Percentage Composition & Empirical Formula
Mole Concept
4
JEE Qs
8%
Hard
75
min
Master the systematic step-by-step procedure for converting percentages to empirical and molecular formulas to avoid calculation errors and secure marks.
🧮 Key Formulas
✅ Key Points for JEE
- 1Percentage composition describes the relative mass of each element in a compound.
- 2Empirical formula represents the simplest whole-number ratio of atoms in a compound; it may or may not be the actual molecular formula.
- 3To find the empirical formula from percentage composition: Assume 100g of compound -> Convert mass to moles for each element -> Divide all mole values by the smallest mole value -> Round to nearest whole number or multiply by a small integer to get whole numbers.
- 4The molecular formula is a whole-number multiple (n) of the empirical formula. 'n' is found by dividing the experimentally determined molar mass by the calculated empirical formula mass.
- 5When solving, assume the total mass of the compound is 100g if percentages are given, simplifying mass-to-mole conversions.
⚠️ Common Mistakes
- ✕Incorrectly rounding off mole ratios, especially when they are close to 0.5 or 0.33, indicating a need to multiply by a factor (e.g., 2 or 3) rather than rounding.
- ✕Confusing empirical formula with molecular formula and forgetting to calculate 'n' to find the molecular formula when molar mass is provided.
- ✕Errors in calculating atomic/molar masses, which propagate through all subsequent calculations.
- ✕Not dividing by the smallest mole value to obtain the simplest ratio.
📝 Practice Questions
See allQ71.Xg of benzoic acid on reaction with aq NaHCO3 released CO2 that occupied 11.2 L volume at STP. X is _____ g.
Q60.The elemental composition of a compound is 54.2%C, 9.2%H and 36.6%O. If the molar mass of the compound is 132 g mol−1 , the molecular formula of the compound is : [Given : The relative atomic mass of C : H : O = 12 : 1 : 16 ] (1) C4H9O3 (2) C6H12O6 (3) C4H8O2 (4) C6H12O3
Q67.Match the LIST-I with LIST-II Choose the correct answer from the options given below: (1) A-II, B-I, C-III, D-IV (2) A-II, B-III, C-I, D-IV (3) A-IV, B-I, C-III, D-II (4) A-IV, B-III, C-I, D-II
Q73.When 81.0 g of aluminium is allowed to react with 128.0 g of oxygen gas, the mass of aluminium oxide produced in grams is_______ - (Nearest integer) Given : Molar mass of Al is 27.0 g mol−1 Molar mass of O is 16.0 g mol−1
Q72.Consider the following reaction occurring in the blast furnace: Fe3O4( s) + 4CO(g) →3Fe(l) + 4CO2( g) ' x ' kg of iron is produced when 2.32 × 103 kgFe3O4 and 2.8 × 102 kgCO are brought together in the furnace. The value of ' x ' is _____. (nearest integer) Given: molar mass of Fe3O4 = 232 g mol−1 molar mass of CO = 28 g mol−1 molar mass of Fe = 56 g mol−1} Q73. 37.8 g N2O5 was taken in a 1 L reaction vessel and allowed to undergo the following reaction at 500 K 2 N2O5( g) ⇌2 N2O4( g) + O2( g) The total pressure at equilibrium was found to be 18.65 bar. Then, Kp = ______ ×10−2 [nearest integer] Assume N2O5 to behave ideally under these conditions. Given: R = 0.082 bar Lmol−1 K−1
Q31.Combustion of glucose (C6H12O6) produces CO2 and water. The amount of oxygen (in g) required for the complete combustion of 900 g of glucose is : [Molar mass of glucose in gmol−1 = 180 ] (1) 480 (2) 800 (3) 960 (4) 32
NCERT Chapters
- Class 11 Chemistry Ch 1: Some Basic Concepts of Chemistry