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JF Tutorial: Mole Calculations

Some Mathematical Functions 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. JF Tutorial: Mole Calculations. Shane Plunkett plunkes@tcd.ie. Recommended reading

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JF Tutorial: Mole Calculations

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  1. Some Mathematical Functions • 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 JF Tutorial: Mole Calculations Shane Plunkett plunkes@tcd.ie • Recommended reading • T.R. Dickson, Introduction to Chemistry, 8th Ed., Wiley, Chapters 2 & 4 • M.S. Silberberg, Chemistry, The Molecular Nature of Matter and Change, • 3rd Ed., Chapter 3 • P. Atkins & L. Jones, Molecules, Matter and Change, 3rd Ed., Chapter 2 • Multiple choice tests: http://www.mhhe.com/silberberg3

  2. Scientific Notation • Makes it easier to deal with large numbers, especially concentrations • Written as A ×10b, where A is a decimal number and b is a whole number Carrying out Calculations 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 × 1023 In chemistry, must deal with several mathematical functions. • Calculators: • Sharp & Casio Type in 6.022 • Press the exponential function [EXP] • Key in 23

  3. Questions How would you write the following: 7.84 × 108 2.3 × 10-4 9.22 × 106 1.5 × 10-8 (a) 784000000 (b) 0.00023 (c) 9220000 (d) 0.000000015 Calculate the following: 1.13 × 1011 (a) (1.38 × 104) × (8.21 × 106) 2.05 × 10-3 (b) (8.56 × 10-8) × (2.39 × 104)

  4. Common Decimal Prefixes

  5. 2. Logarithms • Makes dealing with a wide range of numbers more convenient, especially pH • 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 • 10n = x • written as: log10x = 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: log101000 = 3 103 = 1000

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

  7. Natural logarithms • Natural log of x is denoted ln x • 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 • lnx or logex = n or en = 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. Questions What is the natural log of: • 50 • 1.25 × 105 • 2.36 × 10-3 • 8.98 × 1013 ln 50 ln 1.25x105 ln 2.36x10-3 ln 8.98x1013 3.91 11.74 -6.05 32.13

  9. 3. Graphs • Experimental data often represented in graph form, especially in straight lines • 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 Slope Intercept x-axis

  10. Sign of slope tells you the direction of the 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: Example Given the (x, y) coordinates (2, 4) and (5, 9), find the slope of the line containing these two points. x1 = 2 y1 = 4 x2 = 5 y2 = 9 Sub into above relation: m = 9 – 4 = 5 = 1.67 5 – 2 3

  11. 4. Quadratic Equations • May be encountered when dealing with concentrations • Involve x2 (x-squared terms) • Take the form ax2 + bx + c = 0 • Can be solved by: • this expression finds the roots or the solution for x of the quadratic equation

  12. Example: Find the roots of the equation x2 – 6x + 8 = 0 ax2 + bx + c = 0 a = 1 b = -6 c = 8 x = x = x = = = 2 or = 4 Therefore, x =

  13. Question x = [H3O+] The following quadratic equation has been given Solve for x. 2.4x2 + 1.5x – 3.6 = 0 You have been asked to calculate the concentration of [H3O+] ions in a chemical reaction.

  14. Therefore or 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. Important formulae so far… y = mx + c Graphs….. ax2 + bx + c = 0 Quadratic equations…..

  16. Calculations: The Mole • Stoichiometry is the study of quantitative aspects of chemical formulas and reactions • 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 ×1023 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 1.9926×10-23g. No. of carbon-12 atoms = atomic mass (g) mass of one atom (g) = 12g _ 1.9926×10-23g = 6.022 ×1023 atoms

  17. Other definitions of the Mole • One mole contains Avogadro’s Number (6.022 x 1023) • 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 • 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

  18. One mole of any object always means 6.022 × 1023units of those objects.For example, 1 mol of H2Ocontains 6.022 × 1023 molecules 1 mol of NaCl contains 6.022 × 1023 formula units Calculating the number of particles 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 (NA) Therefore No. of atoms = (0.450mol) × (6.022 × 1023) = 2.7 × 1023 atoms

  19. Example No. of molecules = no. of moles × Avogadro’s number (NA) = 4mol × (6.022 × 1023 mol-1) How many molecules are there in 4 moles of hydrogen peroxide (H2O2)? = 24 ×1023 molecules = 2.4 × 1024 molecules Questions How many atoms are there in 7.2 moles of gold (Au)? Answer: 4.3 × 1024 atoms The visible universe is estimated to contain 1022 stars. How many moles of stars are there? Answer: 1022 stars = 1022 = 0.17 mol. 6.022×1023

  20. Calculating the mass of one molecule Step 1: Calculate the molar mass of water Example: What is the mass of one molecule of water? Molar mass of water = (2 × atomic mass H) + (1 × atomic mass O) Molar mass H2O = (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 6.022×1023mol-1 = 2.992×10-23g Note: Always check the units you have in your answer to ensure you are correct

  21. Example Step 1: Calculate the molar mass 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 2 Nitrogen atoms 8 Hydrogen atoms 1 Carbon atom 3 Oxygen atoms Calculate the mass of one molecule of ammonium carbonate [(NH4)2CO3] Total = 96.09 gmol-1 Step 2: Employ Avogadro’s Number, NA Mass of one molecule = 96.09 gmol-1 . 6.022×1023mol-1 = 1.59 × 10-22g Questions Calculate the mass of one molecule of: • Ethanoic acid (CH3COOH) • Methane (CH4) • Potassium dichromate (K2Cr2O7) 9.96 × 10-23 g 2.66 × 10-23 g 4.89 × 10-22 g

