A mole is a specific amount of 6x1023 particles of an element or molecule. Moles allow conversions between mass, number of particles, volume, and concentration. Balanced chemical equations provide mole ratios to solve multi-step stoichiometry problems involving excess reactants or product yields. Empirical formulas give the simplest whole number ratio of elements in a molecule.

1. Stoichiometry
Studentgrinds.ie lesson 2

2. Moles
● A mole is just a specific amount of something. More specifically, it’s 6×1023 of
anything.
● So, one mole of carbon means you have 6×1023 carbon atoms. In one mole of
CO2 means you have 6×1023 molecules of CO2.
● It’s important because moles are like a currency exchange office. It allows you to
convert one piece of information about a chemical to another, by converting to
moles first. There are 4 properties of any chemical that we can use to calculate
how many moles we have of it, or use moles to work out how much of it we have:
mass, number of molecules, gas volume or concentration.
● Moles = how much we have / how much is in one mole
● How much we have = moles x how much is one mole
You will see these 2
formulae written in
different ways over the
next few pages

3. Moles and Mass
● Molecular weight: the molecular weight of something tells you the mass of one
mole of it (how much it weighs, in grams). This is also called the ‘molar mass’
● The molecular weight of an element is its mass number, so you can read it right
off the periodic table. (So, one mole of C = 12g).
● The molecular weight of a molecule is the sum of the mass numbers of all the
atoms in it. (So, one mole of CO2 is C+O+O, or 12+12+16 +44g).
● Moles=mass (g) /molecular weight
● Mass (g)=moles x molecular weight

4. Moles and Molecules
● One mole of absolutely anything contains 6×1023 molecules (or atoms). So, in
one mole of F, there are 6×1023 F atoms; in one mole of H2O there are 6×1023
H2O atoms. The chemical itself does not matter! Do not use the periodic table
here!
● Moles=number of molecules / 6×1023
● Number of molecules=moles x 6×1023

5. Moles and Gas Volume
● One mole of absolutely any gas ( at standard temperature and pressure) will
have the following volume (size): 22,400cm3 . This is the same as 22,400mL, or
just 22.4L.
● Moles= volume (cm3 ) / 22,400
● Volume (cm3 ) = moles x 22,400.

6. Moles and Concentration
● Concentration means measuring how much of a chemical is dissolved in a
solution. We usually measure this as how many moles of the chemical are
dissolved in 1L of the solution.
● This method of measuring concentration is called moles per litre (it’s also called
molarity).
● Moles per litre=moles/litres
● Moles=moles per litre x litres
● Another way we can measure concentration is grams per L (g/L).
● Grams per litre=grams / litres.
● We can convert between these two types of concentrations using the following
2 formulae.
● Grams per litre=moles per litre x molecular weight
● Moles per litre=grams per litre/molecular weight

7. 2-part questions
● Before going into these, make sure you’re confident changing to moles and from
moles for any chemical.
● What these questions tell you about a chemical is its mass, number of
molecules, volume as a gas or concentration, and you must work out one of the
others for it.
● These questions have the same 2 steps every single time.
● 1. Convert what they’ve given you into moles.
● 2. Convert from moles to what the question is asking you.
Example: You have 12g of NaOH. How many molecules are in this sample?
1. Molecular weight of NaOH is 23+16+1=40. So to work out moles we do 12/40=0.3
2. To convert from moles to number of molecules we do 0.3 x 6×1023 = 1.8 x 1023 . So,
the answer is just 1.8 x 1023 molecules of NaOH.

8. 3-part questions
● In this case they give you information about one of the chemicals (either a reactant or a
product) and want you to work out information about another one.
● You must have a balanced equation for the reaction before you can do these!
● The three steps are:
● 1. Convert what they’ve given you into moles
● 2. Use the ratio in the chemical equation to work out how many moles you have of the other
chemicals.
● 3. Convert from moles into what the question is asking you, for these other chemicals.
● Example: In the following reaction, 15g of HCl are added. What is the volume of CO2 gas
produced? 2HCL + Na2CO3 CO2 + H2O + 2NaCl.
1. Change the HCl to moles. 15/36.5 = 0.41.
2. The ratio here between HCl and CO2 is 2:1. So if we have 0.41 moles of HCl, we should have 0.41 / 2 = 0.205 moles of CO2.
3. Convert moles of CO2 to gas volume. 0.205 x 22,400 = 4,592cm3

9. When one reactant is in abundance
● Sometimes we end up using more of one reactant than we really need.
● Some questions can ask you to work out which reactant is in abundance (which
one we have too much of) and by how much.
● There are 3 steps to this.
● 1. Work out how many moles you have of each reactant.
● 2. Use the balanced equation to work out how many moles you should have of
each reactant.
● 3. Compare these two values for either one of the reactants. If we have more
than we should, it’s in abundance; if we have less than we should, then the other
one is in abundance.
● Example: In the following reaction, we have 10g of C2H4 and 15g of O2. Which
reactant is in abundance? C2H4 + 3O2 2H2O + 2CO2
● 1. We have 10/28 = 0.35 moles of C2H4 and 15/32 = 0.47 moles of o2.
● 2.The ration between these reactants is 1:3. So, if we have 0.35 moles of C2H4, we should have 0.35 x 3 = 1.05
moles of O2.
● 3. However, we only have 0.47 moles of O2. We have less than we should, so, it is the other reactant, C2H4, which
is in excess.

10. The percentage yield
● This is kind of like the opposite to the last type of question. Instead of comparing
the moles we have to the moles we should have of a reactant, we compare the
mass we have and the mass we should have of a product.
● There are 2 steps to this.
● 1. Work out the mass we should have of the product, based on the information
they’ve given us. The way you do this is explained in the slide on ‘3-part
questions’.
● 2. Use the following formula: actual yield (g) / theoretical yield (g) x 100.
● Example: In the following reaction, we added 10g of C2H4 and produced 11g of
CO2. calculate the percentage yield of CO2. C2H4 + 3O2 2H2O + 2CO2
● 1. The ‘theoretical yield’ of CO2 is 30.8g. (Revise the slide on ‘3-part questions’ to see how I’ve worked this out).
● 2. Actual yield of CO2 is 11g, and the theoretical yield is 30.8g. So, the percentage yield of CO2 is 11 / 30.8 x 100 =
35.7%.

11. The Empirical Formula
● There are three ways we can describe any molecule. These are called the
molecular, the empirical and the structural formula.
● Molecular formula: the exact amount of atoms of each element in a molecule.
● Empirical formula: the simplest ratio between the number of atoms of each
element in a molecule.
● Structural formula: the actual structural layout of the different atoms in a
molecule. (This is only important in hydrocarbons and organic chemistry, not in
stoichiometry).
● These are the 3 ways to describe the same molecule, C2H6:
● Molecular formula: C2H6
● Empirical formula: CH3 H H
● Structural formula: H-C-C-H
● H H

12. Summary
● Understand what a mole is
● Convert moles to mass, number of molecules, gas volume and concentration
● Use mole ratios to work with chemical equations
● Calculate which reactant is in abundance, and the percentage yield of products
● Understand a compound’s molecular formula, empirical formula and structural formula