calculation Questions deducing molecular formula from empirical formula using moles problem solving gcse A level chemistry igcse KS4 science GCE AS A2 O Level practice revision exercises

HOME PAGE * KS3 SCIENCES * GCSE BIOLOGY CHEMISTRY PHYSICS * ADVANCED LEVEL CHEMISTRY

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

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

8. Using moles to calculate empirical formula and deduce molecular formula

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.

Spotted any careless error? EMAIL query ? comment or request a type of GCSE calculation not covered?

Self-assessment Quizzes on determining empirical formula and molecular formula

type in answer QUIZ or multiple choice QUIZ

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.

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 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.

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

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 (A_r) 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 (A_r = 23.0)	Sulphur S (A_r= 32.0)	Comments and tips
masses	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
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 (A_r = 27.0)	Oxygen O (A_r= 16.0)	Comments and tips
masses	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
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

TOP OF PAGE

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 (A_r = 56.0)	Oxygen O (A_r= 16.0)	Comments and tips
masses	1.448g	2.001 - 1.448 = 0.553g	not the real atom ratio
moles (mass in g / A_r)	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 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 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.

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.

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

RATIOS ...	carbon (A_r = 12.0)	hydrogen O (A_r= 1.0)	Comments and tips
masses	81.8	18.2	not the real atom ratio
moles (mass in g / A_r)	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.

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

RATIOS ...	carbon (A_r = 12.0)	hydrogen O (A_r= 1.0)	Comments and tips
masses	83.72	16.28	not the real atom ratio
moles (mass in g / A_r)	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!

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 (A_r = 12.0)	Hydrogen (A_r= 1.0)	Oxygen (A_r = 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 / A_r)	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	can now divide by smallest ratio number
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!

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 (A_r = 12)	Hydrogen (A_r= 1)	Chlorine (A_r = 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
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	can now divide by smallest ratio number
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 or calculation above 1,2-dichloroethane and 1,1-dichloroethane, two possible atom arrangements of the same molecular formula.

Example 8.10:

TOP OF PAGE

Self-assessment Quizzes on determining empirical formula and molecular formula

type in answer QUIZ or multiple choice QUIZ

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

Doc

OTHER CALCULATION PAGES

the interconversion of empirical formula and molecular formula examples explained, empirical formula calculations using moles how to deduce molecular formula Revision KS4 Science revising empirical formula calculations using moles how to deduce molecular formula Additional Science Triple Award Science Separate Sciences Courses aid to empirical formula calculations using moles how to deduce molecular formula textbook revision GCSE/IGCSE/O level Chemistry empirical formula calculations using moles how to deduce molecular formula Information Study Notes for revising for AQA GCSE Science empirical formula calculations using moles how to deduce molecular formula, Edexcel GCSE Science/IGCSE Chemistry empirical formula calculations using moles how to deduce molecular formula & OCR 21st Century Science, OCR Gateway Science empirical formula calculations using moles how to deduce molecular formula WJEC gcse science chemistry empirical formula calculations using moles how to deduce molecular formula CEA/CEA gcse science chemistry O Level Chemistry (revise courses equal to US grade 8, grade 9 grade 10 empirical formula calculations using moles how to deduce molecular formula) A level Revision notes for GCE Advanced Subsidiary Level empirical formula calculations using moles how to deduce molecular formula AS Advanced Level A2 IB Revising empirical formula calculations using moles how to deduce molecular formula AQA GCE Chemistry OCR GCE Chemistry empirical formula calculations using moles how to deduce molecular formula Edexcel GCE Chemistry Salters Chemistry empirical formula calculations using moles how to deduce molecular formula CIE Chemistry empirical formula calculations using moles how to deduce molecular formula, WJEC GCE AS A2 Chemistry empirical formula calculations using moles how to deduce molecular formula, CCEA/CEA GCE AS A2 Chemistry revising empirical formula calculations using moles how to deduce molecular formula courses for pre-university students (equal to US grade 11 and grade 12 and AP Honours/honors level empirical formula calculations using moles how to deduce molecular formula revision guide to empirical formula calculations using moles how to deduce molecular formula How to convert from empirical formula to molecular formula, how to calculate molecular formula from reacting masses or percentage composition of the compound gcse chemistry revision free detailed notes on deducing molecular formula to help revise igcse chemistry igcse chemistry revision notes on deducing molecular formula O level chemistry revision free detailed notes on deducing molecular formula to help revise gcse chemistry free detailed notes on deducing molecular formula to help revise O level chemistry free online website to help revise deducing molecular formula for gcse chemistry free online website to help revise deducing molecular formula for igcse chemistry free online website to help revise O level deducing molecular formula chemistry how to succeed in questions on deducing molecular formula for gcse chemistry how to succeed at igcse chemistry how to succeed at O level chemistry a good website for free questions on deducing molecular formula to help to pass gcse chemistry questions on deducing molecular formula a good website for free help to pass igcse chemistry with revision notes on deducing molecular formula a good website for free help to pass O level chemistry

TOP OF PAGE

Website content © Dr Phil Brown 2000+. All copyrights reserved on revision notes, images, quizzes, worksheets etc. Copying of website material is NOT permitted. Exam revision summaries & references to science course specifications are unofficial.