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GCSE-A level chemistry calculations: calculating empirical/molecular formula via moles

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Deducing empirical formula & molecular formula using the mole concept

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

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

Quantitative chemistry calculations 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. Online practice exam chemistry CALCULATIONS and solved problems for KS4 Science GCSE/IGCSE CHEMISTRY and basic starter chemical calculations for A level AS/A2/IB courses. These revision notes and practice questions on how to work out empirical formula and molecular formula using moles with worked examples should prove useful for the new AQA, Edexcel and OCR GCSE (9–1) chemistry science courses.

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Self-assessment Quizzes on determining empirical formula and molecular formula

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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 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 - it shows how many of each different type of atoms is present in one molecule.

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.

Therefore, 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 where molecular formula = empirical formula

e.g. sodium sulfate Na2SO4  and  propane C3H8

You cannot 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

Suppose you start with a molecular structure like dodecane.

If you count the atoms you find the molecular formula is C12H26.

BUT the simplest ratio formula, that is the empirical formula, is 'half' of the molecular formula i.e. C6H13.

(c) doc b

The molecule on the left is an ester called ethyl ethanoate.

From the diagram you deduce that the molecular formula is C4H8O2.

From the molecular formula you deduce that the empirical formula is C2H4O.

AND don't forget to be able to think, calculate and deduce the other way round e.g.

(i) Suppose a hydrocarbon molecule has an empirical formula of C2H5 and a molecular mass of 58 (C = 12, H = 1).

Deduce its molecular formula. The empirical formula mass = (2 x 12) + 5 = 29.

Dividing 58 by 29 gives 2. So the molecular formula is 2 x the empirical formula = C4H10

(ii) Suppose a molecule has an empirical formula of simply CH, but a molecular mass of 78 (C = 12, H = 1).

The empirical formula mass is 12 + 1 = 13. Therefore 78/13 = 6, so the molecular formula is 6 x CH = C6H6

Note the simple numerical factors connecting the empirical formula and molecular formula

empirical formula x n  = molecular formula

where n = 1, 2, 3 etc. If n = 1 then they are the same formula!

BUT remember, if you have to suggest a structure for a molecule you must know at least its molecular formula because where the empirical formula and molecular formula are different, you need extra information to deduce the molecular formula from the empirical formula (examples set out further on down the page).

Simple empirical formula calculations NOT using moles were covered in section 5.

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)

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.

RATIOS ... Sodium Na (Ar = 23.0) Sulphur S (Ar = 32.0) Comments and tips
masses 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
atom ratio = simplest whole number mole 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 integer ratio = 2 : 1, so empirical formula for sodium sulphide = Na2S

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.

RATIOS ... Aluminium Al (Ar = 27.0) Oxygen O (Ar = 16.0) Comments and tips
masses 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
atom ratio = simplest whole number mole 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 integer ratio is 2 : 3, so empirical formula for aluminium oxide = Al2O3

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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.

RATIOS ... iron (Ar = 56.0) Oxygen O (Ar = 16.0) Comments and tips
masses 1.448g 2.001 - 1.448 = 0.553g not the real atom ratio
moles (mass in g / Ar) 1.448 / 56 = 0.0259 mol 0.553 / 16 = 0.0346 mol can now divide by smallest ratio number or scale up by x factor to get simplest whole number ratio, in this case you have to make a reasonable judgement as to the values of the integers
atom ratio = simplest whole number mole ratio by trial and error 0.0259 / 0.0259 = 1

(then x 3 = 3)

3

you arrive at this

 0.0346 / 0.0259 = 1.336

(then x 3 = 4.008)

~4

stage by trial and error!
therefore the simplest integer ratio = 3 : 4, so empirical formula for the iron oxide = Fe3O4

 

Example 8.4

A compound of contains 43.4% sodium, 11.3 % carbon and 45.3 % of oxygen.

Atomic masses: Na = 23.0;  C = 12.0;  O = 16

(a) Using the % composition by mass, calculate a mole ratio for the three constituent elements

mole ratio: Na = 43.4/23.0 = 1.887;  C = 11.3/12.0 = 0.942;  O 45.3/16 = 2.831

Na : C : O is 1.887 : 0.942 : 2.831

(b) Deduce the empirical formula of the compound

Take the above ratio and divide by the smallest number gives

1.887/0.942 : 0.942/0.942 : 2.832/0.942

gives 2.003 : 1.0 : 3.006

within a reasonable experimental error, this is pretty clearly a whole number ratio for Na : C : O of 2 : 1 : 3

Therefore the empirical formula of the compound is Na2CO3

 

Example 8.5

A copper ore compound contains 51.5% copper, 9.7% carbon and an undetermined % of oxygen.

Atomic masses: Cu = 63.5;  C = 12.0;  O = 16

(a) Calculate the % oxygen in the compound.

