A Kilogram of Salt Holds Far More Particles Than a Kilogram of Sugar

A shopkeeper sells rice by the kilogram, not by counting grains. Chemists do the same with atoms and molecules — they weigh them out in moles, a fixed, enormous number of particles.

This part covers the laws of chemical combination, the mole concept, percentage composition and formulae, and stoichiometry with concentrations. Take $N_A=6.022\times 10^{23}$ mol${}^{-1}$.

Mass is conserved in reactions, a compound always has the same elements in the same mass ratio, fixed masses of one element combine with masses of another in simple whole-number ratios, gases react in simple volume ratios, and equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.

- Conservation of mass: $2.4$ g of magnesium burns with $1.6$ g of oxygen to give exactly $4.0$ g of magnesium oxide
- Definite proportions: water is always $1:8$ hydrogen to oxygen by mass
- Multiple proportions: $12$ g of carbon combines with $16$ g of oxygen in CO and $32$ g in CO${}_2$ — a ratio of $1:2$
- Gay-Lussac's law: $1$ volume of H${}_2$ and $1$ volume of Cl${}_2$ give $2$ volumes of HCl
- Avogadro's law: equal gas volumes hold equal numbers of molecules at the same temperature and pressure

Dalton's atomic theory explains these: matter is made of atoms, atoms of one element are identical in mass, and atoms combine in fixed whole-number ratios — so compositions are fixed and ratios are simple.

An everyday example. A candle seems to lose mass as it burns, but the "missing" mass has left as carbon dioxide and water vapour; in a sealed jar the total stays the same.

The substance. Atoms are now known to be divisible, but the idea that they combine in whole-number ratios still holds.

**Average atomic mass weights each isotope's mass by its abundance, molecular or formula mass adds the atomic masses in the formula, and one mole contains $6.022\times 10^{23}$ particles with a mass in grams equal to the molar mass.**

$$n=\frac{m}{M}=\frac{N}{N_A}$$

Worked example 1 — average atomic mass. Chlorine is $75.77\%$ chlorine-35 ($34.97$ u) and $24.23\%$ chlorine-37 ($36.97$ u):

$$0.7577\times 34.97+0.2423\times 36.97\approx 35.45\text{ u}$$

Worked example 2 — molecular and formula masses.

- Glucose C${}_6$H${}_{12}$O${}_6$: $6(12.01)+12(1.008)+6(16.00)\approx 180.2$ u
- Sodium chloride: formula mass $22.99+35.45=58.44$ u

Worked example 3 — moles and particles. $36$ g of water with $M=18$ g/mol:

$$n=\frac{36}{18}=2\text{ mol}N=2\times 6.022\times 10^{23}\approx 1.20\times 10^{24}\text{ molecules}$$

That is $2.41\times 10^{24}$ hydrogen atoms.

An everyday example. Buying dal by the kilogram is like chemistry's mole — mass stands in for counting grains you could never count one by one.

The substance. "Formula mass" is used for ionic solids like NaCl, which are not made of separate molecules.

**Percentage composition gives each element's share of the molar mass; dividing each percentage by the atomic mass gives mole ratios, the simplest whole-number ratio is the empirical formula, and multiplying it by $n=\frac{\text{molar mass}}{\text{empirical formula mass}}$ gives the molecular formula.

Worked example 1 — urea.** CO(NH${}_2$)${}_2$ has molar mass $60.06$ g/mol and contains $28.02$ g of nitrogen per mole:

$$\%N=\frac{28.02}{60.06}\times 100\approx 46.7\%$$

Worked example 2 — from analysis to formula. A compound has $40.0\%$ C, $6.7\%$ H and $53.3\%$ O, and molar mass $180$ g/mol. Take $100$ g:

- C: $\frac{40.0}{12.01}\approx 3.33$ mol
- H: $\frac{6.7}{1.008}\approx 6.65$ mol
- O: $\frac{53.3}{16.00}\approx 3.33$ mol

Dividing by $3.33$ gives $1:2:1$, so the empirical formula is **CH${}_2$O**, with mass $30$.

$$n=\frac{180}{30}=6\Rightarrow {\text{molecular formula C}}_6{\text{H}}_{12}{\text{O}}_6$$

An everyday example. Farmers choose urea as a nitrogen fertiliser partly because nearly half of its mass is nitrogen.

The substance. Different compounds can share an empirical formula — CH${}_2$O is also the empirical formula of acetic acid, C${}_2$H${}_4$O${}_2$.

A balanced equation gives mole ratios; the reactant that runs out first is the limiting reagent and fixes the amount of product; and concentration can be given as mass per cent, mole fraction, molarity (mol per litre of solution) or molality (mol per kg of solvent).

Worked example 1 — limiting reagent. N${}_2$ + 3H${}_2$ $\to$ 2NH${}_3$, starting with $28$ g of N${}_2$ and $10$ g of H${}_2$.

- N${}_2$: $1$ mol, which needs $3$ mol ($6$ g) of H${}_2$
- H${}_2$ available: $5$ mol — more than enough

So **N${}_2$ is limiting**: $2$ mol ($34$ g) of NH${}_3$ forms, and $4$ g of H${}_2$ is left over.

Worked example 2 — concentrations.

- Mass per cent: $2$ g NaCl in $18$ g water $=\frac{2}{20}\times 100=10\%$
- Molarity: $4$ g NaOH ($0.1$ mol) in $250$ mL of solution $=\frac{0.1}{0.250}=0.4$ M
- Molality and mole fraction: $18$ g glucose ($0.1$ mol) in $90$ g water ($5$ mol) gives $m=\frac{0.1}{0.090}\approx 1.11$ mol/kg and $x=\frac{0.1}{5.1}\approx 0.0196$

An everyday example. Mixing an ORS packet into exactly one litre of water follows a concentration recipe — too little water makes it too strong.

The substance. Molality does not change with temperature, but molarity does, because a solution's volume expands when warmed.