AQA AS Chemistry 1.2 Formulae, Equations & Calculations PDF

Head to www.savemyexams.com for more awesome resources AQA AS Chemistry Your notes 1.2 Formulae, Equations & Calculations Contents 1.2.1 Relative Atomic Mass & Relative Molecular Mass 1.2.2 Empirical & Mole...

Head to www.savemyexams.com for more awesome resources AQA AS Chemistry Your notes 1.2 Formulae, Equations & Calculations Contents 1.2.1 Relative Atomic Mass & Relative Molecular Mass 1.2.2 Empirical & Molecular Formula 1.2.3 Balanced Equations 1.2.4 Reaction Yields 1.2.5 Atom Economy 1.2.6 Hydrated Salts Page 1 of 21 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources 1.2.1 Relative Atomic Mass & Relative Molecular Mass Your notes Relative Atomic Mass Definition What is Relative Atomic Mass? Atomic Mass Unit The mass of a single atom is so small that it is impossible to weigh it directly Atomic masses are therefore defined in terms of a standard atom which is called the unified atomic mass unit This unified atomic mass is defined as one-twelfth of the mass of a carbon-12 isotope The symbol for the unified atomic mass is u (often Da, Dalton, is used as well) 1 u = 1.66 x 10-27 kg Relative Atomic Mass Definition The relative atomic mass (Ar) of an element is the ratio of the average mass of the atoms of an element to the unified atomic mass unit The relative atomic mass is determined by using the average mass of the isotopes of a particular element The Ar has no units as it is a ratio and the units cancel each other out Relative atomic mass of X = a vera g e ma s s of on e atom of X on e twelfth of th e ma s s of on e carbon − 12 atom Relative isotopic mass The relative isotopic mass is the mass of a particular atom of an isotope compared to the value of the unified atomic mass unit Atoms of the same element with a different number of neutrons are called isotopes Isotopes are represented by writing the mass number as 20Ne, or neon-20 or Ne-20 To calculate the average atomic mass of an element the percentage abundance is taken into account Multiply the atomic mass by the percentage abundance for each isotope and add them all together Divide by 100 to get average relative atomic mass This is known as the weighted average of the masses of the isotopes Relative molecular mass, Mr Page 2 of 21 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources The relative molecular mass (Mr) is the ratio of weighted average mass of a molecule of a molecular compound to the unified atomic mass unit The Mr has no units Your notes Mr = weighte d a vera g e ma s s of mole cule s in a give n sa mple of a mole cular compound unifie d atomic ma s s unit The Mr can be found by adding up the relative atomic masses of all atoms present in one molecule When calculating the Mr the simplest formula for the compound is used, also known as the formula unit Eg. silicon dioxide has a giant covalent structure, however the simplest formula (the formula unit) is SiO2 Example Mr calculations Substance Atoms present Mr Hydrogen 2xH (2 x 1.0) = 2.0 H2 Water (2 x H) + (1 x O) (2 x 1.0) + (1 x 16.0) = 18.0 H2 O Potassium carbonate (2 x 39.1) + (1 x 12.0) (2 x K) + (1 x C) + (3 x O) K2CO3 + (3 x 16.0) = 138.2 Calcium hydroxide (1 x 40.1) + (2 x 16.0) (1 x Ca) + (2 x O) + (2 x H) Ca(OH)2 + (2 x 1.0) = 74.1 Ammonium sulfate (2 x 14.0) + (8 x 1.0) + (1 x 32.1) + (4 x 16.0) (2 x N) + (8 x H) + (1 x S) + (4 x O) (NH4)2SO4 = 132.1 Relative formula mass, Mr The relative formula mass (Mr) is used for compounds containing ions It has the same units and is calculated in the same way as the relative molecular mass In the table above, the Mr for potassium carbonate, calcium hydroxide and ammonium sulfates are relative formula masses Page 3 of 21 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources 1.2.2 Empirical & Molecular Formula Your notes Empirical & Molecular Formulae The molecular formula is the formula that shows the number and type of each atom in a molecule E.g. the molecular formula of ethanoic acid is C2H4O2 The empirical formula is the simplest whole number ratio of the elements present in one molecule or formula unit of the compound E.g. the empirical formula of ethanoic acid is CH2O Organic molecules often have different empirical and molecular formulae Simple inorganic molecules however have often similar empirical and molecular formulae Ionic compounds always have similar empirical and molecular formulae Page 4 of 21 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Empirical & Molecular Formulae Calculations Empirical formula Your notes Empirical formula is the simplest whole number ratio of the elements present in one molecule or formula unit of the compound It is calculated from knowledge of the ratio of masses of each element in the compound The empirical formula can be found by determining the mass of each element present in a sample of the compound It can also be deduced from data that gives the percentage compositions by mass of the elements in a compound Worked example Calculating empirical formula from mass Determine the empirical formula of a compound that contains 2.72 g of carbon and 7.28 g of oxygen. Answer: Elements Carbon Oxygen Mass of each element 2.72 7.28 (g) Atomic mass 12.0 16.0 Moles = mass / Ar 2. 