This document discusses the mole concept and stoichiometry calculations. It defines key terms like the mole, molar mass, empirical and molecular formulas, and limiting reagents. It also describes how to use balanced chemical equations and molar relationships to calculate amounts of reactants and products involved in chemical reactions. Experimental procedures are provided for determining relative molecular masses and identifying unknown metals based on stoichiometric calculations.

1. AMOUNT OF SUBSTANCE
Module 2
Chapter 3

2. AMOUNT OF SUBSTANCE AND THE
MOLE
The mole
• Chemists use moles to measure the number of particles in a substance (symbol - n and unit
– mol)
• One mole is the amount of a substance that contains 6.02×10 23 particles (Avogadr’s
constant and the number of particles in each mole of carbon-12)
• One mole of an element is equal to (the element’s mass number)g
• 1 mol of H = 1 mol of hydrogen atoms
• 1 mol of H 2 = 1 mol of hydrogen molecules

3. AMOUNT OF SUBSTANCE AD THE
MOLE
Molar Mass
• The mass per mol of a substance (units – g mol -1)
• Add up all atomic masses in the substance
Amount (n) = mass / molar mass

4. DETERMINATION OF FORMULAE
Molecular Formulae
• Some compounds are made up of small elements called molecules (2+ atoms held together
by covalent bonds)
• The molecular formula is the exact number of atoms of each element in a molecule
• Some elements exist as molecules (e.g H 2 , N 2 , O 2) so are shown in equations as their
molecular formula

5. DETERMINATION OF FORMULAE
Empirical Formula
• The simplest whole number ratio of atoms of each element in a compound
• This is important do substances that do not exist as molecules, including metals, some non
metals and ionic compounds)
• These form giant crystalline structures of atoms or ions
• It is the ratio of atoms or ions in a structure and will always be the same

6. DETERMINATION OF FORMULAE
Relative molecular mass
• Compares the mass of a molecule with the mass of an atom of carbon-12
• Add relative atomic masses of elements making up molecule
Relative formula mass
• Compares the mass of a formula unit with the mass of an atom of carbon-12
• Add relative atomic masses of elements in empirical formula
• Analysis – investigating the chemical composition of a substance

7. DETERMINATION OF FORMULAE
Empirical Formula from mass
1. Convert mass into moles (n=m/M)
2. Simplify (divide by smallest)
3. Write empirical formula
Determinatio of molecular formula
1. Convert % into moles (n=m/M)
2. Simplify (divide by smallest)
3. Write relative mass of empirical
4. Divide molecular mass by answer to part 3
5. Write molecular formula

8. DETERMINATION OF FORMULAE
Hydrated salts
• Many coloured salts are hydrated (water is a part of their chrystalline structure)
• This water is known as water of crystallization
• When blue crystals of hydrated copper (||) sulfate are heated, the bonds holding the water
within the crystal are broken, leaving behind white anhydrous copper (||) sulfate
• Without water the crystalline structure is lost leaving behind white powder, it is difficult to
remove all water (so powder is very light blue)
Hydrated salt -> anhydrous salt + water

9. MOLES AND VOLUMES
Volume
• Cubic centimetre (cm 3) = millitre (ml)
• Cubic decimated (dm 3) = litre (l)
Moles and solutions
• The concentration of a solution is the amount of salute, in moles, dissolved in each dm 3 of
solution (unit – mol dm -3)
Concentration = Volume / Amount (n)

10. MOLES AND VOLUME
Standard solution
• A solution of a known concentration
• Made by dissolving an exact mass of the solute in a solveand making up the
solution to an exact volume
Moles and gas volume
• Difficult to measure mass of a gas but easy to measure volume
• At the same pressure and temperature equal volumes of different gases contain the
same number of molecules

11. MOLES AND VOLUMES
Molar gas volume
• Volume per mol of gas at a stated temperature and pressure
• RTP – 20°C and 101kPa, 1 mol of gas has a volume of 24dm 3 (molar gas volume =
24 dm 3 mol -1)
Amount (n) = Volume / molar gas volume

12. MOLES AND VOLUMES
Ideal gas equation
• Assumptions: random motion, elastic collisions, negligible size, no inyermolecular
forces
Pressure × volume = amount (n) × ideal gas
constant × temperature
• Ideal gas constant = 8.314 J mol -1 k –1
• Volume: m 3, Temperature: K, Amount (n): mol, Pressure: Pa

13. MOLES AND VOLUMES
Conversions
• Cm 3 to m 3 × 10 -6
• Dm 3 to m 3 × 10 -3
• °C to K + 273
• KPa to Pa × 10 3

14. MOLES AND VOLUMES
Experiment – Finding a relative molecular mass
1. Get a compound which is liquid at room temperature but has a boiling point below 100°C
so it vaporises
2. Add a sample of the volatile liquid to a small syringe via a needle and weigh
3. Inject sample into a gas syringe through the self sealing rubber cap
4. Reweigh syringe to find mass of volatile liquid added to gas syringe
5. Place gas syringe in a boiling water bath (100°C), the liquid should vaporize forming a gas
6. Record pressure
7. Convert all quantities to math ideal gas equation
8. Use equation to calculate unknown
9. Find molar mass

15. REACTING QUANTITIES
Stoichiometry
• In a balanced equation, the balancing numbers give the ratio of the amount, in moles, of
each substance. This ratio is called the stoichiometry of the reaction.
• Chemists use balanced equations to find: the quantities of reactant required to prepare a
desired quantity of a product, the quantities of products that should be formed from certain
quantities of reactants
Quantities from amounts and equations
1. Work out the amount in moles of whatever you can
2. Use the equation to work out the amount in moles of an unknown chemical
3. Work out unknown information required

16. REACTING QUANTITIES
Experiment – Identifying an unknown metal
1. Attach a gas syringe to a flask
2. Weigh a sample of the unknown metal and place it in the flask
3. Add 25 cm 3 1 mol dm -3 into the flask and put in a bung
4. Measure the maximum volume of gas in the syringe
5. Calculate mols of gas created
6. Determine mols of unknown metal
7. Work out unknown information

17. REACTING QUANTITIES
Percentage yield
• Reasons for avtual yield not equal to theoretical: reaction may not have gown
into completion, Side reactions have taken place alongside main reaction,
purificstion resulted in loss
• Percentage yield is never 100%
Percentage yield = (actual yield / theoretical
yield) × 100

18. REACTING QUANTITIES
Limiting reagent
• The reactant which is not in excess so Is used up, stopping the reaction
• The reactant with more moles is the limiting raegent
Atom economy
• A measure of how well atoms have been utilised
• High a to economy: few waste products, more sustainable
Atom economy = (Mr of desired / Mr of total) ×
100