Stoichiometry and reacting masses

A chemical equation is a symbolic summary of a reaction. The reactants are written on the left and the products on the right, joined by an arrow.

For the equation to obey the conservation of mass, it must be balanced: the number of atoms of every element must be the same on both sides. You balance by adjusting the coefficients (the big numbers in front of formulae), never by changing a chemical formula (e.g. you may not turn $\text{O}_2$ into $\text{O}_3$ to balance oxygen).

A reliable method:

• Write the correct formulae of all reactants and products.
• Balance the most complicated species or those appearing in only one place first.
• Balance free elements (like $\text{O}_2$, $\text{H}_2$) last.
• Check every element, then check that the coefficients are in the lowest whole-number ratio.

Example: $\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}$.

For ionic equations the total charge must also balance on each side. State symbols $(s)$, $(l)$, $(g)$ and $(aq)$ describe the physical state of each species and should be added when the question asks for them.

Once an equation is balanced, the coefficients give the mole ratio in which substances react and form. This ratio is the link between any two species.

A general method for a reacting-mass calculation:

• Convert the given quantity to moles, using $n = \dfrac{m}{M}$, $n = CV$, or $n = \dfrac{V}{V_m}$ as appropriate.
• Use the mole ratio from the balanced equation to find the moles of the species you want.
• Convert that amount back to the quantity asked for (mass, volume, or concentration).

For example, in $\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$, every $1$ mol of $\text{N}_2$ reacts with $3$ mol of $\text{H}_2$ to make $2$ mol of $\text{NH}_3$.

For gases at the same temperature and pressure, Avogadro's law means volume ratios equal mole ratios, so you can use the coefficients directly with gas volumes. The mole ratio is the single most reusable tool in stoichiometry, so always write and balance the equation before doing any arithmetic.

When the amounts of two or more reactants are given, they are rarely in the exact stoichiometric ratio. The limiting reactant runs out first and determines the amount of product; the excess reactant has some left over.

To identify the limiting reactant:

• Convert each reactant to moles.
• Divide each amount by its coefficient in the balanced equation.
• The smallest of these values belongs to the limiting reactant.

All yield calculations must then be based on the limiting reactant only, not the reactant in excess.

To find how much excess reactant remains:

• Use the mole ratio to find how much excess reactant is used up by the limiting reactant.
• Subtract that from the starting amount of the excess reactant.

A common check: if you accidentally base a product calculation on the excess reactant, you will over-predict the yield. Always confirm which reactant is limiting first. Using excess of a cheap or easily removed reactant is a common strategy to ensure an expensive reactant is fully converted.

The theoretical yield is the maximum amount of product predicted by stoichiometry from the limiting reactant, assuming the reaction goes to completion with no losses.

In practice the actual (experimental) yield is almost always smaller, because of:

• Side reactions producing other products.
• Reversible reactions that do not go to completion.
• Losses during filtering, transferring, or purifying the product.
• Impure or incompletely reacted starting materials.

The percentage yield measures efficiency:

$$\% \text{ yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100$$

Make sure the actual and theoretical yields are in the same units (both in grams, or both in moles) before dividing.

Note that percentage yield is not the same as atom economy: a reaction can have a high percentage yield but a low atom economy (or vice versa). Percentage yield describes how much of the possible product was actually obtained; it says nothing about how much waste the equation itself produces. This formula is not in the data booklet, so memorise it.

Atom economy measures how much of the reactant mass ends up in the desired product, rather than in by-products or waste. It is a central idea in green chemistry.

$$\% \text{ atom economy} = \frac{\text{molar mass of desired product}}{\text{molar mass of all reactants}} \times 100$$

This formula is given in the data booklet (R2.1). Key points when using it:

• Multiply each molar mass by its coefficient in the balanced equation.
• The denominator is the total molar mass of all reactants used.
• A reaction with only one product (an addition or combination reaction) has $100\%$ atom economy, because all reactant atoms appear in the product.
• Reactions that release small molecules (e.g. $\text{H}_2\text{O}$ or $\text{CO}_2$) as by-products have lower atom economy.

High atom economy means less waste, lower disposal costs, and more efficient use of raw materials. It differs from percentage yield: atom economy is fixed by the equation, while percentage yield depends on the practical conditions of a particular experiment.