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| Formulae for calculating Mr, Moles & Mass | |
| Relative atomic mass (Mr) | Weighted average mass of atom of element __________________________________________ 1/12 mass of Carbon 12 (12C =1) |
| Explain the meaning of the numbers in the above graphic. | 6 = Atomic number = Number of Protons. 12.011 = Mass number (Ar) = Protons+Neutrons (not an exact number due to weighted averaging of Isotopes of element). 2 4 = number of electrons in 1st & 2nd shell. |
| Mass number (Ar) | = Protons + Neutrons (different for each isotope) |
| Mr = Relative formula mass = relative molecular mass | = the sum of all the atomic masses for all the atoms in a given formula |
| For an isotope, define the numbers in the adjacent graphic | 12 = Mass number (Ar) = exact number of Protons + Neutrons for the isotope. 6 = Atomic/proton number. |
| Molecular/formula mass calculation Example: The diatomic molecules of the elements hydrogen H2 and chlorine Cl2 | Formula masses, RMM or Mr: relative molecular mass for hydrogen H2 = 2 x 1 = 2 relative molecular mass for chlorine Cl2 = 2 x 35.5 = 71 respectively. |
| Molecular/formula mass calculation Example: The compound water H2O | Relative atomic masses are H=1 and O=16 relative molecular mass or Mr = (1x2) + 16 = 18 (molecular mass of water). |
| Law of conservation of mass calculation Example: Magnesium + Oxygen ==> Magnesium oxide 2Mg + O2 ==> 2MgO (atomic masses required: Mg=24 and O=16) | The mass changes in the reaction are: (2 x 24) + (2 x 16) = 2 x (24 + 16) 48 + 32 = 2 x 40 and so 80 mass units of reactants produces 80 mass units of products. You can work with any mass units such as g, kg or tonne (1 tonne = 1000 kg), as long as you use the same units for all the masses involved. |
| Law of conservation of mass calculation Example: When limestone (calcium carbonate) is strongly heated, it undergoes thermal decomposition to form lime (calcium oxide) and carbon dioxide gas. CaCO3 ==> CaO + CO2 (relative atomic masses: Ca = 40, C = 12 and O = 16) Calculate the mass of calcium oxide and the mass of carbon dioxide formed by decomposing 50 tonnes of calcium carbonate. | (40 + 12 + 3x16) ==> (40 + 16) + (12 + 2x16) 100 ==> 56 + 44 scaling down by a factor of two gives 50 ==> 28 + 22 so decomposing 50 tonnes of limestone produces 28 tonnes of lime and 22 tonnes of carbon dioxide gas. |
| Formulae for Hydrochloric Acid, Nitric Acid and Sulphuric Acid | HCl, HNO3, H2SO4 |
| Diatomic molecules of elements | H2, O2, N2, - F2, Cl2, Br2, I2. |
| Formuae for Methane and Ammonia | CH4 and NH3 |
| Calculation of % composition Example: Calculate the % of copper in copper sulphate, CuSO4 Relative atomic masses: Cu = 64, S = 32 and O = 16 | Relative formula mass = 64 + 32 + (4x16) = 160 only one copper atom of relative atomic mass 64 % Cu = 100 x 64 / 160 = 40% copper by mass in the compound. |
| Calculate the % of water in hydrated magnesium sulphate MgSO4.7H2O Relative atomic masses: Mg = 24, S = 32, O = 16 and H = 1 | Relative formula mass = 24 + 32 + (4 x 16) + [7 x (1 + 1 + 16)] = 246 7 x 18 = 126 is the mass of water so % water = 100 x 126 / 246 = 51.2 % H2O |
| The empirical formula of a compound is? | The simplest whole number ratio of atoms present in a compound. e.g. butane molecular formula C4H10, empirical formula C2H5 |
| Empirical formula calculation Example: The empirical formula of a lead oxide. It is found that 207g of lead combined with oxygen to form 239g of a lead oxide. Work out the formula of the lead oxide. (Relative atomic masses: Pb = 207 and O = 16) | Work out the amount of oxygen combined with the lead. By simple logic from the law of conservation of mass, this is 239 - 207 = 32g In atomic ratio terms, the 207 is equivalent to 1 atom of lead and the 32 is equivalent to 2 atoms of oxygen (1 x 207 to 2 x 16), so the formula is simply PbO2 |
| Empirical formula calculation Example: The analysis of sodium sulphate, calculating its empirical formula from the % composition by mass. On analysis, the salt sodium sulfate was found to contain 32.4% sodium, 22.5% sulphur and 45.1% oxygen (atomic masses: Na = 23, S = 32 and O = 16 | |
| The formulae to calculate Molarity (concentration) of a solution | |
| The formulae to calculate Moles and Volume of a gas. | Volume = Moles x 24,000 cm3 |
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