1. 1 JF Tutorial: Mole Calculations Shane Plunkett plunkes@tcd.ie 1.Some Mathematical Functions 2.What is a mole? Avogadro’s Number Converting between moles and mass Calculating mass % from a chemical formula Determining empirical and molecular formulae from mass Recommended reading T.R. Dickson, Introduction to Chemistry, 8 th Ed., Wiley, Chapters 2 & 4 M.S. Silberberg, Chemistry, The Molecular Nature of Matter and Change, 3 rd Ed., Chapter 3 P. Atkins & L. Jones, Molecules, Matter and Change, 3 rd Ed., Chapter 2 Multiple choice tests: http://www.mhhe.com/silberberg3

2. 2 Carrying out Calculations In chemistry, must deal with several mathematical functions. 1.Scientific Notation Makes it easier to deal with large numbers, especially concentrations Written as A ×10 b, where A is a decimal number and b is a whole number Example: Avogadro’s number 602 213 670 000 000 000 000 000 It is very inconvenient to write this. Instead, use scientific notation: 6.022 × 10 23 Calculators: Sharp & CasioType in 6.022 Press the exponential function [EXP] Key in 23

3. 3 Questions (a) 784000000 (b) 0.00023 (c) 9220000 (d) 0.000000015 How would you write the following: 7.84 × 10 8 2.3 × 10 -4 9.22 × 10 6 1.5 × 10 -8 (a) (1.38 × 10 4 ) × (8.21 × 10 6 ) Calculate the following: 1.13 × 10 11 (b) (8.56 × 10 -8 ) × (2.39 × 10 4 ) 2.05 × 10 -3

4. 4 Common Decimal Prefixes PrefixSymbolNumberWord Exponential Notation teraT1,000,000,000,000trillion 10 12 gigaG1,000,000,000billion 10 9 MegaM1,000,000million 10 6 kilok1,000thousand 10 3 hectoh100hundred 10 2 decada10ten 10 1 decid0.1tenth 10 -1 centic0.01hundredth 10 -2 millim0.001thousandth 10 -3 micro0.000001millionth 10 -6 nanon0.000000001billionth 10 -9 picop0.000000000001trillionth 10 -12 femtof0.000000000000001quadrillionth 10 -15

5. 5 2. Logarithms Makes dealing with a wide range of numbers more convenient, especially pH Makes dealing with a wide range of numbers more convenient, especially pH Two types: common logarithms and natural logarithms Two types: common logarithms and natural logarithms Common Logarithms Common log of x is denoted log x gives the power to which 10 must be raised to equal x 10 n = x written as: log 10 x = n (base 10 is not always specified) Example: The common log of 1000 is 3, i.e. 10 must be raised to the power of 3 to get 1000 Written as: log 10 1000 = 3 10 3 = 1000

6. 6 Calculators Sharp:Press the [LOG] function Sharp:Press the [LOG] function Type the number Hit answer Casio:Key in the number Casio:Key in the number Press the [LOG] function Questions (a)10 (b)1,000,000 (c)0.001 (d)853 1 6 -3 2.931 Calculate the common logarithms of the following: log 10 log 1000000 log 0.001 log 853

7. 7 Natural logarithms Natural log of x is denoted ln x Natural log of x is denoted ln x the difference here is, instead of base 10, we have base e (where e = 2.71828) the difference here is, instead of base 10, we have base e (where e = 2.71828) Gives the power to which e must be raised to equal x Gives the power to which e must be raised to equal x lnx or log e x = n or e n = x lnx or log e x = n or e n = x Example The natural log of 10 is 2.303, i.e. e must be raised to the power of 2.303 to get 10 Calculators Sharp:Press the [ln] function Enter the number and hit answer Casio:Enter the number Press the [ln] function

8. 8 Questions What is the natural log of: (a)50 (b)1.25 × 10 5 (c)2.36 × 10 -3 (d)8.98 × 10 13 3.91 11.74 -6.05 32.13 ln 50 ln 1.25x10 5 ln 2.36x10 -3 ln 8.98x10 13

9. 9 3. Graphs Experimental data often represented in graph form, especially in straight lines Experimental data often represented in graph form, especially in straight lines Equation of straight line given by Equation of straight line given by y = mx + c where x and y are the axes values m is the slope of the graph c is the intercept of the plot y- axis x-axis Slope Intercept

