Ratios, Proportion & Percentages — Maths for General Chemistry

Scaling, sharing and proportion

The everyday arithmetic of the whole wet lab

Two quantities are in direct proportion when one doubles as the other doubles (mass and moles), and in inverse proportion when one halves as the other doubles (Boyle's law, pV = constant).

direct: y = k x inverse: y = \frac{k}{x}

A percentage is just a ratio expressed out of 100. It appears in chemistry as percentage by mass (composition), percentage purity, and percentage yield — three very different ideas that all share the same arithmetic: (part / whole) × 100.

The dilution equation C₁V₁ = C₂V₂ is proportion in disguise: when you add solvent the moles of solute do not change, so the product of concentration and volume is conserved. The mole ratio from a balanced equation works the same way — coefficients are a ratio you can use as a unit factor.

Empirical and molecular formulae are found by converting mass percentages to moles and then finding the simplest whole-number ratio — the most common ratio problem in introductory chemistry.

When two reagents are mixed, the one that is completely consumed first is the limiting reagent. All other quantities (theoretical yield, excess reagent remaining) are calculated from the moles of the limiting reagent alone. A common error is to compare masses instead of moles — always convert to moles before applying a mole ratio. The percentage yield is always ≤ 100 % for a real experiment; a value above 100 % signals an error, often a sample that is not pure or not completely dry.

Trace-level concentrations are often expressed in ppm (parts per million) or ppb (parts per billion) by mass or by volume. For dilute aqueous solutions where density ≈ 1 g mL⁻¹, 1 ppm ≈ 1 mg L⁻¹ and 1 ppb ≈ 1 µg L⁻¹. The conversion to molarity requires dividing by the molar mass: $c (M) = \frac{ppm \times 1 0 ^{- 3}}{M ( g mol ^{- 1} )}$ .

Common pitfalls

Comparing masses instead of moles to find the limiting reagent. The reagent present in smaller mass is not necessarily limiting — you must convert to moles and apply the stoichiometric ratio first.
Inverting a ratio. The mole ratio of A to B from a balanced equation is coefficient(A) : coefficient(B), not the inverse. Writing the unit factor upside-down is the most common algebra error in stoichiometry.
Percentage yield greater than 100 %. This indicates the product is impure (contains solvent, unreacted reagents, or by-products) or that the actual yield was mis-measured. It is never physically possible to create more product than the theoretical yield.
Confusing empirical and molecular formulas. The empirical formula gives the simplest whole-number ratio of atoms. The molecular formula is a whole-number multiple of it — you need the molar mass to find which multiple.

Going deeper

In analytical chemistry, ppm and ppb concentrations arise for trace metals, pollutants, and pharmaceuticals. A serial dilution strategy — repeated dilutions each by a fixed factor — lets you span many orders of magnitude from a single concentrated stock. Each step multiplies the total dilution factor; after $n$ steps of factor $f$ , the final concentration is $C_{0} / f^{n}$ . Uncertainties accumulate with each pipetting step, so the minimum number of steps consistent with the required accuracy is preferred.

Key ratio formulas

Dilution

C_{1} V_{1} = C_{2} V_{2}

— moles are conserved; volume and concentration trade off.

% composition

% X = \frac{n _{X} M _{X}}{M _{compound}} \times 100

% yield

% yield = \frac{m _{actual}}{m _{theoretical}} \times 100

Mole ratio

From

a A + b B \to c C

, the mole ratio A:B = a:b — a direct proportion.

Where you'll use it

Making standard solutions, percentage yield, percentage composition, empirical formulae, and reading mole ratios straight off a balanced equation. Try the dilution solver and composition tools.

Dilution solver

Slide the stock concentration, volume taken, and target concentration — C₁V₁ = C₂V₂ finds the final volume.

Set the stock, the volume taken and the target — solve for the final volume

2.0 M

10 mL taken

→

0.50 M

40 mL total

C_{1} V_{1} = C_{2} V_{2} \Rightarrow V_{2} = \frac{C _{1} V _{1}}{C _{2}}

C₁ stock concentration2.0 M

V₁ volume of stock10 mL

C₂ target concentration0.50 M

final volume V₂40.0 mL

add solvent30.0 mL

Moles of solute don't change on dilution — only the volume does. That single idea is the whole equation.

