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Some Basic Concepts of Chemistry is where the mole comes in, and the mole is behind nearly every numerical in physical chemistry. These notes cover significant figures, the laws of chemical combination, atomic and molar mass, empirical formulas, limiting reagent problems and concentration terms, each with a worked example.
The one idea to hold on to
Equations are balanced in moles, not grams. Whatever the question gives you — mass, number of particles or gas volume — convert it to moles first, do the chemistry in moles, and convert back at the end.
1. Significant figures
| Rule | Example | Significant figures |
|---|---|---|
| All non-zero digits count | 285 cm | 3 |
| Zeros between non-zero digits count | 2.005 | 4 |
| Zeros before the first non-zero digit do not count | 0.0034 | 2 |
| Zeros at the end count only if there is a decimal point | 2.500 / 100 | 4 / 1 |
| Exact numbers (counted items, defined values) have unlimited significant figures | 20 students | Unlimited |
- Multiplying or dividing: the answer keeps as many significant figures as the least precise value.
- Adding or subtracting: the answer keeps as many decimal places as the value with the fewest decimal places.
- To show that trailing zeros are significant, use scientific notation: $1.00 \times 10^2$ has three significant figures.
2. Laws of chemical combination
| Law | Proposed by | What it says |
|---|---|---|
| Conservation of mass | Antoine Lavoisier | Matter is neither created nor destroyed in a chemical reaction. |
| Definite proportions | Joseph Proust | A compound always contains the same elements in the same proportion by mass. |
| Multiple proportions | John Dalton | When two elements form more than one compound, the masses of one that combine with a fixed mass of the other are in a small whole-number ratio. |
| Gaseous volumes | Joseph Gay-Lussac | Reacting gases, and gaseous products, are in a simple whole-number ratio by volume at the same temperature and pressure. |
| Avogadro's law | Amedeo Avogadro | Equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules. |
Worked example: multiple proportions
Hydrogen and oxygen form water (H₂O) and hydrogen peroxide (H₂O₂). For the same 2 g of hydrogen, 16 g of oxygen combine in water and 32 g in hydrogen peroxide. The oxygen masses are in the ratio $16 : 32 = 1 : 2$, a small whole-number ratio.
3. Atomic mass and average atomic mass
Atomic masses are measured in unified atomic mass units. One unit is exactly one-twelfth of the mass of a carbon-12 atom:
$$1\ \text{u} = 1.66056 \times 10^{-24}\ \text{g}$$Most elements exist as a mix of isotopes, so the atomic mass in the periodic table is a weighted average of the isotope masses.
Worked example: chlorine
Chlorine is 75.77% ³⁵Cl (34.9689 u) and 24.23% ³⁷Cl (36.9659 u).
Average atomic mass $= (0.7577 \times 34.9689) + (0.2423 \times 36.9659) = 26.4961 + 8.9568 = 35.45\ \text{u}$
Molecular mass is the sum of atomic masses in a molecule: for CO₂, $12.0 + 2(16.0) = 44.0\ \text{u}$. For ionic solids such as NaCl, which have no discrete molecules, the same sum is called the formula mass.
4. The mole
One mole contains exactly $6.02214076 \times 10^{23}$ particles — the Avogadro constant, $N_A$. The mass of one mole of a substance in grams, the molar mass $M$, is numerically equal to its atomic, molecular or formula mass in u.
$$n = \frac{m}{M} = \frac{N}{N_A} = \frac{V_{\text{gas}}}{V_m}$$For the molar volume of a gas, $V_m$, 22.4 L is the value at 273 K and 1 atm, which most NEET questions use. At 273.15 K and 1 bar — the current IUPAC standard — it is 22.7 L. Use whichever the question states.
Worked example
How many moles, molecules and atoms are there in 22 g of CO₂?
Moles: $n = \dfrac{22}{44} = 0.5\ \text{mol}$
Molecules: $0.5 \times 6.022 \times 10^{23} = 3.011 \times 10^{23}$
Atoms: each CO₂ molecule has 3 atoms, so $3 \times 3.011 \times 10^{23} = 9.033 \times 10^{23}$ atoms, of which $6.022 \times 10^{23}$ are oxygen.
5. Percentage composition and empirical formula
The empirical formula gives the simplest whole-number ratio of atoms. The molecular formula gives the actual numbers, and is a whole-number multiple of the empirical formula.
- Take a 100 g sample, so each percentage becomes a mass in grams.
- Divide each mass by its atomic mass to get moles.
