S1.4 Counting particles by mass: The moleIB Chemistry HL: Revision notes
Section 1
The mole and the Avogadro constant
The mole (mol) is the SI unit of amount of substance, n. One mole contains exactly the number of elementary entities given by the Avogadro constant, Nₐ = 6.02 × 10²³ mol⁻¹ (in the data booklet). An elementary entity can be an atom, molecule, ion, electron or a specified group of particles, so you must always say what you are counting.
Number of entities N = n × Nₐ. One mole of CaCO₃ contains 6.02 × 10²³ formula units, but 2 × 6.02 × 10²³ ions and 3 × 6.02 × 10²³ oxygen atoms.
Multiplying by Nₐ gives the number of formula units; multiply again by the number of atoms or ions per formula unit when the question asks for those.
Section 2
Relative masses and molar mass
Masses of atoms are compared on a scale relative to carbon-12, which is defined as exactly 12. Relative atomic mass, Aᵣ, is the weighted mean mass of an element's atoms relative to 1/12 of the mass of a ¹²C atom; relative formula mass, Mᵣ, is the sum of the Aᵣ values in a formula. Both have no units.
Molar mass, M, is numerically equal to Mᵣ but has units of g mol⁻¹. For example Mᵣ(CaCO₃) = 40.08 + 12.01 + 3(16.00) = 100.09, so M = 100.09 g mol⁻¹.
The key relationship is n = m ÷ M (mass in grams).
Use the Aᵣ values from the data booklet to two decimal places and convert mg or kg to g before using n = m/M.
Section 3
Empirical and molecular formulas
The empirical formula is the simplest whole-number ratio of atoms of each element in a compound. The molecular formula is the actual number of atoms of each element in one molecule; it is a whole-number multiple of the empirical formula.
From percentage composition: assume 100 g, divide each mass by Aᵣ, then divide by the smallest answer. If a ratio is close to x.5 or x.33, multiply all values by 2 or 3.
To get the molecular formula: divide the molar mass by the empirical formula mass and multiply the empirical formula by the answer. Ethyl hexanoate: empirical C₄H₈O (72.12), M = 144, so C₈H₁₆O₂.
You can also reverse the process: % by mass of element = (number of atoms × Aᵣ ÷ Mᵣ) × 100.
Rounding 1.5 to 2 (or 1.33 to 1) gives the wrong formula. Only round values that are within about 0.1 of a whole number.
Section 4
Molar concentration
Molar concentration, c, is the amount of solute per unit volume of solution: c = n ÷ V, with c in mol dm⁻³ and V in dm³ (1 dm³ = 1000 cm³). Square brackets show concentration: [NaCl] = 0.100 mol dm⁻³.
In a titration, use c × V for the solution of known concentration, the mole ratio from the equation, then c = n ÷ V for the unknown. If only a portion (aliquot) of a larger volume was titrated, scale up, e.g. × 10 for 25.00 cm³ taken from 250.0 cm³.
Convert cm³ to dm³ by dividing by 1000 before using c = n/V.
Section 5
Avogadro's law and reacting gas volumes
Avogadro's law: equal volumes of all gases measured at the same temperature and pressure contain equal numbers of molecules. So for gases at the same conditions, the ratio of volumes equals the mole ratio in the equation.
Example: 2H₂(g) + O₂(g) → 2H₂O(g). 40 cm³ of H₂ reacts with 20 cm³ of O₂ to give 40 cm³ of steam.
In combustion problems, cool to room temperature so water is liquid (negligible volume); aqueous NaOH absorbs CO₂, and the gas left is excess O₂. For CₓHᵧ + (x + y/4)O₂ → xCO₂ + (y/2)H₂O, compare volumes to find x and y.
Volume ratios only apply to gases. Liquids and solids in the equation do not follow Avogadro's law.
Must know
- n = m ÷ M; N = n × Nₐ; c = n ÷ V (V in dm³).
- Say which entity you are counting: atoms, molecules, ions or formula units.
- Aᵣ and Mᵣ are relative to ¹²C and have no units; M has units g mol⁻¹.
- % composition → divide by Aᵣ → divide by smallest → empirical formula; M ÷ empirical mass → molecular formula.
- Gases at the same T and p: volume ratio = mole ratio.
That's the notes covered.
Carry on to the next subtopic.