Percent Composition Formula: Definition, Questions and Examples

Edited By Shivani Poonia | Updated on Aug 22, 2025 06:01 PM IST

To determine the molecular formula of a compound, we begin by measuring the mass or mass percentage of each element present in the compound. The composition is generally expressed as the mass percentage composition. The mass percentage gives the mass of each element expressed as the percentage of the total mass. In other words, it gives the number of grams of the element present in 100 g of the compound.

This Story also Contains

Percentage Composition:
Equivalent Weight:
Some Solved Examples
Summary

Percent Composition Formula

Percentage Composition:

The percentage of any element or constituent in a compound is the number of parts by mass of that element or constituent present in 100 parts by mass of the compound.

Mass $\%$ of an element $=\frac{\text { Mass of that element in one mole of the compound }}{\text { Molar mass of the compound }} \times 100$

Let us take an example of water (H₂O), it contains hydrogen and oxygen, and the percentage composition of both these elements can be calculated as follows:

The molar mass of water = 18.02 g

Mass $\%$ of Hydrogen $=\frac{2 \times 1.008}{18.02} \times 100=11.18 \%$

Mass $\%$ of Oxygen $=\frac{16.00}{18.02} \times 100=88.79 \%$

One can check the purity of a given sample by analyzing percentage composition.

Equivalent Weight:

Equivalent weight of any substance is that weight which react or liberate 1g of hydrogen (11.2 litre of H₂ at STP) or 8g of oxygen or 35.5g of chlorine or 80g of bromine or 127g of iodine or 108g of silver or 1 mole of electron.
Equivalent weight is a number and when it is denoted in grams, it is called gram equivalent.
It depends upon the nature of the chemical reaction in which the substance takes part

How To Find Equivalent Weight:

Equivalent Weight $=\frac{\text { Molecular weight }}{n-\text { factor }(x)}$

n-Factor or Valence Factor:

It calculates the molar ratio of the species taking part in reactions that are, reactants. The reciprocal of the n-factor 's ratio of the reactants represents the molar ratio of the reactants. For example, If A (having n-factor = a) reacts with B (having n-factor = b) then its n-factor's ratio is a: b, so the molar ratio of A to B is b: a.
It can be represented as follows: $\begin{array}{ll}\mathrm{bA} \quad+ & \mathrm{aB} \rightarrow \text { Product } \\ (\mathrm{n}-\text { factor }=\mathrm{a}) & (\mathrm{n}-\text { factor }=\mathrm{a})\end{array}$

Calculation of n-Factor

Before calculating the n-factor of any of the reactants in a given chemical reaction we must have a clear idea about the type of reaction. The reaction may be any of these types:

(i) Acid-base or neutralization reaction

(li) Redox reaction

Acid-Base or Neutralization Reactions:
As we know according to the Arrhenius concept, "An acid provides H⁺ ion(s) while a base provides OH^- ion(s) in neutralization these H⁺ and OH^- ion/ions combines together".
The number of H⁺ ion(s) and OH^- ion(s) represent the n-factor for acid and base respectively, that is, basicity and acidity respectively.
Example,

$\mathrm{HCl} \rightarrow \mathrm{H}^{+}+\mathrm{Cl}^{-}$ $(\mathrm{n}=1)$ that is, monobasic acid

$$
\mathrm{H}_2 \mathrm{SO}_4 \rightarrow 2 \mathrm{H}^{+}+\mathrm{SO}_4^{2-}
$$

$(\mathrm{n}=2)$ that is, dibasic acid
Redox Reactions
These reactions involve oxidation and reduction simultaneously. Here the exchange of electrons occurs. To find the n-factor for Oxidizing or agent we must find out the change in the oxidation state of these species.

You will be learning the following in detail in the chapter of redox. For now, just look at the definition. Sufficient questions will be practiced later.

