TheIBAxolotl
Balancing equations
A chemical equation must have the same number of each type of atom on both sides (conservation of mass), and the same total charge for ionic equations. Change only the coefficients, never the subscripts in formulas. Add state symbols: (s), (l), (g), (aq).
Example: C\(_3\)H\(_8\)(g) + 5O\(_2\)(g) → 3CO\(_2\)(g) + 4H\(_2\)O(l). Balance C first, then H, then O last (it appears in several products).
Mole ratios and stoichiometry
The coefficients give the mole ratio. The standard method is:
- Convert the known quantity to moles (\(n = m/M\), \(n = cV\) or \(n = V/22.7\)).
- Use the mole ratio from the equation.
- Convert moles of the target back to the quantity asked for.
For gases, the coefficients also give volume ratios at the same temperature and pressure.
Limiting and excess reactants
The limiting reactant is completely used up and decides how much product forms. The other is in excess. To find it, divide the moles of each reactant by its coefficient: the smaller result is limiting.
Worked example
2.43 g of magnesium is added to 50.0 cm\(^3\) of 2.00 mol dm\(^{-3}\) hydrochloric acid:
\[\text{Mg(s)} + 2\text{HCl(aq)} \rightarrow \text{MgCl}_2\text{(aq)} + \text{H}_2\text{(g)}\]
- \(n(\text{Mg}) = 2.43 / 24.31 = 0.100\) mol
- \(n(\text{HCl}) = 2.00 \times 0.0500 = 0.100\) mol
- Mg needs \(2 \times 0.100 = 0.200\) mol HCl, but only 0.100 mol is present, so HCl is limiting.
- \(n(\text{H}_2) = 0.100 / 2 = 0.0500\) mol
- \(V(\text{H}_2)\) at STP \(= 0.0500 \times 22.7 = 1.14\) dm\(^3\)
- Mg left over: \(0.100 - 0.0500 = 0.0500\) mol \(= 1.22\) g
Percentage yield
\[\%\ \text{yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\]
The theoretical yield is calculated from the limiting reactant. If only 0.980 dm\(^3\) of H\(_2\) was collected in the example above: \(0.980 / 1.135 \times 100 = 86.3\%\).
Reasons for yields below 100%: incomplete or reversible reaction, side reactions, loss of product during transfer, filtration or purification. A yield above 100% usually means the product is impure or still wet.
Atom economy and green chemistry
\[\text{atom economy} = \frac{M(\text{desired product})}{\sum M(\text{all reactants})} \times 100\]
Use the coefficients from the balanced equation.
Worked example
Fermentation: C\(_6\)H\(_{12}\)O\(_6\) → 2C\(_2\)H\(_5\)OH + 2CO\(_2\)
- \(M(\text{glucose}) = 180.16\) g mol\(^{-1}\); \(2 \times M(\text{ethanol}) = 2 \times 46.07 = 92.14\) g mol\(^{-1}\)
- Atom economy \(= 92.14 / 180.16 \times 100 = 51.1\%\)
By contrast, hydration of ethene (C\(_2\)H\(_4\) + H\(_2\)O → C\(_2\)H\(_5\)OH) has 100% atom economy, because every reactant atom ends up in the product. Addition reactions always score 100%; substitution and elimination reactions produce by-products.
Green chemistry aims for high atom economy, less waste, safer solvents, lower energy use and renewable feedstocks. Yield and atom economy are different: a reaction can have a high yield but low atom economy, and vice versa.
Common mistakes
- Changing subscripts to balance an equation.
- Picking the limiting reactant by comparing masses instead of moles adjusted for coefficients.
- Calculating theoretical yield from the reactant in excess.
- Leaving out coefficients in atom economy.
- Confusing atom economy with percentage yield.
Exam tips
- Lay out calculations in a clear line: moles given → ratio → moles required → answer with units.
- In multi-step problems, keep unrounded values in the calculator.
- Explain a low yield with a specific, practical reason rather than "human error".
- Check that a percentage yield is below 100% before moving on.