10. 10 Sign of slope tells you the direction of the line Sign of slope tells you the direction of the line Magnitude of slope tells you steepness of line Magnitude of slope tells you steepness of line Slope found by taking two x values and the two corresponding y values and substituting these into the following relation: Slope found by taking two x values and the two corresponding y values and substituting these into the following relation: Example Given the (x, y) coordinates (2, 4) and (5, 9), find the slope of the line containing these two points. x 1 = 2y 1 = 4x 2 = 5y 2 = 9 Sub into above relation: m = 9 – 4= 5= 1.67 5 – 2 3

11. 11 4. Quadratic Equations May be encountered when dealing with concentrations May be encountered when dealing with concentrations Involve x 2 (x-squared terms) Involve x 2 (x-squared terms) Take the form ax 2 + bx + c = 0 Take the form ax 2 + bx + c = 0 Can be solved by: Can be solved by: this expression finds the roots or the solution for x of the quadratic equation

12. 12 Find the roots of the equation x 2 – 6x + 8 = 0 Example: a = 1b = -6c = 8 x = = Therefore, x = or = 4 = 2 ax 2 + bx + c = 0

13. 13 Question You have been asked to calculate the concentration of [H 3 O + ] ions in a chemical reaction. x = [H 3 O + ] The following quadratic equation has been given Solve for x. 2.4x 2 + 1.5x – 3.6 = 0

14. 14 Thereforeor x = 0.95 or x = -1.58 Because we are dealing with concentrations, a negative value will not make sense. Therefore, we report the positive x value, 0.95, as our answer. Never round up this number!

15. 15 Important formulae so far… y = mx + c ax 2 + bx + c = 0 Graphs….. Quadratic equations…..

16. 16 Calculations: The Mole Stoichiometry is the study of quantitative aspects of chemical Stoichiometry is the study of quantitative aspects of chemical formulas and reactions Mole: SI unit of the amount of a substance Mole: SI unit of the amount of a substance Definition:A mole is the number of atoms in exactly 12g of the carbon-12 isotope This number is called Avogadro’s number and is given by 6.022 ×10 23 The mole is NOT just a counting unit, like the dozen, which specifies only the number of objects. The definition of a mole specifies the number of objects in a fixed mass of substance. Mass spectrometry tells us that the mass of a carbon-12 atom is 1.9926×10 -23 g. No. of carbon-12 atoms = atomic mass (g) mass of one atom (g) = 12g _ 1.9926×10 -23 g = 6.022 ×10 23 atoms

17. 17 Other definitions of the Mole One mole contains Avogadro’s Number (6.022 x 10 23 ) One mole contains Avogadro’s Number (6.022 x 10 23 ) A mole is the amount of a substance of a system which contains as many elementary entities as there are atoms in 0.012kg (or12g) of Carbon- 12 A mole is the amount of a substance of a system which contains as many elementary entities as there are atoms in 0.012kg (or12g) of Carbon- 12 A mole is that quantity of a substance whose mass in grams is the same as its formula weight A mole is that quantity of a substance whose mass in grams is the same as its formula weight E.g. Fe55.85 E.g. Fe55.85 Iron has an atomic mass or 55.85g mol -1, so one mole of iron has a mass or 55.85g Iron has an atomic mass or 55.85g mol -1, so one mole of iron has a mass or 55.85g

18. 18 One mole of any object always means 6.022 × 10 23 units of those objects. For example,1 mol of H 2 O contains 6.022 × 10 23 molecules 1 mol of NaCl contains 6.022 × 10 23 formula units Avogadro’s number is used to convert between the number of moles and the number of atoms, ions or molecules. Example 0.450mol of iron contains how many atoms? Number of atoms = number of moles × Avogadro’s number (N A ) Therefore No. of atoms = (0.450mol) × (6.022 × 10 23 ) = 2.7 × 10 23 atoms Calculating the number of particles

19. 19 Example How many molecules are there in 4 moles of hydrogen peroxide (H 2 O 2 )? No. of molecules = no. of moles × Avogadro’s number (N A ) = 4mol × (6.022 × 10 23 mol -1 ) = 24 ×10 23 molecules = 2.4 × 10 24 molecules Questions How many atoms are there in 7.2 moles of gold (Au)? Answer: 4.3 × 10 24 atoms The visible universe is estimated to contain 10 22 stars. How many moles of stars are there? Answer: 10 22 stars = 10 22 = 0.17 mol. 6.022×10 23

