Laws of chemical combination
Law | Statement |
|---|---|
Conservation of mass (Lavoisier) | Matter is neither created nor destroyed in a chemical reaction. |
Definite proportions (Proust) | A given compound always contains the same elements in the same proportion by mass. |
Multiple proportions (Dalton) | When two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in a simple whole-number ratio. |
Gay-Lussac's law of gaseous volumes | Gases react in simple whole-number ratios by volume, at the same temperature and pressure. |
Avogadro's law | Equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules. |
Example of multiple proportions: in H₂O and H₂O₂, the masses of oxygen combining with 2 g of hydrogen are 16 g and 32 g, a simple ratio of 1 : 2.
Atomic and molecular mass
Masses of atoms are expressed in unified atomic mass units (u). One u is exactly 1/12 of the mass of a carbon-12 atom.
The average atomic mass accounts for natural isotopes. Chlorine is about 75% ³⁵Cl and 25% ³⁷Cl, so its average atomic mass is 0.75 × 35 + 0.25 × 37 = 35.5 u.
The molecular mass is the sum of the atomic masses in the molecule. For H₂O: 2 × 1.008 + 16.00 = 18.016 u.
For ionic solids, which do not exist as discrete molecules, the same sum is called the formula mass. For NaCl it is 58.5 u.
The mole
One mole contains exactly 6.022 140 76 × 10²³ elementary entities (atoms, molecules, ions, …). This number is the Avogadro constant, N_A.
The molar mass is the mass of one mole, in g/mol. It is numerically equal to the atomic or molecular mass in u: water has molar mass 18 g/mol.
number of moles n = mass / molar mass = number of particles / N_A
At STP (273.15 K, 1 bar), one mole of an ideal gas occupies 22.7 L. (Older books use 22.4 L, which applies at 1 atm.)
Example. How many molecules are there in 9 g of water?
n = 9 / 18 = 0.5 mol. Molecules = 0.5 × 6.022 × 10²³ = 3.011 × 10²³.
Percentage composition and empirical formula
The empirical formula gives the simplest whole-number ratio of atoms. The molecular formula gives the actual numbers of atoms, and is a whole-number multiple of the empirical formula.
Example. A compound contains 40.0% C, 6.7% H and 53.3% O by mass. Its molar mass is 180 g/mol. Find its formulae.
Take 100 g: 40.0 g C, 6.7 g H, 53.3 g O.
Moles: C 40.0/12 = 3.33, H 6.7/1 = 6.7, O 53.3/16 = 3.33.
Divide by the smallest (3.33): C 1, H 2, O 1. The empirical formula is CH₂O (mass 30).
n = 180 / 30 = 6, so the molecular formula is C₆H₁₂O₆ (glucose).
Stoichiometry
A balanced equation gives the mole ratio in which substances react.
CH₄ + 2O₂ → CO₂ + 2H₂O
So 1 mol of methane (16 g) reacts with 2 mol of oxygen (64 g) to give 1 mol of CO₂ (44 g) and 2 mol of water (36 g). Mass is conserved: 16 + 64 = 44 + 36.
Limiting reagent
When reactants are not present in the exact ratio, the one that runs out first is the limiting reagent. It decides how much product forms.
Example. 3 mol of H₂ react with 2 mol of N₂: N₂ + 3H₂ → 2NH₃.
2 mol of N₂ would need 6 mol of H₂, but only 3 mol are available. So H₂ is limiting, only 1 mol of N₂ reacts, and 2 mol of NH₃ form. 1 mol of N₂ is left over.
Concentration of solutions
Term | Definition |
|---|---|
Mass percent | (mass of solute / mass of solution) × 100 |
Mole fraction (x) | moles of a component / total moles |
Molarity (M) | moles of solute per litre of solution |
Molality (m) | moles of solute per kilogram of solvent |
Molality does not change with temperature, because it is based on mass. Molarity does change, because the volume of a solution changes with temperature.
Example. 4 g of NaOH is dissolved to make 250 mL of solution. n = 4/40 = 0.1 mol, so the molarity is 0.1 / 0.250 = 0.4 M.