Chemical Formulas
Main objectives for this chapter:
• Empirical and molecular formulas.
• Structural formulas. Simple examples
• Calculations of empirical formulas, given the percentage
composition by mass.
• Calculation of empirical formulas, given the masses of
reactants and products.HIGHER LEVEL only
• Calculation of molecular formulas, given the empirical
formulas and the relative molecular masses (examples
should include simple biological substances, such as
glucose and urea).
• Percentage composition by mass. Calculations
Structural formulas, molecular
formulas, empirical formulas
Structural Formulas
Def: The structural formula of a compound tells you the
arrangement of the atoms in a molecule of the compound
H
The structural formula of ethane:
H
H
H
The structural formula of ethene:
H
H
H
H
H
H
Molecular Formula
• Molecular formula tells you the number of
atoms of each element present in a
molecule of a compound
Name
Methane
Structural Formula Molecular Formula
H
H C H
H
Carbon dioxide
O C O
H
Water
H
O
CH4
CO2
H2O
Empirical Formula
• An empirical formula shows only the ratios in which
different atoms are present in a molecule of a
compound
For glucose C6H12O6 is the
molecular formula
The empirical formula is CH2O
Once you know the structural formula you can work out
the molecular formula and the empirical formula
H
• The structural formula of ethane is :
H
H
H
H
• The molecular formula is: C2H6
• The empirical formula is : CH3
H
Formula of ionic compounds
• Ionic compounds are giant
structures.
• There can be any number of
ions in an ionic crystal - but
always a definite ratio of ions.
- +
-+ + -+
- + + -- +
-+
+ - + -+ + -+
+
Sodium chloride
A 1:1 ratio
Name
Sodium chloride
Ratio
1:1
Formula
NaCl
Magnesium chloride
1:2
MgCl2
Aluminium chloride
1:3
AlCl3
Aluminium Oxide
2:3
Al2O3
The structural formula of ethene is:
H
H
• Find the
(i) molecular formula
formula= C2H4
(ii) empirical formula
formula= CH2
H
H
Calculating Empirical Formulas
Method 1when give &
mass of
element
Finding the empirical formula of a compound
when the percentage composition by mass
of each element in the compound
1. Divide the percentage of each element by
its atomic mass.
2. Then find the simplest whole number
ratio between the elements in that compound
•
•
A compound contains 40% sulfur and 60 % oxygen. What is its empirical
formula?
Element
%
moles
Ratio
sulfur
40
40/ 32 = 1.25
1
oxygen
60
60/ 16 = 3.75
3
Therefore the empirical formula of this compound is SO3.
Q282
Finding the empirical formula of a compound
when the percentage composition by mass
of each element in the compound
1. Divide the percentage of each element by
its atomic mass.
2. Then find the simplest whole number
ratio between the elements in that compound
•
•
A compound contains 48.8% carbon and 13.5% hydrogen and 37.7%
nitrogen respectively by mass. Determine the empirical formula of the
compound.
Element
%
%/ Ar
Carbon
48.8
48.8/ 12 =
1.51
4.066666667
Hydrogen
13.5
13.5 / 1=
13.5
Nitrogen
37.7
37,5/14=
1
2.678571429
Therefore the empirical formula of this compound is
Ratio
5.04
Q284
Finding
empirical
formula
Finding the empirical formula of a compound
when the percentage composition by mass
of each element in the compound
1. Divide the percentage of each element by
its atomic mass.
2. Then find the simplest whole number
ratio between the elements in that compound
•
•
A compound contains 64.9% carbon and 13.5% hydrogen and 21.6 %
oxygen respectively by mass. Determine the empirical formula of the
compound.
Element
%
%/ Ar
Ratio
Carbon
64.9
64.9/ 12 =
5.4083
4.006
Hydrogen
13.5
13.5 / 1=
13.5
10
Oxygen
21.6
21.6 /16=
1.35
1
Therefore the empirical formula of this compound is C4H10O
Conservation of Mass in a reaction
• During chemical reactions the same atoms are present
before and after reaction. They have just joined up in
different ways.
