Topic 1- Formulae, equations and amount of

substance
Topic 1- Formulae, equations and amount of
substance
Find the definitions of the following terms:
• Atom
• Element
• Ion
• Compound
• Empirical formula
• Molecular formula
Topic 1- Formulae, equations and amount of
substance
• Atom: smallest part of an element that has the properties of that element
• Element: A most simple substance that cannot chemically be broken down into
any simpler substance
• Ion: a particle of one or more atoms joined together that carries a positive or a
negative charge
• Molecule = a particle in which 2 or more atoms have combined
• Compound: a substance in which 2 or more elements have combined
• Empirical formula: shows the smallest whole number ratio of the atoms of each
element in a compound
• Molecular formula: shows the actual number of the atoms of each element in a
compound
Topic 1- Formulae, equations and amount of
substance
• A mole = the amount of substance that contains the same number of
particles as the number of carbon atoms in exactly 12g of the 12C
isotope

• Or also 6.02 x 1023 particles which could be atoms, molecules, ions,

electrons, …

• Molar mass = the mass of 1 mole of any substance expressed in g/mol

Topic 1- Formulae, equations and amount of
substance – Balancing symbol equations
Watch the video clip if forgotten how to it
Topic 1- Formulae, equations and amount of
substance – Ionic equations
Practice,
practice,
practice!!
Topic 1- Formulae, equations and amount of
substance
Topic 1- Formulae, equations and amount
of substance
Topic 1- Formulae, equations and amount of
substance
Topic 1- Formulae, equations and amount of
substance
Topic 1- Formulae, equations and amount of
substance
Topic 1- Formulae, equations and amount of
substance – More definitions
 
• Relative atomic mass of an element is the weighted average mass of all isotopes of the element
compared to 1/12th of the mass of one 12C atom which has a mass of 12.

• Relative molecular mass is the average mass (allowing for all the isotopes of the elements in the
compound) of a molecule compared to 1/12 th of the mass of one 12C atom.

• Relative formula mass is used instead of relative molecular used for ionic compounds and giant covalent
substances.
 
• Molar mass is the mass per mole of a substance in g mol-1
• Parts per million (ppm) used to indicate concentration e.g. including gases in the atmosphere or
dissolved in liquids.
Topic 1- Formulae, equations and amount of substance – More definitions

Relative molecular mass ( Mr )

 
To be able to calculate the Mr correctly, the formula of the compound must be correct and the relative atomic
masses (including 2 decimals) of the elements in the compound need to be used.
 
Examples:
 relative molecular mass of H2O is 18.02 because (2 x 1.01) + (1 x 16.00) = 18.02

 relative molecular mass of H2SO4 is 98.08 because (2 x 1.01) + 32.06 + (4 x 16.00) = 98.08

Relative formula mass (used for ionic compounds and giant covalent substances 

Examples:
 relative formula mass of NaOH is 40 because 22.99 + 16.00 + 1.01 = 40.00
 relative formula mass of Al2(SO4)3 is because (2 x 26.92 ) + (3 x 32.06) + (12 x 16.00) = 342.02
Topic 1- Formulae, equations and amount of
substance – More definitions
Topic 1- Formulae, equations and amount of
substance – Why the mole?
Example: How much iron oxide is needed to produce 28g of iron during the
reduction of iron oxide by carbon monoxide?

Fe2O3 + 3CO 2Fe + 3CO2

Topic 1- Formulae, equations and amount of
substance – What is the mole, the central concept
• As particles such as atoms, ions and molecules are extremely small chemists
measure amounts of substance using a fixed amount called the MOLE.

• The fixed amount is referred to as Avogadro’s constant which has a value of

6.02 x 1023 mol-1
• So 1 mole of a substance = 6.02 x 1023 units of that substance; these units can be
atoms, molecules, ions, electrons or formula units.

• 1 mole of atoms = 1 mole x 6.02 x 1023 mol-1 = 6.02 x 1023 atoms. We can use the
mole concept for any kind of particle e. atoms, molecules, formula units in case of
ionic compounds or complex ions, electrons, ..
Topic 1- Formulae, equations and amount of
substance – molar mass
• How do we measure 1 mole of a substance?

• The mass of 1 mole of a substance = molar mass which is measured in g

mol-1.

• The molar mass is the mass of 6.02 x 1023 units of a substance; it is the
relative atomic, molecular or formula mass but expressed in grams as
the molar mass is an absolute mass.