- Divide every mole value by the smallest one to get the simplest ratio.
- If needed, multiply by a small whole number to clear fractions.
- Molecular formula $= n \times$ empirical formula, where $n = \dfrac{\text{molar mass}}{\text{empirical formula mass}}$.
Worked example
A compound contains 4.07% hydrogen, 24.27% carbon and 71.65% chlorine. Its molar mass is 98.96 g/mol. Find its empirical and molecular formulas.
| Element | Mass in 100 g | Moles | Ratio |
|---|---|---|---|
| C | 24.27 g | $24.27 \div 12.01 = 2.021$ | 1 |
| H | 4.07 g | $4.07 \div 1.008 = 4.04$ | 2 |
| Cl | 71.65 g | $71.65 \div 35.453 = 2.021$ | 1 |
Empirical formula: CH₂Cl, with empirical formula mass $12.01 + 2.016 + 35.453 = 49.48$.
$n = \dfrac{98.96}{49.48} = 2$, so the molecular formula is C₂H₄Cl₂.
6. Stoichiometry and the limiting reagent
The coefficients of a balanced equation give the mole ratio in which substances react. When reactants are not in that exact ratio, one runs out first. That is the limiting reagent, and it alone decides how much product forms.
- Balance the equation.
- Convert every reactant to moles.
- Divide each by its coefficient. The smallest result is the limiting reagent.
- Use the limiting reagent's moles and the mole ratio to find the product.
Worked example
50.0 kg of N₂ and 10.0 kg of H₂ react to form ammonia: $\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$. How much ammonia forms, and what is left over?
Moles of N₂ $= \dfrac{50\,000}{28.0} = 1785.7\ \text{mol}$. Moles of H₂ $= \dfrac{10\,000}{2.016} = 4960.3\ \text{mol}$.
All the N₂ would need $3 \times 1785.7 = 5357.1\ \text{mol}$ of H₂. Only 4960.3 mol is available, so H₂ is the limiting reagent.
NH₃ formed $= 4960.3 \times \tfrac{2}{3} = 3306.9\ \text{mol} = 3306.9 \times 17.0\ \text{g} \approx 56.2\ \text{kg}$.
N₂ used $= 4960.3 \div 3 = 1653.4\ \text{mol}$, so N₂ left over $= 132.3\ \text{mol} \approx 3.70\ \text{kg}$.
7. Concentration terms
| Term | Formula | Changes with temperature? |
|---|---|---|
| Mass percent | $\dfrac{\text{mass of solute}}{\text{mass of solution}} \times 100$ | No |
| Mole fraction | $x_A = \dfrac{n_A}{n_A + n_B}$ (all mole fractions add to 1) | No |
| Molarity, $M$ | $\dfrac{\text{moles of solute}}{\text{volume of solution in L}}$ | Yes, because volume changes |
| Molality, $m$ | $\dfrac{\text{moles of solute}}{\text{mass of solvent in kg}}$ | No |
Worked examples
Molarity: 4 g of NaOH (molar mass 40 g/mol) is dissolved to make 250 mL of solution. $n = 0.1\ \text{mol}$, so $M = \dfrac{0.1}{0.250} = 0.4\ \text{mol L}^{-1}$.
Molality: 4 g of NaOH is dissolved in 500 g of water. $m = \dfrac{0.1}{0.500} = 0.2\ \text{mol kg}^{-1}$.
Mole fraction: 1 mol of ethanol is mixed with 3 mol of water. $x_{\text{ethanol}} = \dfrac{1}{1+3} = 0.25$.
Mistakes that cost marks
- Comparing reactant masses instead of moles to find the limiting reagent.
- Skipping the balancing step before using mole ratios.
- Using millilitres in molarity. Convert to litres first.
- Dividing by the mass of the solution in molality. It is the mass of the solvent, in kg.
- Mixing up molecules and atoms: 1 mol of CO₂ has $N_A$ molecules but $3N_A$ atoms.
- Rounding early in multi-step problems, so the final answer lands between options.
- Using 22.4 L when the question specifies 1 bar, or 22.7 L when it specifies 1 atm.
60-second recap
- $n = \dfrac{m}{M} = \dfrac{N}{N_A}$, with $N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}$.
- Average atomic mass is the abundance-weighted mean of isotope masses.
- Empirical formula: percentages → moles → divide by the smallest → whole numbers.
- The limiting reagent is the reactant with the smallest moles ÷ coefficient.
- Molarity depends on temperature; molality and mole fraction do not.