Percentage Composition, Empirical & Molecular Formula | Chemistry Class 11 - NEET PDF Download

What is Percentage Composition?

The percentage composition of a compound is the expression of the relative amounts of each element present in the compound, given as the percentage by mass of each element with respect to the total mass of the compound. To find it, we determine the mass contributed by each element in a defined amount of the compound (commonly 1 mole), divide each element's mass by the total mass, and multiply by 100 to obtain the percentage by mass (mass percent).

• Mass percentage composition (mass percent, w/w %): mass of component / total mass of compound × 100.
• The percent composition is widely used in chemical analysis and is essential for deducing empirical formulas from experimental data.

Different Compounds

Mass Percentage Composition 

To calculate mass percent of elements from a chemical formula:

• Write the chemical formula of one mole of the compound and find the number of atoms of each element in one formula unit.
• Multiply the number of atoms of each element by its atomic mass (from the periodic table) to obtain the mass contributed by that element in one mole of the compound.
• Sum the masses of all elements to get the molar mass of the compound (mass of 1 mole).
• Mass fraction of an element = (mass of that element in 1 mole) / (molar mass of compound).
• Mass % of an element = mass fraction × 100.

Importance of Mass Percentage

• Mass percentage is indispensable for empirical formula determination from experimental analysis.
• The empirical formula (simplest whole-number ratio of atoms) is often deduced from the percentage composition; from the empirical formula and molar mass one can obtain the molecular formula.

Determining Mass Percent from a Chemical Formula 

Consider glucose with formula C₆H₁₂O₆. Calculate the mass percentage of each element.

Solve by calculating masses in one mole of glucose:
Mass contributed by carbon = 6 × 12.01 g = 72.06 g.
Mass contributed by hydrogen = 12 × 1.008 g = 12.096 g.
Mass contributed by oxygen = 6 × 16.00 g = 96.00 g.
Molar mass of glucose = 72.06 + 12.096 + 96.00 = 180.156 g (rounded to 180.16 g).
Mass fraction of C = 72.06 / 180.16 = 0.4000 → mass % C = 40.00%.
Mass fraction of H = 12.096 / 180.16 = 0.06714 → mass % H = 6.714%.
Mass fraction of O = 96.00 / 180.16 = 0.53286 → mass % O = 53.286%.

Note: Despite C and O having the same number of atoms in glucose, O contributes a larger mass percent because its atomic mass (16.00) is greater than that of C (12.01).

Solved Example 

Q. Find the percent composition of each element in water.
Solution.

Empirical Formula

The empirical formula of a compound expresses the simplest whole-number ratio of atoms of each element present in the compound. It does not necessarily equal the true molecular formula (actual number of atoms in a molecule); several different molecules can have the same empirical formula.

Examples: C₆H₁₂O₆, CH₃COOH and HCHO have the same empirical formula CH₂O.

Structure of Glucose

Procedure to determine empirical formula from percentage composition

• Write the elements present and their atomic masses.
• Assume 100 g of compound so that percent values become grams of each element.
• Convert mass of each element to moles by dividing by atomic mass.
• Divide all mole values by the smallest mole value to obtain simplest ratio.
• If any ratio is not a whole number, multiply all ratios by an appropriate integer to get whole numbers.
• Write the empirical formula using the whole-number ratio as subscripts.

Example 1. A compound contains C = 71.23%, H = 12.95% and O = 15.81%. What is the empirical formula of the compound?
Solution.

Example 2. The simplest formula of a compound containing 50% of element X (Atomic mass = 10) and 50% of the element Y (Atomic mass = 20) is: 
(A) XY 
(B) X₂Y
(C) XY₂ 
(D) X₂Y_3
Solution.

Molecular Formula

The molecular formula shows the actual number of atoms of each element in a molecule and may be the same as the empirical formula or a whole-number multiple of it.

• Molecular formula = (empirical formula)n, where n is a positive integer.
• n = (molar mass of compound) / (empirical formula mass).
• If the vapour density (VD) is known, molecular mass can be found using: Molecular mass = 2 × Vapour density.

Examples: Butene (C₄H₈) has empirical formula CH₂ (four times CH₂). Ethene (C₂H₄) has empirical formula CH₂ (twice CH₂). Some compounds have identical empirical and molecular formulas.

How to find molecular formula

• Calculate empirical formula mass (sum of atomic masses in empirical formula).
• Divide the given molar mass by the empirical formula mass to obtain n.
• Molecular formula = empirical formula multiplied by n.

Example 3. A compound contains 4.07 % hydrogen, 24.27 % carbon, and 71.65 % chlorine. Its molar mass is 98.96 g. What are its empirical and molecular formulas?
Solution.

Example 4. A compound of carbon, hydrogen, and nitrogen contains these elements in the ratio 9:1:3.5. Calculate the empirical formula. If its molecular mass is 108, what is the molecular formula? 
Solution.

_{Example 5. 2.38 gm of uranium was heated strongly in a current of air. The resulting oxide weighed 2.806 g. Determine the empirical formula of the oxide. (At. mass U = 238; O = 16).}
Solution. 

Example 6. Chemical analysis of a carbon compound gave the following percentage composition by weight of the elements present. Carbon 10.06%, hydrogen 0.84%, chlorine 89.10%. Calculate the empirical formula of the compound.
Solution.

Example 7. A carbon compound on analysis gave the following percentage composition. Carbon 14.5%, hydrogen 1.8%, chlorine 64.46%, oxygen 19.24%. Calculate the empirical formula of the compound.
Solution. 

Example 8. The empirical formula of a compound is CH₂O. Its molecular weight is 90. Calculate the molecular formula of the compound. (Atomic weights C = 12, H = 1, O = 16)
Solution. Empirical formula mass = 12 + 2×1 + 16 = 30 g.

n = molecular mass / empirical mass = 90 / 30 = 3.

Molecular formula = (CH₂O)₃ = C₃H₆O₃.

Relation: Molecular formula = Empirical formula × n.

Density and Vapour Density (Brief)

Absolute (mass) density = mass / volume.

Relative density = density of substance / density of a reference substance.

Specific gravity = density of substance / density of H₂O at 4 °C.

Vapour density (VD) is defined for a gas as its density relative to hydrogen (H2) under the same conditions:

Practice Questions with Answers

Q.1. A crystalline hydrated salt, on being rendered anhydrous, loses 45.6% of its mass. The percentage composition of the anhydrous salt is: Al = 10.5%, K = 15.1%, S =24.8% and oxygen = 49.6%. Find the empirical formula of the anhydrous and crystalline hydrated salt. [K = 39; Al = 27; S = 32; O = 16; H = 1]
Ans. The empirical formula of anhydrous salt = KAlS₂O₈
Hydrated salt composition: % anhydrous part = 54.4% and % H₂O = 45.6%
The empirical formula of hydrated salt = KAlS₂O₈·12H₂O

Q.2. A colorless crystalline compound has the following percentage composition: Sulphur 24.24%, nitrogen 21.21%, hydrogen 6.06% and the rest is oxygen. Determine the empirical formula of the compound. If the molecular mass is 132, what is the molecular formula of the compound? Name the compound if it is found to be sulphite.
Ans. Empirical formula: SN₂H₈O₄
Molecular formula = SN₂H₈O₄
Name: Ammonium sulphite: (NH₄)₂SO₃

Q.3. A gaseous hydrocarbon contains 85.7% carbon and 14.3% hydrogen. 1 litre of the hydrocarbon weighs 1.26 g at NTP. Determine the molecular formula of the hydrocarbon.
Ans.  Empirical Formula = CH₂  
Result : (CH₂)x₂=C₂H₄