Units, SI & Dimensional Analysis — Maths for General Chemistry

Units carry the meaning

Track them and most chemistry calculations check themselves

Every physical quantity is a number × a unit. The SI system builds everything from seven base units — for chemistry the key ones are the mole (amount of substance), kilogram (mass), metre (length), kelvin (temperature), and second (time). All other units — litres, joules, pascals — are derived by combining these.

The unit-factor method (dimensional analysis) treats units as algebra: multiply by fractions equal to one, arranged so the unwanted units cancel, leaving only the target unit. If the units do not come out right, the setup is wrong — no arithmetic is needed to spot the error.

5.0 g \times \frac{1 mol}{40 g} = 0.125 mol

In chemistry, concentrations appear as $mol dm^{- 3}$ (= mol L⁻¹ = M). Rates are in $mol dm^{- 3} s^{- 1}$ . The gas constant is $R = 8.314 J K^{- 1} mol^{- 1}$ — notice how the units tell you exactly which quantities to multiply and divide when using the ideal-gas law.

SI prefixes extend the range without changing the base unit. Nano (n, 10⁻⁹) is used for bond lengths in nanometres; micro (µ, 10⁻⁶) for enzyme concentrations in µM; milli (m, 10⁻³) for mmol in biochemical work; and kilo (k, 10³) for kilojoules in thermochemistry.

Temperature is the unit that trips students most often. The SI unit is the kelvin (K), not the degree Celsius. The conversion is exact: $T / K = θ /^{\circ} C + 273.15$ . The ideal-gas law, the Arrhenius equation, and the Boltzmann factor all require absolute temperature in kelvin — plugging in a Celsius value gives a wildly wrong answer. Note also that a temperature difference has the same size in K and in °C: Δ $T = 5 K$ and Δθ = 5 °C are identical, so heat-capacity calculations that only need a ΔT are unaffected by the choice of scale.

Volume is another frequent source of errors. The SI base unit of volume is the cubic metre (m³), but the litre (L) and cubic decimetre (dm³) are identical: $1 L = 1 dm^{3} = 1 0^{- 3} m^{3}$ . When a prefix is cubed, the exponent triples: 1 cm = 10⁻² m, so 1 cm³ = (10⁻²)³ m³ = 10⁻⁶ m³. Similarly, to derive the units of the ideal-gas constant $R$ start from $p V = n RT$ : $R = p V / (n T)$ , so $[R] = \frac{Pa \cdot m ^{3}}{mol \cdot K} = J mol^{- 1} K^{- 1}$ because 1 Pa · m³ = 1 J.

Common pitfalls

Using °C instead of K in gas-law or Arrhenius calculations. Always convert first: $T = θ + 273.15$ . A room temperature of 25 °C → 298.15 K, not 25 K.
Forgetting to cube a prefix when converting volume. 1 dm³ ≠ 10⁻¹ m³; it is $(1 0^{- 1})^{3} = 1 0^{- 3} m^{3}$ . Similarly 1 cm³ = 10⁻⁶ m³.
Mixing up mol dm⁻³ and mol m⁻³. The ideal-gas law with p in Pa and V in m³ gives n in mol; divide by m³ for mol m⁻³. Divide by dm³ for the everyday mol dm⁻³. The two differ by a factor of 1000.
Inverting the unit factor. The unit you want to cancel must go on the bottom of the fraction. Writing $5 g \times \frac{40 g}{1 mol}$ gives g² mol⁻¹, not mol — always check your units cancel.

Going deeper

The seven SI base units (metre, kilogram, second, ampere, kelvin, mole, candela) were redefined in 2019 in terms of fixed values of fundamental constants — the Boltzmann constant, Avogadro constant, elementary charge, etc. This means the kilogram is no longer the mass of a physical artefact but is instead defined exactly via $h = 6.62607015 \times 1 0^{- 34} J s$ . For chemistry the practical impact is negligible, but the conceptual shift is profound: every unit is now anchored to a constant of nature rather than to a human-made object.

Key relationships to memorise

Moles from mass

n = \frac{m}{M}

— mass (g) divided by molar mass (g mol⁻¹).

Concentration

c = \frac{n}{V}

— moles divided by volume in dm³ gives mol dm⁻³.

Ideal-gas law

p V = n RT

— pressure in Pa, V in m³, T in K, R = 8.314 J K⁻¹ mol⁻¹.

