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5-Formulae, Equations and
Substances
Formulae
Molar Mass, Mr
Empirical Formula
Calculation from Percentage Composition
Calculation from Masses on Combustion
Molecular Formula from Empirical Formula
Calculation of Molar Mass
Ionic Compounds
Cations
Anions
Chemical Equations
State Symbols
Solubility Rules
Ionic Equations
Mole Calculations
Ideal Gas Calculation
Water of Crystallisation Calculations
Mole to Number of Particles
Calculations from Chemical Equations
Mass and Volume Calculations
Mass to Mass Calculations
Volume of Gas Calculations
5-Formulae, Equations and Substances 1
, Mass to Volume of Gas Calculations
Volume of Gas to Volume of Another Gas
Percentage Yield
Atom Economy
Percentage Composition
Limiting Reagent
Concentration
Titrations
Back Titrations
Minimising Errors in Titrations
Errors due to Apparatus
Errors due to Poor Technique
Formulae
What is the Amount of Substances measured in?
Moles
The molecular formula of a compound shows the number of
atoms of each element in one molecule of the substance.
E.g. the molecular formula of glucose is C6 H12 O6 .
Molar Mass, Mr
Molar mass is the mass of 1 mole of the substance. The units
are g mol− 1 .
The molar mass is found by adding the relative atomic masses of all the atoms
present in the formula.
E.g. glucose C6 H12 O6 - carbon = 12, hydrogen = 1 and oxygen = 16.
Molar mass of glucose = (6x12) + (12x1) + (6x16) = 180 g mol− 1 .
Empirical Formula
What is the Empirical Formula?
The simplest whole number ratio of the atoms of each element in the substance.
5-Formulae, Equations and Substances 2
, E.g. the empirical formula of glucose is CH2 O.
Calculation from Percentage Composition
The mass of an element can be found from the mass of a substance. The values
can be converted to percentage by mass. This can be used to find the empirical
formula.
1. Divide each percentage by the RAM of the element - this is the number of moles
of each element in 100g of the compound.
2. Divide these results by the smallest value obtained - this gives the simple mole
ratio.
3. If the ratio are not integers, then times the number by 2 or 3.
Example: A compound contained 34.3% sodium, 17.9% carbon and 47.8%
oxygen by mass.
Empirical Formula from %
Element % Divide by RAM Divide by the smallest Ratio
Sodium 34.3 34.3/23 = 1.49 1.49/1.49 = 1 1
Carbon 17.9 17.9/12 = 1.49 1.49/149 = 2 1
Oxygen 47.8 47.8/16 = 2.99 2.99/1.49 = 2.01 = 2 2
Calculation from Masses on Combustion
The masses of carbon dioxide and water produced when a known mass of a
substance is burnt are measured.
The mass of carbon in carbon dioxide is calculated as mass of CO2 x 12/44.
The mass of hydrogen is mass of water x 2/18.
The percentage of carbon and hydrogen are calculated and any oxygen in the
compound is found by subtraction.
Example: 2.22g of an organic compound X containing carbon, hydrogen and
oxygen was burnt in excess air and 3.26g of carbon dioxide and 1.33g of water
were obtained. Calculate the empirical formula of compound X.
Mass of carbon = 3.26 x 12/44 = 0.889g
5-Formulae, Equations and Substances 3
, Mass of hydrogen = 1.33 x 2/18 = 0.148g
Mass of oxygen = 2.22 - 0.889 - 0.148 = 1.18g
% carbon = 0.889 x 100/2.22 = 40%
% hydrogen = 0.148 x 100/2.22 = 6.67%
% oxygen = 1.18 x 100/2.22 = 53.2%
Empirical Formula from Combustion
Element % Divide by RAM Divide by the smallest Ratio
C 40 40/12 = 3.3 3.3/3.3 = 1 1
H 6.67 6.67/1 = 6.67 6.67/3.3 = 2 2
O 53.2 53.2/16 = 3.3 3.3/3.3 = 1 1
The empirical formula of X is CH2 O.
