} = \frac{0.5}{0.25} = 2.0 \text{ mol/dm}^3$$
Gas Volumes and the Molar Volume
At room temperature and pressure (RTP: 20 °C, 1 atmosphere), one mole of any gas occupies the same volume: 24 dm³ (or 24,000 cm³). This is called the molar volume.
To calculate the volume of a gas at RTP: $$\text{Volume (dm}^3\text{)} = \text{moles} \times 24$$ To calculate moles from volume: $$\text{Moles} = \frac{\text{volume (dm}^3\text{)}}{24}$$
Example: Calculate the volume of 0.3 mol of carbon dioxide at RTP.
$$\text{Volume} = 0.3 \times 24 = 7.2 \text{ dm}^3$$