Quantitative Aspects of Chemical Change — Complete Notes

What you'll learn

• Use the mole concept and Avogadro's number to count particles.
• Convert between mass, moles and number of particles.
• Use molar mass and concentration in calculations.
• Balance equations and solve stoichiometry problems (mass–mass and solution).
• Identify the limiting reagent and calculate percentage yield and purity.
• Determine empirical and molecular formulae.

The mole and Avogadro's number

A mole (mol) is the amount of substance that contains the same number of particles as there are atoms in 12 g of carbon-12. That number is Avogadro's number:

N_A = 6,022 × 10²³ particles·mol⁻¹

The particles can be atoms, molecules, ions or formula units. The mole is simply a chemist's counting unit, just as a dozen means 12.

Key formulas

Molar mass M is the mass of one mole of a substance, in g·mol⁻¹. It is numerically equal to the relative formula mass. Example: M(NaCl) = 23 + 35,5 = 58,5 g·mol⁻¹.

Note: 1 dm³ = 1 litre = 1 000 cm³. To convert cm³ to dm³, divide by 1 000.

Stoichiometry

A balanced equation gives the mole ratio in which substances react. For example:

N₂(g) + 3H₂(g) → 2NH₃(g)

means 1 mol N₂ reacts with 3 mol H₂ to form 2 mol NH₃. The coefficients tell you the ratio, not the masses.

Steps for a stoichiometry calculation:

1. Write and balance the equation.
2. Convert the given quantity to moles.
3. Use the mole ratio from the balanced equation.
4. Convert moles of the required substance to the quantity asked for (mass, volume, concentration).

Limiting reagent

When reactants are not present in the exact ratio of the balanced equation, one runs out first — the limiting reagent. It determines the maximum amount of product formed. The other reactant is said to be in excess.

To find it: calculate the moles of product that each reactant could form; the smallest value comes from the limiting reagent.

Percentage yield and purity

Percentage yield = (actual yield / theoretical yield) × 100

Percentage purity = (mass of pure substance / mass of sample) × 100

The theoretical yield is calculated from the limiting reagent. The actual yield is what you really obtain in the lab, and it is always less than or equal to the theoretical yield.

Empirical and molecular formula

• Empirical formula = the simplest whole-number ratio of atoms in a compound.
• Molecular formula = the actual number of atoms; it is a whole-number multiple of the empirical formula.

From percentage composition: divide each percentage by the element's atomic mass, then divide all the answers by the smallest one, and round to whole numbers. Then M_molecular / M_empirical gives the multiplier.

Worked example 1 — mass, moles and particles

Calculate the mass of 0,50 mol of sodium carbonate, Na₂CO₃, and the number of formula units present.

Step 1 — Molar mass:

M(Na₂CO₃) = 2(23) + 12 + 3(16) = 46 + 12 + 48 = 106 g·mol⁻¹

Step 2 — Mass:

m = n M = 0,50 × 106 = 53 g

Step 3 — Number of formula units:

N = n N_A = 0,50 × 6,022 × 10²³ = 3,011 × 10²³ ≈ 3,0 × 10²³ formula units

Worked example 2 — solution stoichiometry

25,0 cm³ of hydrochloric acid of concentration 0,100 mol·dm⁻³ exactly neutralises a sodium hydroxide solution. Calculate the mass of NaOH that reacted.

Equation: HCl + NaOH → NaCl + H₂O (mole ratio 1 : 1).

Step 1 — Convert the volume to dm³:

V = 25,0 cm³ = 0,0250 dm³

Step 2 — Moles of HCl:

n = c V = 0,100 × 0,0250 = 2,50 × 10⁻³ mol

Step 3 — Mole ratio HCl : NaOH = 1 : 1, so:

n(NaOH) = 2,50 × 10⁻³ mol

Step 4 — Molar mass of NaOH:

M(NaOH) = 23 + 16 + 1 = 40 g·mol⁻¹

Step 5 — Mass:

m = n M = 2,50 × 10⁻³ × 40 = 0,100 g

Worked example 3 — limiting reagent and percentage yield

2,0 mol of nitrogen gas reacts with 4,0 mol of hydrogen gas:

N₂(g) + 3H₂(g) → 2NH₃(g)

(a) Identify the limiting reagent. (b) Calculate the mass of ammonia that should form. (c) If only 38,0 g of ammonia is obtained, calculate the percentage yield.

(a) H₂ needed for 2,0 mol N₂ = 3 × 2,0 = 6,0 mol, but only 4,0 mol is available → H₂ is the limiting reagent. Equivalently, from N₂: n(NH₃) = 2 × 2,0 = 4,0 mol; from H₂: n(NH₃) = (2/3) × 4,0 = 2,67 mol. The smallest value comes from H₂.

