A-Level Chemistry AQA Notes

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Relative Masses

Relative isotopic mass is the mass of an atom of an isotope compared with 1/12th of the mass of an atom of carbon-12. For an isotope, the relative isotopic mass = its mass number.
Relative atomic mass is the ratio of the average mass of an atom of an element to 1/12th of the mass of an atom of carbon-12.
Relative molecular mass is the ratio of the average mass of a molecule of an element or compound to 1/12th of the mass of an atom of carbon-12.
Relative formula mass is similar to relative molecular mass but applies to ionic compounds.

The Mole & Concentration

The mole is the unit used to quantify the amount of a substance. It can be applied to any amount of chemical species, including atoms, electrons, molecules and ions
A mole is the amount of substance that contains the same number of atoms or particles as 12 g of carbon-12.
The number of particles in 12g of ^12C is the Avogadro constant of 6.022 x 10^23 mol-^1.

n is the number of moles (mol)
m is the mass (g)
M is the molar mass (g mol-^1)
The concentration of a solution is the amount of solute present in a known volume of solution.

c is the concentration (mol dm-^3)
n is the number of moles in solution (mol)
V is the volume (dm^3)
Remember:
- 1 dm^3 = 1000 cm^3
- 1 m^3 = 1000dm^3

Gas Equations

One mole of any gas under standard conditions will occupy the same volume.
The molar gas volume is 24 dm^-3 mol^-1 under standard conditions of 298 K and 100 kPa
The number of moles of gas can be calculated using the equation:

n is the number of moles of gas (mol)
V is volume (dm^-3)

In an ideal gas the assumptions are made that:
- Intermolecular forces between the gas particles are negligible
- The volume of the particles themselves, relative to the volume of their container, is negligible
The ideal gas equation is:

p is pressure (Pa)
V is volume (m^3)
n is the number of moles (mol)
R is the gas constant (8.314 JK^-1)
T is temperature (K)

Empirical & Molecular Formula

The empirical formula is the simplest whole-number ratio of atoms of each element present in a compound.
The empirical formula can be calculated from the composition by mass or percentage by mass. e.g 6.2 g of P is combined with O2 to form 14.2 g of phosphorous oxide. Calculate the empirical formula of the compound.

mass of O2: 14.2 g - 6.2 g = 8 g

number of moles of each element:

Divide through by the smallest number of moles to get the whole number ratio:

Empirical formula: P2O5

The molecular formula gives the number and type of atoms of each element in a molecule. It is made up of a whole number of empirical units.
The molecular formula can be determined using the empirical formula and relative molecular mass of the molecule. e.g. Determine the molecular formula of a compound with empirical formula CH2 and a relative molecular mass of 224.

Relative molecular mass of the empirical formula:

C H2
12 + (1 × 2) = 14

Divide the relative molecular mass by that of the empirical formula:
224/14=16

Molecular formula:
C16H32

Balanced Equations

When a chemical reaction occurs, no atoms are created or destroyed. The atoms in the reactants rearrange to form the products.
In a balanced equation, there is the same number of atoms of each element in both the reactants and products.
State symbols are written after every species to indicate the physical state
- Solid (s)
- Liquid (l)
- Gaseous (g)
- Aqueous (aq) - dissolved in water

Ionic equations can be written for any reaction involving ions in solution, where only the reacting ions and the products they form are included.
Spectator ions are ions that do not take part in the overall reaction and are found in both the reactants and products
The net ionic equation shows only the ions directly involved in the reaction (removing spectator ions).

e.g.

NaCl (aq) + AgNO3 (aq) → AgCl (s) + NaNO3 (aq)

Na+ (aq) + Cl- (aq) + Ag+ (aq) + NO3- (aq)→ AgCl (s) + Na+ (aq) + No3- (aq)

Net ionic equation: Cl- (aq) + Ag+ (aq)→ AgCl (s)

Stoichiometry expresses the molar ratios between reactants and products in a reaction. It is used to determine the quantity of products or reactants required or produced by a known reaction.

Atom Economy & Percentage Yield

Atom economy is a theoretical measure of the proportion of atoms from the reactants that form the desired product. In order to calculate it a balanced chemical equation is required.

Maximising atom economy has important economic, ethical and environmental advantages:
- More sustainable (uses fewer raw materials)
- Minimises chemical waste
- Maximises efficiency
- Less money is spent on separation processes

The limiting reagent is the reagent not in excess. It dictates the theoretical yield and the amount of product actually formed.
Percentage yield is a measure of the percentage of reactants that have been converted into the desired product. It gives a measure of the efficiency of a reaction route.
The percentage yield is reduced by the formation of unwanted by-products, any reactant that remains unreacted, or product that cannot be extracted from the reaction vessel.

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