The percent composition (percentage composition) of a compound is a relative measure of the mass (or weight) of each different element present in the compound.

To calculate the percent composition (percentage composition) of a compound:

Calculate the total mass of each element present in the molecular formula^{1} of the compound

total mass of element present = number of atoms of this element x relative atomic mass of element^{2}

add together all the total masses for all the elements present in the molecule

Calculate the percentage compositon : % by mass (or % by weight) of element

% by mass(element)

=

total mass of element present in compound relative molecular mass of compound

x 100

When the % by mass for every element in the compound is added together, the result will be 100%^{4}

Percent Composition Concepts

The diagram on the right shows a box containing 3 balls:

one red ball shown as o

two black balls shown as o

A red ball has a mass of 16 g
A black ball has a mass of 1 g
What percentage of the total mass of all the balls in the box is due to the black balls?

percent by mass(black balls)

=

combined mass of all black balls total mass of all balls in box

x 100

=

1 + 1 16 + 1 + 1

x 100

=

2 18

x 100

=

11 %

What percentage of the total mass of all the balls in the box is due to the red ball?
100% by mass = % by mass of black balls + % by mass of red ball
100% = 11% + % by mass of red ball
% by mass of red ball = 100 - 11 = 89%

o

o

o

Chemists often think of atoms and molecules as really tiny balls, and refer to this as the particle theory of matter.
If each ball in the box represents an atom making up a water molecule, H_{2}O, then the diagram on the right shows a box containing a molecule of water in which

the red ball o is an oxygen atom

each black ball o is a hydrogen atom

each diagonal line ( \ and / ) represents a chemical bond between an oxygen atom and a hydrogen atom

A water molecule, H_{2}O, is made up of 1 oxygen atom and 2 hydrogen atoms.
Using the Periodic Table you will find:
relative atomic mass of an oxygen atom is 16
relative atomic mass of a hydrogen atom is 1.0
What percentage of the total mass of the water molecule is due to hydrogen atoms?

percent by mass(hydrogen)

=

combined mass of all hydrogen atoms total mass of all atoms in molecule

x 100

=

1.0 + 1.0 16 + 1.0 + 1.0

x 100

=

2.0 18

x 100

=

11 %

What percentage of the mass of a water molecule is due to the oxygen atom?
100% by mass = % by mass of all hydrogen atoms + % by mass of the oxygen atom
100% = 11% + % by mass of the oxygen atom
% by mass of the oxygen atom = 100 - 11 = 89%

o

o

\

/

o

For a water molecule:

percent by mass(hydrogen)

=

combined mass of all hydrogen atoms total mass of all atoms in molecule

x 100

percent by mass(oxygen)

=

combined mass of all oxygen atoms total mass of all atoms in molecule

x 100

% by mass of oxygen + % by mass of hydrogen = 100 % (% by mass of water molecule)

Equations (formulae, or, expressions) for Percentage Composition

For any element present in any molecule:

percent by mass(element)

=

combined mass of all atoms of this element total mass of all atoms of all elements in molecule

x 100

And, if you add together the % by mass of every element present in the compound the result will be 100%

relative molecular mass of compound M_{r}(X_{a}Y_{b}Z_{c})

= a x M_{r}(X) + b x M_{r}(Y) + c x M_{r}(Z)

% by mass(X) =

a x M_{r}(X) a x M_{r}(X) + b x M_{r}(Y) + c x M_{r}(Z)

x 100

% by mass(Y) =

b x M_{r}(Y) a x M_{r}(X) + b x M_{r}(Y) + c x M_{r}(Z)

x 100

% by mass(Z) =

c x M_{r}(Z) a x M_{r}(X) + b x M_{r}(Y) + c x M_{r}(Z)

x 100

% by mass(X) + % by mass(Y) + % by mass(Z) = 100%

Examples

Example 1

Calculate the percent by mass (weight) of sodium (Na) and chlorine (Cl) in sodium chloride (NaCl)

What is the question asking you to do?
(i) Calculate the % by mass of Na in NaCl
(ii)Calculate the % by mass of Cl in NaCl

What information (data) has been given in the question?
symbol for sodium: Na
symbol for chlorine: Cl
molecular formula for sodium chloride: NaCl
which tells us:
number of atoms of Na = 1
number of atoms of Cl = 1

What is the relationship between the % by mass of an element and the compound?

