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)