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1 Book 2 - Amount of Substance ...................The MOLE!!! 2 Valency This is a number assigned to each element that indicates the number of electrons that it has lost/gained in order to become stable. Valencies are used to write chemical formulae and to determine the ions formed Sodium tends to lose electron where possible. We know this because sodium is in group so has electron in its outer shell. We would say that sodium has a VALENCY of 1 Valencies of elements in groups Group 1 2 3 4 5 6 7 Number of outer shell electrons 1 2 3 4 5 6 7 Valency Ion formed Name of ion 1 2 3 4 3 2 1 +1 +2 +3 -3 -2 -1 Sodium Magnesium Aluminium Nitride Sulphide chloride Group 8 elements are called the noble gases. They do not react with other elements as they have a full octet Valencies of elements not in groups (transition metals) Metal Copper Zinc Iron Manganese Chromium Silver Valency 1 or 2 2 2 or 3 2, 4 or 7 3 or 6 1 Ion formed/ name Cu+, copper (I), Cu2+, copper (II) Zn2+, Zinc Fe2+, Iron (II), Fe3+, Iron (III) Mn2+, Manganese (II) Cr3+, Chromium (III) Ag+, Silver Learn these!! Complex ions contain more than one element, you will probably be familiar with some of these. You must learn the formula, charge & name of the ion. 3 Name formula Valency Carbonate Hydrogencarbonate Sulphate Sulphite Hydrogensulphate Nitrate Hydroxide Dichromate Manganate Ammonium Phosphate CO32HCO3SO42SO32HSO4NO3OHCr2O72MnO4NH4+ PO43- 2 1 2 2 1 1 1 2 1 1 3 complete the following table Name iodide Suphate sulphide aluminium dichromate Formula of ion ISO42S2Al3+ Cr2O72- Valency 1 2 2 3 2 Name Formula of ion magnesium Mg2+ ammonium NH4+ nitride N3hydroxide OHIron (II) Fe3+ Valency 2 1 3 1 3 When we know what ion an element forms we can write equations to show the process. These are called half ionic equations e.g. magnesium Mg Mg2+ + 2ewrite ½ eqns to show the following forming ions a) Lithium Li b) Chlorine Cl2 + c) Sulphur S + Li+ 2e2e- e- + 2Cl- S2- Reading and writing chemical formulae 4 Naming compounds Rules 1. metal name goes first and doesn’t change 2. non-metal elements become ‘ides’ to show that they are negative ions e.g. chloride 3. ate means oxygen is present e.g. sulphate 4. ite also means oxygen present but LESS than ate 5. when non-metals form positive ions they are called ‘ium’ e.g. ammonium 6. if a metal has more than one valency it has a Roman numeral in the name of the compound to show which ion is present. e.g. Cu2O = copper (I) oxide Reading Formula K2O CaSO4 Mg(NO3)2 CuSO4 .5H2O Na2CO3.10H2O Name of compound Potassium oxide Calcium sulphate Magnesium nitrate Copper (II) sulphate Sodium carbonate Composition 2K and 1 O 1 Ca, 1 S and 4 O 1 Mg, 2N and 6 O 1 Cu, 1S 9O and 10 H 2 Na, 1C, 13 O and 20 H Writing There are two methods, one involves using valencies, the other - ions. Example: magnesium chloride Valency method ion method Mg Cl 2 1 swap = MgCl2 Mg2+ Cl- to be neutral need 2 Cltherefore = MgCl2 5 Complete the table Name Strontium fluoride formula Sr F 2 1 Na SO4 1 2 Si Cl 4 1 Al CO3 3 2 NH4 PO4 1 3 Cu O 2 2 Sodium sulphate Silicon chloride Aluminium carbonate Ammonium phosphate Copper (II) oxide Formula CsF CaCO3 FeI3 