For Redox Reactions:
E = (Molecular weight) / (Change in oxidation number),

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x= change in oxidation state
For Example, for KMnO₄
(a) In acidic medium: E = M/5

$2 \mathrm{KMnO}_4^{+7}+3 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{K}_2 \mathrm{SO}_4+2 \mathrm{MnSO}_4^{+2}+3 \mathrm{H}_2 \mathrm{O}+5[\mathrm{O}]$
5 unit change in oxidation number.

(b) In basic medium: E = M/1

$2 \mathrm{KMnO}_4^{+7}+2 \mathrm{KOH} \rightarrow 2 \mathrm{~K}_2 \stackrel{+6}{\mathrm{MnO}_4}+\mathrm{H}_2 \mathrm{O}+[\mathrm{O}]$
one unit change in oxidation number

(c) In neutral medium: E = M/3

$2 \mathrm{KMnO}_4^{+7}+\mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{KOH}+2 \stackrel{+4}{\mathrm{MnO}_2}+3[\mathrm{O}]$
3 unit change in oxidation number

Formulae for calculation of Equivalent Weight:

For Acids:
E = (Molecular weight) / (Protocity or Basicity of Acid), x= number of furnishable protons
For Example, for H₃PO₄, E = M/3
For H₂SO₄ , E =M/2
For Bases:
E = (Molecular weight) / (Acidity or number of OH^- ions),x= number of furnishable OH^- ions
For Example, for Ca(OH)₂, E = M/2
For Al(OH)₃, E =M/3
For Ions:
E = (Molecular weight) / (Charge on ion), x= charge on ion
For Example, for SO₄^2-, E = M/2
For PO₄^3-, E = M/3
For Compounds:
E = (Molecular weight) / (total positive charge or negative charge present in compound),

x= total positive charge or negative charge present in the compound
For Example, for CaCO₃, E = M/2
For AlCl₃, E =M/3

For Acidic Salt:
E = (Molecular weight) / (Number of replaceable H-atoms)
For example, for H₃PO₄

$\begin{aligned} & 2 \mathrm{NaOH}+\mathrm{NaH}_2 \mathrm{PO}_4 \rightarrow \mathrm{Na}_3 \mathrm{PO}_4+2 \mathrm{H}_2 \mathrm{O} \\ & \mathrm{E}=\mathrm{M} / 2\end{aligned}$

Metal displacement method
E₁ / E₂ = W₁/ W₂

Also Read:

Recommended topic video on (Percent Composition Formula)

Some Solved Examples

Que 1: Find the equivalent mass of H₃PO₄ in the reaction:

$\mathrm{Ca}(\mathrm{OH})_2+\mathrm{H}_3 \mathrm{PO}_4 \rightarrow \mathrm{CaHPO}_4+2 \mathrm{H}_2 \mathrm{O}$

1) 55

2) 43

3) 49

4) 37

Solution

In this given reaction only two hydrogen atoms are replaced so their equivalent mass will be given as follows:

Equivalent mass of H₃PO₄ = (Molecular Mass of H₃PO₄ ) / 2

=98 / 2

= 49

Hence, the answer is option (3).

Que 2: 74.5 g of metallic chloride contains 35.5 g of chlorine. The equivalent weight of the metal is

1) 19.5

2) 35.5

3) 39

4) 78

Solution

74.5 g of metallic chloride contains 35.5 g of chlorine.

Therefore, the weight of another element present in the chloride is

= Weight of Metal Chloride - Weight of Chloride

= 74.5 - 35.5 = 39 g

Thus, 39 g of the metal combines with 35.5 g of Chlorine

Mole of Chloride present = weight/molar mass = 35.5/35.5 = 1 mole

So, one mole of Cl is present and we know the charge on Cl is (-1).

So, MCl configuration will be there where M-metal.

MCl = M⁺+ Cl^-

So, one mole of Metal will be present and the charge would be (+1) on metal.

Molecular weight = mass of 1 mol metal = 39 g

Equivalent Weight = Molecular weight/ Valency.

Equivalent Weight = 39/1 =39

hence, the equivalent weight of the metal is 39.