• You should be able to calculate the molar mass of many different

substances.
Topic 1- Formulae, equations and amount of
substance - mole

• How many moles is 6.02 x 1021 particles?

• How many particles is 0.05 moles?

Mole = number of particles /6.02 x 1023

Topic 1- Formulae, equations and amount of
substance – How to take 1 mole? Molar mass
Topic 1- Formulae, equations and amount of
substance – Molar mass
Topic 1- Formulae, equations and amount of substance - mole

The mole is a very useful concept in quantitative chemistry.

We should be able to calculate it starting from amounts and be to use ratio’s.

V = molar volume
M = molar mass;
V = actual volume
m = mass
Topic 1- Formulae, equations and amount of substance - mole
Topic 1- Formulae, equations and amount of
substance – Molar mass and moles
Topic 1- Formulae, equations and amount of
substance – Molar mass and moles
Topic 1- Formulae, equations and amount of
substance – Molar mass and moles
Topic 1- Formulae, equations and amount of substance – molar mass and moles
Topic 1- Formulae, equations and amount of
substance – Molar mass and moles - answers
Topic 1- Formulae, equations and amount of
substance – Let’s use the mole
Using molar ratios in compounds or formula units:
 
Examples:
In 1 mole of Fe2O3 there are 2 moles of iron ions and 3 moles of oxygen ions.

In 3 moles of Na2CO3 there are 3 moles of CO32- ions and 6 moles of Na+ ions.

In 2 moles of (NH4)3PO4 there are 6 moles of NH4+ ions and 24 moles of hydrogen atoms.

Exercises
 
1. If there are 2 x 1022 C-atoms in a sample of ethane, C2H6, how many hydrogen atoms are there in that same
sample?
2. If you have 4.5 x 1022 atoms of hydrogen in a sample of ammonia, NH3, how many molecules of ammonia
do you have?
3. How many moles of oxide ions are there in 2 moles of Al2O3?
4. How many moles of NO3- ions are there in 1 mole of Ga(NO3)3?
5. What is the total number of atoms in 0.05 moles of NH4NO3?
 Topic 1- Formulae, equations and amount of
substance - mole

1. Find the mass of 1 molecule of ethanol,

C2H5OH.

2. Find the number of carbon atoms in 0.1

mole of C2H5OH

3. Which sample contains the greatest

number of particles?
5g of NH3 or 5g of H2O or 5g of CaO?
Topic 1- Formulae, equations and amount of
substance – Let’s use the mole
Topic 1- Formulae, equations and amount of
substance - mole
Topic 1- Formulae, equations and amount of substance - mole

Answer = 0.1 x 6.02 x 1023 = 6.02 x 1022 molecules of C2H5OH x 2 C atoms molecule-1 = 1.204 x 1023 atoms of carbon.

NH3
Topic 1-
Formulae,
equations and
amount of
substance -
mole
Topic 1- Formulae, equations and amount of substance - mole

• Empirical formula = the simplest whole number ratio of the atoms present
in a compound. (obtained from experimentation)

• Molecular formula = shows the actual number of atoms (or ions) of each
element present in the molecule (or formula unit) of a compound.

•For each of the molecular formula below, find the empirical formula.
a) H2O b) H2O2 c) C6H6 d) C2H6 e) C2H4 f) C4H8O2 g) NH3 h) Fe2O3 I) Al2Cl6
Topic 1- Formulae, equations and amount of substance –
Determining empirical formula

What information do we need so that we can determine the empirical

formula?
Topic 1- Formulae, equations and amount of substance –
Determining empirical formula

What information do we need so that we can determine the empirical formula?

Determining experimentally the empirical
formula of magnesium oxide.

Mass of magnesium (g) 0.0245 What is the mass of

Mass of magnesium oxide (g) (1st reading) 0.0411 oxygen that reacted with
Mass of magnesium oxide (g) (2nd reading) 0.0411 magnesium? Heat it to
constant mass
Topic 1- Formulae, equations and amount of
substance – Determining empirical formula

elements Mg O
mass (in g or %) 0.0245 0.0166

moles
Simple ratio
Topic 1- Formulae, equations and amount of
substance – Determining empirical formula

elements Mg O
mass (in g or %) 0.0245 0.0166

molar mass/relative 24 16
atomic mass
number of moles 0.00102 0.00103
most simple molar ratio 0.00102/0.00102 = 1 0.00103/0.00102 = 1
1:1
empirical formula MgO
Topic 1- Formulae, equations and amount of
substance – Determining empirical formula
elements in compound carbon hydrogen
 
mass in grams (or in %) 47.98 (74.9%) 16.11 (25.1%)
Empirical formula – Rounding up/down
• Elements X Y
moles ratio 1 1.25