20. 20 Calculating the mass of one molecule Example: What is the mass of one molecule of water? Step 1: Calculate the molar mass of water Molar mass of water = (2 × atomic mass H) + (1 × atomic mass O) Molar mass H 2 O = (2 × 1.008g mol -1 ) + (1 ×16.000g mol -1 ) = 18.00 g mol -1 Step 2: Employ Avogadro’s number Mass of one molecule = Molar mass Avogadro’s no. = 18.00g mol -1 6.022×10 23 mol -1 = 2.992×10 -23 g Note:Always check the units you have in your answer to ensure you are correct

21. 21 Example Calculate the mass of one molecule of ammonium carbonate [(NH 4 ) 2 CO 3 ] Step 1: Calculate the molar mass 2 Nitrogen atoms 8 Hydrogen atoms 1 Carbon atom 3 Oxygen atoms 2 × 14.01gmol -1 8 × 1.008 gmol -1 1 ×12.01gmol -1 3 × 16.00 gmol -1 = 28.02 gmol -1 = 8.064 gmol -1 = 12.01 gmol -1 = 48.00 gmol -1 Total = 96.09 gmol -1 Step 2: Employ Avogadro’s Number, N A Mass of one molecule = 96.09 gmol -1. 6.022×10 23 mol -1 Questions Calculate the mass of one molecule of: = 1.59 × 10 -22 g (a)Ethanoic acid (CH 3 COOH) (b)Methane (CH 4 ) (c)Potassium dichromate (K 2 Cr 2 O 7 ) 9.96 × 10 -23 g 2.66 × 10 -23 g 4.89 × 10 -22 g

22. 22 Converting between mass and moles In the lab, we measure the mass of our reactants in grams using a balance. However, when these react they do so in a ratio of moles. Therefore, we need to convert between the mass we measure and the number of moles we require. The expression relating mass and number of moles is: Mass of sample (g) = no. of moles (mol) × molar mass (gmol -1 ) Example Calculate the mass in grams in 0.75mol of sodium hydroxide, NaOH Step 1: Find the molar mass of the compound Na:22.99 gmol -1 O:16.00 gmol -1 H:1.008 gmol -1 M r :40.00 gmol -1 Step 2: Substitute into the above expression Mass of sample = 0.75mol × 40.00 gmol -1 = 30g

23. 23 Questions Calculate the mass in grams present in: (a) 0.57mol of potassium permanganate (KMnO 4 ) Answer: Molar mass KMnO 4 = 158.03 gmol -1 Mass in grams = 0.57mol × 158.03 gmol -1 = 90.07 g (b) 1.16mol of oxalic acid (H 2 C 2 O 4 ) Answer: Molar mass H 2 C 2 O 4 = 90.04 gmol -1 Mass in grams = 1.16mol × 90.04 gmol -1 = 104.44 g (c) 2.36mol of calcium hydroxide (Ca(OH) 2 ) Answer: Molar mass Ca(OH) 2 = 74.1 gmol -1 Mass in grams = 2.36mol × 74.1 gmol -1 = 174.87 g

24. 24 Converting between moles and mass Number of moles = mass of sample (g) molar mass (gmol -1 ) molar mass (gmol -1 ) Example Convert 25.0g of KMnO 4 to moles Step 1: Calculate the molar mass K Mn O 1 × 39.10 gmol -1 1 × 54.93 gmol -1 4 × 16.00 gmol -1 39.10 gmol -1 54.93 gmol -1 64.00 gmol -1 M r = 158.03 gmol -1 Step 2: Substitute into above expression No. of moles = 25.0g. 158.03gmol -1 = 0.158 mol

25. 25 Questions Calculate the number of moles in: (a) 1.00g of water (H 2 O) Answer: Molar mass water = 18.02 gmol -1 1.00g H 2 O = 0.055mol (b) 3.0g of carbon dioxide (CO 2 ) Answer: Molar mass carbon dioxide = 44 gmol -1 3.0g CO 2 = 0.068mol (c) 500g of sucrose (C 12 H 22 O 11 ) Answer: Molar mass sucrose = 342.30 gmol -1 500g C 12 H 22 O 11 = 1.46mol (d) 2.00g of silver chloride (AgCl) Answer: Molar mass silver chloride = 143.38 gmol -1 2.00g AgCl = 0.014mol