CuSO4 NaHCO3 FeO = SrF2 = Na2SO4 = SiCl4 = Al2(CO3)3 = (NH4)3PO4 = CuO name Caesium fluoride Calcium carbonate Iron (III) iodide Copper (II) sulphate Sodium hydrogen carbonate Iron (II) oxide WRITING EQUATIONS Chemical equations must be BALANCED and often include state symbols (g) (l) (s) (aq) Gas e.g. H2, N2, O2, CO2, Cl2, F2 and group 8 Liquid H2O, Hg, Br2 Solid all metals, C , S, P, I2, Si Aqueous (dissolved in water), all acids and alkalis, most salts, Write and balance the following equations and add state symbols:Magnesium + hydrochloric acid magnesium chloride + hydrogen Mg(s) + nitrogen N2(g) + 2HCl(aq) + H2(g) hydrogen MgCl2(aq) NH3(g) + H2(g) ammonia (NH3) sodium hydroxide + sulphuric acid sodium sulphate NaOH(aq) + H2SO4(aq) Na2SO4(aq) + water H2O(l) 6 THE MOLE! An amount of substance is the quantity whose unit is the mole the amount of any substance containing as many particles as there are carbon atoms in exactly 12g of carbon-12 A mole is Symbol = n A mole is an actual amount, it is 6.02 x 1023. (Compare: a dozen = 12, a score = 20, a pair = 2) This is called the Avogadro constant. the number of atoms per The Avagadro constant, NA, is mole of the carbon-12 isotope. So one mole of magnesium contains 6.02 x10 23 magnesium atoms. Two moles of magnesium would contain 1.2. x 1024 Mg atoms BEWARE! Because oxygen exists as O2 if we have one mole of O2 we would have 2 x 6.02 x 1023 = 1.2 x 1024 atoms Questions: No. of atoms = How many atoms are there in a) 3 moles of Na moles of x 6.02 x 1023 3 x 6.02 x 1023 = 1.81 x 1024 atoms b) 0.25 moles of Al 0.25 x 6.02 x 1023 = 1.51 x 1023 c) 2 moles of H2 2 x 2 x 6.02 x 1023 = 2.41 x 1024 Relative molecular mass, Mr is the average mass of a molecule compared to carbon-12 This is calculated using relative atomic masses. It is used for simple molecular (covalent) molecules e.g. CO2 Relative formula mass is also calculated using Ar 7 This is used for giant molecular (covalent) molecules. e.g. diamond Calculating moles: The relationship between moles, mass and Mr is used continuously in chemical calculations in grams Moles = (n) mass Mr Calculated from the periodic table Calculate the Mr of a) H2SO4 (2 x 1) + 32.1 + (4 x 16) = 98.1 b) Mg(OH)2 24.3 + {(16 + 1) x 2] = 58.3 Calculate the mass of a) 6 moles of rubidium 6 x 85.5 = 513g mass = moles x molar mass b) 0.25 moles of CO2 0.25 x 44 = 11g c) 0.01 moles of H2SO4 0.01 x 98.1 = 0.981g how many moles are present in a) 1.2g of Mg n = 1.2/24.3 = 0.0494 b) 30g of C n = 30/12 = 2.5 c) 6.7g of PCl5 Mr PCl5 = 208.5 n = 6.7/208.5 = 0.0321 Molar mass Molar mass is the mass of one mole of a substance, it is the same as Mr except that Mr does NOT have units, molar mass has units of g mol-1 (grams per mole) Molar mass is the mass per mole of a substance Units are g mol-1 (g/mol) 8 To calculate the molar mass we use the relative atomic masses of the atoms from the periodic table. Using moles in calculations Steps: 1. calculate moles of the reactant 2. use the balanced eqn to deduce moles of product that should be made (theory) 3. calculate the mass of product that should be made (theory) Zinc reacts with hydrochloric acid producing zinc chloride and hydrogen gas. Write a balanced