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===About Empirical Formulae===
 
===About Empirical Formulae===
 
The '''empirical formula''' of a [[compound]] may not be the same as the [[Chemical Formula|chemical formula]]:
 
The '''empirical formula''' of a [[compound]] may not be the same as the [[Chemical Formula|chemical formula]]:
−
*Ethane
+
*[[Ethane]] - [[Chemical Formula]] C<sub>2</sub>H<sub>6</sub>, '''Empirical Formula''' CH<sub>3</sub>
−
: [[Chemical Formula]] C<sub>2</sub>H<sub>6</sub>
+
*[[Ethene]] - [[Chemical Formula]] C<sub>2</sub>H<sub>4</sub>, '''Empirical Formula''' CH<sub>2</sub>
−
: '''Empirical Formula''': CH<sub>3</sub>
+
*[[Propene]] - [[Chemical Formula]] C<sub>3</sub>H<sub>6</sub>, '''Empirical Formula''' CH<sub>2</sub>
−
*Ethene
+
*[[Glucose]] - [[Chemical Formula]] C<sub>6</sub>H<sub>12</sub>O<sub>6</sub>, '''Empirical Formula''' CH<sub>2</sub>O
−
: [[Chemical Formula]]: C<sub>2</sub>H<sub>4</sub>
+
*[[Lactic Acid]] - [[Chemical Formula]] C<sub>3</sub>H<sub>6</sub>O<sub>3</sub>, '''Empirical Formula''' CH<sub>2</sub>O
−
: '''Empirical Formula''': CH<sub>2</sub>
 
  
 
: '''Empirical formulae''' are calculated from the amount of [[atom]]s in a [[Chemical Reaction|chemical reaction]].
 
: '''Empirical formulae''' are calculated from the amount of [[atom]]s in a [[Chemical Reaction|chemical reaction]].
 
: The number of [[atom]]s can be found if you know the [[mass]] of different [[element]]s and the [[Relative Atomic Mass|relative atomic mass]] of the [[element]]s in the [[Chemical Reaction|reaction]].
 