Hence, the answer is the option (3).

Que 3: The oxide of a metal has 32% oxygen. Its equivalent weight would be

1) 17

2) 34

3) 32

4) 8

Solution

100g of the oxide contains 68g of metal and 32g of oxygen

$\therefore$ 8g of oxygen will be present with 17g of the metal.

$\therefore$ The equivalent weight of the metal is 17.

Hence, the answer is the option (1).

Que 4: What is the equivalent weight of Sulphuric Acid (H₂SO₄) when it behaves as a diprotic acid?

1) 98

2) 49

3) 24.5

4) 12.25

Solution

n-factor of H₂SO₄ = 2

$\therefore$ Equivalent weight $=\frac{98}{2}=49$

Hence, the answer is the option (2)

Que 5. A compound possesses 8% sulfur by mass. The least molecular mass is

1) 200

2) 400

3) 155

4) 355

Solution

Let, the molar mass of the compound be M.

The compound contains 8% by weight of Sulphur, mathematically it implies

$\frac{8}{100} \times M=32$

$\Rightarrow M=400 \mathrm{~g}$

Hence, the answer is the option (2).

Practice more Questions from the link given below:

Percentage Composition And Equivalent Weight

MOLE CONCEPT AND MOLAR MASS

Summary

Percentage composition is very important in chemistry because it helps in determining proportion of each element within a compound.This information is vital for understanding a compound's properties, identifying it, and calculating its empirical and molecular formulas.It’s also a analytical educational tool—serving as a bridge between chemical formulas and real-world applications. Its ability to authenticate purity and accuracy strengthens both lab precision and industrial credibility.

Frequently Asked Questions (FAQs)

1. How does percent composition relate to molecular formula?

The percent composition can help determine the empirical formula, which can then be used to find the molecular formula if the molar mass is known. The molecular formula is a whole number multiple of the empirical formula.

2. How does percent composition relate to empirical formulas?

Percent composition can be used to determine the empirical formula of a compound. By converting percentages to masses and then to moles, you can find the simplest whole-number ratio of atoms, which is the empirical formula.

3. Can percent composition be used to identify an unknown compound?

Yes, percent composition can be a valuable tool in identifying unknown compounds, especially when combined with other analytical techniques. It provides information about the relative amounts of elements present.

4. What's the relationship between percent composition and stoichiometry?

Percent composition is crucial in stoichiometry as it helps determine the amount of each element in a compound, which is necessary for balancing chemical equations and calculating reactant or product quantities.

5. Can two different compounds have the same percent composition?

Yes, it's possible for different compounds to have the same percent composition. These are called isomers - compounds with the same formula but different structural arrangements.

6. What is percent composition in chemistry?

Percent composition is the proportion of each element present in a compound, expressed as a percentage of the total mass. It tells us how much of each element contributes to the overall mass of the compound.

7. Why is percent composition important in chemistry?

Percent composition is crucial because it allows us to determine the relative amounts of elements in a compound, helps in identifying unknown substances, and is useful in calculating empirical formulas and stoichiometry problems.

8. How do you calculate the percent composition of an element in a compound?

To calculate the percent composition of an element, divide the mass of that element in the compound by the total mass of the compound, then multiply by 100. The formula is: (mass of element / total mass of compound) × 100.

9. Can percent composition be greater than 100%?

No, percent composition cannot be greater than 100%. Since it represents the proportion of each element in a compound, the sum of all percentages must equal 100%.

10. Does the percent composition of a compound change if you have different amounts of the same compound?

No, the percent composition remains constant regardless of the amount of compound. It represents the ratio of elements, which stays the same whether you have a small or large sample of the pure compound.

11. How does percent composition affect a compound's properties?

The percent composition directly influences a compound's properties, including its reactivity, melting point, boiling point, and other physical and chemical characteristics.

12. How does percent composition relate to the concept of limiting reagents?

While percent composition doesn't directly determine the limiting reagent, understanding the composition of reactants helps in calculating the amounts available for reaction, which is crucial in identifying the limiting reagent.