72 = 0.227 7. 28 = 0.455 12. 0 16. 0 Ratio (divide by smallest value) 0. 227 0. 455 =1 =2 0. 227 0. 227 So, the empirical formula of the compound is CO2 The above example shows how to calculate empirical formula from the mass of each element present in the compound The example below shows how to calculate the empirical formula from percentage composition Page 5 of 21 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Worked example Your notes Calculating empirical formula from percentage Determine the empirical formula of a hydrocarbon that contains 90.0% carbon and 10.0% hydrogen. Answer: Elements Carbon Hydrogen Mass of each element 90.0 10.0 (g) Atomic mass 12.0 1.0 Moles = mass / Ar 90. 0 10. 0 = 7.5 = 10.0 12. 0 1. 0 Ratio (divide by smallest value) 7. 5 10. 0 =1 = 1.33 7. 5 7. 5 Convert to whole number ratio 1x3=3 1.33 x 3 = 4 (x3 for this example) So, the empirical formula of the compound is C3H4 Molecular formula The molecular formula gives the exact numbers of atoms of each element present in the formula of the compound The molecular formula can be found by dividing the relative formula mass of the molecular formula by the relative formula mass of the empirical formula Multiply the number of each element present in the empirical formula by this number to find the molecular formula Page 6 of 21 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Worked example Your notes Calculating molecular formula The empirical formula of X is C4H10S and the relative molecular mass of X is 180 What is the molecular formula of X? (Ar data: C = 12, H = 1, S = 32) Answer: Step 1: Calculate relative mass of the empirical formula Relative empirical mass = (C x 4) + (H x 10) + (S x 1) Relative empirical mass = (12 x 4) + (1 x 10) + (32 x 1) Relative empirical mass = 90 Step 2: Divide relative formula mass of X by relative empirical mass Ratio between Mr of X and the Mr of the empirical formula = 180/90 Ratio between Mr of X and the Mr of the empirical formula = 2 Step 3: Multiply each number of elements by 2 (C4 x 2) + (H10 x 2) + (S x 2) = (C8) + (H20) + (S2) Molecular Formula of X is C8H20S2 Page 7 of 21 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Worked example Your notes Calculating empirical formula and molecular formula Analysis of a compound X shows that it contains 24.2 % by mass of carbon, 4.1 % by mass of hydrogen and 71.7% by mass of chlorine. Calculate the empirical formula of X. Use this empirical formula and the relative molecular mass of X (Mr = 99.0) to calculate the molecular formula of X. Answer: Elements Carbon Hydrogen Chlorine Value (g or %) 24.2 4.1 71.7 Atomic mass 12.0 1.0 35.5 24. 2 4. 1 71. 7 Moles = mass / Ar = 2.02 = 4.1 = 2.02 12. 0 1. 0 35. 5 2. 02 4. 1 2. 02 Ratio (divide by smallest) =1 =2 =1 2. 02 2. 02 2. 02 So, the empirical formula of compound X is CH2Cl The relative formula mass of the empirical formula is: Relative formula mass = (1 x C) + (2 x H) + (1 x Cl) Relative formula mass = (1 x 12.0) + (2 x 1.0) + (2 x 35.5) Relative formula mass = 49.5 Divide the relative formula mass of X by the relative formula mass of the empirical formula Ratio between Mr of X and the Mr of the empirical formula = 99.0/45.9 Ratio between Mr of X and the Mr of the empirical formula = 2 Multiply each number of elements by 2 (C1 x 2) + (H2 x 2) + (Cl1 x 2) = (C2) + (H4) + (Cl2) The molecular formula of X is C2H4Cl2 Page 8 of 21 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources 1.2.3 Balanced Equations Your notes Deducing Formulae of Compounds Ionic compounds are formed from a metal and a non-metal bonded together Ionic compounds are electrically neutral; the positive charges equal the negative charges Charges on positive ions All metals form positive ions There are also some non-metal positive ions, such as ammonium, NH4+, and hydrogen, H+ The metals in Group 1, Group 2 and Group 3 (13) have a charge of 1+ and 2+ and 3+ respectively The charge on the ions of the transition elements