symbol equation, taking care with the formula of zinc chloride. Zn + 2HCl ZnCl2 + H2 The equation tells us that: one mole of zinc reacts with 2 moles of HCl mole ratio is 1 : 2 : 1 : 1 so 1 mole of zinc makes 1 mole of ZnCl2 If 6.5g of zinc are used, how many moles of HCl are needed to react with n of Zn = it? As ratio is 1 : 2 Zn : HCl 6.5 65.4 = 0.0994 ( to 3 s.f.) n of HCl = 2 x 0.0994 = 0.199 How many moles of zinc chloride should be made? As ratio Zn : ZnCl2 1 : 1 n of ZnCl2 = 0.0994 What mass of zinc chloride should we make (in theory) in this reaction? Mass = n x Mr = 0.0994 x [65.4 + (2 x 35.5)] = 13.6g If only 7.3g of zinc chloride is ACTUALLY made, what is the percentage yield? % yield = 7.3 x 100 = 53.7% 13.6 9 Percentage yield calculations given in question % Yield = actual mass theoretical mass x 100 the one you calculate Questions 1. when 20g of calcium carbonate are thermally decomposed, only 4g of calcium oxide are actually made. Calculate % yield. CaCO3 CaO + CO2 Step 1 n of CaCO3 = step 2 ratio is 1 : 1 so moles of CaO step 3 theoretical mass CaO step 4 20 = 0.200 (3s.f.) 100.1 = = = 0.200 0.2 x 56.1 11.2g % yield = 4 x 100 11.2 = 35.7% 2. when 30g of chlorine react directly with excess iron, 25g of the product iron (III) chloride is actually made. Write a balanced symbol eqn and calc % yield. 2Fe + 3Cl2 2FeCl3 n of Cl2 = 30 = 0.423 mol 70 n of FeCl3 = 0.423 x 2/3 theoretical mass = % yield = 25 45.8 0.282 x 100 x = = 0.282 mol 162.3 54.9% = 45.8g 10 MOLES AND GASES The ideal gas equation Volume Pressure Moles Temp PV = nRT 1000 cm3 R = Universal gas constant 8.31jk-1mol-1 dm3 11 When using the ideal gas equation we must use certain units of volume, pressure and temperature we call these SI units Temperature: Kelvin C o (+273) k Pressure: Pa (Pascals) 100kpa = 1000,00pa Kpa x103 pa Volume: m3 cm3 x10-3 dm3 x10-3 m3 1. Rearrange the ideal gas equation to find Volume V = nRT P 2. Rearrange the ideal gas equation to find Pressure P = nRT V 12 3. Rearrange the ideal gas equation to find Moles n = PV RT 4. Rearrange the ideal gas equation to find Temperature T = PV nR 13 Question: Potassium Nitrate decomposes on strong heating forming Oxygen and Potassium Nitrite as the only products a) A 1.00g sample of KNO3 (Mr 101.1) was heated strongly until fully decomposed. Calculate the number of moles of KNO3 in the sample b) At 25OC and 100kpa the oxygen gas produced in the decomposition occupied a volume of 12.6cm3. State the ideal gas equation and use it to calculate the number of moles of O2 produced. (R = 8.31J k-1mol-1) 1. n KNO3 = 1.00 / 101.1 = 9.89X10-3 2. Convert units (k, Pa and m3) 25OC + 273 = 298k 12.6cm3 x10-6 = 1.26x10-5m3 100kpa x103 = 100,000pa 4.Apply the ideal gas equation (rearranged) n = PV RT 100,000pa X 1.26x10-5m3 = 5.088x10-4 8.31JK-1mol-1 X 298k C) Calculate the % yield of Oxygen in the reaction KNO3KNO2 + ½ O2 % yield = Actual moles / Theoretical moles x 100 5.088x10-4 / 100 = 10.3% 4.945x10-3 (9.89x10-3 / 2 due to 1:0.5 ratio) x 14 MOLES AND CONCENTRATION The concentration of a solution is the amount of solute, in mol, dissolved in 1 dm3 (1000cm3) of solution. The units are mol dm-3 ( mol/dm3) We can calculate the concentration of any solution if we know number of moles (n) volume in dm3 concentration moles C = n v volume in dm3 [ n = c x v ] Acids are said to be concentrated or dilute. one in which a large amount of acid is in each dm3 of the solution. A concentrated acid is one in which a small amount of acid is in each dm of the solution. A dilute acid is 3 Calculate the concentration of the following: a) 9.4g of NaCl dissolved into 500cm3 of water. 