13. Can percent composition help in predicting chemical reactions?

Yes, percent composition can help predict reactions by providing information about the relative amounts of elements available for reaction. This is particularly useful in combustion analysis and other stoichiometric calculations.

14. Can percent composition be used to distinguish between allotropes?

No, percent composition alone cannot distinguish between allotropes (different forms of the same element) because they have the same elemental composition. Other analytical methods are needed.

15. How does percent composition relate to molarity?

Percent composition can be used to calculate the mass of solute in a solution, which, combined with the solution's volume, allows for the calculation of molarity (moles of solute per liter of solution).

16. Can percent composition be used to determine the strength of chemical bonds?

Percent composition alone cannot determine bond strength. While it provides information about elemental ratios, bond strength depends on factors like electronegativity and bond type, which aren't reflected in percent composition.

17. What's the difference between percent composition by mass and by atoms?

Percent composition by mass refers to the percentage of each element's mass in a compound, while percent composition by atoms refers to the percentage of atoms of each element. They often differ due to variations in atomic masses.

18. How does the periodic table help in calculating percent composition?

The periodic table provides atomic masses of elements, which are essential for calculating the mass of each element in a compound and subsequently determining its percent composition.

19. Why might experimental percent composition differ from theoretical calculations?

Experimental results may differ due to impurities in the sample, measurement errors, or incomplete reactions. Theoretical calculations assume a pure, ideal compound.

20. Can you have a negative percent composition?

No, percent composition cannot be negative. It represents the proportion of an element in a compound and must always be a positive value between 0 and 100%.

21. How does percent composition relate to the law of definite proportions?

The law of definite proportions states that a chemical compound always contains the same proportion of elements by mass. Percent composition is a direct application of this law, expressing these proportions as percentages.

22. Why is water's percent composition not 50% hydrogen and 50% oxygen?

Water's percent composition isn't 50-50 because oxygen atoms are much heavier than hydrogen atoms. The actual percent composition of water is about 11.1% hydrogen and 88.9% oxygen by mass.

23. How does percent composition help in determining the purity of a sample?

By comparing the experimental percent composition of a sample to the theoretical values for a pure compound, we can assess the sample's purity. Significant deviations may indicate impurities or a different compound altogether.

24. Can percent composition be used for mixtures?

While percent composition typically refers to pure compounds, a similar concept can be applied to mixtures. In this case, it would represent the percentage of each component in the mixture by mass.

25. How does percent composition relate to mole fraction?

Both percent composition and mole fraction describe the relative amounts of components in a substance, but percent composition is based on mass, while mole fraction is based on the number of moles.

26. Why is it important to specify "by mass" when discussing percent composition?

Specifying "by mass" is crucial because percent composition can also be calculated by atoms or molecules. "By mass" clarifies that we're dealing with mass percentages, which is the most common form.

27. How does percent composition change in a hydrated compound?

In a hydrated compound, the percent composition changes because water molecules are included in the total mass. This results in a lower percentage for each element compared to the anhydrous form.

28. Why might a chemist need to convert between percent composition and mass?

Converting between percent composition and mass is often necessary in practical applications, such as determining how much of an element is present in a given mass of compound, or vice versa.

29. How does percent composition relate to the concept of yield in chemical reactions?

Percent composition is useful in calculating theoretical yields in chemical reactions. By knowing the composition of reactants and products, we can determine the expected amount of product and compare it to the actual yield.

30. Why is it important to round percent composition results appropriately?

Appropriate rounding in percent composition calculations is crucial to reflect the precision of the original measurements and avoid implying a level of accuracy that isn't justified by the data.

31. How does percent composition relate to the concept of significant figures?

The number of significant figures in percent composition calculations should reflect the precision of the original measurements. This ensures that the final result doesn't imply more precision than is actually present in the data.

