go to the AUS-e-TUTE homepage

Percent Composition of Compounds Calculations Chemistry Tutorial

Key Concepts

Please do not block ads on this website.
No ads = no money for us = no free stuff for you!

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
o       o
    o    

Each ball is weighed and it is found that :

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
  ×   100
  =     1 + 1    
16 + 1 + 1
  ×   100
  =   2  
18
  ×   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%

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, H2O, then the diagram below 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
o       o
  \   /  
    o    

A water molecule, H2O, is made up of 1 oxygen atom and 2 hydrogen atoms.

Using the Periodic Table you will find:

relative atomic mass (atomic weight) of an oxygen atom is 16

relative atomic mass (atomic weight) 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
  ×   100
  =     1.0 + 1.0    
16 + 1.0 + 1.0
  ×   100
  =   2.0  
18
  ×   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%

Do you know this?

Join AUS-e-TUTE!

Play the game now!

Mathematical Equations (mathematical expressions) for Percentage Composition Calculations

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
  ×   100

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

So, for a compound with the formula XaYbZc we can write mathematical equations (expressions) to find the percentage by mass of elements X, Y and Z in the compound as shown in the table below:

elementnumber of
atoms of element
relative atomic mass of
element from Periodic Table
total mass
of element
XaMr(X)a × Mr(X)
YbMr(Y)b × Mr(Y)
ZcMr(Z)c × Mr(Z)

relative molecular mass of compound =
Mr(XaYbZc)
a × Mr(X) + b × Mr(Y) + c × Mr(Z)

% by mass(X) =
                a × Mr(X)                
a × Mr(X) + b × Mr(Y) + c × Mr(Z)
× 100
% by mass(Y) =
                b × Mr(Y)                
a × Mr(X) + b × Mr(Y) + c × Mr(Z)
× 100
% by mass(Z) =
                c × Mr(Z)                
a × Mr(X) + b × Mr(Y) + c × Mr(Z)
× 100

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

Do you understand this?

Join AUS-e-TUTE!

Take the test now!

Worked Examples of Percent Composition Calculations

Example 1:

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

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 (Na2SO4).

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 [(NH4)3PO4]

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%

Can you apply this?

Join AUS-e-TUTE!

Take the exam now!

Problem Solving: Percentage Composition Calculations

The problem:

Jackie the Geologist has just discovered a new iron ore deposit.
Jackie took a sample of this iron 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?

Solving the problem:

Using the StoPGoPS model for problem solving:

STOP!
State the question. What is the question asking you to do?

Determine the molecular formula of the compound making up the ore

PAUSE!
Plan. 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%

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:

Mr(iron)

Mr(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 Mr(compound) =

using %(oxygen) to find Mr(compound) =

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

since Mr(compound) calculated using %(iron) = Mr(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:

Mr(iron) = 55.85

Mr(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: FeaOb

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 × Mr(Fe)/Mr(FeaOb) × 100

69.94 = 55.85a/Mr(FeaOb) × 100

69.94/100 = 55.85a/Mr(FeaOb) × 100/100

0.6994 = 55.85a/Mr(FeaOb)

% by mass(oxygen) = b × Mr(O)/Mr(FeaOb) × 100

30.06 = 16.00b/Mr(FeaOb) × 100

30.06/100 = 16.00b/Mr(FeaOb) × 100/100

0.3006 = 16.00b/Mr(FeaOb)

Step 5: Rearrange each equation above in order to find relative molecular mass of compound

using %(iron) to find Mr(FeaOb):

0.6994 × Mr(FeaOb) = 55.85a/Mr(FeaOb) × Mr(FeaOb)

0.6994 × Mr(FeaOb) = 55.85a

0.6994/0.6994 × Mr(FeaOb) = 55.85a/0.6994

Mr(FeaOb) = 55.85a/0.6994 = 79.85a

using %(oxygen) to find Mr(FeaOb):

0.3006 × Mr(FeaOb) = 16.00b/Mr(FeaOb) × Mr(FeaOb)

0.3006 × Mr(FeaOb) = 16.00b

0.3006/0.3006 × Mr(FeaOb) = 16.00b/0.3006

Mr(FeaOb) = 16.00b/0.3006 = 53.23b

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

Mr(FeaOb) calculated using %(iron) = Mr(FeaOb) 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 FeO1.5

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 FeO3/2,

so we can clear the fraction by multiplying the number of oxygen atoms and the number of iron atoms by 2:

F1 × 2O3/2 × 2 which is Fe2O3

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:

Mr(Fe2O3) = 2 × 55.85 + 3 × 16.00 = 111.7 + 48.00 = 159.7

% by mass(Fe) = (2 × 55.85)/159.7 × 100 = 69.94%

% by mass(O) = (3 × 16.00)/159.7 × 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. What is a possible molecular formula for the compound in the ore sample?

Fe2O3


Footnotes:

(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%.