• Elements X Y Z
1.33 2 1

• Elements X Y Z
1. 1.67 1
Topic 1- Formulae, equations and amount of
substance – Determining empirical formula
elements in compound carbon hydrogen
 
mass in grams (or in %) 47.98 (74.9%) 16.11 (25.1%)
     
number of moles 47.98/12 = 3.99 mole 16.11/1 = 16.02 moles
(74.9/12 = 6.24 moles) (25.1/1 = 25.1 moles)
 
ratio of moles 4 (6.25) 16 (25)
most simple ratio of moles 4/4 = 1 (6.25/6.25 = 1) 16/4 = 4 (25/6.25 = 4)
 
empirical formula CH4
Empirical formula of simple compounds
• A student carried out an experiment to calculate the empirical
formula of an oxide of iron. She burnt 3.808 g of iron until it has all
reacted. She found that the mass of the product was 5.440 g.
• Calculate the empirical formula. Watch out for the rounding off.
Topic 1- Formulae, equations and amount of
substance – Determining molecular formula
There are many possible molecular formulae that can be derived from one empirical formula.
However, if the relative molecular/formula is known, the molecular formula can be obtained.

Example: Calculate the molecular formula of a compound with a relative molecular mass of 84 and an
empirical formula of CH2.

Answer:

• mass empirical formula: 14g

• relative molecular mass of formula: 84
• whole number ratio (n) of molar mass/empirical mass = 84 g/14g = 6
• molecular formula = n x empirical formula: (CH2 ) x 6 = C6H12
Topic 1- Formulae, equations and amount of substance – Determining molecular formula

• Example: Naphthalene, best known as ‘mothballs’, is composed of

carbon (93.71%) and hydrogen (6.29%). If the molar mass of the
compound is 128g, what is the molecular formula of naphthalene?
 Topic 1- Formulae, equations and amount of substance – Determining molecular formula

 Example: Naphthalene, best known as ‘mothballs’, is composed of carbon (93.71%)

and hydrogen (6.29%). If the molar mass of the compound is 128g, what is the
molecular formula of naphthalene?
  carbon hydrogen
mass of element 93.71 6.29
number of moles
 
most simple ratio
 
lowest whole number ratio
empirical formula:
ratio molecular formula
/empirical formula:
molecular formula = 2 x C5H4 =
 Topic 1- Formulae, equations and amount of substance – Determining molecular formula

 Example: Naphthalene, best known as ‘mothballs’, is composed of carbon (93.71%)

and hydrogen (6.29%). If the molar mass of the compound is 128g, what is the
molecular formula of naphthalene?
  carbon hydrogen
mass of element 93.71 6.29
number of moles 93.71/12 = 7.8 6.29/ 1 = 6.29
  DO NOT ROUND UP OR DOWN
most simple ratio 7.8/ 6.29 = 1.25 6.29/6.29 = 1
  AGAIN DO NOT ROUND UP OR DOWN BUT FIND LOWEST
WHOLE NUMBER RATIO
lowest whole number ratio 5 4
empirical formula: C 5 H4
ratio molecular formula 128g/ 64g = 2  
/empirical formula:  
molecular formula = 2 x C5H4 = C10H8
Answer to slide 16
• 5.440 – 3.808 = 1.632

• Fe O
• 3.808/56 1.632/16
• 0.068 0.102
• 0.068/0.068 0.102/0.068
• 1 1.5
• 2 3 A= Fe2O3
 Topic 1- Formulae, equations and amount of
substance – Determining molecular formula
For each of the following substances, calculate the empirical formula and the molecular formula

