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Doc Brown's Chemistry - GCSE/IGCSE/GCE (basic A level) O Level  Online Chemical Calculations

study examples carefully8. Using moles to calculate empirical formula and deduce molecular formulastudy examples carefully

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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study examples carefully8. Using moles to calculate empirical formula deduce molecular formulastudy examples carefully

  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.

Do not confuse with molecular formula which depicts the actual total numbers of each atom in a molecule.

The molecular formula and empirical formula can be different or the same.

They are the same if the molecular formula cannot be simplified on a whole number basis.

Examples where molecular formula = empirical formula

e.g. sodium sulfate Na2SO4  and  propane C3H8

You can simplify the atomic ratios 2 : 1 : 4 or 3 : 8 to smaller whole number (integer) ratios

Examples of where molecular formula and empirical formula are different e.g.

butane molecular formula C4H10, empirical formula C2H5

numerically, the empirical formula of butane is 'half' of its molecular formula

4 : 10 ==> 2 : 5

glucose molecular formula C6H12O6, empirical formula CH2O

numerically, the empirical formula of glucose is '1/6th' of the molecular formula

6 : 12 : 6 ==> 1 : 2 : 1

 

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.

RATIOS ... Sodium Na (Ar = 23) Sulphur S (Ar = 32) Comments and tips
Reacting mass 1.15g 0.80g not the real atom ratio
moles (mass in g / Ar) 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

 


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.

RATIOS ... Aluminium Al (Ar = 27) Oxygen O (Ar = 16) Comments and tips
Reacting mass 1.35g 2.55 - 1.35 = 1.2g not the real atom ratio
moles (mass in g / Ar) 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

 


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)

RATIOS ... Carbon (Ar = 12) Hydrogen (Ar = 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 / Ar) 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
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:

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OTHER CALCULATION PAGES

  1. What is relative atomic mass?, relative isotopic mass and calculating relative atomic mass

  2. Calculating relative formula/molecular mass of a compound or element molecule

  3. Law of Conservation of Mass and simple reacting mass calculations

  4. Composition by percentage mass of elements in a compound

  5. Empirical formula and formula mass of a compound from reacting masses (easy start, not using moles)

  6. Reacting mass ratio calculations of reactants and products from equations (NOT using moles) and brief mention of actual percent % yield and theoretical yield, atom economy and formula mass determination

  7. Introducing moles: The connection between moles, mass and formula mass - the basis of reacting mole ratio calculations (relating reacting masses and formula mass)

  8. Using moles to calculate empirical formula and deduce molecular formula of a compound/molecule (starting with reacting masses or % composition) (this page)

  9. Moles and the molar volume of a gas, Avogadro's Law

  10. Reacting gas volume ratios, Avogadro's Law and Gay-Lussac's Law (ratio of gaseous reactants-products)

  11. Molarity, volumes and solution concentrations (and diagrams of apparatus)

  12. How to do volumetric titration calculations e.g. acid-alkali titrations (and diagrams of apparatus)

  13. Electrolysis products calculations (negative cathode and positive anode products)

  14. Other calculations e.g. % purity, % percentage & theoretical yield, volumetric titration apparatus, dilution of solutions (and diagrams of apparatus), water of crystallisation, quantity of reactants required, atom economy

  15. Energy transfers in physical/chemical changes, exothermic/endothermic reactions

  16. Gas calculations involving PVT relationships, Boyle's and Charles Laws

  17. Radioactivity & half-life calculations including dating materials


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