Quantitative Chemistry: Moles, Mass and Yield
What Is a Mole?
A mole is the chemist's counting unit. Just as a dozen always means 12, one mole always contains \(6.02 \times 10^{23}\) particles — whether those particles are atoms, molecules or ions. This enormous number is called Avogadro's constant (\(N_A\)).
The key link between moles and mass is the molar mass. For any element, the molar mass in grams equals its relative atomic mass (\(A_r\)). For compounds, add up the \(A_r\) values of all the atoms in the formula to get the relative formula mass (\(M_r\)).
\[\text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}\] One mole of any element has a mass equal to its relative atomic mass in grams. For CO₂: \(M_r = 12 + (2 \times 16) = 44\,\text{g/mol}\).
Q: Calculate the number of moles in 22 g of carbon dioxide (\(CO_2\), \(M_r = 44\)).
Step 1 — Find \(M_r\): \(12 + (2 \times 16) = 44\,\text{g/mol}\).
Step 2 — Apply the formula: \(\text{moles} = \frac{22}{44} = 0.5\,\text{mol}\).
Q: What mass of sodium hydroxide (NaOH, \(M_r = 40\)) contains 0.25 mol?
Rearrange: \(\text{mass} = \text{moles} \times M_r = 0.25 \times 40 = 10\,\text{g}\).
Using Moles in Balanced Equations
A balanced chemical equation shows the molar ratio of reactants to products. For example:
$$2\,Mg + O_2 \rightarrow 2\,MgO$$
The numbers in front (coefficients) tell you that 2 moles of magnesium react with 1 mole of oxygen to produce 2 moles of magnesium oxide. To find the mass of a product from a known mass of reactant: (1) calculate moles of reactant, (2) use the ratio to find moles of product, (3) convert moles of product to mass.
Q: What mass of MgO is produced when 12 g of Mg burns completely? (\(A_r\): Mg = 24, O = 16)
Step 1 — Moles of Mg: \(\frac{12}{24} = 0.5\,\text{mol}\).
Step 2 — Ratio from equation: 2 mol Mg : 2 mol MgO → 1:1 ratio, so moles of MgO = 0.5 mol.
Step 3 — Mass of MgO: \(M_r(\text{MgO}) = 24 + 16 = 40\,\text{g/mol}\). Mass \(= 0.5 \times 40 = 20\,\text{g}\).
Limiting Reactants
In most reactions, one reactant runs out before the others. The limiting reactant determines the maximum amount of product. To identify it: calculate the moles of each reactant and compare to the stoichiometric ratio. Whichever reactant would be exhausted first is the limiting reactant; the other is in excess.
The limiting reactant is the one that runs out first. The amount of product formed is always calculated from the moles of the limiting reactant, not the excess reactant.
Percentage Yield
In practice, reactions rarely produce the theoretical maximum amount of product. Some product may be lost during transfer between vessels, some reactions may not go to completion, or unwanted side reactions may occur. The percentage yield compares the actual yield to the theoretical maximum:
$$\text{percentage yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100$$
Industrial chemists aim to maximise percentage yield to reduce waste and cost. Factors lowering yield include incomplete reactions, reversible reactions that reach equilibrium before completion, and losses during filtration or evaporation.
Q: A reaction has a theoretical yield of 50 g of aspirin. The chemist isolates 38 g. Calculate the percentage yield.
\(\text{percentage yield} = \frac{38}{50} \times 100 = 76\%\)
In an industrial context, a 76% yield would be assessed against the cost of raw materials and any waste disposal requirements before deciding if the process is commercially viable.
Atom Economy
Atom economy measures how efficiently a reaction converts reactants into the desired product. Unlike percentage yield (which measures what you got versus what you could get), atom economy is a property of the reaction equation itself:
$$\text{atom economy} = \frac{\text{sum of } M_r \text{ of desired products}}{\text{sum of } M_r \text{ of all reactants}} \times 100$$
A reaction where all atoms end up in the desired product has an atom economy of 100%. Addition reactions (where everything combines into one product) are ideal. Reactions that produce many by-products have low atom economy and generate chemical waste — an important issue in green chemistry and industrial sustainability.
Atom economy = (\(M_r\) of useful products ÷ \(M_r\) of all reactants) × 100. High atom economy is desirable — less waste, lower cost, more sustainable. A 100% atom economy means every atom of reactant ends up in the desired product.