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- 1. INTRODUCTION TO THE MOLE (chemistry)
- 2. WHAT IS A MOLE? • The mole is a unit that is used to quantitatively measures the amount of substance. • 1 mol (of any substance) contains 6.022 x 1023 particles (or molecules). Avogadro’s number
- 3. MOLAR MASS • Defined as the mass (g) of a substance per mol (mol-1) = g/mol • Molar mass of any substance can be calculated from the periodic table (moles x molecular weight) • e.g. molar mass of H2O = (2 x 1.01) + 16.00 = 18.02 g/mol
- 4. EXAMPLE CALCULATION • Calculate the molar mass of the following compounds (using BOS periodic table https://www.boardofstudies.nsw.edu.au/syllabus_hsc/pdf_doc/chemistry-data-sheet.pdf): • Copper (I) sulfate 1. Write out the chemical formula (REMEMBER: net charge of compound must be zero) • Copper (I) ionises into Cu+ in solution and sulfate ion (SO4 2-) has a net charge of -2 in solution. • Sum of charges = (+1) + (-2) = -1 • In order to balance the charges, there must be 2 Cu+ ions for every SO4 2- ion. With this, sum of charges = 2(1) + (-2) = 0 • Cu2SO4 , since 2 copper ions for every sulfate ion 2. Calculate the molar mass using the periodic table and chemical formula • Molar mass (Cu2SO4) = 2(63.55) + 32.07 + 4(16.00) = 223.15 g/mol
- 5. TRY THIS! • Now try calculating the molar mass of these on your own: • Magnesium hydroxide • Iron (III) oxide (ferric oxide) • Calcium carbonate
- 6. CHECK YOUR ANSWERS
- 7. REARRANGING THE FORMULA • Molar mass = m / n • To make m the subject: • Flip the equation so m is on the left-hand side: • m / n = Molar mass • Have to get rid of / n on left-hand side, so we multiply both sides by n: • m x n / n = Molar mass x n • Any number divided by itself equals 1, therefore n/n=1 : • m x 1 = Molar mass x n • Any number multiplied by 1 is equal to itself, so: • m = Molar mass x n
- 8. EXAMPLE CALCULATION • Now that we can rearrange the molar mass formula to make m the subject, we can calculate the mass of any substance, given the molar mass and moles. • Calculate the mass of 0.05 mol of iron (III) oxide 1. Write the chemical formula • Fe2O3 2. Calculate the molar mass from the periodic table • Molar mass (Fe2O3) = 2(55.85) + 3(16.00) = 159.7 g/mol 3. Substitute data into the formula to find the mass (m) • Molar mass = 159.7 g/mol, n = 0.05 mol m = Molar mass x n = 159.7 x 0.05 = 7.985 g 4. Round off to the least number of significant figures • m (Fe2O3) = 8.00 g (3 sig. fig.)
- 9. TRY THIS! • Now try answering these on your own: • Calculate the mass (in g) of: • 60 mol of calcium carbonate • 0.0045 mol of sodium bicarbonate • 50 µmol of sodium chloride • Give all answers to 2 significant figures with scientific notation
- 10. CHECK YOUR ANSWERS • 60 mol of calcium carbonate (CaCO3) • m = 60 x 100.09 = 6.0 x 103 g • 0.0045 mol of sodium bicarbonate (NaHCO3) • m = 0.0045 x 84.008 = 3.8 x 10-1 g • 50 µmol of sodium chloride (NaCl) • m = 50 x 10-6 x 58.44 = 2.9 x 10-3 g
- 11. REARRANGING THE FORMULA • Molar mass = m / n • To make n the subject: • Flip equation so n is on the left: • m / n = Molar mass • Divide both sides by m: • m/m / n = Molar mass / m • m/m = 1: • 1 / n = Molar mass / m • Take the reciprocal of both sides: • n = m / Molar mass
- 12. EXAMPLE CALCULATION • Now that we can rearrange the molar mass formula to make n the subject, we can calculate the moles of any substance, given the molar mass and mass. • Calculate the number of moles in 45.7 kg of iron (III) oxide 1. Write the chemical formula • Fe2O3 2. Calculate the molar mass from the periodic table • Molar mass (Fe2O3) = 2(55.85) + 3(16.00) = 159.7 g/mol 3. Substitute data into the formula to find the number of moles (n) • Molar mass = 159.7 g/mol, m = 45.7 x 103 g n = m / Molar mass = 45700 / 159.7 = 286.16155 mol 4. Round off to the least number of significant figures • n (Fe2O3) = 286 mol (3 sig. fig.)
