Structure 1.4 Flashcards

6.02 × 10²³ particles per mole (in the IB data booklet; defined in SI as exactly 6.02214076 × 10²³ mol⁻¹). Historically, it was the number of atoms in exactly 12 g of carbon-12.

n = m / M, where n = moles, m = mass in grams, M = molar mass in g/mol.

The mass of one mole of a substance, in g/mol. Numerically equal to the relative molecular/formula mass.

22.7 dm³ mol⁻¹ (at 273.15 K and 100 kPa). This applies to all ideal gases.

n = V / 22.7, where V is volume in dm³ at STP.

Number of particles = \(n \times 6.02 \times 10^{23}\) (moles times Avogadro's constant).

The simplest whole-number ratio of atoms of each element in a compound. E.g. CH2O for glucose.

The actual number of atoms of each element in one molecule. E.g. C6H12O6 for glucose.

Divide the molar mass of the compound by the molar mass of the empirical formula. Multiply the empirical formula by this whole number.

The mass of each element in a compound expressed as a percentage of the total molar mass.

The reactant that is completely consumed first in a reaction, determining the maximum amount of product that can form.

Calculate moles of each reactant, divide by the stoichiometric coefficient, and the smallest value identifies the limiting reagent.

c = n / V, where c = concentration in mol dm⁻³, n = moles, V = volume in dm³.

Divide by 1000. So 250 cm³ = 0.250 dm³.

A solution of known, precisely determined concentration, prepared by dissolving an accurate mass of solute and making up to an exact volume of solution in a volumetric flask.

c1V1 = c2V2, where c1 and V1 are initial concentration and volume, c2 and V2 are final.

The number in front of a formula in a balanced equation, showing the molar ratio of reactants and products.

Percentage yield = (actual yield / theoretical yield) x 100%. In practice it is often less than 100% due to side reactions, mechanical transfer losses, or incomplete reactions (apparent yields > 100% usually indicate wet or impure product).