9.4 = 0.161 mol 58.5 b) 10g of KOH dissolved into 2dm3 water n = 10 = 0.178 56.1 calc moles Calculate conc. C = n/v c = n = 0.161 = 0.322 mol n = mass v 0.500 Mr c = 0.178 = 0.0890 mol dm 2 What mass of solid must be added to make the following? a) 200cm3 of a 0.1 mol dm-3 solution of NaOH n = 0.1 x 200/1000 = 0.0200 mol mass = 0.0200 x 40 = 0.8g b) 750cm3 of a 2 mol dm-3 solution of Na2CO3.10H2O -3 n = c Mass Calc moles x v/1000 Calc mass = n x Mr 15 n = 2 x 750 / 1000 = 1.5 mol [Mr for Na2 CO3.10H2O = (23 x 2) + 12 + (16 x 3) + (10 x 18) = 286 ] mass = 1.5 x 286 = 429g TITRATIONS There are several types of titration; we will only focus on ACID-BASE titration’s. Titration’s are used to find the concentration of an acid or alkali (base) An indicator is used to show the ‘end-point’ which is when the acid has reacted with all of the alkali Acid name Formula Hydrochloric acid HCl Sulphuric acid H2SO4 Nitric acid HNO3 Note: all acids and alkalis are given the state symbol (aq). The salts made when these react are also usually (aq). Example 25cm3 of 0.05mol dm-3 NaOH was added to a conical flask. This required exactly 17.5cm3 of HCl to react with it. Calculate the concentration of the HCl used. Steps 1. write a balanced symbol eqn 2. calculate moles of the KNOWN (i.e. the thing we know conc. and vol for) 3. use the stoichiometry of the eqn to DEDUCE the number of moles of UNKNOWN used 4. calc the conc. of the unknown calculation: 1. NaOH + HCl NaCl + H2O 2. n of NaOH = 0.05 x 25/1000 = 1.25 x 10-3 ( 0.00125) 3. as ratio 1 : 1 n of HCl = 1.25 x 10-3 4. conc of HCl = 1.25 x 10-3/ 0.0175 = 0.0714 mol dm-3 16 Questions: 1. 25cm3 of 0.2moldm-3 NaOH reacts with 11.2cm3 of H2SO4. Calculate the conc. of the sulphuric acid. (work out the formula of the salt carefully) 2. A student has 50cm3of 0.01 moldm-3 KOH, she wishes to exactly neutralise this with 0.1 moldm-3 HNO3. What volume of the acid will she need? Answer on the back of a page. 1. 2NaOH + 25 cm3 0.2 mol dm-3 H2SO4 11.2 cm3 ? n of NaOH = 0.2 x n of H2SO4 25 = 1000 KOH + 50 cm3 0.01 mol dm-3 + 2H2O 5.0 x 10-3 ( 0.00500) ratio 2 : 1 = 0.00500 = 2 conc H2SO4 = 0.00250 0.0112 2. Na2SO4 0.00250 = 0.223 mol dm-3 HNO3 KNO3 ? 0.100 mol dm-3 + H2O n of KOH = 0.01 x 50 = 5.0 x 10 –4 (0.000500) 1000 n of HNO3 = 5.0 x 10-4 due to 1.1 ratio vol = 0.00050 = 0.1 5 x 10-3 dm-3 (0.00500) or 5cm3 17 Calculating formulae: Copper has a valency of 1 or 2. Use the following data to calculate the correct formula for an oxide made when copper is heated in air. 4.3g of copper was heated in air and 4.85g of solid oxide was formed. Steps: 1. 2. 