can vary which is why Roman numerals are often used to indicate their charge Roman numerals are used in some compounds formed from transition elements to show the charge (or oxidation state) of metal ions E.g. in copper (II) oxide, the copper ion has a charge of 2+ whereas in copper (I) nitrate, the copper has a charge of 1+ Non-metal ions The non-metals in group 15 to 17 have a negative charge and have the suffix ‘ide’ E.g. nitride, chloride, bromide, iodide Elements in group 17 gain 1 electron so have a 1- charge, eg. Br- Elements in group 16 gain 2 electrons so have a 2- charge, eg. O2- Elements in group 15 gain 3 electrons so have a 3- charge, eg. N3- There are also polyatomic negative ions, which are negative ions made up of more than one type of atom Page 9 of 21 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Your notes The charges of simple ions depend on their position in the Periodic Table Formulae of Ionic Compounds Table Ion Formula Silver(I) Ag+ Ammonium NH4+ Zinc(II) Zn2+ Hydroxide OH– Nitrate NO3– Sulfate SO42– Carbonate CO32– Hydrogen carbonate HCO3– Phosphate PO43– Page 10 of 21 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Worked example Your notes Formulae Determine the formulae of the following ionic compounds 1. Magnesium chloride 2. Iron(III) oxide 3. Aluminium nitrate Answer 1: Magnesium chloride Magnesium is in Group 2 so has a charge of 2+ Chlorine is in group 17 so has a charge of 1- Magnesium needs two chloride ions for each magnesium ion to be balanced so the formula is MgCl2 Answer 2: Iron (III) oxide The Roman numeral states that iron has a charge of 3+ Oxygen is in group 16 so has a charge of 2- The charges need to be equal so 2 iron ions to 3 oxide ions will balance electrically, so the formula is Fe2O3 Answer 3: Aluminum nitrate Aluminium is in group 13 so has a charge of 3+ Nitrate is a polyatomic ion and has a charge of 1- The polyatomic ion needs to be placed in a bracket if more than 1 is needed The formula of aluminium nitrate is Al(NO3)3 Examiner Tip Remember: Polyatomic ions are ions that contain more than one type of element, such as OH- Page 11 of 21 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Balancing Equations A symbol equation is a shorthand way of describing a chemical reaction using chemical symbols to Your notes show the number and type of each atom in the reactants and products A word equation is a longer way of describing a chemical reaction using only words to show the reactants and products Balancing equations During chemical reactions, atoms cannot be created or destroyed The number of each atom on each side of the reaction must therefore be the same E.g. the reaction needs to be balanced When balancing equations remember: Not to change any of the formulae To put the numbers used to balance the equation in front of the formulae To balance firstly the carbon, then the hydrogen and finally the oxygen in combustion reactions of organic compounds When balancing equations follow the following the steps: Write the formulae of the reactants and products Count the numbers of atoms in each reactant and product Balance the atoms one at a time until all the atoms are balanced Use appropriate state symbols in the equation The physical state of reactants and products in a chemical reaction is specified by using state symbols (s) solid (l) liquid (g) gas (aq) aqueous Page 12 of 21 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Ionic Equations Ionic equations Your notes In aqueous solutions ionic compounds dissociate into their ions Many chemical reactions in aqueous solutions involve ionic compounds, however only some of the ions in solution take part in the reactions The ions that do not take part in the reaction are called spectator ions An ionic equation shows only the ions or other particles taking part in a reaction, and not the spectator ions Worked example Balance the following equation: magnesium + oxygen → magnesium oxide Answer: Step 1: Write out the symbol equation showing reactants and products Mg + O2 → MgO Step 2: Count the number of atoms in each reactant and product Mg O Reactants 1 2 Products 1 1 Step 3: Balance the atoms one at a time until all the atoms are balanced 2Mg + O2 → 2MgO This is now showing that 2 moles of magnesium react with 1 mole of oxygen to form 2 moles of magnesium oxide Step 4: Use appropriate state symbols in the fully balanced equation 2Mg (s) + O2 (g) → 2MgO (s) Page 13 of 21 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Worked example Your notes 1. Balance the following equation: zinc + copper(II) sulfate → zinc(II) sulfate + copper 2. Write the ionic equation for the above reaction. Answer 1: Step 1: To balance the equation, write out the symbol equation showing reactants and products Zn + CuSO4 → ZnSO4 + Cu Step 2: Count the number of atoms in each reactant and product. The equation is already balanced Zn Cu S O Reactants 1 1 1 4 Products 1 1 1 4 Step 3: Use appropriate state symbols in the equation Zn (s) + CuSO4 (aq) → ZnSO4 (aq) + Cu (s) Answer 2: Step 1: The full chemical equation for the reaction is Zn (s) + CuSO4 (aq) → ZnSO4 (aq) + Cu (s) Step 2: Break down reactants into their respective ions Zn (s) + Cu2+ SO42- (aq) → Zn2+SO42- (aq) + Cu (s) Step 3: Cancel the spectator ions on both sides to give the ionic equation Zn (s) + Cu2+SO42- (aq) → Zn2+SO42- (aq) + Cu (s) Zn (s) + Cu2+(aq) → Zn2+ (aq) + Cu (s) Page 14 of 21 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources 1.2.4 Reaction Yields Your notes Percentage Yield In a lot of reactions, not all reactants react to form products which can be due to several factors: Other reactions take place simultaneously The reaction does not go to completion Reactants or products are lost to the atmosphere The percentage yield shows how much of a particular product you get from the reactants compared to the maximum theoretical amount that you can get: percentage yield = actual yield theoretical yield Where actual yield is the number of moles or mass of product obtained experimentally The predicted yield is the number of moles or mass obtained by calculation You will often have to use the following equation to work out the reacting masses, to calculate the predicted yield number of mol = mass of a substance in grams g ( ) molar mass g mol − 1 ( ) It is important to be clear about the type of particle you are referring to when dealing with moles Eg. 1 mole of CaF2 contains one mole of CaF2 formula units, but one mole of Ca2+ and two moles of F- ions Page 15 of 21 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Worked example Your notes Calculate % yield using moles In an experiment to displace copper from copper(II) sulfate, 6.54 g of zinc was added to an excess of copper(II) sulfate solution. The copper was filtered off, washed and dried. The mass of copper obtained was 4.80 g. Calculate the percentage yield of copper. Answer: 1. Write the balanced symbol equation: Zn (s) + CuSO4 (aq) → ZnSO4 (aq) + Cu (s) 2. Calculate the number of moles of zinc: 6. 54 g n(Zn) = = 0.10 moles 65. 4 g mol −1 3. Deduce the number of moles of copper, using the balanced chemical equation: 1 mole of zinc forms 1 mole of copper The ratio is 1 : 1 Therefore, n(Cu) = 0.10 moles 4. Calculate the maximum mass (theoretical yield) of copper: Mass = mol x Mr Mass = 0.10 mol x 63.5 g mol-1 Mass = 6.35 g 5. Calculate the percentage yield of copper: 4. 80 g Percentage yield = x 100 = 75.6 % 6. 35 g Page 16 of 21 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Limiting & Excess Reagents Limiting & Excess reagents Your notes Sometimes, there is an excess of one or more of the reactants (excess reagent) The reactant which is not in excess is called the limiting reagent To determine which reactant is limiting: The number of moles of each reactant should be calculated The ratio of the reactants shown in the equation should be taken into account e.g. 2Na + S → Na2S Here, the ratio of Na : S is 2 : 1, and this should be taken into account when doing calculations Once all of one reactant has been used up, the reaction will stop, even if there are moles of the other reactant(s) leftover The reactant leftover is in excess, the reactant which causes the reaction to stop once it is used up is the limiting reagent Worked example Excess & limiting reagent 9.2 g of sodium is reacted with 8.0 g of sulfur to produce sodium sulfide, Na2S. Which reactant is in excess and which is the limiting reactant? Answer Step 1: Calculate the moles of each reactant 9. 2 g mol(Na) = = 0.40 mol 23 g mol −1 8. 