(a) Using the % composition by mass, calculate a mole ratio for the three constituent elements

(b) Deduce the empirical formula of the compound

Do this one for yourself and you should get CuCO3

Examples 8.1 to 8.5 involved ionic compounds, so no molecular formula or molecular mass were involved.

However examples 8.6 to 8.9 involve covalent molecules and both empirical formula and molecular formula, and the connection between the two.

You are likely to encounter these sort of problems when studying advanced level organic chemistry.

Relative atomic masses: C = 12.0;  H = 1.0;  O = 16.0;  Cl = 35.5

Empirical formula and molecular formula calculation Example 8.6 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 - from the molecular ion peak.

Treat the percentages as if they were masses in grams, and it all works out fine.

RATIOS ... carbon (Ar = 12.0) hydrogen (Ar = 1.0) Comments and tips
masses 81.8 18.2 not the real atom ratio
moles (mass in g / Ar) 81.8 / 12 = 6.817 18.2 / 1 = 18.2 can now divide by smallest ratio number or scale up by x factor to get simplest whole number ratio in this case you have to make a reasonable judgement as to the values of the integers
atom ratio = simplest whole number mole ratio by trial and error 6.817/6.817 = 1.0

1.0 x 2 = 2

1.0 x 3 = 3.0

you arrive at this

 18.2/6.817 = 2.67

2.670 x 2 = 5.34

2.670 x 3 = 8.01 ~8.0

stage by trial and error!
therefore the simplest integer ratio = 3 : 8, so empirical formula for the hydrocarbon = C3H8

The empirical formula mass = (3 x 12) + 8 = 44

This equals the molecular mass, therefore the molecular formula is also C3H8

 

Empirical formula and molecular formula calculation Example 8.7 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 - from the molecular ion peak.

Again, treat the percentages as if they were masses in grams, and it all works out fine.

RATIOS ... carbon (Ar = 12.0) hydrogen (Ar = 1.0) Comments and tips
masses 83.72 16.28 not the real atom ratio
moles (mass in g / Ar) 83.72 / 12 = 6.977 16.28 / 1 = 16.28 can now divide by smallest ratio number or scale up by x factor to get simplest whole number ratio, a bit fiddly to find the ratio on this one to deduce the empirical formula
atom ratio = simplest whole number mole ratio by trial and error 6.977/6.977 = 1.0

1.0 x 2 = 2.0

1.0 x 3 = 3.0

you arrive at this

 16.28/6.977 = 2.333

2.333 x 2 = 4.667

2.333 x 3 = 7.00

stage by trial and error!
therefore the simplest integer ratio = 3 : 7, so empirical formula for the hydrocarbon = C3H7

The empirical formula mass = (3 x 12) + 7 = 43

BUT the molecular mass is double this, so the molecular formula must be double the empirical formula

Therefore the molecular formula is C6H14 of which there are many structural isomers!

You would need more information to deduce the structural formula of this hydrocarbon.

 

Empirical formula and molecular formula calculation Example 8.8 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.

RATIOS ... Carbon (Ar = 12.0) Hydrogen (Ar = 1.0) Oxygen (Ar = 16.0) Comments and tips
masses 40.00 6.67 53.33 just think of it as based on 100g
molar ratio (mass in g / Ar)  40.00 / 12 =  3.333 mol 6.67 / 1 =  6.67 mol 53.33 / 16.0 = 3.333 mol can now divide by smallest ratio number
atom ratio = simplest whole number mole ratio by trial and error 3.333/3.333 = 1.0 6.67/3.333 ~2.0 3.333/3.333 = 1.0
therefore the simplest integer ratio = 1 : 2 : 1, so empirical formula for the sugar = CH2O

The empirical formula mass = 12 + 2 + 16 = 30

BUT the molecular mass is 5 x  this (150/30), so the molecular formula must be 5 x the empirical formula

Therefore the molecular formula is C5H10O5 , a pentose, of which there are many structural isomers!

You would need more information to deduce the structural formula of this sugar molecule.

 

Empirical formula and molecular formula calculation Example 8.9 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.

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
atom ratio = simplest whole number mole ratio by trial and error 2.02 / 2.02 = 1 4.04 / 2.02 = 2 2.02 / 2.02 = 1
therefore the simplest atomic ratio = 1 : 2 : 1, so 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

two possible structures, which cannot be distinguished by the data given or calculation above

1,2-dichloroethane and 1,1-dichloroethane, two possible atom arrangements of the same molecular formula known as structural positional isomers, because one of the chlorine atoms can be in two different positions.

 

Example 8.10:

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Self-assessment Quizzes on determining empirical formula and molecular formula

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Above is typical periodic table used in GCSE science-chemistry specifications in doing reacting gas volume ratio calculations, and I've 'usually' used these values in my exemplar calculations to cover most syllabuses

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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 acid-alkali titration calculations, diagrams of apparatus, details of procedures

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

  14. Other calculations e.g. % purity, % percentage & theoretical yield, 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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