26. 26 Important formulae so far…. No. of carbon-12 atoms = atomic mass (g) mass of one atom (g) No. of atoms = No. of moles × Avogadro’s number (N A ) No. of molecules = No. of moles × Avogadro’s number (N A ) Mass of one molecule = Molar mass Avogadro’s no. Mass of sample (g) = no. of moles (mol) × molar mass (gmol -1 ) Number of moles = mass of sample (g) molar mass (gmol-1) Defining the mole: Calculating the number of atoms or molecules, given the number of moles: Most important equation: Calculating the mass of an individual molecule:

27. 27 Calculating mass percentage from a chemical formula Many of the elements in the periodic table of the elements occur in combination with other elements to form compounds. A chemical formula of a compound tells you the composition of that compound in terms of the number of atoms of each element present. The mass percentage composition allows you to determine the fraction of the total mass each element contributes to the compound. Example Ammonium nitrate (NH 4 NO 3 ) is an important compound in the fertiliser industry. What is the mass % composition of ammonium nitrate? Step 1: Calculate the molar mass of ammonium nitrate Molar mass NH 4 NO 3 = 80.05 gmol -1 Two N atoms: 28.016 gmol -1 Four H atoms: 4.032 gmol -1 Three O atoms: 48.00 gmol -1

28. 28 Step 2: Determine the mass % composition for each element Nitrogen: 28.016g N in one mol of ammonium nitrate Mass fraction of N = 28.016g 80.05g Mass % composition of N = 28.016g × 100% 80.05g = 34.99% ≈ 35% Hydrogen: 4.032g H in one mol of ammonium nitrate Mass fraction of N = 4.032g 80.05g Mass % composition of H = 4.032g × 100% 80.05g = 5.04% ≈ 5% Oxygen: 48.00g O in one mol of ammonium nitrate As above, the mass % composition of O is found to be 60%

29. 29 Therefore, the mass % composition of ammonium nitrate (NH 4 NO 3 ) is: % Nitrogen:35% % Hydrogen:5% % Oxygen:60% To check your answer, make sure it adds up to 100% Question What is the mass % composition of C 12 H 22 O 11 ? Answer: % Carbon:42.1% % Hydrogen:6.5% % Oxygen:51.4%

30. 30 Determining empirical formula from mass The empirical formula of a compound tells you the relative number of atoms of each element present in that compound. It gives you the simplest ratio of the elements in the compound. For example, the empirical formula of glucose (C 6 H 12 O 6 ) is CH 2 O, giving the C:H:O ratio of 1:2:1 If you know the mass % composition and the molar mass of elements present in a compound, you can work out the empirical formula Example What is the empirical formula of a compound which has a mass % composition of 50.05% S and 49.95% O? Step 1: Find the atomic masses of the elements present Sulfur (S) : 32.066 gmol -1 Oxygen (O) : 16.000 gmol -1

31. 31 Step 2: Determine the number of moles of each element present Since we are dealing with percentages, we can express the mass % as grams if we assume we have 100g of the compound. grams if we assume we have 100g of the compound. Therefore, 100g of our compound contains 50.05g of sulfur and 49.95g of oxygen. Convert number of grams to number of moles Number of mol Sulfur = mass of sulfur in sample (g) atomic mass of sulfur (gmol -1 ) = 50.05g. 32.066 gmol -1 = 1.56 mol Similarly, the no. of mol of Oxygen is found to be 3.12mol Step 3: Determining the ratios of elements Sulfur: 1.56mol Oxygen: 3.12mol Ratio 1.56 : 3.12 Ratio must be in whole numbers. Here we must divide across by 1.56 Therefore, we have a ratio of 1:2 giving an empirical formula of SO 2

32. 32 Question Determine the empirical formula of a compound that contains 27.3 mass% Carbon and 72.7 mass% Oxygen. Answer: No. of mol Carbon = 2.27mol No. of mol Oxygen = 4.54mol Ratio1:2Empirical formula CO 2 Monosodium glutamate (MSG) has the following mass percentage composition: 35.51% C, 4.77 % H, 37.85% O, 8.29% N, and 13.60% Na. What is its molecular formula if its molar mass is 169 gmol -1 ? Answer: C 5 H 8 O 4 NNa