0 mol(S) = = 0.25 mol 32 g mol −1 Step 2: Write the balanced equation and determine the molar ratio 2Na + S → Na2S The molar ratio of Na: Na2S is 2:1 Step 3: Compare the moles and determine the limiting reagent So, to react completely 0.40 moles of Na require 0.20 moles of S Since there are 0.25 moles of S, then S is in excess Na is therefore the limiting reactant. Once all of the Na has been used up, the reaction will stop, even though there is S left Page 17 of 21 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources 1.2.5 Atom Economy Your notes Atom Economy The atom economy of a reaction shows how many of the atoms used in the reaction become the desired product The rest of the atoms or mass is wasted It is found directly from the balanced equation by calculating the Mr of the desired product percentage atom economy = molecular mass of desired product × 100 sum of molecular masses of all reactants In addition reactions, the atom economy will always be 100%, because all of the atoms are used to make the desired product Whenever there is only one product, the atom economy will always be 100% For example, in the reaction between ethene and bromine: CH2=CH2 + Br2 → CH2BrCH2Br The atom economy could also be calculated using mass, instead or Mr In this case, you would divide the mass of the desired product formed by the total mass of all reactants, and then multiply by 100 Efficient processes have high atom economies and are important to sustainable development They use fewer resources Create less waste Worked example Qualitative atom economy Ethanol can be produced by various reactions, such as: Hydration of ethene: C2H4 + H2O → C2H5OH Substitution of bromoethane: C2H5Br + NaOH → C2H5OH + NaBr Explain which reaction has a higher atom economy. Answer: The hydration of ethene has a higher atom economy The atom economy is 100 % This is because all of the reactants are converted into products Whereas the substitution of bromoethane produces NaBr as a waste product Page 18 of 21 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Worked example Your notes Quantitative atom economy The blast furnace uses carbon monoxide to reduce iron(III) oxide to iron. Fe2O3 + 3CO → 2Fe + 3CO2 Calculate the atom economy for this reaction, assuming that iron is the desired product. Answer: 1. Write the equation: Atom economy = molecular mass of desired product × 100 sum of molecular masses of all reactants 2. Calculate the relevant atomic / molecular masses: Fe2O3 = (2 x 55.8) + (3 x 16.0) = 159.6 CO = (1 x 12.0) + (1 x 16.0) = 28.0 Fe = 55.8 The Mr of CO2 is not required as it is not the desired product 3. Substitute values and evaluate: 2 × 55. 8 Atom economy = x 100 159. 6 + ( 3 × 28. 0 ) Atom economy = 45.8 % Examiner Tip Careful: Sometimes a question may ask you to show your working when calculating atom economy. In this case, even if it is an addition reaction and it is obvious that the atom economy is 100%, you will still need to show your working. Page 19 of 21 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources 1.2.6 Hydrated Salts Your notes Water of Crystallisation Water of crystallisation is when some compounds can form crystals which have water as part of their structure A compound that contains water of crystallisation is called a hydrated compound The water of crystallisation is separated from the main formula by a dot when writing the chemical formula of hydrated compounds E.g. hydrated copper(II) sulfate is CuSO4∙5H2O A compound which doesn’t contain water of crystallisation is called an anhydrous compound E.g. anhydrous copper(II) sulfate is CuSO4 A compound can be hydrated to different degrees E.g. cobalt(II) chloride can be hydrated by six or two water molecules CoCl2 ∙6H2O or CoCl2 ∙2H2O The conversion of anhydrous compounds to hydrated compounds is reversible by heating the hydrated salt: Hydrated: CuSO4 5H2O ⇌ CuSO4 + 5H2O :Anhydrous The degree of hydration can be calculated from experimental results: The mass of the hydrated salt must be measured before heating The salt is then heated until it reaches a constant mass The two mass values can be used to calculate the number of moles of water in the hydrated salt - known as the water of crystallisation Page 20 of 21 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Worked example Your notes Calculating water of crystallisation 11.25 g of hydrated copper sulfate, CuSO4.xH2O, is heated until a constant mass of 7.19 g. Calculate the formula of the hydrated copper(II) sulfate. Ar (Cu) = 63.5 Ar (S) = 3. Ar (O) = 16 Ar (H) = 1 Answer: 1. Salt and water CuSO4 H 2O 11.25 - 7.19 2. Value 7.19 = 4.06 63.5 + 32 + (16 x 4) (1 x 2) + 16 3. Mr = 159.5 = 18 mass 7. 19 4. 06 4. Moles = = 0.045 = 0.226 Mr 159. 5 18 0. 045 0. 226 5. Salt : water ratio =1 =5 0. 045 0. 045 The formula is 6. Formula of hydrated salt CuSO4 5H2O Examiner Tip A water of crystallisation calculation can be completed in a similar fashion to an empirical formula calculation Instead of elements, you start with the salt and water Instead of dividing by atomic masses, you divide by molecular / formula masses The rest of the calculation works the same way as the empirical formula calculation Page 21 of 21 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers