Calculations 8 Deducing Empirical and molecular formula using moles

The basis of this method, is that a mole of defined species has the same number of defined 'particles' in it. So calculating the mole ratio of atoms in the combination, gives you the actual atomic ratio in the compound. This ratio is then expressed in the simplest whole number atomic ratio, which is the empirical formula. (it means the formula 'from experiment', see examples 1 and 2). (see section 3. for some simpler examples not using moles)

However, there is one problem to resolve for covalent molecular compounds. The molecular formula is the summary of all the atoms in one individual molecule - so the molecular formula might not be the same as the empirical formula! In order to deduce the molecular formula from the empirical formula, you ALSO need to know the molecular mass of the molecule from another data source. See example 3, which also illustrates the calculation using % element composition AND the method is no different for more than two elements!

Examples of where the empirical formula is the same as the molecular formula ...

water H₂O, methane CH₄, propane C₃H₈ (these molecular formula cannot be 'simplified')

Examples of where the molecular formula is different from the empirical formula in () ...

e.g. ethane C₂H₆ (CH₃), phosphorus(V) oxide P₄O₁₀ (P₂O₅), benzene C₆H₆ (CH)

Three examples are set out below to illustrate all the situations. The relative atomic masses of the elements (A_r) are given in the tabular format method of solving the problem.

Example 8.1: 1.15g of sodium reacted with 0.8g of sulphur. Calculate the empirical formula of sodium sulphide.

RATIOS ...	Sodium Na (A_r = 23)	Sulphur S (A_r= 32)	Comments and tips
Reacting mass	1.15g	0.80g	not the real atom ratio
moles (mass in g / A_r)	1.15 / 23 = 0.05 mol	0.8 / 32 = 0.025 mol	can now divide by smallest ratio number or scale up by x factor to get simplest whole number ratio
simplest whole number ratio by trial and error	0.05 / 0.025 = 2 or 0.05 x 40 = 2	0.025 / 0.025 = 1 or 0.025 x 40 = 1
therefore the simplest ratio = empirical formula for sodium sulphide = Na₂S

Example 8.2: 1.35g of aluminium was heated in oxygen until there was no further gain in weight. The white oxide ash formed weighed 2.55g. Deduce the empirical formula of aluminium oxide. Note: to get the mass of oxygen reacting, all you have to do is to subtract the mass of metal from the mass of the oxide formed.

RATIOS ...	Aluminium Al (A_r = 27)	Oxygen O (A_r= 16)	Comments and tips
Reacting mass	1.35g	2.55 - 1.35 = 1.2g	not the real atom ratio
moles (mass in g / A_r)	1.35 / 27 = 0.05 mol	1.2 / 16 = 0.075 mol	can now divide by smallest ratio number or scale up by x factor to get simplest whole number ratio
simplest whole number ratio by trial and error	0.05 / 0.05 = 1 (then x 2 = 2) or 0.05 x 40 = 2	0.075 / 0.05 = 1.5 (then x 2 = 3) or 0.075 x 40 = 3
therefore the simplest ratio = empirical therefore the formula for aluminium oxide = Al₂O₃

Example 8.3: A chlorinated hydrocarbon compound when analysed, consisted of 24.24% carbon, 4.04% hydrogen, 71.72% chlorine. The molecular mass was found to be 99 from another experiment. Deduce the empirical and molecular formula. (you can 'treat' the %'s as if they were grams, and it all works out like examples 1 and 2)

RATIOS ...	Carbon (A_r = 12)	Hydrogen (A_r= 1)	Chlorine (Ar = 35.5)	Comments and tips
Reacting mass or % mass	24.24	4.04	71.72	just think of it as based on 100g
molar ratio (mass in g / A_r)	24.24 / 12 = 2.02 mol	4.04 / 1 = 4.04 mol	71.72 / 35.5 = 2.02 mol	can now divide by smallest ratio number
simplest whole number ratio	2.02 / 2.02 = 1	4.04 / 2.02 = 2	2.02 / 2.02 = 1	can now divide by smallest ratio number
therefore the simplest ratio = empirical formula for the chlorinated hydrocarbon = CH₂Cl BUT the molecular mass is 99, and the empirical formula mass is 49.5 (12+2+35.5) AND 99 / 49.5 = 2, and so the molecular formula must be 2 x CH₂Cl = C₂H₄Cl₂

Example 8.4: