A laboratory technician checks the formulae and relative formula masses of several compounds.
The atom count for one molecule of compound R is recorded in the table.
| Element | Number of atoms |
|---|---|
| Carbon | 9 |
| Hydrogen | 18 |
| Oxygen | 3 |
Write the molecular formula and the empirical formula of compound R.
Compound S is a liquid used as a solvent. Its empirical formula is CH2Cl\mathrm{CH_{2}Cl}CH2Cl and its Mr M_r\,Mr is 99.0.
Deduce the molecular formula of S.
Relative atomic masses (ArA_rAr): C = 12.0, H = 1.0, Cl = 35.5
A sulfate of metal M has the formula MSO4\mathrm{MSO_{4}}MSO4 and an Mr M_r\,Mr of 151.9.
| Metal | ArA_rAr |
|---|---|
| Magnesium | 24.3 |
| Iron | 55.8 |
| Copper | 63.5 |
| Zinc | 65.4 |
Relative atomic masses (ArA_rAr): O = 16.0, S = 32.1
Identify metal M.
The same metal forms a chloride with the formula MCl2\mathrm{MCl_{2}}MCl2.
Calculate the Mr M_r\,Mr of MCl2\mathrm{MCl_{2}}MCl2.
Relative atomic mass (ArA_rAr): Cl = 35.5
A technician models a reaction used to remove carbon monoxide from an exhaust stream.
2CO+O2→2CO2 \mathrm{2CO + O_{2} \rightarrow 2CO_{2}} 2CO+O2→2CO2A student says, "The total for the reactants is (2×12.0)+16.0+32.0=72.0(2 \times 12.0) + 16.0 + 32.0 = 72.0(2×12.0)+16.0+32.0=72.0. The total for the products is (2×12.0)+(2×16.0)=56.0(2 \times 12.0) + (2 \times 16.0) = 56.0(2×12.0)+(2×16.0)=56.0, so the equation is not balanced."
Evaluate the student's statement.
Relative atomic masses (ArA_rAr): C = 12.0, O = 16.0
102 exam-style questions on OCR GCSE Chemistry 2.1 Purity and separating mixtures, covering 2.1.1 Pure substances and melting point data, 2.1.2 Relative formula mass and empirical formula, 2.1.3 Formulations as useful mixtures, 2.1.4 Techniques for separating mixtures, and 2.1.5 Chromatography and Rf values. Each one has a worked solution and a mark scheme showing where the marks go.