- 13. TRY THIS! • Now try answering these on your own: • Calculate the moles in: • 2.1 g of barium sulfate • 0.5 g of cobalt (II) phosphate • 1.68 tonnes of lead • Give all answers to 2 significant figures with scientific notation
- 14. CHECK YOUR ANSWERS • 2.1 g of barium sulfate (BaSO4) • n = 2.1 / 233.43 = 9.0 x 10-3 mol • 0.5 g of cobalt (II) phosphate (Co3(PO4)2) • n = 0.5 / 366.74 = 1.4 x 10-3 mol • 1.68 tonnes of lead • n = 1.68 x 106 / 207.2 = 8.1 x 103 mol
- 15. HOW CAN WE USE MOLAR MASS? • Molar mass = m (mass) / n (moles) • We can rearrange the formula above to find the mass of substance, given the number of moles • Likewise, we can find the moles of a substance, given the mass • We can calculate the concentration of any substance: • Concentration is the amount of substance (n) dissolved in solution per given volume (L) • M (molar concentration) = n / V (volume in litres) • M is measured in mol/L
- 16. EXAMPLE CALCULATION • Dissolving 20.0 g of sodium chloride in 2.5 L of water will give a salt solution of what concentration (in mol/L)? 1. Calculate the molar mass of sodium chloride • (using this periodic table: https://www.boardofstudies.nsw.edu.au/syllabus_hsc/pdf_doc/chemistry-data-sheet.pdf) molar mass (NaCl) = 22.99 + 35.45 = 58.44 g/mol 2. Find out how many moles of sodium chloride has been dissolved • Molar mass = m / n, therefore rearranging the formula gives n = m / Molar mass n (NaCl) = 20.0g / 58.44 g mol-1 = 0.3422313 mol 3. Divide by the volume to give molarity • M = n / V M = 0.3422313 mol / 2.5 L = 0.13689254 mol/L 4. Round off to the least number of significant figures given in question • 20.0 is to 3 sig. fig., 2.5 is to 2 sig. fig., therefore round final answer to 2 sig. fig. M (NaCl) = 0.1 mol/L or 0.1 M
- 17. TRY THIS! • Now try calculating the molarity of the following on your own (assuming complete dissociation of ions): • 10 g of copper (II) sulfate in 800 mL solution • 0.05 g of barium hydroxide in 10 mL solution • Extension: 25 mL of 0.5600 M sulfuric acid diluted to a total volume of 100 mL
- 18. CHECK YOUR ANSWERS • 10 g of copper (II) sulfate in 800 mL solution • 0.1 M • 0.05 g of barium hydroxide in 10.0 mL solution • 0.03 M • 25.0 mL of 0.5600 M sulfuric acid diluted to a total volume of 100 mL • Find the dilution factor = final volume / initial volume: • 25.0 / 100 = ¼ • Multiply the dilution factor by the original concentration (give answer to 3 sig. fig.): • ¼ x 0.5600 = 0.14 M
- 19. SUMMARY • Moles are a unit of measurement for the amount of substance, specifically 6.022 x 1023 molecules • So, molar mass is the mass (g) of 6.022 x 1023 molecules of substance (or 1 mol) • Atomic mass of any element on the periodic table is equivalent to its molar mass • e.g. Hydrogen has atomic mass of 1.01 amu, therefore it has a molar mass of 1.01 g/mol • Molar mass (g/mol) = m (g) / n (mol) • Concentration is the amount of substance dissolved per given volume • Concentration or Molarity (M) = n (mol) / V (L)