: The number of [[atom]]s can be found if you know the [[mass]] of different [[element]]s and the [[Relative Atomic Mass|relative atomic mass]] of the [[element]]s in the [[Chemical Reaction|reaction]].
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===Finding the Empirical Formula===
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{| class="wikitable"
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| style="height:20px; width:200px; text-align:center;" |'''Number of Atoms'''
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| style="height:20px; width:200px; text-align:center;" |'''Skeletal Diagram'''
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| style="height:20px; width:200px; text-align:center;" |'''Ball and Stick Model'''
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|-
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| style="height:20px; width:200px; text-align:center;" |100 [[atom]]s of [[Hydrogen]] [[Chemical Reaction|react]] completely with 50 [[atom]]s of [[Oxygen]].
 +
|[[File:SkeletalFormulaButane.png|center|200px]]
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|[[File:BallandStickButan12diol.png|center|200px]]
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|-
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| style="height:20px; width:200px; text-align:left;" |
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The [[ratio]] of [[Hydrogen]] [[atom]]s to [[Oxygen]] [[atom]]s
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H:O
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100:50
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2:1
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So the '''empirical formula''' is H<sub>2</sub>O
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| style="height:20px; width:200px; text-align:left;" |
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In this [[diagram]] there are 4 [[Carbon]] [[atom]]s, 10 [[Hydrogen]] [[atom]]s.
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The [[ratio]] of [[atom]]s is:
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C:H
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4:10
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2:5
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So the '''empirical formula''' is C<sub>2</sub>H<sub>5</sub>
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| style="height:20px; width:200px; text-align:left;" |
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In this [[diagram]] there are 4 [[Carbon]] [[atom]]s, 10 [[Hydrogen]] [[atom]]s and 2 [[Oxygen]] [[atom]]s.
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 +
The [[ratio]] of [[atom]]s is:
 +
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C:H:O
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4:10:2
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2:5:1
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So the '''empirical formula''' is C<sub>2</sub>H<sub>5</sub>O
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|}
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{| class="wikitable"
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|-
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| style="height:20px; width:200px; text-align:center;" |64g of [[Oxygen]] is found to [[Chemical Reaction|react]] completely with 8g of [[Hydrogen]]. Find the '''empirical formula''' for the [[product]].
 +
| style="height:20px; width:200px; text-align:center;" |60g of [[Carbon]] is found to [[Chemical Reaction|react]] completely with 160g of [[Oxygen]]. Find the '''empirical formula''' for the [[product]].
 +
| style="height:20px; width:200px; text-align:center;" |416g of [[Sulphur]] is found to [[Chemical Reaction|react]] completely with [[Oxygen]] to produce 832g of [[product]]. Find the '''empirical formula''' for the [[product]].
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|-
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| style="height:20px; width:200px; text-align:left;" |State the [[mass]] of all [[reactant]]s.
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[[Mass]] of [[Oxygen]] = 64g
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[[Mass]] of [[Hydrogen]] = 8g
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| style="height:20px; width:200px; text-align:left;" |State the [[mass]] of all [[reactant]]s.
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[[Mass]] of [[Carbon]] = 60g
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[[Mass]] of [[Oxygen]] = 416g
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| style="height:20px; width:200px; text-align:left;" |State the [[mass]] of all [[reactant]]s.
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[[Mass]] of [[Sulphur]] = 416g
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[[Mass]] of [[Oxygen]] = 416g
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|-
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| style="height:20px; width:200px; text-align:left;" |Find the number of [[mole]]s of each [[element]].
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Number of [[Mole]]s = ([[Mass]]/[[Relative Atomic Mass]])
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[[Relative Atomic Mass]] of [[Oxygen]] = 16g
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[[Mole]]s of [[Oxygen]] = 64/16
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[[Mole]]s of [[Oxygen]] = 4 mole
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[[Relative Atomic Mass]] of [[Hydrogen]] = 1g
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[[Mole]]s of [[Hydrogen]] = 8/1
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[[Mole]]s of [[Hydrogen]] = 8 mole
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| style="height:20px; width:200px; text-align:left;" |Find the number of [[mole]]s of each [[element]].
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Number of [[Mole]]s = ([[Mass]]/[[Relative Atomic Mass]])
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[[Relative Atomic Mass]] of [[Carbon]] = 12g
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[[Mole]]s of [[Carbon]] = 60/12
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[[Mole]]s of [[Carbon]] = 5 mole
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[[Mass]] of [[Oxygen]] = 160g
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[[Relative Atomic Mass]] of [[Oxygen]] = 16g
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[[Mole]]s of [[Oxygen]] = 160/16
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[[Mole]]s of [[Oxygen]] = 10 mole
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| style="height:20px; width:200px; text-align:left;" |Find the number of [[mole]]s of each [[element]].
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Number of [[Mole]]s = ([[Mass]]/[[Relative Atomic Mass]])
 +
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[[Relative Atomic Mass]] of [[Sulphur]] = 32g
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[[Mole]]s of [[Sulphur]] = 416/32
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[[Mole]]s of [[Sulphur]] = 32 mole
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[[Relative Atomic Mass]] of [[Oxygen]] = 16g
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[[Mole]]s of [[Oxygen]] = 416/16
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[[Mole]]s of [[Oxygen]] = 26 mole
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|-
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| style="height:20px; width:200px; text-align:left;" |Find the [[Ratio]]:
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[[Hydrogen]]:[[Oxygen]]
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8:4
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2:1
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| style="height:20px; width:200px; text-align:left;" |Find the [[Ratio]]:
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[[Carbon]]:[[Oxygen]]
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5:10
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1:2
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| style="height:20px; width:200px; text-align:left;" |Find the [[Ratio]]:
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[[Sulphur]]:[[Oxygen]]
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13:26
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1:2
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|-
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| style="height:20px; width:200px; text-align:left;" |The '''empirical formula''' of the [[product]] is H<sub>2</sub>O
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| style="height:20px; width:200px; text-align:left;" |The '''empirical formula''' of the [[product]] is CO<sub>2</sub>
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| style="height:20px; width:200px; text-align:left;" |The '''empirical formula''' of the [[product]] is SO<sub>2</sub>
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|}
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===References===
 +
====AQA====
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 +
:[https://www.amazon.co.uk/gp/product/0008158762/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=0008158762&linkCode=as2&tag=nrjc-21&linkId=a0fffa35b3ea49a63404f6704e0df7cc ''Formula, formulae; empirical, pages 62-3, 90-1, 252, GCSE Chemistry; Student Book, Collins, AQA '']
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====Edexcel====
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 +
:[https://www.amazon.co.uk/gp/product/1292120193/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1292120193&linkCode=as2&tag=nrjc-21&linkId=572df39392fb4200db8391d98ae6314e ''Empirical formula, page 217, GCSE Combined Science, Pearson Edexcel '']
 +
:[https://www.amazon.co.uk/gp/product/1292120215/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1292120215&linkCode=as2&tag=nrjc-21&linkId=8f96ddb76196848bafdb124354e4cf77 ''Empirical formulae, page 73, GCSE Chemistry, Pearson, Edexcel '']
 +
:[https://www.amazon.co.uk/gp/product/1782948147/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1782948147&linkCode=as2&tag=nrjc-21&linkId=f63dcd8345f4e49c717b39a228a36c7c ''Empirical formulae, pages 71, 72, 77, 78, GCSE Chemistry, CGP, Edexcel  '']
 +
:[https://www.amazon.co.uk/gp/product/1782945725/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1782945725&linkCode=as2&tag=nrjc-21&linkId=694be7494de75af3349537d34e13f7f0 ''Empirical formulas, pages 27, 30, GCSE Chemistry; The Revision Guide, CGP, Edexcel '']
 +
:[https://www.amazon.co.uk/gp/product/1782945741/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1782945741&linkCode=as2&tag=nrjc-21&linkId=30da4f2178da182547b62a7329d13b57 ''Empirical formulas, pages 90, 93, GCSE Combined Science; The Revision Guide, CGP, Edexcel '']
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====OCR====
 +
:[https://www.amazon.co.uk/gp/product/0198359829/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=0198359829&linkCode=as2&tag=nrjc-21&linkId=90e8d7b4f039d53035238fa0320fe00b ''Empirical formulae, page 38, Gateway GCSE Chemistry, Oxford, OCR  '']
 +
:[https://www.amazon.co.uk/gp/product/1782945695/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1782945695&linkCode=as2&tag=nrjc-21&linkId=ceafcc80bcad6b6754ee97a0c7ceea53 ''Empirical formulas, page 105, Gateway GCSE Combined Science; The Revision Guide, CGP, OCR '']
 +
:[https://www.amazon.co.uk/gp/product/1782945679/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1782945679&linkCode=as2&tag=nrjc-21&linkId=a2db42f7b4bdf10cafaafa3bb9120940 ''Empirical formulas, page 32, Gateway GCSE Chemistry; The Revision Guide, CGP, OCR '']

