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Quantitative Analysis A.S. 2.1 (Chemistry 91161) Year 12 4 Internal Credits The Mole and Molar Mass The amount (n) of a substance that contains 23 6.02 X 10 particles is called a mole The Mole • A counting unit • Similar to a dozen, except instead of 12, it’s 602 billion trillion 602,000,000,000,000,000,000,000 • 1 mole = 6.02 X 1023 particles • Avagadro’s number • Enough soft drink cans to cover the surface of the earth to a depth of over 300 Km. The Mole • 1 dozen cars = 12 cars • 1 mole of cars = 6.02 X 1023 cars • 1 dozen Al atoms = 12 Al atoms • 1 mole of Al atoms = 6.02 X 1023 atoms Note that the NUMBER is always the same, but the MASS is very different! Mole can be abbreviated, but only to mol!!! Learning Check 1. Number of atoms in 0.500 mole of Al a) 500 Al atoms b) 6.02 x 1023 Al atoms c) 3.01 x 1023 Al atoms 2.Number of moles of S in 1.8 x 1024 S atoms a) 1.0 mole S atoms b) 3.0 mole S atoms c) 1.1 x 1048 mole S atoms Molecular Mass (M) • The Mass of 1 mole of a substance (in grams) • Equal to the atomic mass number 1 mole of C atoms = 12.0 g 1 mole of Mg atoms = 24.3 g 1 mole of Cu atoms = 63.5 g • Unit gmol-1 • 7gmol-1 means 1 mole of the substance would weigh 7 g Molecular Mass of Molecules and Compounds Mass (in grams) of 1 mole is equal to the sum of the atomic masses e.g. What is the Molar mass of CaCl2 1 mole Ca x 40.1 g/mol + 2 moles Cl x 35.5 g/mol = 111.1 g/mol CaCl2 You try 1 mole of N2O4 = 92.0 g/mol Learning Check! A. Molar Mass of K2O = 94 ? Grams/mole B. Molar Mass of antacid Al(OH)3 ? gmol-1 = 78 C. Prozac, C17H18F3NO, is a widely used antidepressant that inhibits the uptake of serotonin by the brain. Find its molar mass. 309 gmol-1 Balancing Equations • 2Ca(s) + O2(g) 2CaO(s) + energy • This means that 2 mols of Ca atoms react with every 1 mol of Oxygen molecules to produce 2 mols of CaO and “n” mols of energy is released. • If 4 mols of Ca was burnt how many mols of oxygen would be required and how much energy is produced? • How many mols of oxygen are reacted when 6 mols of CaO are produced? • If 40g of Ca are reacted what mass of CaO is produced? • What is the mole ratio of ions formed when MgCl2 dissolves? Use a balanced equation to calculate quantities • If 50g of ethane burns what is the mass of water produced? • 2C2H6 + 7O2 4CO2 + 6H2O • M(C2H6) = 30 • n=m/M =50/30 = 1.67 moles ethane • C2H6:3H2O Mole ratio STOICHIOMETRY • 1.67 x 3 = 5 moles water m • M(H2O)= 18 • m = nxM = 5x18 = 90g water n M Converting Moles (n), Mass (g) and Molecular Mass (M) n = m/M m n M Aluminum is often used for the structure of light-weight bicycle frames. How many grams of Al are in 3.00 moles of Al? m=nxM = 3 x 27 = 81g Al m Learning Check! n M The artificial sweetener aspartame (Nutra-Sweet) formula C14H18N2O5 is used to sweeten diet foods, coffee and soft drinks. How many moles of aspartame are present in 225 g of aspartame? 0.77 moles Learning Check! How many atoms of K are present in 78.4 g of K? 2.01 x 6.02 x 1023 = 12.1 x 1023 What is the mass (in grams) of 1.20 X 1024 molecules of glucose (C6H12O6)? n=1.20 X 1024 / 6.02 x 1023 =2 moles M= 180g/mole so m=nxM =2x180 = 360g How many Na+ in 25g of Na2CO3? M(Na2CO3)= 106g/mol N=m/M = 25/106 = 0.236moles X 2 = 0.472 moles Percent Composition Gives the breakdown of mass between different atoms e.g. Find the percentage of Carbon in Carbon monoxide M(C) =12 M(O) =16 Percentage(C) = M(C) /M(CO) x 100 = 42.86% What is the percent carbon in C5H8NO4 (the glutamic acid used to make MSG monosodium glutamate), a compound used to flavour foods and tenderise meats? a) 8.2 %C b) 27.8 %C c) 41.1 %C Types of Formulas Empirical Formula The formula of a compound that has the