% by mass(element)

=

sum of relative atomic masses of all atoms of this element relative molecular mass of compound

x 100

Calculate the sum of the relative atomic masses of all Na atoms present in the compound:
Using the Periodic Table : relative atomic mass(Na) = M_{r}(Na) = 22.99
1 Na atom is present in the chemical formula NaCl
sum of relative atomic masses of all Na atoms present in NaCl = 1 x 22.99 = 22.99

Calculate the sum of the relative atomic masses of all Cl atoms present in the compound:
Using the Periodic Table : relative atomic mass(Cl) = M_{r}(Cl) = 35.45
1 Cl is present in the formula NaCl
sum of relative atomic masses of all Cl atoms present in NaCl = 1 x 35.45 = 35.45

Calculate the relative molecular mass of NaCl:
M_{r}(NaCl) = 1 x M_{r}(Na) + 1 x M_{r}(Cl)
M_{r}(NaCl) = 1 x 22.99 + 1 x 35.45 = 58.44

Calculate the percent by mass (weight) of Na in NaCl:

% by mass(Na)

=

sum of relative atomic masses of all atoms of Na relative molecular mass of NaCl

x 100

% by mass(Na)

=

22.99 58.44

x 100

%Na = 39.34%

Calculate the percent by mass (weight) of Cl in NaCl:

% by mass(Na)

=

sum of relative atomic masses of all atoms of Cl relative molecular mass of NaCl

x 100

% by mass(Cl)

=

35.45 58.44

x 100

%Cl = 60.66%

The answers above are probably correct if %Na + %Cl = 100%, that is,
39.34% + 60.66% = 100%

Example 2

Calculate the percent by mass (weight) of each element present in sodium sulfate (Na_{2}SO_{4}).

What is the question asking you to do?
(i) Calculate % by mass(Na) in Na_{2}SO_{4} (ii) Calculate % by mass(S) in Na_{2}SO_{4} (iii) Calculate % by mass(O) in Na_{2}SO_{4}

What information (data) has been given in the question?
molecular formula for sodium sulfate: Na_{2}SO_{4} which tells us the number of atoms of each element present in the compound:
there are 2 atoms of Na in Na_{2}SO_{4} there is 1 atom of S in Na_{2}SO_{4} there are 4 atoms of O in Na_{2}SO_{4}

What is the relationship between the % by mass of an element and the compound?

% by mass(element)

=

sum of relative atomic masses of all atoms of this element relative molecular mass of compound

x 100

Calculate the sum of the relative atomic masses of all Na atoms present in the compound:
Using the Periodic Table : relative atomic mass(Na) = M_{r}(Na) = 22.99
2 Na atoms are present in the chemical formula Na_{2}SO_{4} sum of relative atomic masses of all Na atoms present in Na_{2}SO_{4} = 2 x 22.99 = 45.98

Calculate the sum of the relative atomic masses of all S atoms present in the compound:
Using the Periodic Table : relative atomic mass(S) = M_{r}(S) = 32.07
1 S atom are present in the chemical formula Na_{2}SO_{4} sum of relative atomic masses of all S atoms present in Na_{2}SO_{4} = 1 x 32.07 = 32.07

Calculate the sum of the relative atomic masses of all O atoms present in the compound:
Using the Periodic Table : relative atomic mass(O) = M_{r}(O) = 16.00
4 O atoms are present in the chemical formula Na_{2}SO_{4} sum of relative atomic masses of all O atoms present in Na_{2}SO_{4} = 4 x 16.00 = 64.00

