Mission 1: Can burning 1 kg of charcoal really make more than 3 kg of carbon dioxide?

What an equation tells you

• Atoms are rearranged, never made or destroyed: each element balances.
• Coefficients give the ratio of amounts in moles, not masses.
• CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l): 1 mol : 2 mol : 1 mol : 2 mol.
• A coefficient multiplies the whole formula: 2O₂ is four O atoms.
• Total mass is conserved; the total amount in moles need not be.

Deducing an equation

• Write correct formulae first: from ion charges, with H₂, O₂, N₂ and Cl₂ as molecules.
• Balance by changing coefficients only, never subscripts.
• Add state symbols for the reaction conditions: (s), (l), (g), (aq).
• (aq) = dissolved in water. A precipitate is (s); water formed at room temperature is (l); a gas given off is (g).
• Sodium in water: 2Na(s) + 2H₂O(l) → 2NaOH(aq) + H₂(g).

Three steps for any reacting quantity

• Step 1: given quantity → amount in moles.
• Step 2: multiply by the mole ratio (wanted coefficient ÷ given coefficient).
• Step 3: moles → the quantity asked for: mass, gas volume or concentration.
• For masses, n = m/M, with A_r from the data booklet to two decimal places.
• M is for one formula unit. The coefficient is used only in step 2.

Worked example: reacting masses

• CaCO₃(s) → CaO(s) + CO₂(g). What mass of CaO forms from 25.0 g of CaCO₃?
• M(CaCO₃) = 40.08 + 12.01 + 3(16.00) = 100.09 g mol⁻¹.
• n(CaCO₃) = 25.0 ÷ 100.09 = 0.2498 mol.
• Ratio 1 : 1, so n(CaO) = 0.2498 mol.
• m(CaO) = 0.2498 × 56.08 = 14.0 g.

Mission 2: How can a few drops of indicator reveal how much acid is in vinegar?

Three ways to moles (from Structure 1.4 and 1.5)

• From a mass: n = m/M.
• From a solution: n = CV, with V in dm³ (cm³ ÷ 1000).
• From a gas at STP: n = V ÷ V_m, with V_m = 22.7 dm³ mol⁻¹ (data booklet).
• From a gas at other conditions: PV = nRT.
• After the mole ratio, go back the same way.

Worked example: mass to gas volume

• Zn(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂(g), with the acid in excess.
• What volume of H₂ at STP forms from 1.30 g of Zn?
• n(Zn) = 1.30 ÷ 65.38 = 0.01988 mol. Ratio 1 : 1, so n(H₂) = 0.01988 mol.
• V = 0.01988 × 22.7 = 0.451 dm³ (451 cm³).
• Only gases have this molar volume; never apply 22.7 to a solid or liquid.
• Gas to gas at the same T and p: volume ratio = mole ratio, with no V_m step.

Titration

• A pipette delivers a fixed volume of one solution into a conical flask.
• A burette adds the other until the indicator changes colour: the end-point.
• The volume added from the burette is the titre.
• Repeat until titres are concordant (differ by 0.10 cm³ or less).
• Use the mean of the concordant titres.
• One solution is a standard solution: its concentration is known accurately.

Worked example: titration calculation

• 25.0 cm³ of NaOH(aq) needs a mean titre of 21.40 cm³ of 0.0500 mol dm⁻³ H₂SO₄.
• 2NaOH(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + 2H₂O(l).
• n(H₂SO₄) = 0.0500 × 0.02140 = 1.070 × 10⁻³ mol.
• n(NaOH) = 2 × 1.070 × 10⁻³ = 2.140 × 10⁻³ mol.
• [NaOH] = 2.140 × 10⁻³ ÷ 0.0250 = 0.0856 mol dm⁻³.

Mission 3: Mix 10 g of hydrogen with 10 g of oxygen. Which runs out first?

Limiting and excess

• The limiting reactant is used up first, and the reaction stops.
• It fixes the maximum amount of product.
• The other reactant is in excess: some is left over.
• 2CO + O₂ → 2CO₂, starting with 6 CO and 4 O₂: 6 CO₂ form and 1 O₂ is left.