• Because of this the total mass of reactants is always equal
to the total mass of products. (Law of Conservation of
Mass)
Reaction
but no
mass change
Conservation of Mass
Gas given off.
HCl
Mg
Mass of
chemicals in flask
decreases
11.71
Same reaction in
sealed container:
No change in
mass
Method 2- when given
mass of element in
compound
You must know the masses of all of the elements in the compound.
You might have to work this out... REMEMBER
SUM OF MASS OF REACTANTS = SUM OF MASS OF PRODUCTS
Example
• When 3.175g of copper reacts with chlorine gas 6.725g of copper
chloride is formed. Find the empirical formula of the copper chloride
Copper + Chlorine gas
3.175g
?
3.55g
Copper Chloride
6.725g
Element
mass
moles
Ratio
Copper
3.175g
0.05
1
Chlorine
3.55g
0.1
2
The empirical formula is CuCl2
Method 2- when given
mass of element in
compound
Example
• When`1.44g of Magnesium was completely burned in oxygen it
resulted in the formation of 2.40g of magnesium oxide. Find the
empirical formula of magnesium oxide
Magnesium + Oxygen
0.96g?
1.44g
Magnesium oxide
2.40g
Element
mass
moles
Ratio
Magnesium
1.44g
0.06
1
Oxygen
0.96g
0.06
1
The empirical formula is MgO
Method 2- when given
mass of element in
compound
Q288
• When`2.07g of lead reacts with iodine, 4.61g of lead iodide was
formed. Find the empirical formula of lead iodide
Lead + Iodine
2.07g
2.54 ?
Lead iodide
4.61g
Element
mass
moles
Ratio
Lead
2.07g
0.01
1
Iodine
2.54
0.02
2
The empirical formula is PbI2
Method 2- when given
mass of element in
compound
Q289
• When`3.94g of hydrated copper (II) sulfate was heated, 2.52g of
anhydrous salt remained. Calculate the formula of the hydrated salt.
Hydrated copper(II) sulfate
3.94g
Anhydrous copper sulphate + water
2.52g
1.42g
Group
mass
moles
Ratio
Water
1.42
0.078888888
5
Copper sulfate
2.52
0.015799373
1
The empirical formula is CuSO4 (H20)5
Q290
Method 2- when given mass of element in compound
• 9.76g of a metal forms 20.9g of its oxide whose formula is M20.
Calculate the relative molecular mass of the metal.
Metal + Oxygen
9.76g
11.14g
Metal oxide
20.90g
Group
mass
moles
Ratio
Metal
9.76
?1.3925
2
oxygen
11.14
0.69625
1
(1.3925/ 9.76= 7.008976661
Mass of one mole of the metal is 7.009g. This is the relative molecular
mass
Calculating Molecular Formulas
Calculation of the moleuclar
formula:
• You need:
1. The empirical formula
2. The molecular mass (can work out using
the periodic table)
Calculating
molecular mass
You need:
1. The empirical formula
2. The relative molecular mass
• Urea is used as a fertiliser and an animal feed. It has a
relative molecular mass of 60 and is composed of
46.66% N, 26.66%O, 20%C and 6.66% H. Determine
the molecular formula of urea
1. The empirical formula N2OCH4
Mass according to EF = 2(14) +16 +12 +4(1) = 60
Element
%
moles
ratio
nitrogen
46.66
3.33
2
oxygen
26.66
1.66
1
Carbon
20
1.66
1
Hydrogen
6.66
6.66
4
2. Relative molecular mass of urea = 60
Molecular formula N2OCH4
3. Empirical formula = molecular formula
284. Error – see
286
You need:
1. The empirical formula
2. The relative molecular mass
• An alcohol was found on analysis to contain 64.9%
carbon, 13.5% hydrogen and 21.6% oxygen. If the
relative molecular mass of the alcohol is 74 show that
the molecular formula is C4H10O
1. The empirical formula C4H10 O
Mass according to EF = 4(12) +16 +10(1) = 74
Element
%
moles
ratio
Carbon
64.9
5.408333333 4
oxygen
21.6
1.35
1
Hydrogen
13.5
13.5
10
2. Relative molecular mass of urea = 74
3. Empirical formula = molecular formula
C4H10 O
285. Calculating
molecular mass
