Number of moles from mass

Turning a mass into a number of moles is the single most-used calculation in SPM Chemistry. It is the gateway to stoichiometry, titration and reacting-mass problems, so our online one-to-one teachers make sure the routine is second nature before you meet the harder questions that build on it.

When you use it

Reach for this whenever a question gives you a mass in grams (or kilograms) and asks for moles, or when you need moles as the first step before using a chemical equation. It pairs with the mole-to-mass rearrangement, m = n x M, which you use in reverse.

The formula

n = m / M

• n = number of moles, in mol
• m = mass, in grams (g)
• M = molar mass, in grams per mole (g mol⁻¹)

The molar mass M is numerically equal to the relative atomic mass (for an element) or the relative molecular / formula mass (for a compound), but it carries the unit g mol⁻¹.

Units

Mass must be in grams. If a question gives kilograms, multiply by 1000 first. The answer, n, is in mol.

Rearranging the relationship

The same triangle gives you three questions from one relationship. If you know moles and molar mass, the mass is m = n x M. If you know mass and moles, the molar mass is M = m / n, which is how you identify an unknown element. Being able to move between these three forms confidently is exactly what separates a full-mark script from a half-finished one, because SPM rarely asks the question in the direction you first practised.

Worked example 1 (easy), sodium hydroxide

Find the number of moles in 8 g of sodium hydroxide, NaOH. (Ar: Na = 23, O = 16, H = 1.)

• Step 1, molar mass: M(NaOH) = 23 + 16 + 1 = 40 g mol⁻¹.
• Step 2, apply n = m / M: n = 8 / 40 = 0.2 mol.

Worked example 2 (medium), carbon dioxide

Find the number of moles in 11 g of carbon dioxide, CO₂. (Ar: C = 12, O = 16.)

• Step 1, molar mass: M(CO₂) = 12 + (2 x 16) = 44 g mol⁻¹.
• Step 2, n = m / M = 11 / 44 = 0.25 mol.

Worked example 3 (SPM level), sodium carbonate

Find the number of moles in 5.3 g of sodium carbonate, Na₂CO₃. (Ar: Na = 23, C = 12, O = 16.)

• Step 1, molar mass: M(Na₂CO₃) = (2 x 23) + 12 + (3 x 16) = 46 + 12 + 48 = 106 g mol⁻¹.
• Step 2, n = m / M = 5.3 / 106 = 0.05 mol.

This small, deliberately “awkward” number is typical of SPM: it looks tricky but drops out cleanly once the molar mass is right, which is exactly why the molar mass step matters so much.

Common traps

• Using the wrong molar mass, the commonest cause of a wrong answer. Build M carefully before dividing.
• Leaving the mass in kilograms. Convert to grams first (x 1000).
• Turning the formula upside down and writing M / m. Remember n = m / M.
• Forgetting the unit mol on the answer, or writing g mol⁻¹ on the number of moles.
• Rounding the molar mass (for example 44 to 40), which throws the whole answer off.

How we help

Our teachers give you a mixed set, elements, simple compounds, then bracketed and hydrated formulae, so n = m / M becomes automatic and you never freeze on the molar-mass step. Lessons are one-to-one and taught in English, from RM50 per hour, with a paid one-hour trial so you can test the approach first. Once moles-from-mass is secure, the reverse (mass from moles) and every reacting-mass question become straightforward, because they are all built on this one relationship.