Calculate the relative molecular mass (M_{r}) of Na_{2}SO_{4}:
M_{r}(Na_{2}SO_{4}) = 2 x M_{r}(Na) + 1 x M_{r}(S) + 4 x M_{r}(O)
M_{r}(Na_{2}SO_{4}) =(2 x 22.99) + (1 x 32.07) + (4 x 16.00)
= 45.98 + 32.07 + 64.00 = 142.05
= 142.05

Calculate the percent by mass (weight) of Na in Na_{2}SO_{4}:

% by mass(Na)

=

sum of relative atomic masses of all atoms of Na relative molecular mass of Na_{2}SO_{4}

x 100

% by mass(Na)

=

45.98 142.05

x 100

% by mass(Na) = 32.37%

Calculate the percent by mass (weight) of S present in Na_{2}SO_{4}:

% by mass(S)

=

sum of relative atomic masses of all atoms of S relative molecular mass of Na_{2}SO_{4}

x 100

% by mass(Na)

=

32.07 142.05

x 100

% by mass(S) = 22.58%

Calculate the percent by mass (weight) of O in Na_{2}SO_{4}:

% by mass(O)

=

sum of relative atomic masses of all atoms of O relative molecular mass of Na_{2}SO_{4}

x 100

% by mass(O)

=

64.00 142.05

x 100

% by mass(O) = 45.05%

The answers above are probably correct if %Na + %S + %O = 100%, that is,
32.37% + 22.58% + 45.05% = 100%

Example 3

Calculate the percent by mass (weight) of each element present in ammonium phosphate [(NH_{4})_{3}PO_{4}]

What have you been asked to do?
(i) Calculate % by mass(N) in (NH_{4})_{3}PO_{4} (ii) Calculate % by mass(H) in (NH_{4})_{3}PO_{4} (iii) Calculate % by mass(P) in (NH_{4})_{3}PO_{4} (iv) Calculate % by mass(O) in (NH_{4})_{3}PO_{4}

What information (data) has been given in the question?
molecular formula for ammonium phosphate, (NH_{4})_{3}PO_{4} which tells us the number of atoms of each element present in the compound:
3 atoms of N
12 atoms of H
1 atom of P
4 atoms of O

What is the relationship between the % by mass of an element and the compound?

% by mass(element)

=

sum of relative atomic masses of all atoms of this element relative molecular mass of compound

x 100

Calculate the sum of the relative atomic masses of all N atoms present in the compound:
Using the Periodic Table : relative atomic mass(N) = M_{r}(N) = 14.01
3 N atoms are present in the chemical formula (NH_{4})_{3}PO_{4} sum of relative atomic masses of all N atoms present in (NH_{4})_{3}PO_{4} = 3 x 14.01 = 42.03

Calculate the sum of the relative atomic masses of all H atoms present in the compound:
Using the Periodic Table : relative atomic mass(H) = M_{r}(H) = 1.008
12 H atoms are present in the chemical formula (NH_{4})_{3}PO_{4} sum of relative atomic masses of all H atoms present in (NH_{4})_{3}PO_{4} = 12 x 1.008 = 12.096

Calculate the sum of the relative atomic masses of all P atoms present in the compound:
Using the Periodic Table : relative atomic mass(P) = M_{r}(P) = 30.97
1 P atom is present in the chemical formula (NH_{4})_{3}PO_{4} sum of relative atomic masses of all P atoms present in (NH_{4})_{3}PO_{4} = 1 x 30.97 = 30.97

Calculate the sum of the relative atomic masses of all O atoms present in the compound:
Using the Periodic Table : relative atomic mass(O) = M_{r}(O) = 16.00
4 O atoms are present in the chemical formula (NH_{4})_{3}PO_{4} sum of relative atomic masses of all O atoms present in (NH_{4})_{3}PO_{4} = 4 x 16.00 = 64.00

Calculate the relative molecular mass (M_{r}) of (NH_{4})_{3}PO_{4}:
M_{r}((NH_{4})_{3}PO_{4}) = 3 x M_{r}(N) + 12 x M_{r}(H) + 1 x M_{r}(P) + 4 x M_{r}(O)
= 3 x 14.01 + 12 x 1.008 + 30.97 + 4 x 16.00
= 42.03 + 12.096 + 30.97 + 64.00
= 149.096

