TheIBAxolotl
The mole
Atoms are far too small to count individually, so chemists count them in batches. One mole is the amount of substance containing \(6.02 \times 10^{23}\) particles (the Avogadro constant, \(N_A\)). The "particles" can be atoms, molecules, ions or electrons, so always say which.
\[N = n \times N_A\]
Example: 0.20 mol of NH\(_3\) contains \(0.20 \times 6.02 \times 10^{23} = 1.2 \times 10^{23}\) molecules, but each molecule has 4 atoms, so there are \(4.8 \times 10^{23}\) atoms.
Relative mass and molar mass
- Relative formula mass, \(M_r\): sum of the \(A_r\) values in a formula (no units).
- Molar mass, M: mass of one mole, numerically equal to \(M_r\) but in g mol\(^{-1}\).
\[n = \frac{m}{M}\]
For CaCO\(_3\): \(M = 40.08 + 12.01 + 3(16.00) = 100.09\ \text{g mol}^{-1}\). So 0.0500 mol has a mass of \(0.0500 \times 100.09 = 5.00\) g.
Empirical and molecular formulas
- Empirical formula: simplest whole-number ratio of atoms.
- Molecular formula: actual number of atoms in one molecule; a whole-number multiple of the empirical formula.
Method: mass (or %) → divide by \(A_r\) → divide by the smallest → round or scale to whole numbers.
Worked example
A compound contains 40.0% C, 6.71% H and 53.3% O by mass. Its molar mass is 180 g mol\(^{-1}\).
| C | H | O | |
|---|---|---|---|
| mass in 100 g | 40.0 | 6.71 | 53.3 |
| ÷ \(A_r\) | 40.0/12.01 = 3.33 | 6.71/1.01 = 6.64 | 53.3/16.00 = 3.33 |
| ÷ smallest | 1.00 | 1.99 | 1.00 |
Empirical formula: CH\(_2\)O, with \(M = 30.03\) g mol\(^{-1}\). Multiplier \(= 180 / 30.03 = 5.99 \approx 6\), so the molecular formula is C\(_6\)H\(_{12}\)O\(_6\).
If a ratio comes out near 1.5 or 1.33, multiply all values by 2 or 3; do not round 1.5 to 2.
Molar concentration
\[c = \frac{n}{V}\]
with c in mol dm\(^{-3}\) and V in dm\(^3\) (1 dm\(^3\) = 1000 cm\(^3\)). Mass concentration in g dm\(^{-3}\) = \(c \times M\).
Worked example
4.00 g of NaOH (\(M = 40.00\) g mol\(^{-1}\)) is dissolved to make 250.0 cm\(^3\) of solution.
- \(n = 4.00 / 40.00 = 0.100\) mol
- \(V = 250.0 / 1000 = 0.2500\) dm\(^3\)
- \(c = 0.100 / 0.2500 = 0.400\) mol dm\(^{-3}\)
A standard solution is one whose concentration is known accurately, made using a volumetric flask.
Avogadro's law
Equal volumes of any gases at the same temperature and pressure contain equal numbers of particles. So, for gases, volume ratios equal mole ratios. If 2 dm\(^3\) of H\(_2\) reacts with O\(_2\) to make water vapour, it needs 1 dm\(^3\) of O\(_2\) (2 : 1 in the equation).
Common mistakes
- Using cm\(^3\) in \(c = n/V\) without dividing by 1000.
- Confusing molecules with atoms when using \(N_A\).
- Rounding mole ratios too early in empirical formula work.
- Writing \(M_r\) with units, or molar mass without them.
- Applying Avogadro's law to solids or liquids; it only works for gases.
Exam tips
- Write the unit next to every number; most errors show up as mismatched units.
- Keep at least one extra significant figure in intermediate steps, then round the final answer to match the data (usually 3 s.f.).
- For hydrated salts, treat water as a "component" and find the ratio of salt to H\(_2\)O moles.
- In multiple-choice gas questions, compare moles directly from volumes; no calculation of mass is needed.