Worked example 1Making 250 mL of 0.100 M HCl from 2.00 M stock

How many millilitres of $2.00 M$ HCl stock solution are needed to prepare $250.0 mL$ of $0.100 M$ HCl?

Worked example 2Serial dilution — two-step

You have $1.00 M$ NaCl and need $0.010 M$ for an experiment, but you only have 10 mL pipettes and 100 mL flasks. Describe a two-step serial dilution.

Percent composition

See how the atoms divide up by mass — the percentages must always add to 100.

Percentage by mass of each element

Glucose · C₆H₁₂O₆

C 40%

O 53.3%

C40.0%

H6.7%

O53.3%

%element = (mass of that element in one mole ÷ molar mass) × 100. The percentages must add to 100 — a quick check on your arithmetic.

Worked example 3Percentage composition of urea

Calculate the percentage by mass of each element in urea, $CO(NH_{2})_{2}$ . (C = 12.01, H = 1.008, N = 14.01, O = 16.00 g mol⁻¹)

Worked examples

Empirical formulae, mole ratios and percentage yield — the core ratio calculations of chemistry.

Worked example 4Empirical formula from combustion data

Combustion of $0.1152 g$ of a compound containing only C and H gives $0.3168 g$ CO₂ and $0.1944 g$ H₂O. Find the empirical formula. (M: CO₂ = 44.01, H₂O = 18.02, C = 12.01, H = 1.008)

Worked example 5Mole ratio in a stoichiometry problem

In the reaction $2 Al + 3 Cl_{2} \to 2 AlCl_{3}$ , how many grams of AlCl₃ form from $5.40 g$ of Al? (Al = 26.98, AlCl₃ = 133.33 g mol⁻¹)

Worked example 6Percentage yield from a limiting reagent

In the esterification $CH_{3} COOH + C_{2} H_{5} OH ⇌ CH_{3} COOC_{2} H_{5} + H_{2} O$ , you start with $6.00 g$ acetic acid (M = 60.05 g mol⁻¹) and excess ethanol. The theoretical yield of ethyl acetate (M = 88.11 g mol⁻¹) is calculated; you actually isolate $6.92 g$ . Find the theoretical yield and percentage yield.

Worked example 7Limiting reagent with two weighed reactants

$3.00 g$ of iron(III) chloride FeCl₃ (M = 162.2 g mol⁻¹) is mixed with $2.00 g$ of sodium hydroxide NaOH (M = 40.00 g mol⁻¹) in the reaction $FeCl_{3} + 3 NaOH \to Fe(OH)_{3} + 3 NaCl$ . Identify the limiting reagent, calculate the theoretical yield of Fe(OH)₃ (M = 106.87 g mol⁻¹), and find the mass of excess reagent remaining.

Worked example 8Molecular formula from empirical formula and molar mass

A compound has empirical formula $CH_{2} O$ (M_emp = 30.03 g mol⁻¹). A mass spectrometry experiment gives a molar mass of $180.2 g mol^{- 1}$ . Find the molecular formula.

Worked example 9ppm to molarity conversion for a trace metal

A drinking-water sample contains $0.050 ppm$ lead (Pb, M = 207.2 g mol⁻¹) by mass. Assuming the water density is 1.00 g mL⁻¹, calculate the molar concentration of Pb in mol dm⁻³. Is it above the WHO guideline of $1.4 \times 1 0^{- 7} mol dm^{- 3}$ ?

ChallengeLimiting reagent + % yield + purity combined

Aspirin (acetylsalicylic acid, M = 180.16 g mol⁻¹) is synthesised from salicylic acid (M = 138.12 g mol⁻¹) and excess acetic anhydride (M = 102.09 g mol⁻¹): $C_{7} H_{6} O_{3} + (CH_{3} CO)_{2} O \to C_{9} H_{8} O_{4} + CH_{3} COOH$ . You use $5.00 g$ of salicylic acid and isolate $4.12 g$ of crude product. Titration shows the product is 94.3 % pure aspirin. (a) Find the theoretical yield. (b) Find the actual yield of pure aspirin. (c) Calculate the % yield.

Check yourself

Four quick questions on dilutions, proportions, composition and yield.

Question 1 of 4 · Score 0

You dilute 10 mL of 2.0 M stock to a final volume of 100 mL. What is the new concentration?

C_{1} V_{1} = C_{2} V_{2}

Choose an answer.