# Percent Composition (Percentage Composition)

## Key Concepts

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 formula1 of the compound
total mass of element present = number of atoms of this element x relative atomic mass of element2

• Calculate the relative molecular mass, Mr, of the compound3
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 compoundrelative 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 ballstotal 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, H2O, 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, H2O, 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 atomstotal 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 atomstotal mass of all atoms in molecule x   100 percent by mass(oxygen) = combined mass of all oxygen atomstotal 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%

For a compound with the formula XaYbZc:

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

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

% by mass(X) =
 a x Mr(X)                 a x Mr(X) + b x Mr(Y) + c x Mr(Z) x 100
% by mass(Y) =
 b x Mr(Y)                 a x Mr(X) + b x Mr(Y) + c x Mr(Z) x 100
% by mass(Z) =
 c x Mr(Z)                 a x Mr(X) + b x Mr(Y) + c x Mr(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)
1. 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

2. 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

3. 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 elementrelative molecular mass of compound x   100

4. Calculate the sum of the relative atomic masses of all Na atoms present in the compound:
Using the Periodic Table : relative atomic mass(Na) = Mr(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

5. Calculate the sum of the relative atomic masses of all Cl atoms present in the compound:
Using the Periodic Table : relative atomic mass(Cl) = Mr(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

6. Calculate the relative molecular mass of NaCl:
Mr(NaCl) = 1 x Mr(Na) + 1 x Mr(Cl)
Mr(NaCl) = 1 x 22.99 + 1 x 35.45 = 58.44

7. Calculate the percent by mass (weight) of Na in NaCl:
 % by mass(Na) = sum of relative atomic masses of all atoms of Narelative molecular mass of NaCl x   100 % by mass(Na) = 22.9958.44 x   100

%Na = 39.34%

8. Calculate the percent by mass (weight) of Cl in NaCl:
 % by mass(Na) = sum of relative atomic masses of all atoms of Clrelative molecular mass of NaCl x   100 % by mass(Cl) = 35.4558.44 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 (Na2SO4).
1. What is the question asking you to do?
(i) Calculate % by mass(Na) in Na2SO4
(ii) Calculate % by mass(S) in Na2SO4
(iii) Calculate % by mass(O) in Na2SO4

2. What information (data) has been given in the question?
molecular formula for sodium sulfate: Na2SO4
which tells us the number of atoms of each element present in the compound:
there are 2 atoms of Na in Na2SO4
there is 1 atom of S in Na2SO4
there are 4 atoms of O in Na2SO4

3. 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 elementrelative molecular mass of compound x   100

4. Calculate the sum of the relative atomic masses of all Na atoms present in the compound:
Using the Periodic Table : relative atomic mass(Na) = Mr(Na) = 22.99
2 Na atoms are present in the chemical formula Na2SO4
sum of relative atomic masses of all Na atoms present in Na2SO4 = 2 x 22.99 = 45.98

5. Calculate the sum of the relative atomic masses of all S atoms present in the compound:
Using the Periodic Table : relative atomic mass(S) = Mr(S) = 32.07
1 S atom are present in the chemical formula Na2SO4
sum of relative atomic masses of all S atoms present in Na2SO4 = 1 x 32.07 = 32.07

6. Calculate the sum of the relative atomic masses of all O atoms present in the compound:
Using the Periodic Table : relative atomic mass(O) = Mr(O) = 16.00
4 O atoms are present in the chemical formula Na2SO4
sum of relative atomic masses of all O atoms present in Na2SO4 = 4 x 16.00 = 64.00

7. Calculate the relative molecular mass (Mr) of Na2SO4:
Mr(Na2SO4) = 2 x Mr(Na) + 1 x Mr(S) + 4 x Mr(O)
Mr(Na2SO4) =(2 x 22.99) + (1 x 32.07) + (4 x 16.00)
= 45.98 + 32.07 + 64.00 = 142.05
= 142.05

8. Calculate the percent by mass (weight) of Na in Na2SO4:
 % by mass(Na) = sum of relative atomic masses of all atoms of Narelative molecular mass of Na2SO4 x   100 % by mass(Na) = 45.98142.05 x   100

% by mass(Na) = 32.37%

9. Calculate the percent by mass (weight) of S present in Na2SO4:
 % by mass(S) = sum of relative atomic masses of all atoms of Srelative molecular mass of Na2SO4 x   100 % by mass(Na) = 32.07142.05 x   100

% by mass(S) = 22.58%

10. Calculate the percent by mass (weight) of O in Na2SO4:
 % by mass(O) = sum of relative atomic masses of all atoms of Orelative molecular mass of Na2SO4 x   100 % by mass(O) = 64.00142.05 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 [(NH4)3PO4]
1. What have you been asked to do?
(i) Calculate % by mass(N) in (NH4)3PO4
(ii) Calculate % by mass(H) in (NH4)3PO4
(iii) Calculate % by mass(P) in (NH4)3PO4
(iv) Calculate % by mass(O) in (NH4)3PO4

2. What information (data) has been given in the question?
molecular formula for ammonium phosphate, (NH4)3PO4
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

3. 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 elementrelative molecular mass of compound x   100

4. Calculate the sum of the relative atomic masses of all N atoms present in the compound:
Using the Periodic Table : relative atomic mass(N) = Mr(N) = 14.01
3 N atoms are present in the chemical formula (NH4)3PO4
sum of relative atomic masses of all N atoms present in (NH4)3PO4 = 3 x 14.01 = 42.03

5. Calculate the sum of the relative atomic masses of all H atoms present in the compound:
Using the Periodic Table : relative atomic mass(H) = Mr(H) = 1.008
12 H atoms are present in the chemical formula (NH4)3PO4
sum of relative atomic masses of all H atoms present in (NH4)3PO4 = 12 x 1.008 = 12.096

6. Calculate the sum of the relative atomic masses of all P atoms present in the compound:
Using the Periodic Table : relative atomic mass(P) = Mr(P) = 30.97
1 P atom is present in the chemical formula (NH4)3PO4
sum of relative atomic masses of all P atoms present in (NH4)3PO4 = 1 x 30.97 = 30.97

7. Calculate the sum of the relative atomic masses of all O atoms present in the compound:
Using the Periodic Table : relative atomic mass(O) = Mr(O) = 16.00
4 O atoms are present in the chemical formula (NH4)3PO4
sum of relative atomic masses of all O atoms present in (NH4)3PO4 = 4 x 16.00 = 64.00

8. Calculate the relative molecular mass (Mr) of (NH4)3PO4:
Mr((NH4)3PO4) = 3 x Mr(N) + 12 x Mr(H) + 1 x Mr(P) + 4 x Mr(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

9. Calculate the percent by mass of N present in (NH4)3PO4
 % by mass(N) = sum of relative atomic masses of all atoms of Nrelative molecular mass of (NH4)3PO4 x   100 % by mass(N) = 42.03149.096 x   100

% by mass(N) = 28.19%

10. Calculate the percent by mass of H present in (NH4)3PO4
 % by mass(H) = sum of relative atomic masses of all atoms of Hrelative molecular mass of (NH4)3PO4 x   100 % by mass(H) = 12.096149.096 x   100

% by mass(H) = 8.11%

11. Calculate the percent by mass of P present in (NH4)3PO4
 % by mass(P) = sum of relative atomic masses of all atoms of Prelative molecular mass of (NH4)3PO4 x   100 % by mass(P) = 30.97149.096 x   100

% by mass(P) = 20.77%

12. Calculate the percent by mass of O present in (NH4)3PO4
 % by mass(O) = sum of relative atomic masses of all atoms of Orelative molecular mass of (NH4)3PO4 x   100 % by mass(O) = 64.00149.096 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?

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 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:     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 x Mr(Fe)/Mr(FeaOb) x 100                     69.94 = 55.85a/Mr(FeaOb) x 100                     69.94/100 = 55.85a/Mr(FeaOb) x 100/100                     0.6994 = 55.85a/Mr(FeaOb)     % by mass(oxygen) =b x Mr(O)/Mr(FeaOb) x 100                     30.06 = 16.00b/Mr(FeaOb) x 100                     30.06/100 = 16.00b/Mr(FeaOb) x 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 x Mr(FeaOb) = 55.85a/Mr(FeaOb) x Mr(FeaOb)         0.6994 x Mr(FeaOb) = 55.85a         0.6994/0.6994 x Mr(FeaOb) = 55.85a/0.6994         Mr(FeaOb) = 55.85a/0.6994 = 79.85a     using %(oxygen) to find Mr(FeaOb):         0.3006 x Mr(FeaOb) = 16.00b/Mr(FeaOb) x Mr(FeaOb)         0.3006 x Mr(FeaOb) = 16.00b         0.3006/0.3006 x 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 x 2O3/2 x 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 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 Fe2O3

What would you like to do now?

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

2We 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.

3We 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.

4There 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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