32. Can percent composition be used to determine the formula of a polymer?

While percent composition can provide information about the relative amounts of elements in a polymer, it alone is not sufficient to determine the complete formula. Additional information about the polymer's structure and repeating units is needed.

33. How does percent composition relate to the concept of molecular mass?

Percent composition is directly related to molecular mass. The mass of each element contributing to the percent composition must sum to the total molecular mass of the compound.

34. Why might the experimental percent composition of a synthesized compound differ from the theoretical value?

Experimental percent composition might differ due to impurities in reactants, incomplete reactions, side reactions, or loss of product during isolation and purification steps.

35. How does percent composition relate to the concept of empirical formula mass?

The empirical formula mass can be calculated from percent composition by assuming 100 grams of the compound and converting the percentages to grams, then to moles, to find the simplest whole number ratio of atoms.

36. Can percent composition be used to distinguish between ionic and covalent compounds?

Percent composition alone cannot distinguish between ionic and covalent compounds. While it provides information about elemental ratios, it doesn't indicate the nature of the chemical bonding.

37. How does percent composition relate to the concept of formula weight?

Formula weight is the sum of the atomic weights of all atoms in a formula unit. Percent composition can be calculated from the formula weight by dividing the mass of each element by the total formula weight and multiplying by 100.

38. Why is it important to consider significant figures in percent composition calculations?

Significant figures in percent composition calculations ensure that the precision of the result matches the precision of the input data, preventing false implications of greater accuracy than the measurements allow.

39. How does percent composition relate to the concept of molar mass?

Molar mass is the mass of one mole of a substance. Percent composition can be calculated from molar mass by dividing the mass of each element in one mole by the total molar mass and multiplying by 100.

40. Can percent composition be used to determine the degree of unsaturation in organic compounds?

While percent composition provides information about elemental ratios, it alone is not sufficient to determine the degree of unsaturation. Additional information about the molecular formula or structure is needed.

41. How does percent composition relate to the concept of stoichiometric coefficients?

Percent composition can help in determining empirical formulas, which are often equivalent to or multiples of the stoichiometric coefficients in balanced chemical equations.

42. Why might the percent composition of a compound change when it undergoes a chemical reaction?

The percent composition changes during a chemical reaction because the elements are rearranged to form new compounds with different ratios of elements, resulting in different mass percentages.

43. How does percent composition relate to the concept of mass spectrometry?

Mass spectrometry can provide information about the relative abundance of different isotopes of an element, which can affect the average atomic mass used in percent composition calculations.

44. Can percent composition be used to determine the oxidation state of elements in a compound?

While percent composition provides information about elemental ratios, it alone cannot determine oxidation states. Additional information about the compound's structure or properties is needed.

45. How does percent composition relate to the concept of molecular geometry?

Percent composition doesn't directly relate to molecular geometry. While it provides information about elemental ratios, it doesn't indicate how atoms are arranged in space.

46. Why is it important to consider the law of conservation of mass when working with percent composition?

The law of conservation of mass ensures that the total mass of elements in reactants equals the total mass in products. This principle underlies percent composition calculations in chemical reactions.

47. How does percent composition relate to the concept of isotopes?

Isotopes of an element have the same number of protons but different numbers of neutrons, affecting their atomic mass. The average atomic mass used in percent composition calculations takes into account the natural abundance of isotopes.

48. How does percent composition relate to the concept of chemical equilibrium?

While percent composition doesn't directly relate to equilibrium, understanding the composition of reactants and products is crucial in equilibrium calculations, especially when dealing with limiting reagents or reaction yields.

49. Why might the percent composition of a compound in its gas phase differ from its solid phase?

The percent composition of a pure compound remains the same regardless of its physical state. However, if the gas phase involves decomposition or a chemical change, the composition would differ from the solid phase.

50. How does percent composition relate to the concept of colligative properties?

While percent composition doesn't directly determine colligative properties, it's useful in calculating the amount of solute present in a solution, which is crucial for determining properties like boiling point elevation or freezing point depression.