C04a) CHEMISCAL CALCULATIONS
Conservation of mass and balanced symbol equations
Conservation of mass | No atoms are lost or made during a chemical reaction | Mass of the products equals the mass of the reactants. |
Relative formula mass (Mr)
Mass changes when a reactant or product is a gas
Moles (HT only)
Amounts of substances in equations (HT only)
Using moles to balance equations (HT only)
Limiting reactants
(HT only)
Balanced symbol equations | Represent chemical reactions and have the same number of atoms of each element on both sides of the equation | H2 + Cl2 🡪 2HCl Subscript numbers show the number of atoms of the element to its left. Normal script numbers show the number of molecules. |
Subscript Normal script
Mr | The sum of the relative atomic masses of the atoms in the numbers shown in the formula | The sum of the Mr of the reactants in the quantities shown equals the sum of the Mr of the products in the quantities shown. | 2Mg + O2 🡪 2MgO
48g + 32g = 80g 80g = 80g |
Mass appears to increase during a reaction | One of the reactants is a gas | Magnesium + oxygen 🡪 magnesium oxide |
Mass appears to decrease during a reaction | One of the products is a gas and has escaped | Calcium carbonate 🡪 carbon dioxide + calcium oxide |
Chemical amounts are measured in moles (mol) | Mass of one mole of a substance in grams = relative formula mass | One mole of H2O = 18g (1 + 1 + 16) One mole of Mg = 24g |
Avogadro constant | One mole of any substance will contain the same number of particles, atoms, molecules or ions. | 6.02 x 1023 per mole
One mole of H2O will contain 6.02 x 1023 molecules One mole of NaCl will contain 6.02 x 1023 Na+ ions |
Number of moles = mass (g) or mass (g) Ar Mr | How many moles of sulfuric acid molecules are there in 4.7g of sulfuric acid (H2SO4)? Give your answer to 1 significant figure. 4.7 = 0.05 mol 98 (Mr of H2SO4) |
The reactant that is completely used up | Limits the amount of product that is made | Less moles of product are made. |
Concentration of solutions
The balancing numbers in a symbol equation can be calculated from the masses of reactants and products | Convert the masses in grams to amounts in moles and convert the number of moles to simple whole number ratios. |
Chemical equations show the number of moles reacting and the number of moles made | Mg + 2HCl 🡪 MgCl2 + H2 One mole of magnesium reacts with two moles of hydrochloric acid to make one mole of magnesium chloride and one mole of hydrogen | If you have a 60g of Mg, what mass of HCl do you need to convert it to MgCl2? Ar : Mg =24 so mass of 1 mole of Mg = 24g Mr : HCl (1 + 35.5) so mass of 1 mole of HCl = 36.5g So 60g of Mg is 60/24 = 2.5 moles Balanced symbol equation tells us that for every one mole of Mg, you need two moles of HCl to react with it. So you need 2.5x2 = 5 moles of HCl You will need 5 x 36.5g of HCl= 182.5g |
Measured in mass per given volume of solution (g/dm3) | Conc. = mass (g) . volume (dm3) | HT only Greater mass = higher concentration. Greater volume = lower concentration. |
Chemical measurements
Whenever a measurement is taken, there is always some uncertainty about the result obtained | Can determine whether the mean value falls within the range of uncertainty of the result |
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C04b) CHEMICAL CALCULATIONS
Yield is the amount of product obtained | It is not always possible to obtain the calculated amount of a product | The reaction may not go to completion because it is reversible. |
Some of the product may be lost when it is separated from the reaction mixture. | ||
Some of the reactants may react in ways different to the expected reaction. |
Percentage yield
Atom economy
Using concentrations of solutions in mol/dm3
(HT only, chemistry only)
Use of amount of substance in relation to volumes of gases
(HT only, chemistry only)
Percentage yield is comparing the amount of product obtained as a percentage of the maximum theoretical amount | % Yield = Mass of product made x 100 Max. theoretical mass | A piece of sodium metal is heated in chlorine gas. A maximum theoretical mass of 10g for sodium chloride was calculated, but the actual yield was only 8g. Calculate the percentage yield.
Percentage yield = 8/10 x 100 =80% |
HT only: 200g of calcium carbonate is heated. It decomposes to make calcium oxide and carbon dioxide. Calculate the theoretical mass of calcium oxide made.
CaCO3 🡪 CaO + CO2 Mr of CaCO3 = 40 + 12 + (16x3) = 100 Mr of CaO = 40 + 16 = 56 100g of CaCO3 would make 56 g of CaO So 200g would make 112g |
A measure of the amount of starting materials that end up as useful products | Atom economy = Relative formula mass of desired product from equation x 100 Sum of relative formula mass of all reactants from equation | High atom economy is important or sustainable development and economic reasons |
Calculate the atom economy for making hydrogen by reacting zinc with hydrochloric acid: Zn + 2HCl → ZnCl2 + H2 Mr of H2  = 1 + 1 = 2 Mr of Zn + 2HCl = 65 + 1 + 1 + 35.5 + 35.5 = 138 Atom economy = 2∕138  × 100 = 2∕138 × 100 = 1.45% This method is unlikely to be chosen as it has a low atom economy. |
Concentration of a solution is the amount of solute per volume of solution | Concentration  = amount (mol) (mol/dm3) volume (dm3) | What is the concentration of a solution that has 35.0g of solute in 0.5dm3 of solution? 35/0.5 = 70 g/dm3 |
Titration | If the volumes of two solutions that react completely are known and the concentrations of one solution is known, the concentration of the other solution can be calculated. | 2NaOH(aq) + H2SO4(aq)→ Na2S04(aq) + 2H2O(l) It takes 12.20cm3 of sulfuric acid to neutralise 24.00cm3 of sodium hydroxide solution, which has a concentration of 0.50mol/dm3. Calculate the concentration of the sulfuric acid in mol/dm3: 0.5 mol/dm3 x (24/1000) dm3 = 0.012 mol of NaOH The equation shows that 2 mol of NaOH reacts with 1 mol of H2SO4, so the number of moles in 12.20cm3 of sulfuric acid is (0.012/2) = 0.006 mol of sulfuric acid Calculate the concentration of sulfuric acid in mol/ dm3 0.006 mol x (1000/12.2) dm3 =0.49mol/dm3 |
Calculate the concentration of sulfuric acid in g/ dm3: H2SO4 = (2x1) + 32 + (4x16) = 98g 0.49 x 98g = 48.2g/dm3 |
Equal amounts of moles or gases occupy the same volume under the same conditions of temperature and pressure | The volume of one mole of any gas at room temperature and pressure (20°C and 1 atmospheric pressure) is 24 dm3 | No. of moles of gas = vol of gas (dm3) 24dm3 |
What is the volume of 11.6 g of butane (C4H10) gas at RTP? Mr : (4 x 12) + (10 x 1) =Â 58 11.6/58 =Â 0.20 mol Volume = 0.20 x 24 =Â 4.8 dm3 |
6g of a hydrocarbon gas had a volume of 4.8 dm3. Calculate its molecular mass. 1 mole = 24 dm3, so 4.8/24 = 0.2 mol Mr = 6 / 0.2 = 30 If 6g = 0.2 mol, 1 mol equals 30 g |