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These notes were written for the old IB syllabus (2009). The new IB syllabus for first examinations 2016 can be accessed by clicking the link below.

### Formulae

1.2.1 Define the term molar mass (M) and calculate the mass of one mole of a species.

#### Molar mass

This is the mass in grams that is numerically equal to the relative molecular mass expressed in grams. The relative molecular mass is obtained by adding the atomic masses of the

 Example: Find the molar mass of sulphuric acid The formula is H2SO4 atomic masses H=1, S=32, O=16 Sum = (2x1) + 32 + (4x16) = 98 Therefore the molar mass = 98g

1.2.2: Distinguish between atomic mass, molecular mass and formula mass. The term molar mass (in g mol-1) can be used for all of these.

#### Atomic mass

This is the mass of the atom of a specific element. It is usually expressed in atomic mass units (a.m.u.)

#### Molecular mass

Is the sum of the component atoms within one molecule of the compound. This refers only to simple molecular compounds. For example water consists of H2O molecules and has a formula mass = 2 + 16 = 18

#### Formula mass

This term is used when the substance is not simple molecular, such as silicon dioxide. The giant structure of silicon dioxide can be simplified to be represented as the formula SiO2 - in other words there are two oxygen atoms for every silicon atom in the structure. The formula mass is then taken from this simplest formula SiO2

The formula mass can also be applied to ionic substances (which also only exist in giant structure form). The structure can be simplified to give the simplest ratio of ions in the structure which is then taken as the formula. Calcium chloride is a giant structure of calcium ions and chloride ions in a ratio of one calcium to every two chloride ions so the simplest ratio is CaCl2 - this is used to calculate the formula mass: 40 + (2 x 35.5) = 111

1.2.3: Define the terms relative molecular mass (Mr) and relative atomic mass (Ar). The terms have no units.

#### Relative atomic mass, Ar

The relative mass of an atom is its mass compared to the 12C isotope = 12 atomic mass units.

In other words the standard unit is defined by the carbon 12 atom, which is assigned a value of 12 units. Everthing else is measured with respect to this atom and is then described as relative atomic/molecular mass. This has no units as it is a comparative value.

For example, the mass of a magnesium atom is 2 x the mass of a carbon 12 atom and therefore it has a relative mass of 24 (which is 2 x 12)

#### Relative molecular mass, Mr

The relative mass of a molecule is the mass compared to the 12C isotope = 12 atomic mass units.

In other words the standard unit is defined by the carbon 12 atom, which is assigned a value of 12 units. Everthing else is measured with respect to this atom and is then described as relative atomic/molecular mass. This has no units as it is a comparative value.

For example, the mass of a water molecule is 1.5 x the mass of a carbon 12 atom and therefore it has a relative molecular mass of 18 (which is 1.5 x 12)

1.2.4: State the relationship between the amount of substance (in moles) and mass, and carry out calculations involving amount of substance, mass and molar mass.

#### Moles, mass and relative molecular mass

Clearly, if we can calculate the mass of 1 mole of a substance from the relative molecular mass then we can also calculate fractions of a mole or perform the reverse calcuation to find the mass from the number of moles.

This gives the general equation:

 moles = mass Mr

1.2.5: Define the terms empirical formula and molecular formula. The molecular formula is a multiple of the empirical formula.

The empirical formula is the simplest possible ratio of atoms in a compound. There are always a whole number (integer) of empirical formulas in a molecular formula.

The molecular formula of glucose is C6H12O6 and the empirical formula is CH2O. You have to multiply the empirical formula by 6 to get the molecular formula

1.2.6 Determine the empirical formula and/or the molecular formula of a given compound.
Determine the:
- empirical formula from the percentage composition or from other suitable experimental data
- percentage composition from the formula of a compound
- molecular formula when given both the empirical formula and the molar mass

Empirical formula is determined by dividing the percentage composition of each element within a compound by its relative mass.

The percentage composition is found by dividing the total mass of an element within a compound by the relative mass of that compound and mutiplying by 100.

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