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)
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)
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
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).
NaCl (aq) + AgNO3 (aq) → AgCl (s) + NaNO3 (aq)
Na+ (aq) + Cl- (aq) + Ag+ (aq) + NO3- (aq)→ AgCl (s) + Na+ (aq) + No3- (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.