33. 33 Important calculations Calculating mass percentage from a chemical formula Step 1: Calculate the molar mass Step 2: Determine the mass % composition for each element Determining empirical formula from mass Step 1: Find the atomic masses of the elements present Step 2: Determine the number of moles of each element present Step 3: Determining the ratios of elements

34. 34 Molarity Some chemical reactions involve aqueous solutions of reactants The concentration of a solution is the amount of solute present in a given quantity of solvent or solution This concentration may be expressed in terms of molarity (M) or molar concentration: M = Molarity = no. of moles volume in Litres Molarity is the number of moles of solute in 1 Litre (L) of solution

35. 35 What is molarity of an 85.0mL ethanol (C 2 H 5 OH) solution containing 1.77g of ethanol? Step 1: Determine the number of moles of ethanol Example Molar mass of ethanol, C 2 H 5 OH: 2 × carbon atoms 1 × oxygen atom 6 × hydrogen atoms 2 × 12.01 gmol -1 1 × 16.00 gmol -1 6 × 1.008 gmol -1 24.02 gmol -1 16.00 gmol -1 6.048 gmol -1 46.07 gmol -1 No. of moles = mass in g molar mass No. of moles ethanol = 1.77g. 46.07 gmol -1 = 0.038 mol

36. 36 Step 2: Convert to molarity Have 85.0mL ethanol 1 L = 1000mL  Have 0.085 L of ethanol Molarity = no. of moles volume in L = 0.038 mol 0.085 L = 0.45 molL -1 ≡ 0.45 M Questions Calculate the molarities of each of the following solutions: (a) 2.357g of sodium chloride (NaCl) in 75mL solution Answer: 0.5378 M (b) 1.567mol of silver nitrate (AgNO 3 ) in 250mL solution Answer: 6.268 M (c) 10.4g of calcium chloride (CaCl 2 ) in 2.20 × 10 2 mL of solution Answer: 0.426 M

37. 37 Example An antacid tablet is not pure CaCO 3 ; it contains starch, flavouring, etc. If it takes 41.3mL of 0.206 M HCl to react with all the CaCO 3 in one tablet, how many grams of CaCO 3 are in the tablet. You are given the following balanced equation: 2HCl (aq) + CaCO 3(s)  CaCl 2(aq) + H 2 O (l) + CO 2(g) Step 1: Determine the no. of moles of HCl that react Have 0.206 M HCl solution  have 0.206 mol in one litre Have 41.3 mL of HCl solution  have 0.0413 L of HCl solution Molarity = no. of moles volume in L  no. of moles = Molarity × volume in L = 0.206 molL -1 × 0.0413L = 0.0085 mol ≡ 8.5 × 10 -3 mol HCl

38. 38 Step 2: Determine no. of moles of CaCO 3 used in the reaction 2HCl (aq) + CaCO 3(s)  CaCl 2(aq) + H 2 O (l) + CO 2(g) From the balanced equation, we can see that 2 moles of HCl are required to react with one mole of CaCO 3 Therefore, if 8.5 × 10 -3 mol of HCl are present in the reaction, we must have 4.25 × 10 -3 mol of CaCO 3 present. Molar mass of CaCO 3 : 1 × calcium atom 1 × carbon atom 3 × oxygen atoms 1 × 40.08 gmol -1 1 × 12.01 gmol -1 3 × 16.00 gmol -1 40.08 gmol -1 12.01 gmol -1 48.00 gmol -1 100 gmol -1 No. of mols = mass in g molar mass  Mass in g = no. of mols × molar mass = (4.25 × 10 -3 mol) × (100 gmol -1 ) = 0.425 g CaCO 3 present in tablet

39. 39 Questions (a) How many moles of NaCl are present in 25.00mL of 1.85M NaCl (aq) ? Answer: 4.62 × 10 -2 mol NaCl (b) What volume of a 1.25 × 10 -3 M solution of C 6 H 12 O 6(aq) contains 1.44 × 10 -6 mol of glucose? Answer: 1.15 mL (c) If stomach acid, given as 0.1 M HCl, reacts completely with an antacid tablet containing 500mg of CaCO 3, what volume of acid in millilitres will be consumed? The balanced equation is: CaCO 3(s) + 2HCl (aq)  CaCl 2(aq) + CO 2(g) + H 2 O (l) Answer: 100mL acid

40. 40 Important formulae… No. of moles = mass in g molar mass Molarity = no. of moles volume in L Calculating the number of moles: Calculating the molarity or concentration: