SC9 · Calculations involving masses · Topic 1 — Key concepts in chemistry | GCSE Edexcel Separate Sciences

Revise the key ideas

Relative formula mass, Mr, is the sum of relative atomic masses of all atoms in a formula. For molecules it can also be called relative molecular mass. Mr has no unit.
For water, H₂O: Mr = 2 × 1 + 16 = 18, using Ar(H) = 1 and Ar(O) = 16. Subscripts multiply the atom immediately before them.
Brackets multiply a whole group: Ca(OH)₂ has one calcium, two oxygen and two hydrogen atoms; Mr = 40 + 2 × (16 + 1) = 74.
Brackets multiply all atoms inside the group.
Percentage by mass of an element = total Ar contribution from that element ÷ Mr × 100. In water, oxygen contributes 16/18 × 100 ≈ 88.9%.
Use supplied Ar values rather than assuming a different precision from another table. Keep unrounded values during working and round at the end.
Do not confuse formula mass with an atom count: CO₂ contains three atoms, but Mr = 12 + 2 × 16 = 44.

A molecular formula gives actual numbers of atoms in one molecule. An empirical formula gives the simplest whole-number ratio of atoms of each element.
For butane, molecular formula C₄H₁₀ reduces to empirical formula C₂H₅. Some molecular formulae, such as H₂O, are already in the simplest ratio.
To obtain an empirical formula from masses, divide each element's mass by its Ar, then divide all results by the smallest to find the mole ratio.
If a ratio is 1:1.5, multiply all parts by 2 to obtain 2:3. Do not round 1.5 directly to 2 and change the composition.
For 10.0 g Ca and 17.8 g Cl with Ar 40 and 35.5, amounts are 0.25 and about 0.501 mol, approximately 1:2, giving CaCl₂. The supplied masses have rounding uncertainty.
Rounded experimental data support CaCl₂; divide masses by Ar before finding the ratio.
Percentages can be treated as masses in a hypothetical 100 g sample. Use the same divide-by-Ar method rather than using mass percentages directly as atom ratios.
Molecular formula = empirical formula multiplied by Mr(molecule) ÷ Mr(empirical formula). For CH₂ with molecular Mr 42, multiplier 42/14 = 3 and formula C₃H₆.

Weigh a dry crucible and lid, add cleaned magnesium ribbon and reweigh. The difference gives the starting magnesium mass.
Heat with the lid slightly lifted periodically to admit oxygen while limiting loss of magnesium oxide powder. Use tongs and eye protection; do not look directly at burning magnesium.
School practical schematic; teacher supervision, eye protection and tongs are needed.
Allow the crucible to cool before weighing, then repeat heating, cooling and weighing to constant mass. This supports completion of reaction and more reliable weighing.
Final oxide mass minus starting magnesium mass gives the mass of oxygen gained, after subtracting crucible and lid masses correctly.
Divide Mg mass by Ar(Mg) = 24 and O mass by Ar(O) = 16, then simplify the ratio. Ideal magnesium oxide has empirical formula MgO.
Product loss, incomplete oxidation and side reactions can affect the result. Explain the direction of an error using how it changes the calculated oxygen or magnesium amount.

Atoms are rearranged in chemical reactions, not created or destroyed, so total mass is conserved. A balanced equation has the same number of each atom on both sides.
In a closed system, measured total mass before and after a reaction is unchanged, including any precipitate or gas kept inside.
In an open flask, mass can appear to fall when a gas escapes. Burning magnesium can gain mass because oxygen from the air joins the product; neither observation breaks conservation.
For 2Mg + O₂ → 2MgO, formula-mass totals give 48 g magnesium + 32 g oxygen → 80 g magnesium oxide. Coefficients give particle or mole ratios, not equal gram ratios.
Coefficients are mole ratios; masses also depend on Mr.
Scale a reacting-mass ratio proportionally: 12 g Mg gives 20 g MgO if magnesium reacts completely and oxygen is in excess.
The limiting reactant is used up first and limits product formed. An excess reactant has some left over; use the limiting amount in the balanced equation.
To work out the reacting ratio (stoichiometry) from masses, convert each mass to moles and simplify the mole ratio. The masses alone usually do not give the numbers in front of the formulae in the balanced equation.

Concentration in g dm⁻³ = mass of dissolved solute in g ÷ volume of solution in dm³. Use solution volume, not just the starting solvent volume.
1 dm³ = 1000 cm³, so divide cm³ by 1000 before substituting. 250 cm³ is 0.250 dm³.
For 5.0 g solute in 250 cm³ solution: concentration = 5.0 ÷ 0.250 = 20 g dm⁻³.
Convert volume units before dividing.
Rearrange as mass = concentration × volume and volume = mass ÷ concentration, with consistent units.
Dilution adds solvent while solute amount remains unchanged, lowering concentration. If solution volume doubles with no solute lost, concentration halves.

A mole is a way of counting a very large number of particles. One mole contains about 6.02 × 10²³ particles: the Avogadro constant. State which particles you mean, such as atoms, molecules, ions or formula units.
Molar mass has units g mol⁻¹ and is numerically equal to the appropriate relative particle mass for GCSE calculations. One mole of H₂O has mass 18 g, not one molecule's mass.
Amount in mol, n = mass in g ÷ molar mass in g mol⁻¹. For 9 g water, n = 9 ÷ 18 = 0.5 mol.
Mass = amount × molar mass. For 0.25 mol CO₂, mass = 0.25 × 44 = 11 g.
Number of particles = amount × 6.02 × 10²³; amount = particle count ÷ 6.02 × 10²³. For 0.5 mol molecules there are 3.01 × 10²³ molecules.
Reverse each step to work backwards from a particle count.
In 2H₂ + O₂ → 2H₂O, the mole ratio is 2:1:2. Four moles H₂ need two moles O₂ and produce four moles water if reaction is complete.
For a limiting-reactant calculation, convert both masses to moles and compare with the equation ratio before calculating product. Excess material cannot create extra product after the limiting reactant is used.