smallest whole number ratio of the atoms present. e.g. CH3 means for every C there are 3H Molecular Formula The formula that gives the actual number of each kind of atom found in the molecule. e.g. C2H6 means every molecule has 2C and 6H atoms Chemical Formulas of Compounds • Formulas give the relative numbers of atoms or moles of each element in a formula - always a whole number ratio. NO2 2 atoms of O for every 1 atom of N 1 mole of NO2 : 2 moles of O atoms to every 1 mole of N atoms • If we know or can determine the relative number of moles of each element in a compound, we can determine a formula for the compound. Find Empirical Formula given the composition 1. Determine the mass in grams of each element present, if needed. 2. Calculate the number of moles of each n atom. 3. Divide each by the smallest number of moles to obtain the simplest whole number ratio. 4. Write empirical formula and determine its molar mass. m M Learning Check! A sample of a brown gas, a major air pollutant, is found to contain 2.34g N and 5.34g O. Determine the empirical formula for this substance. moles of N = 2.34g of N = 0.167 moles of N 14.01 g/mole moles of O = 5.34g of O = 0.334 moles of O 16.00 g/mole Therefore N0.167O0.334 Formula: N 0.167 O 0.334 N O 2 0.167 0.167 Find Molecular Formula given the molar mass Calculate the molecular formula of a compound with empirical formula CH4 and a molar mass of 64g/mol m n M 1. Identify the empirical formula and calculate its molar mass. CH4 M(CH4) =16 2. Divide the given molar mass by the empirical molar mass. 64/16 = 4 3. Multiply the empirical formula by the answer from step 2. C4H16 Learning Check! A compound has an empirical formula of NO2. The colourless liquid, used in rocket engines has a mass of 92.0 g. What is the molecular formula of this substance? Molecular Formula from % Composition A substance contains 60.80 % Na ; 28.60 % B ; 10.60 % H by mass. Its molar mass is 114g/mole. What is the molecular formula of the substance? HELP! Consider a sample size of 100 grams Determine the number of moles of each atom Determine the simplest whole number ratio Write empirical formula Calculate molar mass of empirical Divide molar mass by empirical molar mass Multiply answer by empirical formula Na3B3H12 Concentration of Solution n c V • n = amount of solute (moles) • c = concentration of solution (moles/litre) • V = volume of solution (litres) Problem 1. Calculate concentration if 0.025 moles of HCl are present in 50ml of solution. • c=n/V • =0.025/0.050 • =0.5 molL-1 Problem 2. Calculate moles of Na2CO3 present in 21mL of 0.3molL-1 solution • n = cV • = 0.3 x 0.021 • = 0.0063 moles n c V 2.4g of sodium carbonate is dissolved in water to make 250ml of solution. Calculate the concentration. M(Na2CO3) = 106gmol-1 • n = m/M • = 2.4/106 • = 0.023 moles m nM • c = n/V • = 0.023/0.250 • = 0.092molL-1 n c V Volumetric Analysis • Acid Base Titrations – Used to determine an unknown concentration ACID-BASE REACTIONS Titrations H2C2O4(aq) + 2 NaOH(aq) ---> Na2C2O4(aq) + 2 H2O(liq) acid base Oxalic acid, H2C2O4 Carry out this neutralisation reaction using a TITRATION. Setup for titrating an acid with a base Titration 1. Use pipette to measure your (unknown concentration) solution into the flask. 2. Add known solution from the buret to find the average titre. 3. Indicator shows when exact stoichiometric reaction has occurred. (Acid = Base) 4. Calculate moles (n) of known solution. 5. Use balanced equation to find the unknown amount of moles. 6. Calculate the concentration of unknown. PROBLEM #1: Standardise a solution of NaOH — i.e., accurately determine its concentration. 