You need:
1. The empirical formula
2. The relative molecular mass
• An organic acid contain 27.6% carbon, 2.2% hydrogen
and 71.1% oxygen. If the relative molecular mass of the
alcohol is 90. Determine the molecular formula
1. The empirical formula CO2 H
Mass according to EF = 12 +2(16) +1 = 45
Element
%
moles
ratio
Carbon
27.6
2.3
1
oxygen
71.1
4.44375
2
Hydrogen
2.2
2.2
1
2. Relative molecular mass of the organic acid = 90
3. Empirical formula x 2 = molecular formula
Molecular formula = C2O4 H2
286. Calculating
molecular mass
You need:
1. The empirical formula
2. The relative molecular mass
• Determine the molecualr formula of a compound whose
composition is carbon 64.8%, hydrogen 13.6%and
oxygen 21.6% and whose relative molecular mass is 74
1. The empirical formula C4H10 O
Mass according to EF = 4(12) +16 +10(1) = 74
Element
%
moles
ratio
Carbon
64.8
5.40
4
oxygen
21.6
1.35
1
Hydrogen
13.5
13.6
10
2. Relative molecular mass of urea = 74
3. Empirical formula = molecular formula
Molecular formula = C4H10 O
Calculating % compositions by
mass
By the end of today’s class you
should be able to:
(iii)Calculate the Percentage composition by
mass of an element in a compound
Percentage composition by mass
Mass % of A in compound =
Mass of A in compound
Relative molecular mass of the compound
x 100
1
What is the percentage by mass
of Fe in Fe2O3 ?
Mass % of A in compound =
Mass of A in compound
Relative molecular mass of the compound
2(56) x 100
160
1
70%
x 100
1
291b)What is the percentage by
mass of nitrogen in NH4NO3 ?
Mass % of A in compound =
Mass of A in compound
Relative molecular mass of the compound
2(14) x 100
80
1
35%
x 100
1
291c) What is the percentage
by mass of carbon in
methylbenzene
?
Mass % of A in compound =
Mass of A in compound
Relative molecular mass of the compound
7(12) x 100
92
1
91.3043478%
x 100
1
291(d)What is the percentage
by mass of N in NO2 ?
Mass % of A in compound =
Mass of A in compound
Relative molecular mass of the compound
14 x 100
46
1
30.4347826%
x 100
1
291(e)What is the percentage by
mass of water in Na2 CO3. 10H20 ?
Mass % of A in compound =
Mass of A in compound
Relative molecular mass of the compound
10(18) x 100
286
1
62.9370629%
x 100
1
291(f)What is the percentage by
mass of Fe in Fe3 O4.?
Mass % of A in compound =
Mass of A in compound
Relative molecular mass of the compound
3(56) x 100
232
1
72.4137931%
x 100
1
Extra questions not on
worksheet
Urea has an empirical formula of CON2H4
and a relative molecular mass of 60.
Find its molecular formula.
The empirical formula of urea is a simple whole number ratio of
the atoms from each element that are present.
•If the atoms in each molecule actually present were CON2H4
then the molecular mass of urea would be:
12 + 16+ 14 +14+ 1+1+1+1 = 60
•We are told the molecular mass of urea is 60.
•molecular formula = empirical formula!
•Molecular formula = CON2H4
Glucose has an empirical formula of CH2O
and a relative molecular mass of 180.
Find its molecular formula.
•If the atoms in each molecule actually present were CH2O then
the molecular mass of glucose would be:
12+ 1+1+16= 30
•But we are told the molecular mass of glucose is 180.
•So the molecular formula is not CH2O!
•30 x n = 180
•n=6
•So the molecular formula = C6H12O6
Q1. Heptane has an empirical formula of C7H16 and a
relative molecular mass of 100.
Find its molecular formula.
• If the atoms in each molecule actually present were
C7H16 then the molecular mass of heptane would be:
(12 X 7) + (1 X 16)= 100
• We are told the molecular mass of Heptane is 100.