3. subtract the mass of the metal from the mass of the solid oxide formed to find mass of oxygen. calculate the number of moles of each element. divide the number of moles of each element by the smallest value. Mass of oxygen 4.85 – 4.3 = 0.55g Cu : O 4.3 : 0.55 63.5 : 16 0.0677 : 0.0344 1.97 : 1 ( Ar) ( smallest) formula = Cu2O Iron has two possible valencies. If 7.21g of iron was heated in air the solid made weighed 10.29g. Calculate the formula of the iron oxide. Mass of oxygen = 10.29 – 7.21 = 3.08g Fe 7.21 55.8 : O 3.08 16 0.129 : 0.193 1 : 1.5 2 : 3 formula = Fe2O3 18 Calculating Stoichiometric relationships. Sodium hydroxide, NaOH, reacts with an acid to form a salt and water. It is known that the acid is either sulphuric acid or hydrochloric acid. Using the information below, calculate the stoichiometric relationship between the acid and NaOH and therefore identify the acid used. Concentration/mol dm Volume/cm3 Moles of NaOH = 0.01 Moles of acid = -3 NaOH 0.01 25 Acid 0.005 25 x 25 / 1000 0.005 x 25 / 1000 = = 2.50 x 10-4 1.25 x 10-4 Ratio of NaOH : acid is 2.5 x 10-4 : 1.25 x 10-4 2 : 1 so reacting ration is 2 NaOH :1 acid acid must be H2SO4 2NaOH + H2SO4 Na2SO4 + H2O Potassium hydroxide reacts with either nitric acid or phosphoric acid, H3PO4. Use the data in the table below, calculate the stoichiometric relationship between the acid and KOH and therefore identify the acid used. KOH Acid -3 Concentration/mol dm 0.2 0.16 3 Volume/cm 20 25 Moles of KOH = 0.2 Moles of acid = x 20 / 1000 = 4.0 x 10-3 0.16 x 25 / 1000 = 4.0 x 10-3 Ratio of KOH : acid is -3 4.0 x 10 : 4.0 x 10-3 1 : 1 so reacting ration is 1 KOH :1 acid must be HNO3 KOH + HNO3 KNO3 + H 2O 19 EMPIRICAL AND MOLECULAR FORMULAE Empirical formula is the simplest whole number ratio of atoms of each element in a compound. Molecular formula is the actual number of atoms of each element in a compound. Determining the Empirical and Molecular formula. A compound contains 6.75g of Al and 26.63g of Cl. What is its empirical formula? The molar mass of the compound is 267gmol-1 , calculate the molecular formula of this compound. Steps 1. find moles of each element (from mass or from % of element) 2. divide by the smallest number 3. use the whole number ratio to deduce Empirical formula 4. use molar mass/Mr to calc molecular formula Al : 6.75 : 27 : Cl 26.63 35.5 0.25 : 0.25 : 0.750 0.25 1 : 3 smallest Empirical formula = AlCl3 mass = 133.5 Mr = 267 so Ef x 2 = Mf = Al2Cl6 Questions (answer over the page) 1. an organic molecule is found to contain 90.6% C and 9.4% H, calculate the empirical formulae. The Mr of the molecule is 106, use this to deduce the molecular formula. 2. A compound contains only carbon, hydrogen and oxygen. Its composition by mass is 40%C and 6.7% H. calculate the Empirical formula. The molar mass is 180 gmol-1. Use this data to calculate Molecular formula. Answers to questions. 20 ANSWERS 1 C 90.6 12 : : : H 9.4 1 7.55 1 : : 4 : 9.4 1.25 (can’t round up, x 4 to get whole number) 5 empirical formula = C4H5 Er mass = 53 Mr 106 106 / 53 = 2 2. Mf = emp. Formula x 2 =C8H10 C 40 12 : : : H 6.7 1 : : : O 53.3 16 3.33 1 : : 6.7 2 : : 3.33 1 empirical formula = CH2O Er mass = 30 Mr = 180 Mf = emp. Formula x 6 = C6H12O6 21 Compounds containing water Some compounds can exist in both the anhydrous and hydrated forms. For example copper (II) sulphate is a white powder when anhydrous and forms blue crystals when hydrated. The difference between the two forms is due to the presence of water of crystallisation. Water of crystallisation: water molecules which make up the part of the crystal structure of a compound, that can be removed by heating. CuSO4 + 5H2O CuSO4. 