(b) n(NH₃) = (2/3) × 4,0 = 2,67 mol

M(NH₃) = 14 + 3(1) = 17 g·mol⁻¹

theoretical m = n M = 2,67 × 17 = 45,3 g

(c) % yield = (actual / theoretical) × 100 = (38,0 / 45,3) × 100 = 83,9%

Worked example 4 — molecular formula

A compound has an empirical formula of CH₂O and a molar mass of 180 g·mol⁻¹. Determine its molecular formula.

Step 1 — Empirical formula mass:

M(CH₂O) = 12 + 2(1) + 16 = 30 g·mol⁻¹

Step 2 — Multiplier:

n = M_molecular / M_empirical = 180 / 30 = 6

Step 3 — Multiply every subscript by 6:

Molecular formula = C₆H₁₂O₆

Concentration, dilution and standard solutions

Concentration tells you how much solute is dissolved in a given volume of solution. A standard solution is one whose concentration is known accurately.

When a solution is diluted, the amount of solute (in moles) does not change, so:

c₁ V₁ = c₂ V₂

where the subscript 1 refers to before dilution and 2 to after dilution. Adding water increases the volume and lowers the concentration, but the number of moles of solute stays the same.

Example: 50,0 cm³ of 0,20 mol·dm⁻³ NaCl is diluted to 250 cm³; then c₂ = c₁ V₁ / V₂ = (0,20 × 0,0500) / 0,250 = 0,040 mol·dm⁻³.

Common mistakes

• Using grams instead of moles in a mole ratio. The ratio works only in moles.
• Forgetting to convert cm³ to dm³ before using n = c V.
• Not balancing the equation first, which gives the wrong ratio.
• Rounding too early. Keep extra figures and round only at the end.
• Confusing Avogadro's number (particles) with molar mass (grams). They are linked through n.
• Assuming the reactant with the larger mass is in excess — it is the number of moles and the ratio that matter.
• Confusing percentage yield (product obtained) with percentage purity (pure substance in a sample).

Quick summary / cheat sheet

• n = m / M ; N = n N_A ; n = c V ; for a gas at STP, V_m = 22,4 dm³·mol⁻¹.
• N_A = 6,022 × 10²³ mol⁻¹.
• 1 dm³ = 1 000 cm³.
• Balance first, then work in moles, then convert out.
• The limiting reagent gives the theoretical yield.
• % yield = actual / theoretical × 100 ; % purity = pure / sample × 100.
• Empirical formula = simplest ratio; molecular formula = whole-number multiple of it.

Practice questions

1. Calculate the number of moles in 8,8 g of carbon dioxide, CO₂. [2]
2. How many atoms are there in 0,25 mol of helium gas? [2]
3. Calculate the concentration of a solution made by dissolving 5,85 g of NaCl in enough water to make 500 cm³ of solution. [3]
4. 20,0 cm³ of 0,250 mol·dm⁻³ H₂SO₄ reacts completely with NaOH: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the mass of NaOH that reacts. [4]
5. A compound contains 40% carbon, 6,7% hydrogen and 53,3% oxygen by mass. Determine its empirical formula. [4]
6. In the reaction 2H₂ + O₂ → 2H₂O, 3,0 mol H₂ reacts with 1,0 mol O₂. Which is the limiting reagent, and what mass of water forms? [4]
7. The theoretical yield of a reaction is 120 g but only 96 g is obtained. Calculate the percentage yield. [2]

Answers

1. M(CO₂) = 12 + 2(16) = 44 g·mol⁻¹; n = m / M = 8,8 / 44 = 0,20 mol.

2. N = n N_A = 0,25 × 6,022 × 10²³ = 1,51 × 10²³ atoms (helium is monatomic).

3. M(NaCl) = 23 + 35,5 = 58,5 g·mol⁻¹; n = 5,85 / 58,5 = 0,100 mol; V = 500 cm³ = 0,500 dm³; c = n / V = 0,100 / 0,500 = 0,200 mol·dm⁻³.

4. n(H₂SO₄) = c V = 0,250 × 0,0200 = 5,00 × 10⁻³ mol; ratio 1 : 2 → n(NaOH) = 1,00 × 10⁻² mol; M(NaOH) = 40 g·mol⁻¹; m = 1,00 × 10⁻² × 40 = 0,400 g.

5. C: 40 / 12 = 3,33; H: 6,7 / 1 = 6,7; O: 53,3 / 16 = 3,33. Divide by 3,33: C = 1, H = 2, O = 1 → empirical formula CH₂O.

6. From H₂: n(H₂O) = 3,0 mol; from O₂: n(H₂O) = 2 × 1,0 = 2,0 mol. The smaller value comes from O₂, so O₂ is limiting. n(H₂O) = 2,0 mol; M(H₂O) = 18 g·mol⁻¹; m = 2,0 × 18 = 36 g.

7. % yield = (96 / 120) × 100 = 80%.

Official EduRank Team notes. Always cross-check with your textbook and teacher.