Amount of Substance and Composition Formulae

Introduction

Amount-of-substance formulae bridge a sample on a balance and the atoms or molecules that react. The two central conversions are n = m/M and N = nN A. Composition adds a separate ratio: mass fraction is a part divided by the whole mass. Correct use requires naming the counted entity and distinguishing a compound's molar mass from its reaction coefficient.

Core explanation

For a pure substance, n = m/M, where n is amount in mol, m is mass and M is molar mass in g mol⁻¹ or kg mol⁻¹ consistent with the mass. The inverse is m = nM. Molar mass follows from the chemical formula and atomic molar masses. Water's molar mass is about 18.015 g mol⁻¹ because each molecule has two hydrogen atoms and one oxygen atom. This formula mass is not the coefficient of water in a balanced reaction.

Entity count is N = nN A. If the entities are water molecules, 0.5 mol contains 0.5N A molecules. That sample contains 1.0N A hydrogen atoms and 0.5N A oxygen atoms. A statement like “0.5 mol particles” is incomplete if atoms and molecules could both be meant. For ionic substances, specify formula units or individual ions and account for dissociation separately.

Stoichiometric coefficients relate amounts consumed and formed. In 2 H₂ + O₂ → 2 H₂O, n(H₂O)/n(O₂) = 2 for ideal complete reaction with oxygen limiting. The coefficient 2 for water does not double its molar mass. If 1.0 mol H₂ and 1.0 mol O₂ are supplied, hydrogen is limiting; at most 1.0 mol water forms and 0.5 mol oxygen remains. This calculation uses both molar conversion and reaction ratios.

Mass fraction w i = m i/m total is dimensionless and lies from zero to one for nonnegative component masses. Percent by mass is 100w i%. For a pure compound, an element's theoretical mass fraction is its atom count times atomic molar mass divided by compound molar mass. For water, hydrogen accounts for about 2(1.008)/18.015 ≈ 0.112, or 11.2% by mass. Mass fraction is not mole fraction, which counts entities.

Empirical-formula analysis reverses mass fractions. Start with a convenient 100 g hypothetical sample, convert each element's mass to moles, divide by the smallest amount, then seek a simple whole-number ratio. If ratios look like 1:1.5, multiply all by two. Small deviations arise from rounding and experimental uncertainty; forcing an arbitrary ratio without considering precision can produce a false formula.

For mixtures, molar mass may refer to an average and must be defined. An air sample does not have a single molecule type, though its mean molar mass can connect total gas mass to total moles approximately. For a hydrated salt, distinguish the formula of the hydrate from anhydrous salt; ignoring waters of crystallization produces a systematic amount error.

Pure-reagent mass is not automatically usable reacting mass. A 90% pure solid contains only 0.90 times its weighed mass as the target compound if the percentage is by mass and impurities are inert. Likewise, a yield calculation compares actual product with theoretical product based on the limiting reactant, not merely on its weighed mass.

Step-by-step reasoning

Write the chemical formula and specify the entity. Compute M, then convert measured mass to n. For a reaction, apply coefficients to find limiting reagent and product amount. For counts multiply by N A; for composition divide component mass by whole mass. Check that fractions sum to one when all components are included.

Visual explanation

Draw a triangle with mass m, moles n and entity count N. The m-to-n arrow divides by M, and the n-to-N arrow multiplies by N A. A separate branch from m splits into component masses, each divided by total m to form a mass fraction.

Real-world analogy

Counting eggs by the dozen and weighing cartons are different ways to describe a supply. A mole is an extremely large count, and molar mass is the mass per such count. The analogy does not imply molecules are identical to eggs in size or behavior.

Real-world example

A formulation requires 0.10 mol sodium chloride. With M ≈ 58.44 g mol⁻¹, weigh about 5.844 g of pure NaCl. If reagent purity is 98% by mass, the required weighed mass is 5.844/0.98 ≈ 5.96 g, assuming impurities do not contribute NaCl.

Why?

Chemical reactions occur in entity ratios, while laboratory instruments often measure mass. These formulae are the bridge between the two descriptions and underpin concentrations, yields, titrations and material balances.

Common misconception

“A 2 in front of H₂O changes its molar mass” confuses reaction amount with molecular composition. Another error divides a component's moles by total mass and calls the result mass fraction; the numerator and denominator must both be masses.

Worked example

A sample contains 9.00 g water. With M = 18.015 g mol⁻¹, n = 9.00/18.015 ≈ 0.500 mol water molecules. The count is 0.500 × 6.022 × 10²³ ≈ 3.01 × 10²³ molecules. It contains twice as many hydrogen atoms, about 6.02 × 10²³. The units and atom ratio check the result.

Quick check

1. What extra information is required to turn a sample's mass into a molecule count? Answer: Its molar mass and Avogadro's constant, with the counted entity specified.

Exam focus

Write n = m/M and N = nN A with units. Distinguish molar mass from equation coefficients. Define mass fraction as component mass divided by total mass and use limiting-reactant ratios for products.

Advanced insight

The mole is defined by a fixed Avogadro constant, while atomic masses and isotopic composition determine a particular sample's molar mass. A high-precision isotopically enriched sample can have a slightly different molar mass from the conventional tabulated value. Ordinary textbook calculations usually ignore that distinction.

Summary

Mass, moles and entity count are linked by molar mass and N A. Reaction coefficients determine relative amounts, not molecular masses. Mass fractions describe composition on a mass basis and should be kept separate from mole fractions.

Practice questions

1. How many moles are in 11.688 g pure NaCl with M = 58.44 g mol⁻¹? Answer: 0.2000 mol. 2. What is the mass fraction of 2 g solute in 10 g total solution? Answer: 0.20, or 20% by mass. 3. In 2 H₂ + O₂ → 2 H₂O, how much water can 0.5 mol O₂ form with excess H₂? Answer: 1.0 mol H₂O. 4. Does one mole of CO₂ contain one mole or three moles of atoms? Answer: Three moles of atoms: one mole carbon atoms and two moles oxygen atoms.

Sources

- BIPM definition of the mole. - IUPAC Gold Book.