Calculating Yields and Atom Economy

(Note: Both yields must be measured in the exact same units, typically grams or moles.)

Worked Example: Simple Mass Substitution

A student prepares 4.5 g of dry magnesium chloride crystals. The calculated theoretical yield was 5.0 g. Calculate the percentage yield.

Step 1: Identify the actual and theoretical yields.

• Actual yield = $4.5\text{ g}$
• Theoretical yield = $5.0\text{ g}$

Step 2: Substitute the values into the percentage yield formula.

• $\text{Percentage Yield} = (4.5\text{ g} \div 5.0\text{ g}) \times 100$

Step 3: Calculate the final percentage.

• $\text{Percentage Yield} = 90\%$

Calculating Theoretical Yield from Moles

Often, exam questions will not give you the theoretical yield directly. You will need to calculate it yourself using the molar ratio from a balanced chemical equation. This assumes 100% of the limiting reactant is converted into the product.

Worked Example: Multi-step Percentage Yield

4.80 g of magnesium reacts with excess hydrochloric acid to form 14.2 g of magnesium chloride ($MgCl_2$). Calculate the percentage yield. Equation: $Mg + 2HCl \rightarrow MgCl_2 + H_2$ ($A_r: Mg=24, Cl=35.5$)

Step 1: Calculate the moles of the limiting reactant.

• $\text{Moles of Mg} = 4.80\text{ g} \div 24 = 0.20\text{ mol}$

Step 2: Determine the theoretical moles of the product using the molar ratio.

• The ratio of $Mg : MgCl_2$ in the equation is $1:1$.
• Theoretical moles of $MgCl_2$ = $0.20\text{ mol}$

Step 3: Calculate the theoretical mass of the product.

• Relative formula mass ($M_r$) of $MgCl_2$ = $24 + (2 \times 35.5) = 95$
• $\text{Theoretical mass} = 0.20\text{ mol} \times 95 = 19.0\text{ g}$

Step 4: Substitute into the percentage yield formula.

• $\text{Percentage Yield} = (14.2\text{ g} \div 19.0\text{ g}) \times 100$

Step 5: Calculate the final answer.

• $\text{Percentage Yield} = 74.7\%$ (to 3 s.f.)

What is Atom Economy?

Even if a reaction has a 100% percentage yield, it might still generate massive amounts of waste if the desired chemical is produced alongside heavy, unwanted by-products.

• Atom economy measures the proportion of the mass of all starting materials that ends up in the desired product.
• Unlike percentage yield (which relies on experimental data), atom economy is a strictly theoretical calculation based entirely on the balanced equation.
• According to the Law of Conservation of Mass, the total mass of the reactants always equals the total mass of the products.
• Only reactions that produce a single product (like addition reactions) can achieve a 100% atom economy.

Calculating Atom Economy

To calculate atom economy, you must use the relative formula masses of the substances involved.

Worked Example: Atom Economy

Iron is extracted from iron(III) oxide by heating it with carbon monoxide. Calculate the atom economy for the production of iron in this reaction. Equation: $Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2$ ($A_r: Fe=56, O=16, C=12$)

Step 1: Calculate the total $M_r$ of the desired product, including the balancing numbers.

• The desired product is $2Fe$.
• $\text{Total } M_r = 2 \times 56 = 112$

Step 2: Calculate the sum of the $M_r$ of all reactants.

• $M_r \text{ of } Fe_2O_3 = (2 \times 56) + (3 \times 16) = 112 + 48 = 160$
• $M_r \text{ of } 3CO = 3 \times (12 + 16) = 3 \times 28 = 84$
• $\text{Sum of reactants} = 160 + 84 = 244$

Step 3: Substitute into the formula.

• $\text{Atom Economy} = (112 \div 244) \times 100$

Step 4: Calculate the final percentage.

• $\text{Atom Economy} = 45.9\%$ (to 3 s.f.)

Evaluating Reaction Pathways

Industrial chemists must choose the best reaction pathway (the sequence of steps used to make a chemical). High atom economy is highly desirable for green chemistry and sustainability.

A highly sustainable pathway:

• Conserves valuable raw natural resources.
• Minimises the production of harmful or useless waste.
• Reduces the economic costs associated with buying extra materials and disposing of waste safely.

When evaluating different pathways, chemists look for the route with the highest atom economy and percentage yield. However, a pathway with a lower atom economy might still be chosen if the reaction rate is much faster, or if the "waste" by-products are actually useful chemicals that can be sold for profit.