5 H2O Anhydrous: the compound which remains when the water is removed from a hydrated salt. Means: being without water. Hydrated: a complex molecule which has water molecules incorporated e.g. CuSO4 is anhydrous CuSO4. 5H2O is hydrated as it contains 5 waters of crystallisation. 22 Questions involving water of crystallisation 1. Calculate the percentage mass of water of crystallisation in copper (II) sulphate-5-water [CuSO4. 5 H2O] Mass of CuSO4 . 5H2O = 63.5 + 32.1 + (4 x 16) + 10 + ( 5 x 16) = 249.5 mass of water in molecule = 5 x 18 = 90 % mass of H2O = 90 249.5 x 100 = 36% 2. Weighed samples of the following crystals were heated to drive off the water of crystallisation. When they reached constant mass, the following masses were recorded. Deduce the empirical formulae of the hydrates a) 0.942g of MgSO4. a H2O gave 0.461g of residue mass of water removed= 0.942 – 0.461 = 0.481g MgSO4 : H 2O 0.461 : 0.481 120.4 18 0.00383 0.0267 1 7 formula = MgSO4 . 7 H2O b) 1.124g of CaSO4. b H2O gave 0.889g of residue mass of water removed = 1.124 – 0.889 = 0.235g CaSO4 : H 2O 0.889 : 0.235 136.1 18 0.00653 0.0130 1 2 formula = CaSO4 . 2H2O c) 1.203g of Hg(NO3)2. c H2O gave 1.172g of residue mass of water removed = 1.203 – 1.172 = 0.031g Hg(NO3)2 : H 2O 1.172 : 0.031 325 18 0.00360 0.00172 2 1 1 ½ formula = Hg(NO3)2 . ½ H2O 23 ATOM ECONOMY atom economy = mass of desired product___ x 100 sum of masses of all products (or reactants) Make sure equation is balanced or you will get the wrong answer!!!! Mass of products = Mass of reactants as long as the equation balances. e.g. ethene reacts with steam, the desired product is ethanol. A single product is made in an addition reaction, giving 100% atom economy. e.g. 2-iodobutane undergoes a hydrolysis reaction with aqueous sodium hydroxide; the desired product is butan-2-ol. Write a balanced equation for this reaction and calculate the atom economy. Step 1: calculate the Mass of the desired product Step 2: divide this by the sum of the Mass of all the products CH3CH2CHICH3 + NaOH CH3CH2CHOHCH3 + NaI Mr = 74 Mr = 149.9 Atom economy = 74 x 100 (74 + 149.9) = 33.1% 24 The majority of the starting materials have turned in to waste. Even if the reaction proceeded with a 100% yield, a large proportion of the mass of atoms used would be wasted. A high atom economy benefits society as it reduces the amount of waste produced, and it reduces the cost to companies. Reactions are more sustainable. “Waste products” are the by-products formed during a reaction. Their disposal can be costly. However, some by-products can be sold on or used elsewhere in the chemical plant. NB. Percentage yield and atom economy are not the same thing!! Atom economy tells you the proportion of reactants that is converted into useful products. Percentage yield tells you the efficiency of this i.e of converting reactants into products Do not confuse them!! 25