  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 Mr: 40.00 gmol-1 O: 16.00 gmol-1 H: 1.008 gmol-1 Step 2: Substitute into the above expression Mass of sample = 0.75mol × 40.00 gmol-1 = 30g

  23. Questions (a) 0.57mol of potassium permanganate (KMnO4) Answer: Molar mass KMnO4 = 158.03 gmol-1 Calculate the mass in grams present in: Mass in grams = 0.57mol × 158.03 gmol-1 = 90.07 g (b) 1.16mol of oxalic acid (H2C2O4) Answer: Molar mass H2C2O4 = 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. Converting between moles and mass Example Convert 25.0g of KMnO4 to moles Number of moles = mass of sample (g) molar mass (gmol-1) 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 Mr = 158.03 gmol-1 Step 2: Substitute into above expression 25.0g . 158.03gmol-1 No. of moles = = 0.158 mol

  25. Questions (a) 1.00g of water (H2O) Answer: Molar mass water = 18.02 gmol-1 Calculate the number of moles in: 1.00g H2O = 0.055mol (b) 3.0g of carbon dioxide (CO2) Answer: Molar mass carbon dioxide = 44 gmol-1 3.0g CO2 = 0.068mol (c) 500g of sucrose (C12H22O11) Answer: Molar mass sucrose = 342.30 gmol-1 500g C12H22O11 = 1.46mol (d) 2.00g of silver chloride (AgCl) Answer: Molar mass silver chloride = 143.38 gmol-1 2.00g AgCl = 0.014mol

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

  27. Calculating mass percentage from a chemical formula A chemical formula of a compound tells you the composition of that compound in terms of the number of atoms of each element present. Many of the elements in the periodic table of the elements occur in combination with other elements to form compounds. The mass percentage composition allows you to determine the fraction of the total mass each element contributes to the compound. Example Ammonium nitrate (NH4NO3) 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 Two N atoms: 28.016 gmol-1 Four H atoms: 4.032 gmol-1 Three O atoms: 48.00 gmol-1 Molar mass NH4NO3 = 80.05 gmol-1

  28. Step 2: Determine the mass % composition for each element Mass fraction of N = 28.016g 80.05g Nitrogen: 28.016g N in one mol of ammonium nitrate 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. Therefore, the mass % composition of ammonium nitrate (NH4NO3) 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 C12H22O11? Answer: % Carbon: 42.1% % Hydrogen: 6.5% % Oxygen: 51.4%

  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 (C6H12O6) is CH2O, 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. Step 2: Determine the number of moles of each element present Therefore, 100g of our compound contains 50.05g of sulfur and 49.95g of oxygen. Since we are dealing with percentages, we can express the mass % as grams if we assume we have 100g of the compound. 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 SO2

  32. Question Answer: No. of mol Carbon = 2.27mol No. of mol Oxygen = 4.54mol Ratio 1:2 Empirical formula CO2 Determine the empirical formula of a compound that contains 27.3 mass% Carbon and 72.7 mass% Oxygen. 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: C5H8O4NNa

  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. Molarity The concentration of a solution is the amount of solute present in a given quantity of solvent or solution Some chemical reactions involve aqueous solutions of reactants 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. Example What is molarity of an 85.0mL ethanol (C2H5OH) solution containing 1.77g of ethanol? Molar mass of ethanol, C2H5OH: Step 1: Determine the number of moles of ethanol 2 × carbon atoms 2 × 12.01 gmol-1 24.02 gmol-1 1 × oxygen atom 1 × 16.00 gmol-1 16.00 gmol-1 6 × hydrogen atoms 6 × 1.008 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. Step 2: Convert to molarity 1 L = 1000mL  Have 0.085 L of ethanol Have 85.0mL 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 (AgNO3) in 250mL solution Answer: 6.268 M (c) 10.4g of calcium chloride (CaCl2) in 2.20 × 102 mL of solution Answer: 0.426 M

  37. Example An antacid tablet is not pure CaCO3; it contains starch, flavouring, etc. If it takes 41.3mL of 0.206 M HCl to react with all the CaCO3 in one tablet, how many grams of CaCO3 are in the tablet. You are given the following balanced equation: 2HCl(aq) + CaCO3(s) CaCl2(aq) + H2O(l) + CO2(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. Step 2: Determine no. of moles of CaCO3 used in the reaction From the balanced equation, we can see that 2 moles of HCl are required to react with one mole of CaCO3 Therefore, if 8.5 × 10-3 mol of HCl are present in the reaction, we must have 4.25 × 10-3 mol of CaCO3 present. 2HCl(aq) + CaCO3(s)  CaCl2(aq) + H2O(l) + CO2(g) Molar mass of CaCO3: 40.08 gmol-1 1 × calcium atom 1 × 40.08 gmol-1 12.01 gmol-1 1 × carbon atom 1 × 12.01 gmol-1 3 × oxygen atoms 3 × 16.00 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 CaCO3 present in tablet

  39. Questions Answer: 4.62 × 10-2 mol NaCl (a) How many moles of NaCl are present in 25.00mL of 1.85M NaCl(aq)? (b) What volume of a 1.25 × 10-3 M solution of C6H12O6(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 CaCO3, what volume of acid in millilitres will be consumed? The balanced equation is: CaCO3(s) + 2HCl(aq) CaCl2(aq) + CO2(g) + H2O(l) Answer: 100mL acid

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

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