Molecular Formula from Empirical Formula
If the molar mass is known, the molecular formula can be derived from the
empirical formula.
First the empirical mass is worked out. if this is the same as the molar mass, the
molecular and empirical formula are the same.
If they are not the same, they will be in a simple ratio to each other.
To find the ratio, divide the molar mass by the empirical mass.
E.g. molar mass / empirical mass = 60/30 = 2
Example: A compound contains 38.7% calcium, 20% phosphorous and 41.3%
oxygen by mass. The molar mass of the compound is 310 g mol− 1 . Calculate its
empirical and molecular formula.
Empirical Formula
Divide by Divide by the Multiply by 2 to find the whole
Element
RAM smallest number ratio
38.7/40.1 =
Calcium 0.97/0.65 = 1.5 1.5 x 2 = 3
0.97
5-Formulae, Equations and Substances 4
, Divide by Divide by the Multiply by 2 to find the whole
Element
RAM smallest number ratio
Phosphorous 20/31 = 0.65 0.65/0.65 = 1 1x2=2
Oxygen 41.3/16 = 2.58 2.58/0.65 = 4 4x2=8
The empirical formula is Ca3 P2 O8 .
Empirical mass = (3 x 40.1) + (2 x 31) + (8 x 16) = 310.3
This is numerically the same as the molar mass. Therefore, the molecular
formula is Ca3 P2 O8 .
Calculation of Molar Mass
The Mr of liquids and gases can be found from mass spectra - measuring the
m/z value of the molecular ion peak.
Can also be calculated with the ideal gas equation - pV = nRT
Ionic Compounds
Made up of cations and anions.
Aqueous solutions of metal compounds are ionic.
Cations
A positive ion formed by the loss of
electrons from an atom - metals form
cations.
Roman numerals - charge on the cation.
Anions
A negative ion formed by the gain of electrons - some non-metals form anions.
Polyatomic anions contain more than one element - CO3 2− .
5-Formulae, Equations and Substances 5
5-Formulae, Equations and
Substances
Formulae
Molar Mass, Mr
Empirical Formula
Calculation from Percentage Composition
Calculation from Masses on Combustion
Molecular Formula from Empirical Formula
Calculation of Molar Mass
Ionic Compounds
Cations
Anions
Chemical Equations
State Symbols
Solubility Rules
Ionic Equations
Mole Calculations
Ideal Gas Calculation
Water of Crystallisation Calculations
Mole to Number of Particles
Calculations from Chemical Equations
Mass and Volume Calculations
Mass to Mass Calculations
Volume of Gas Calculations
5-Formulae, Equations and Substances 1
, Mass to Volume of Gas Calculations
Volume of Gas to Volume of Another Gas
Percentage Yield
Atom Economy
Percentage Composition
Limiting Reagent
Concentration
Titrations
Back Titrations
Minimising Errors in Titrations
Errors due to Apparatus
Errors due to Poor Technique
Formulae
What is the Amount of Substances measured in?
Moles
The molecular formula of a compound shows the number of
atoms of each element in one molecule of the substance.
E.g. the molecular formula of glucose is C6 H12 O6 .
Molar Mass, Mr
Molar mass is the mass of 1 mole of the substance. The units
are g mol− 1 .
The molar mass is found by adding the relative atomic masses of all the atoms
present in the formula.
E.g. glucose C6 H12 O6 - carbon = 12, hydrogen = 1 and oxygen = 16.
Molar mass of glucose = (6x12) + (12x1) + (6x16) = 180 g mol− 1 .
Empirical Formula
What is the Empirical Formula?
The simplest whole number ratio of the atoms of each element in the substance.
5-Formulae, Equations and Substances 2
, E.g. the empirical formula of glucose is CH2 O.
Calculation from Percentage Composition
The mass of an element can be found from the mass of a substance. The values
can be converted to percentage by mass. This can be used to find the empirical
formula.
1. Divide each percentage by the RAM of the element - this is the number of moles
of each element in 100g of the compound.