1. Water (one mole = 18g) that contains 11.1% of hydrogen and 88.9% of oxygen.
2. Ammonia (one mole = 17g) that contains 82.4% of nitrogen and 17.6% of hydrogen.
3. Potassium carbonate (one mole = 138g) that contains 56.5% of potassium, 8.7% of carbon and 34.8% of
oxygen.
4. Hydrazine (one mole = 32g) that contains 87.5% of nitrogen and 12.5% of hydrogen.
5. Glucose (one mole = 180g) that contains 40.0% of carbon, 6.67% of hydrogen and 53.3% of oxygen.
6. Ethane (one mole = 30g) that contains 80% of carbon and 20% of hydrogen.
7. Phosphorus(III) chloride (one mole = 137.5g) that contains 22.5% of phosphorus and 77.5% of chlorine.
8. Butane (one mole = 58g) that contains 82.8% carbon and 17.2% of hydrogen.
9. A compound X (one mole = 342g) that contains 15.8% of aluminium, 28.1% of sulfur and 56.1% of oxygen.
The compound forms a white precipitate when it reacts with barium nitrate solution.
Topic 1- Formulae, equations and amount of substance
– Determining formula from combustion analysis
Topic 1- Formulae, equations and amount of substance –
Determining formula from combustion analysis
Topic 1- Formulae, equations and amount of substance
– Determining formula from combustion analysis
Topic 1- Formulae, equations and amount of substance
– Determining formula from combustion analysis
Topic 1- Formulae, equations and amount of substance
– Determining formula from combustion analysis
Topic 1- Formulae, equations and amount of substance
– Determining formula from combustion analysis
Topic 1- Formulae, equations and amount of
substance – Determining formula
Topic 1- Formulae, equations and amount of
substance – Determining formula
Topic 1- Formulae, equations and amount of
substance – Solutions
Topic 1- Formulae, equations and amount of
substance – Solutions – mass concentration
Topic 1- Formulae, equations and amount of
substance – Solutions – mass concentration
Topic 1- Formulae, equations and amount of
substance – Solutions
Topic 1- Formulae, equations and amount of substance –
Solutions
Topic 1- Formulae, equations and amount of
substance – Calculations involving solutions
Topic 1- Formulae, equations and amount of
substance – Calculations involving solutions
Topic 1- Formulae, equations and amount of
substance – Calculations involving solutions
Topic 1- Formulae, equations and amount of
substance – Calculations involving solutions
Topic 1- Formulae, equations and amount of substance – Calculations involving solutions
Topic 1- Formulae, equations and amount of
substance – Calculations involving solutions
Topic 1- Formulae, equations and amount of
substance – Calculations involving solutions
Topic 1- Formulae, equations and amount of
substance – Concentration in ppm
Topic 1- Formulae, equations and amount of
substance – Calculating reacting masses
Conservation of mass
 
When applying the molar mass concept to the equation below the Law of
Conservation of Mass can be proven:

Fe2 O3 (s) + 3CO (g)  2Fe (s) + 3 CO2 (g)

 
1 x 160 g + 3 x 28g 2 x 56 g + 3 x 44 g

244 g 244 g

mass before = mass after the reaction!!!!!

Topic 1- Formulae, equations and amount of
substance – Calculating reacting masses

Example: How much iron oxide is needed to produce 28g of iron during the
reduction of iron oxide by carbon monoxide?

Fe2 O3 (s) + 3CO (g)  2Fe (s) + 3 CO2 (g)

28/56 = 0.5 ratio : 1:2 0.25 mol iron oxide x 160 = 40g
Topic 1- Formulae, equations and amount of
substance – Calculating reacting masses

Example: How much iron can be produced from 80 g of iron oxide using the
process below.

Fe2 O3 (s) + 3CO (g)  2Fe (s) + 3 CO2 (g)

Topic 1- Formulae, equations and amount of
substance – Calculating reacting masses
Example: How much iron oxide is needed to produce 28g of iron during the reduction of
iron oxide by carbon monoxide?
Step 1: write balanced equation
Fe2 O3 (s) + 3CO (g)  2Fe (s) + 3 CO2 (g)
Step 2: convert data in question into moles
 
data in question = 28 g or iron number of moles of iron = 28g / 56g = 0.5 mole of iron
Step 3: find answer in equation i.e. use molar ratios
the equation tells us that to get 2 moles of Fe we need 1 mole of Fe2 O3 (= molar ratio)
 
 0.5 mole of iron needs 0.25 moles of Fe2 O3 .
Step 4: convert answer into relevant units
 mass = number of moles x molar mass = 0.25g x 160 g = 40 g
 