Calculate the percent by mass of N present in (NH_{4})_{3}PO_{4}

% by mass(N)

=

sum of relative atomic masses of all atoms of N relative molecular mass of (NH_{4})_{3}PO_{4}

x 100

% by mass(N)

=

42.03 149.096

x 100

% by mass(N) = 28.19%

Calculate the percent by mass of H present in (NH_{4})_{3}PO_{4}

% by mass(H)

=

sum of relative atomic masses of all atoms of H relative molecular mass of (NH_{4})_{3}PO_{4}

x 100

% by mass(H)

=

12.096 149.096

x 100

% by mass(H) = 8.11%

Calculate the percent by mass of P present in (NH_{4})_{3}PO_{4}

% by mass(P)

=

sum of relative atomic masses of all atoms of P relative molecular mass of (NH_{4})_{3}PO_{4}

x 100

% by mass(P)

=

30.97 149.096

x 100

% by mass(P) = 20.77%

Calculate the percent by mass of O present in (NH_{4})_{3}PO_{4}

% by mass(O)

=

sum of relative atomic masses of all atoms of O relative molecular mass of (NH_{4})_{3}PO_{4}

x 100

% by mass(O)

=

64.00 149.096

x 100

% by mass(O) = 42.93%

The answers above are probably correct if :
%by mass(N) + %by mass(H) + %by mass(P) + %by mass(O) =100%, that is,
28.19% + 8.11% + 20.77% + 42.93% = 100%

Problem Solving : Percentage Composition Calculations

The problem: Jackie the Geologist has just discovered a new ore deposit.
Jackie took a sample of this ore to Chris the Chemist for analysis.
First, Chris purified the sample, removing the mud and other impurities from the sample.
Then Chris analysed the purified ore sample and found it was composed of 69.94% iron and 30.06% oxygen.
What is the most likely formula for the compound that makes up Jackie's ore?

What is the question asking you to do?
Determine the molecular formula of the compound making up the ore

What chemical principle will you need to apply?
Apply stoichoimetry (chemical calculations)

What information (data) have you been given?

compound contains 69.94% iron

compound contains 30.06% oxygen

The compound must be made up of only iron and oxygen since 69.94% + 30.06% = 100%

PAUSE!

Plan.

Step 1: Use the Periodic Table for find the symbols for each element
symbol for iron
symbol for oxygen

Step 2: Use the Periodic Table to find the relative atomic mass of each element:
M_{r}(iron)
M_{r}(oxygen)

Step 3: Write a partial formula for the compound:
number of atoms of iron =
number of atoms of oxygen =

Step 4: Write equations for the relationship between the % by mass of each element in the compound and the relative molecular mass of the compound:
% by mass(iron) =
% by mass(oxygen) =

Step 5: Rearrange each equation above in order to find relative molecular mass of compound
using %(iron) to find M_{r}(compound) =
using %(oxygen) to find M_{r}(compound) =

Step 6: Write an expression for the relationship between %(iron) and %(oxygen)
since M_{r}(compound) calculated using %(iron) = M_{r}(compound) using %(oxygen)

Step 7: Solve the equation in order to find the number of iron atoms and the number of oxygen atoms and write the formula

GO!

Go with the plan.

Step 1: Use the Periodic Table for find the symbols for each element
symbol for iron: Fe symbol for oxygen: O

Step 2: Use the Periodic Table to find the relative atomic mass of each element:
M_{r}(iron) = 55.85 M_{r}(oxygen) = 16.00

Step 3: Write a partial formula for the compound:
number of atoms of Fe (iron) = a number of atoms of O (oxygen) = b partial formula: Fe_{a}O_{b}