Latest revision as of 01:28, 6 December 2019

Contents

Key Stage 4

Meaning

An empirical formula is the simplest ratio of the different types of atom in a compound.

About Empirical Formulae

The empirical formula of a compound may not be the same as the chemical formula:

Empirical formulae are calculated from the amount of atoms in a chemical reaction.
The number of atoms can be found if you know the mass of different elements and the relative atomic mass of the elements in the reaction.

Finding the Empirical Formula

Number of Atoms Skeletal Diagram Ball and Stick Model
100 atoms of Hydrogen react completely with 50 atoms of Oxygen.
 
 

The ratio of Hydrogen atoms to Oxygen atoms

H:O

100:50

2:1

So the empirical formula is H2O

In this diagram there are 4 Carbon atoms, 10 Hydrogen atoms.

The ratio of atoms is:

C:H

4:10

2:5

So the empirical formula is C2H5

In this diagram there are 4 Carbon atoms, 10 Hydrogen atoms and 2 Oxygen atoms.

The ratio of atoms is:

C:H:O

4:10:2

2:5:1

So the empirical formula is C2H5O

64g of Oxygen is found to react completely with 8g of Hydrogen. Find the empirical formula for the product. 60g of Carbon is found to react completely with 160g of Oxygen. Find the empirical formula for the product. 416g of Sulphur is found to react completely with Oxygen to produce 832g of product. Find the empirical formula for the product.
State the mass of all reactants.

Mass of Oxygen = 64g

Mass of Hydrogen = 8g

State the mass of all reactants.

Mass of Carbon = 60g

Mass of Oxygen = 416g

State the mass of all reactants.

Mass of Sulphur = 416g

Mass of Oxygen = 416g


Find the number of moles of each element.

Number of Moles = (Mass/Relative Atomic Mass)

Relative Atomic Mass of Oxygen = 16g

Moles of Oxygen = 64/16

Moles of Oxygen = 4 mole


Relative Atomic Mass of Hydrogen = 1g

Moles of Hydrogen = 8/1

Moles of Hydrogen = 8 mole

Find the number of moles of each element.

Number of Moles = (Mass/Relative Atomic Mass)

Relative Atomic Mass of Carbon = 12g

Moles of Carbon = 60/12

Moles of Carbon = 5 mole

Mass of Oxygen = 160g


Relative Atomic Mass of Oxygen = 16g

Moles of Oxygen = 160/16

Moles of Oxygen = 10 mole

Find the number of moles of each element.

Number of Moles = (Mass/Relative Atomic Mass)

Relative Atomic Mass of Sulphur = 32g

Moles of Sulphur = 416/32

Moles of Sulphur = 32 mole


Relative Atomic Mass of Oxygen = 16g

Moles of Oxygen = 416/16

Moles of Oxygen = 26 mole

Find the Ratio:

Hydrogen:Oxygen

8:4

2:1

Find the Ratio:

Carbon:Oxygen

5:10

1:2

Find the Ratio:

Sulphur:Oxygen

13:26

1:2

The empirical formula of the product is H2O The empirical formula of the product is CO2 The empirical formula of the product is SO2

References

AQA

Formula, formulae; empirical, pages 62-3, 90-1, 252, GCSE Chemistry; Student Book, Collins, AQA

Edexcel

Empirical formula, page 217, GCSE Combined Science, Pearson Edexcel
Empirical formulae, page 73, GCSE Chemistry, Pearson, Edexcel
Empirical formulae, pages 71, 72, 77, 78, GCSE Chemistry, CGP, Edexcel
Empirical formulas, pages 27, 30, GCSE Chemistry; The Revision Guide, CGP, Edexcel
Empirical formulas, pages 90, 93, GCSE Combined Science; The Revision Guide, CGP, Edexcel

OCR

Empirical formulae, page 38, Gateway GCSE Chemistry, Oxford, OCR
Empirical formulas, page 105, Gateway GCSE Combined Science; The Revision Guide, CGP, OCR
Empirical formulas, page 32, Gateway GCSE Chemistry; The Revision Guide, CGP, OCR
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