To get 28g or iron, 40g of iron oxide is needed to get the reaction to go to completion!!!
Topic 1- Formulae, equations and amount of
substance – Calculating reacting masses
Topic 1- Formulae, equations and amount of
substance – Calculating reacting masses
Topic 1- Formulae, equations and amount of
substance – Calculating reacting masses
Topic 1- Formulae, equations and amount of
substance – Calculating reacting masses
Topic 1- Formulae, equations and amount of
substance – Calculating reacting masses
Topic 1- Formulae, equations and amount of
substance – Calculating reacting masses
Topic 1- Formulae, equations and amount of
substance – Percentage yield
Topic 1- Formulae, equations and amount of
substance – Theoretical yield
Topic 1- Formulae, equations and amount of substance
– Theoretical yield
Topic 1- Formulae, equations and amount of
substance – Percentage yield
Topic 1- Formulae, equations and amount of
substance – Percentage yield
Topic 1- Formulae, equations and amount of
substance – Percentage yield
Topic 1- Formulae, equations and amount of
substance – Percentage yield
Topic 1- Formulae, equations and amount of
substance – Atom economy
The concept of atom economy has been introduced to emphasize the need for an
appreciation of sustainability of chemical reactions. Atom economy looks at the
efficiency (like percentage yield) of a chemical reaction. Both the mass of the
atoms of useful product and the total mass are calculated using molar mass and
multiplied by the coefficient as in the balanced equation.

Calculate the atom economy to extract silver according to:

2AgNO3 + Mg  2Ag + Mg(NO3)2

Topic 1- Formulae, equations and amount of
substance – Atom economy
The concept of atom economy has been introduced to emphasize the need for an
appreciation of sustainability of chemical reactions. Atom economy looks at the
efficiency (like percentage yield) of a chemical reaction. Both the mass of the
atoms of useful product and the total mass are calculated using molar mass and
multiplied by the coefficient as in the balanced equation.
Topic 1- Formulae, equations and amount of
substance – percentage yield & atom economy
• Watch power point “IAS Topic 1 percentage yield and atom economy”
for more examples.
 Topic 1- Formulae, equations and amount of
substance – Atom economy
Exercises

Calculate the percentage atom economy of the named product in the reactions
shown below:

1. sodium in the reaction 2NaCl  2Na + Cl2

2. hydrogen in the reaction Zn + 2HCl  ZnCl2 + H2
3. iron in the reaction Fe2O3 + 3CO  2Fe + 3CO2
4. calcium oxide in the reaction CaCO3  CaO + CO2
5. sulfur trioxide in the reaction 2SO2 + O2  2SO3
6. oxygen in the reaction 2H2O2  2H2O + O2
Topic 1- Formulae, equations and amount of
substance – Percentage yield & atom economy
Topic 1- Formulae, equations and amount of
substance – percentage yield & atom economy
Topic 1- Formulae, equations and amount of
substance – Formula and equation
Topic 1- Formulae, equations
and amount of substance –
Formula and equation
Topic 1- Formulae, equations and amount of
substance – Formula and equation
Topic 1- Formulae, equations and amount of substance – Formula and equation
Topic 1- Formulae, equations and amount of
substance – Formula and equation
Topic 1- Formulae, equations and amount of
substance – Exercises
Topic 1- Formulae, equations and amount of
substance – Exercises
Topic 1- Formulae, equations and amount of
substance – Volumes of gases
Topic 1- Formulae, equations and amount of
substance – Volumes of gases
volume of 1 mole of a gas = 2.4 x 10-2 m3 mol-1 or 24 dm3 mol-1
Topic 1- Formulae, equations and amount of
substance – Volumes of gases
Topic 1- Formulae, equations and amount of
substance – Volumes of gases
Topic 1- Formulae, equations and amount of
substance – Volumes of gases
Topic 1- Formulae, equations and amount of
substance – Volumes of gases2
Topic 1- Formulae, equations and amount of substance – Volumes of gases2
Topic 1- Formulae, equations and amount of
substance – Volumes of gases
Topic 1- Formulae, equations and amount of
substance – Volumes of gases
Topic 1- Formulae, equations and amount of
substance – Volumes of gases
Topic 1- Formulae, equations and amount of
substance – Volumes of gases – Core
Practical

Only watch the first experiment

Topic 1- Formulae, equations and amount of
substance – Volumes of gases – Core
Practical
How to reduce loss of gas
Topic 1- Formulae, equations and amount of
substance – gasses in ppm
Do now - pV =nRT
Do Now
Mr = mass/moles

Moles = pV/RT