35 mL of NaOH is pipetted into a flask and neutralised (by titration) with 25.2 ? mL of 0.0998 M HCl. What is the concentration of the NaOH? 2. Find average titre of HCl (L) Titre 1 CONCORDANCE 3 consistent results 2 3 4 Av. Vol. 25.0 25.4 29.4 25.2 25.2 ml PROBLEM #1: Standardise a solution of NaOH — i.e., accurately determine its concentration. 4. Calculate moles (n) of HCl n = cV = 0.0998 x 0.0252 = 0.025 moles HCl used n c V 5. Use balanced equation to find the unknown amount of moles NaOH NaOH + HCl NaCl + H20 1:1 ratio, therefore 0.025 moles NaOH reacts with 0.025 moles HCl PROBLEM #1: Standardise a solution of NaOH — i.e., accurately determine its concentration. 6. Calculate the concentration of NaOH • c=n/V • = 0.025/0.035 • = 0.71molL-1 n c V Your turn 20 mL of Al(OH)3 is neutralised (by titration) with 18.6mL of 0.2 M HCl. What is the concentration of the Al(OH)3? Concentration of Al(OH)3 = 0.062molL-1 PROBLEM 3: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? Add water to the 3.0 M solution to lower its concentration to 0.50 M i.e. Dilute it! But how much water do we add? PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? c • V = Amount (n) of NaOH in original solution (3.0 mol/L)(0.050 L) = 0.15 mol NaOH Amount of NaOH in final solution must also = 0.15 mol NaOH n/C = Volume of final solution (0.15 mol NaOH)/(0.50 mol/L) = 0.30 L or 300 mL n c V PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? Conclusion: add 250 mL of water to 50.0 mL of 3.0 M NaOH to make 300 mL of 0.50 M NaOH. Diluting Solutions No need to convert ml to L if both units are the same Finding the new concentration Your volume X concentration = diluted Total volume concentration e.g. 20ml of 0.5molL-1 is diluted to 100ml. What is the final concentration? 0.02/0.1 X 0.5 =0.1molL-1 Find the original concentration of HCl when 20ml is pipeted into a flask and made up to 250ml giving a final concentration of 4x10-3molL-1 0.05molL-1 Test Yourself • You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. How much water do you need to add? 250ml • What is the concentration of Cu2+ if 20ml of 0.64molL-1 of CuSO4 is diluted to 100ml? 0.128molL-1 • 25ml CuSO4 solution was diluted to 200ml. The new concentration is 7.36x10-4 molL-1. What was the original concentration? 5.89x10-3 molL-1 Density • Convert L g and g L • E.g. Calculate the volume of 6 moles of ethanol given the density of ethanol (CH3CH2OH) is 0.789gml-1 • M = 46gmol-1 • m = Mn = 46 x 6 = 276g • 276g/0.789gml-1 = 349.8ml or 0.350L Apple Juice • 20ml apple juice was titrated against 0.1molL-1 NaOH and the average was 10.36ml. Assuming all apple juice is citric acid and that 1mole reacts with 3 moles of NaOH. Calculate the concentration of citric acid in gL-1. Apple juice must contain between 0.3 and 0.8g per 100ml. Is this considered apple juice or apple drink? • M(citric acid)=192gmol-1 Gravimetric analysis Decomposing of solids • Hydrated magnesium sulphate (MgSO4.XH2O) is heated in a crucible (total mass 37.95g) so the water is lost from the crystals. When it is reweighed it has a constant mass of 36.82g. The crucible weighs 35.75g what is the mass of the water and anhydrous MgSO4? MgSO4 = 1.07g H20 = 1.13g Gravimetric analysis • Calculate X the water of crystallisation in MgSO4.XH2O • M(MgSO4) = 120gmol-1 M(H2O) = 18gmol-1 • Calculate the number of moles of each and the mole ratio. • n(MgSO4) = 1.07/120 = 0.00892moles • n(H2O) = 1.13/18 = 0.0628moles m • 0.0089:0.0628 = 1:7 n M • X=7 Gravimetric analysis • AgNO3 is added to 25ml of water containing Cl- so that the ions are precipitated out. • Ag+(aq) + Cl-(aq) AgCl(s) • They are found to weigh 0.32g. Calculate the concentration of Cl- in solution. • n= m/M = 0.32/143.5 = 0.00223 moles AgCl • Cl- : AgCl so 0.00223 moles Cl• c = n/V = 0.00223/0.025 = 0.089molL-1 m n M