Unit-factor chain

Multiply by as many fractions-equal-to-one as needed; cancel units step by step.

Temperature scales

T / K = θ /^{\circ} C + 273.15

— always convert to K before using in any formula involving RT or k_BT.

Volume cube-prefix rule

1 dm^{3} = 1 0^{- 3} m^{3}

;

1 cm^{3} = 1 0^{- 6} m^{3}

. Cube the prefix factor, not just the number.

Why it matters

If your units don't come out right, your formula is wrong — full stop. Concentration in mol dm⁻³, rate in mol dm⁻³ s⁻¹, the gas constant in J K⁻¹ mol⁻¹: the units tell you exactly how the quantities must combine. Try the converter and prefix explorer.

Unit converter

Switch between common pressure and energy units — the conversion factor is always a ratio of SI equivalents.

Category:

value1.00 atm

From

1.00 atm = 101.3 kPa

Each unit is converted to the SI base (Pa) and back out, so the conversion factor is 101.3. Write conversions as fractions and let the unwanted units cancel — that's the whole trick of dimensional analysis.

Worked example 1Converting pressure for the ideal-gas law

A gas sample has a pressure of $2.50 atm$ . Convert to pascals (Pa) and to kilopascals (kPa), given $1 atm = 101325 Pa$ .

Worked example 2Energy units in thermochemistry

The standard enthalpy of combustion of ethanol is $Δ H^{\circ} = - 1368 kJ mol^{- 1}$ . Express this in kcal mol⁻¹, given $1 cal = 4.184 J$ .

Prefix explorer

Each SI prefix is a power of ten — slide through them and see how the exponent changes.

Slide through the SI prefixes — each step is a factor of a thousand

1 mol = 10⁰ mol

(base) → 1 = 1

Worked example 3Moles from grams via molar mass

A sample of NaOH (molar mass $M = 40.00 g mol^{- 1}$ ) has a mass of $2.400 g$ . Calculate (a) the amount in moles and (b) the amount in millimoles.

Worked examples

Multi-step unit-factor problems — try each before revealing the solution.

Worked example 4Concentration of a solution

$4.90 g$ of sulfuric acid ( $H_{2} SO_{4}$ , molar mass $98.08 g mol^{- 1}$ ) is dissolved to make $500.0 mL$ of solution. Calculate the molar concentration.

Worked example 5Ideal-gas law — finding volume

What volume (in dm³) does $0.250 mol$ of an ideal gas occupy at $25^{\circ} C$ and $100 kPa$ ? ( $R = 8.314 J K^{- 1} mol^{- 1}$ )

Worked example 6Rate law — confirming units of the rate constant

For the second-order reaction $rate = k [A]^{2}$ , the rate is $3.6 \times 1 0^{- 4} mol dm^{- 3} s^{- 1}$ when $[A] = 0.060 mol dm^{- 3}$ . Find $k$ and state its units.

Worked example 7Temperature conversion for the Arrhenius equation

The rate constant of a reaction doubles between 25 °C and 35 °C. To use the Arrhenius equation $ln (k_{2} / k_{1}) = - \frac{E _{a}}{R} (\frac{1}{T _{2}} - \frac{1}{T _{1}})$ , convert both temperatures to kelvin and evaluate the bracket.

Worked example 8Volume unit chain — cm³ → dm³ → m³

A crystal of NaCl has a unit-cell volume of $179.0 \overset{˚}{A}^{3}$ (where $1 \overset{˚}{A} = 1 0^{- 10} m$ ). Express this volume in (a) m³, (b) cm³, and (c) dm³.

ChallengeDeriving R and checking its units from first principles

One mole of an ideal gas occupies $22.414 dm^{3}$ at exactly 0 °C and 101 325 Pa. (a) Calculate $R$ from $p V = n RT$ in SI units. (b) Show that J K⁻¹ mol⁻¹ is consistent with Pa · m³ · K⁻¹ · mol⁻¹. (c) Convert $R$ to dm³ · atm · K⁻¹ · mol⁻¹ (1 atm = 101 325 Pa).

Check yourself

Four quick questions on SI units, prefixes and dimensional analysis.

Question 1 of 4 · Score 0

How many millimoles (mmol) are in 0.25 mol?

Choose an answer.