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8. Using moles to calculate empirical formula and deduce molecular formula
Help for problem solving in doing empirical formula AND molecular formula calculations using moles, using experiment data, making predictions. Practice revision questions on deducing molecular formula from an empirical formula using reacting masses or % mass composition and converting reacting masses to moles. 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 by mass
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 and how to calculate an empirical formula and a molecular formula from reacting masses or % element composition by mass.
For non-molecular compounds (i.e. inorganic ionic compounds), the empirical formula is usually, but not always, quoted as the compound formula.
The relative atomic masses of the elements (Ar) are given in the tabular format method of solving the problem.
Empirical formula calculation Example 8.1: The compound formed between sodium and sulfur
1.15g of sodium reacted with 0.8g of sulphur. Calculate the empirical formula of sodium sulphide.
You convert the masses to moles i.e. mass in g divided by the relative atomic mass.
Since one mole of any defined substance contains the same number of particles (e.g. atoms), it means that the atomic mole ratio is also the actual ratio of atoms in the compound.
The ratio is then expressed as the simplest whole ratio from which the empirical formula is derived.
Apart from a 1, other numbers e.g. 2, 3 etc. should be seen as subscripts in the empirical formula.
Empirical formula calculation Example 8.2 The empirical formula of aluminium oxide
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.
Empirical formula calculation Example 8.3 for an oxide of iron.
1.448g of iron was heated in air in a crucible until no further gain in weight was observed.
The final mass of the iron oxide was found to be 2.001g
Calculate the empirical formula of the iron oxide.
Atomic masses: Fe = 56 and O = 16
The mass of oxygen combined with the iron is deduced by subtracting the original mass of iron from final total mass of iron oxide.
Empirical formula and molecular formula calculation Example 8.4 for a hydrocarbon compound
On analysis a hydrocarbon was found to consist of 81.8% carbon and 18.2% hydrogen.
Molecular ion measurements in a mass spectrometer show that the hydrocarbon has a molecular mass of 44.
Treat the percentages as if they were masses in grams, and it all works out fine.
Empirical formula and molecular formula calculation Example 8.5 for another hydrocarbon compound
On analysis a hydrocarbon was found to consist of 83.72% carbon and 16.28% hydrogen.
Molecular ion measurements in a mass spectrometer show that the hydrocarbon has a molecular mass of 86.
Again, treat the percentages as if they were masses in grams, and it all works out fine.
Empirical formula and molecular formula calculation Example 8.6 for a carbohydrate sugar compound
A carbohydrate compound e.g. a sugar, was found on analysis to contain 40.00% carbon, 6.67% hydrogen and 53.33% oxygen.
The molecular mass was 150. From the information calculate the empirical formula and deduce the molecular formula.
Empirical formula and molecular formula calculation Example 8.7 for a chloroalkane compound
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, i.e. based on a total mass of 100g.
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