Step 4: Write equations for the relationship between the % by mass of each element in the compound and the relative molecular mass of the compound:
% by mass(iron) = a x M_{r}(Fe)/M_{r}(Fe_{a}O_{b}) x 100 69.94 = 55.85a/M_{r}(Fe_{a}O_{b}) x 100 69.94/100 = 55.85a/M_{r}(Fe_{a}O_{b}) x 100/100 0.6994 = 55.85a/M_{r}(Fe_{a}O_{b}) % by mass(oxygen) =b x M_{r}(O)/M_{r}(Fe_{a}O_{b}) x 100 30.06 = 16.00b/M_{r}(Fe_{a}O_{b}) x 100 30.06/100 = 16.00b/M_{r}(Fe_{a}O_{b}) x 100/100 0.3006 = 16.00b/M_{r}(Fe_{a}O_{b})

Step 5: Rearrange each equation above in order to find relative molecular mass of compound
using %(iron) to find M_{r}(Fe_{a}O_{b}):
0.6994 x M_{r}(Fe_{a}O_{b}) = 55.85a/M_{r}(Fe_{a}O_{b}) x M_{r}(Fe_{a}O_{b})
0.6994 x M_{r}(Fe_{a}O_{b}) = 55.85a 0.6994/0.6994 x M_{r}(Fe_{a}O_{b}) = 55.85a/0.6994 M_{r}(Fe_{a}O_{b}) = 55.85a/0.6994 = 79.85a using %(oxygen) to find M_{r}(Fe_{a}O_{b}):
0.3006 x M_{r}(Fe_{a}O_{b}) = 16.00b/M_{r}(Fe_{a}O_{b}) x M_{r}(Fe_{a}O_{b})
0.3006 x M_{r}(Fe_{a}O_{b}) = 16.00b 0.3006/0.3006 x M_{r}(Fe_{a}O_{b}) = 16.00b/0.3006 M_{r}(Fe_{a}O_{b}) = 16.00b/0.3006 = 53.23b

Step 6: Write an expression for the relationship between %(iron) and %(oxygen)

M_{r}(Fe_{a}O_{b}) calculated using %(iron) = M_{r}(Fe_{a}O_{b}) calculated using %(oxygen)

79.85a = 53.23b

Step 7: Solve the equation in order to find the number of iron atoms and the number of oxygen atoms and write the formula
79.85a/53.23 = 53.23b/53.23 1.5a = b So if b =1 then a = 1.5 which would make the formula FeO_{1.5} Chemists prefer not to use decimals, that is fractions of atoms, in a chemical formula,
since, 1.5 represents the fraction ^{3}/_{2}, we could write the formula as FeO_{3/2},
so we can clear the fraction by multiplying the number of oxygen atoms and the number of iron atoms by 2:
F_{1 x 2}O_{3/2 x 2} which is Fe_{2}O_{3}

PAUSE!

Ponder plausability.

Does this solution answer the question that was asked?
Yes, we have determined a possible molecular formula for this compound.

Is the solution reasonable?
Work backwards by calculating the percentage composition of the compound using the chemical formula we have determined:
: M_{r}(Fe_{2}O_{3}) = 2 x 55.85 + 3 x 16.00 = 111.7 + 48.00 = 159.7
% by mass(Fe) = (2 x 55.85)/159.7 x 100 = 69.94%
% by mass(O) = (3 x 16.00)/159.7 x 100 = 30.06%
Since the % by mass(Fe) and the % by mass(O) we calculated using the formula we determined are the same as the % by mass for Fe and O given in the question, we are confident that our solution to the question is plausible.

STOP!

State the solution.

A possible molecular formula for the compound in the ore sample is Fe_{2}O_{3}

^{1}We will be using the term "molecular formula" generically to include ionic compounds as well as covalent compounds.

^{2}We will be using relative atomic mass of elements which you will find in the Periodic Table but you can use molar mass of the element since the molar mass is just the relative atomic mass of the element expressed in grams.

^{3}We will be using relative molecular mass (also known as formula mass, molecular weight, or formula weight), but you can use molar mass instead since molar mass is just the relative molecular mass expressed in grams.

^{4}There may be rounding errors in your calculations so that the result will be extremely close to, but not exactly, 100%. In general you can expect results to be between 99% and 101%.

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