Doc Brown's Chemistry - GCSE/IGCSE/GCE (basic A level) O Level Online Chemical Calculations
8. Using moles to calculate empirical formula and deduce molecular formula
This page describes, and explains, how to use the mole concept, with fully worked out examples, to deduce the empirical formula of a compound from the masses of each element in a sample of the compound or from the % of each element in the compound. Where appropriate, the method of how to deduce the molecular formula of a compound from its empirical formula is further explained.
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8. Using moles to calculate empirical formula deduce molecular formula
of a compound/molecule starting with reacting masses or % composition
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!
The empirical formula of a compound is the simplest whole number ratio of atoms present in a compound. Here, the word 'empirical' means derived from experimental data.
Examples where molecular formula = empirical formula
Examples of where molecular formula and empirical formula are different e.g.
Examples of where the empirical formula is the same as the molecular formula ...
water H2O, methane CH4, pentane C5H12 (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 (Ar) are given in the tabular format method of solving the problem.
Emprical formula calculation Example 8.1:
1.15g of sodium reacted with 0.8g of sulphur. Calculate the empirical formula of sodium sulphide.
Emprical formula calculation 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.
Emprical formula and molecular formula calculation 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)
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