Mathematical Techniques and Unit Conversions

Percentage Yield and Atom Economy

Industrial chemists constantly battle to prevent waste during manufacturing. To measure how successfully a reaction produces a desired product, chemists use two different calculations: yield and economy.

• Theoretical yield is the maximum calculated mass of product possible, assuming 100% of the reactants turn into products.
• Actual yield is the mass of product actually obtained in a real experiment, which is always lower due to side reactions, incomplete reactions, or separation losses.
• Percentage yield compares the actual mass obtained to the theoretical maximum, acting as a measure of reaction effectiveness.
• Atom economy measures the efficiency of a reaction by comparing the mass of the desired product to the total mass of all reactants.

Reactions with high atom economy are highly valued because they produce less waste and use raw materials more sustainably.

Worked Example: Calculating Percentage Yield

A student reacts $5.00\text{ g}$ of calcium carbonate ($CaCO_3$, $M_r=100$) to produce calcium oxide ($CaO$, $M_r=56$). The actual mass of $CaO$ obtained is $2.45\text{ g}$. Calculate the percentage yield.

Step 1: Calculate the moles of the reactant. $\text{moles} = \frac{\text{mass}}{M_r} = \frac{5.00}{100} = 0.050\text{ mol}$

Step 2: Use the mole ratio (1:1) to find the expected moles of product. Expected moles of $CaO = 0.050\text{ mol}$.

Step 3: Calculate the theoretical yield (maximum possible mass). $\text{Theoretical mass} = \text{moles} \times M_r = 0.050 \times 56 = 2.80\text{ g}$

Step 4: Calculate the percentage yield. $\%\text{ Yield} = \left( \frac{2.45}{2.80} \right) \times 100 = 87.5\%$

Final Answer: $87.5\%$

Worked Example: Calculating Atom Economy

Calculate the percentage atom economy for producing iron ($Fe$) in the following reaction: $Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2$. ($A_r$: $Fe=56, O=16, C=12$)

Step 1: Calculate the total $M_r$ of all reactants. $\text{Total } M_r = (2 \times 56 + 3 \times 16) + 3 \times (12 + 16) = 160 + 84 = 244$

Step 2: Calculate the total $M_r$ of the desired product ($2Fe$). $\text{Desired } M_r = 2 \times 56 = 112$

Step 3: Apply the formula. $\text{Atom economy} = \left( \frac{112}{244} \right) \times 100 = 45.901...\%$

Final Answer: $45.9\%$ (to 3 sig figs)

Rearranging Mathematical Equations

In chemistry, rearranging equations like the mole formula or the concentration formula is an essential skill. To move a variable to the other side of an equals sign, you must perform the inverse operation.

Worked Example: Rearranging Concentration Formula

Rearrange the concentration equation ($C = n / V$) to calculate the required volume ($V$) in $\text{dm}^3$.

Step 1: Multiply both sides by the denominator ($V$) to clear the fraction. $C \times V = n$

Step 2: Isolate $V$ by dividing both sides by $C$. $V = \frac{n}{C}$

Unit Conversions and Significant Figures

Failing to convert units is the leading cause of calculation errors in chemistry exams.

• Mass conversions: To convert grams ($g$) to kilograms ($kg$), divide by $1000$. To convert $kg$ to $g$, multiply by $1000$.
• Volume conversions: To convert cubic centimetres ($\text{cm}^3$) to cubic decimetres ($\text{dm}^3$), divide by $1000$.
• Significant figures: Give your final answer to the same number of significant figures as the least precise value provided in the question.

Worked Example: Gram to Kilogram Conversion

Calculate the number of moles in $2.5\text{ kg}$ of Copper(II) Oxide ($CuO$, $M_r=79.5$). Give your answer to 2 significant figures.

Step 1: Convert mass from $kg$ to $g$. $\text{mass} = 2.5 \times 1000 = 2500\text{ g}$

Step 2: Calculate moles. $\text{moles} = \frac{\text{mass}}{M_r} = \frac{2500}{79.5} = 31.446...\text{ mol}$

Final Answer: $31\text{ mol}$ (to 2 sig figs)

Worked Example: Multistep Conversion and Precision

Calculate the concentration in $\text{mol/dm}^3$ when $3.2\text{ g}$ of sodium hydroxide ($NaOH, M_r=40$) is dissolved in $200\text{ cm}^3$ of water.

Step 1: Convert mass to moles: $3.2 / 40 = 0.080\text{ mol}$. Step 2: Convert volume: $200 / 1000 = 0.200\text{ dm}^3$. Step 3: Calculate concentration ($C = n / V$): $0.080 / 0.200 = 0.4\text{ mol/dm}^3$. Step 4: Adjust to 2 sig figs (matching $3.2\text{ g}$).

Final Answer: $0.40\text{ mol/dm}^3$