A chemist is determining the formulae of several substances.
The numbers of atoms in one molecule of compound P are shown in the table.
| Element | Number of atoms |
|---|---|
| Carbon | 6 |
| Hydrogen | 14 |
| Oxygen | 2 |
Write the molecular formula and the empirical formula of compound P.
Measurements give compound Q an empirical formula of NO2\mathrm{NO_{2}}NO2 and an Mr M_r\,Mr of 92.0.
Deduce the molecular formula of Q.
Relative atomic masses (ArA_rAr): N = 14.0, O = 16.0
An unknown metal, M, is present in a compound tested by a chemist. The results are shown in the table.
| Result | Value |
|---|---|
| Formula of the compound | M(NO3)2\mathrm{M(NO_{3})_{2}}M(NO3)2 |
| Relative formula mass of the compound | 189.4 |
| Solid formed when the compound is heated | MO\mathrm{MO}MO |
The equation for the reaction is
2M(NO3)2→2MO+4NO2+O2 \mathrm{2M(NO_{3})_{2} \rightarrow 2MO + 4NO_{2} + O_{2}} 2M(NO3)2→2MO+4NO2+O2| Metal | ArA_rAr |
|---|---|
| Magnesium | 24.3 |
| Calcium | 40.1 |
| Zinc | 65.4 |
| Barium | 137.3 |
Relative atomic masses (ArA_rAr): N = 14.0, O = 16.0
Identify metal M.
Calculate the relative formula mass of the solid formed in the reaction.
Nitrogen reacts with hydrogen.
N2+3H2→2NH3 \mathrm{N_{2} + 3H_{2} \rightarrow 2NH_{3}} N2+3H2→2NH3A student says, "I calculate the total for the products using 2×(14.0+2×1.0)2 \times (14.0 + 2 \times 1.0)2×(14.0+2×1.0). This does not match the total for the reactants, so mass is not conserved."
Evaluate the student's statement.
Relative atomic masses (ArA_rAr): N = 14.0, H = 1.0
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.