Finding the limiting reactant

• Convert every reactant to moles. Masses alone tell you nothing.
• Divide each amount by its coefficient. The smallest result is the limiting reactant.
• Calculate all products from the limiting reactant only.
• Excess left = amount present − amount that reacted.

Worked example

• CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), with 8.00 g of CH₄ and 24.0 g of O₂.
• n(CH₄) = 8.00 ÷ 16.05 = 0.4984 mol; ÷ 1 = 0.4984.
• n(O₂) = 24.0 ÷ 32.00 = 0.750 mol; ÷ 2 = 0.375. Smaller, so O₂ limits.
• n(CO₂) = 0.750 ÷ 2 = 0.375 mol; m = 0.375 × 44.01 = 16.5 g at most.
• CH₄ left = 0.4984 − 0.375 = 0.1234 mol = 1.98 g. The smaller mass was in excess.

Theoretical and experimental yield

• Theoretical yield: the maximum mass of product, calculated from the limiting reactant.
• It assumes complete reaction, no side reactions and no losses.
• Experimental yield: the mass of product actually obtained.
• The experimental yield is normally less than the theoretical yield.

Mission 4: If the equation promises 10 g of product, why do chemists often collect only 7 g?

Percentage yield

• Percentage yield = experimental yield ÷ theoretical yield × 100%.
• Use the same units for both yields, usually grams.
• Product is lost through incomplete reaction, side reactions, transfer and purification.
• A catalyst speeds a reaction up; it does not raise the theoretical yield.
• Above 100% signals an impure product (such as a wet one) or an error.

Worked example: a combined problem

• CuO(s) + H₂SO₄(aq) → CuSO₄(aq) + H₂O(l); crystallizing gives CuSO₄·5H₂O.
• 4.00 g CuO: n = 4.00 ÷ 79.55 = 0.0503 mol. 40.0 cm³ of 1.00 mol dm⁻³ acid: n = 0.0400 mol.
• Ratio 1 : 1, so H₂SO₄ limits. Excess CuO is filtered off.
• Theoretical: 0.0400 mol × 249.72 g mol⁻¹ = 9.99 g of CuSO₄·5H₂O.
• 7.50 g collected: yield = 7.50 ÷ 9.99 × 100 = 75.1%.

Working backwards from a yield

• To make a target mass at a known yield, first find the theoretical mass needed.
• Theoretical = target ÷ (percentage yield ÷ 100). It is always larger than the target.
• Example: 5.00 g of CaO at 80.0% yield needs a theoretical 6.25 g.
• n(CaO) = 6.25 ÷ 56.08 = 0.1114 mol = n(CaCO₃). Mass = 0.1114 × 100.09 = 11.2 g CaCO₃.

Mission 5: Can a reaction with a 100% yield still turn most of its reactants into waste?

Atom economy

• Atom economy = M of desired product ÷ total M of all reactants × 100% (data booklet).
• Multiply each M by its coefficient in the balanced equation.
• Every reactant counts, even water.
• It is fixed by the equation: neither the masses used nor the yield change it.
• Low atom economy means more reactant mass ends as by-products: more waste.

Worked example

• CaCO₃(s) → CaO(s) + CO₂(g), where CaO is the desired product.
• Atom economy = 56.08 ÷ 100.09 × 100 = 56.0%.
• 44% of the reactant mass leaves as CO₂, even at 100% yield.
• An addition reaction with a single product has 100% atom economy.
• Waste can still come from unreacted reactants, solvents and separation.

The big ideas

• Coefficients give mole ratios. Balance with coefficients only, and add state symbols.
• Any quantity → moles → mole ratio → moles → the quantity wanted.
• Titration: n = CV for the known solution, then the mole ratio, then C = n ÷ V.
• Limiting reactant: divide moles by coefficients; the smallest limits and fixes the theoretical yield.
• Percentage yield = experimental ÷ theoretical × 100%.
• Atom economy is set by the equation; the lower it is, the more waste.