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# The mole (chemistry)

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Moles and molar mass
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# The mole (chemistry)

An introduction to the mole and its use in chemistry, including its use to calculate molar mass and concentration

An introduction to the mole and its use in chemistry, including its use to calculate molar mass and concentration

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### The mole (chemistry)

1. 1. INTRODUCTION TO THE MOLE (chemistry)
2. 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. 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. 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. 5. TRY THIS! • Now try calculating the molar mass of these on your own: • Magnesium hydroxide • Iron (III) oxide (ferric oxide) • Calcium carbonate
6. 6. CHECK YOUR ANSWERS
7. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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)