• Molecular formula = empirical formula
• Molecular formula of Heptane = C7H16
Q2. Butanoic acid has an empirical formula of C2H4O and
a relative molecular mass of 88.
Find its molecular formula.
• If the atoms in each molecule actually present were C2H4O2 then the
molecular mass of Butanoic acid would be:
12 + 12 + 1+1+ 1+1 + 16 = 44
• We are told the molecular mass of Butanoic acid is 88.
• Molecular formula x 2 = empirical formula
• Molecular formula of Butanoic acid = C4H8O2
Q3. Fructose has the following composition by mass – 40%
carbon, 6.66% hydrogen, 53.33% oxygen. If the relative
molecular mass of fructose is 180 find its molecular
formula.
First you need to find the empirical formula:
Element
present
Mass
present in
100g of
fructose
Carbon
40g
Hydrogen
6.66g
Oxygen
53.33g
Moles
present in
100g of
fructose
Molar
ratio
Simplest Simplest
ratio
whole
number
ratio
First you need to find the empirical formula:
Element
present
Mass
present in
100g of
fructose
Moles
present in
100g of
fructose
Carbon
40g
40
12
Hydrogen
6.66g
6.66 = 6.67
1
Oxygen
53.33g
53.33 = 3.33
16
Molar
ratio
= 3.33
Moles present = mass in grams
Ar
Simplest Simplest
ratio
whole
number
ratio
First you need to find the empirical formula:
Element
present
Mass
present in
100g of
fructose
Moles
present in
100g of
fructose
Molar
ratio
Carbon
40g
40
12
3.33 =1
3.33
Hydrogen
6.66g
6.66 = 6.67 6.67 =2
1
3.33
Oxygen
53.33g
53.33 = 3.33 3.33 =1
16
3.33
= 3.33
Simplest Simplest
ratio
whole
number
ratio
To get a simple ratio divide each number by the smallest number
present!
First you need to find the empirical formula:
Element
present
Mass
present in
100g of
fructose
Moles
present in
100g of
fructose
Molar
ratio
Carbon
40g
40
12
3.33 =1 1
3.33
1
Hydrogen
6.66g
6.66 = 6.67 6.67 =2 2
1
3.33
2
Oxygen
53.33g
53.33 = 3.33 3.33 =1 1
16
3.33
1
= 3.33
empirical formula = CH2O
Simplest Simplest
ratio
whole
number
ratio
the relative molecular mass of fructose is 180, the
empirical formula is CH2O - find its molecular formula.
• If the atoms in each molecule actually present were
CH2O then the molecular mass of fructose would be:
12 + 1+1+ 16 = 30
• We are told the molecular mass of Fructose is 180.
• Molecular formula x 6 = empirical formula
• Molecular formula of Fructose = C6H12O6
What is the percentage composition by mass of carbon
present in ethanol ( C2H5OH )?
• Moles of carbon present : 2
• Mass of carbon present :
Ar x number of moles = mass present in grams
12 x 2 = 24g
• Total mass of C2H5OH =
12 +12+ 1+1+1+1+1+16+1 = 46
• Percentage of carbon by mass in ethanol =
24 x 100 = 52.17%
46
What is the percentage composition by mass of nitrogen
present in (NH4)2HPO4 ?
• (NH4)2HPO4 = N2H9PO4
• Moles of nitrogen present : 2
• Mass of nitrogen present :
Ar x number of moles = mass present in grams
14 x 2 = 28g
• Total mass of N2H9PO4:
(14 x 2)+ (1 x 9)+ 40 +(16 x4) = 109g
• Percentage of nitrogen by mass in N2H9PO4
28 x 100 = 25.68%
109
What is the percentage composition by mass of sodium in
sodium hydroxide NaOH?
• Moles of sodium present : 1
• Mass of sodium present :
Ar x number of moles = mass present in grams
23 x 1 = 23g
• Total mass of NaOH =
23 + 16 +1 = 40
• Percentage of sodium by mass in sodium hydroxide=
23 x 100 = 57.5%
40