2. Divide these results by the smallest value obtained - this gives the simple mole
ratio.
3. If the ratio are not integers, then times the number by 2 or 3.
Example: A compound contained 34.3% sodium, 17.9% carbon and 47.8%
oxygen by mass.
Empirical Formula from %
Element % Divide by RAM Divide by the smallest Ratio
Sodium 34.3 34.3/23 = 1.49 1.49/1.49 = 1 1
Carbon 17.9 17.9/12 = 1.49 1.49/149 = 2 1
Oxygen 47.8 47.8/16 = 2.99 2.99/1.49 = 2.01 = 2 2
Calculation from Masses on Combustion
The masses of carbon dioxide and water produced when a known mass of a
substance is burnt are measured.
The mass of carbon in carbon dioxide is calculated as mass of CO2 x 12/44.
The mass of hydrogen is mass of water x 2/18.
The percentage of carbon and hydrogen are calculated and any oxygen in the
compound is found by subtraction.
Example: 2.22g of an organic compound X containing carbon, hydrogen and
oxygen was burnt in excess air and 3.26g of carbon dioxide and 1.33g of water
were obtained. Calculate the empirical formula of compound X.
Mass of carbon = 3.26 x 12/44 = 0.889g
5-Formulae, Equations and Substances 3
, Mass of hydrogen = 1.33 x 2/18 = 0.148g
Mass of oxygen = 2.22 - 0.889 - 0.148 = 1.18g
% carbon = 0.889 x 100/2.22 = 40%
% hydrogen = 0.148 x 100/2.22 = 6.67%
% oxygen = 1.18 x 100/2.22 = 53.2%
Empirical Formula from Combustion
Element % Divide by RAM Divide by the smallest Ratio
C 40 40/12 = 3.3 3.3/3.3 = 1 1
H 6.67 6.67/1 = 6.67 6.67/3.3 = 2 2
O 53.2 53.2/16 = 3.3 3.3/3.3 = 1 1
The empirical formula of X is CH2 O.
Molecular Formula from Empirical Formula
If the molar mass is known, the molecular formula can be derived from the
empirical formula.
First the empirical mass is worked out. if this is the same as the molar mass, the
molecular and empirical formula are the same.
If they are not the same, they will be in a simple ratio to each other.
To find the ratio, divide the molar mass by the empirical mass.
E.g. molar mass / empirical mass = 60/30 = 2
Example: A compound contains 38.7% calcium, 20% phosphorous and 41.3%
oxygen by mass. The molar mass of the compound is 310 g mol− 1 . Calculate its
empirical and molecular formula.
Empirical Formula
Divide by Divide by the Multiply by 2 to find the whole
Element
RAM smallest number ratio
38.7/40.1 =
Calcium 0.97/0.65 = 1.5 1.5 x 2 = 3
0.97
5-Formulae, Equations and Substances 4
, Divide by Divide by the Multiply by 2 to find the whole
Element
RAM smallest number ratio
Phosphorous 20/31 = 0.65 0.65/0.65 = 1 1x2=2
Oxygen 41.3/16 = 2.58 2.58/0.65 = 4 4x2=8
The empirical formula is Ca3 P2 O8 .
Empirical mass = (3 x 40.1) + (2 x 31) + (8 x 16) = 310.3
This is numerically the same as the molar mass. Therefore, the molecular
formula is Ca3 P2 O8 .
Calculation of Molar Mass
The Mr of liquids and gases can be found from mass spectra - measuring the
m/z value of the molecular ion peak.
Can also be calculated with the ideal gas equation - pV = nRT
Ionic Compounds
Made up of cations and anions.
Aqueous solutions of metal compounds are ionic.
Cations
A positive ion formed by the loss of
electrons from an atom - metals form
cations.
Roman numerals - charge on the cation.
Anions
A negative ion formed by the gain of electrons - some non-metals form anions.
Polyatomic anions contain more than one element - CO3 2− .
5-Formulae, Equations and Substances 5
