Chemistry · Some Basic Concepts Of Chemistry · NEET
Use 4 easy steps. Step 1: Assume you have 100 g of the compound, so each percent becomes grams (78% C means 78 g C). Step 2: Divide each element's mass by its atomic mass to get moles. Step 3: Divide every mole value by the SMALLEST mole value. Step 4: If the numbers are not whole (like 1.5), multiply all of them by a small number (2, 3...) to make them whole. The final whole numbers are the atom ratio. For NEET, this exact 4-step method solves almost every empirical formula question fast.
Empirical formula = the simplest ratio of atoms. Molecular formula = the actual number of atoms in one molecule. Example: glucose empirical formula is CH2O, but its molecular formula is C6H12O6 (which is CH2O times 6). They are the SAME only when the ratio is already the smallest, like H2O or CO2. Molecular formula = n x (empirical formula).
Dividing by the smallest mole value turns the numbers into a ratio starting from 1. It gives you the simplest comparison between atoms. The smallest element becomes 1, and the others show how many times more of them there are. This is just a shortcut to reach the simplest whole-number ratio quickly.
First find n, using this formula: n = (molar mass of compound) / (empirical formula mass). Then multiply every atom count in the empirical formula by n. Example: empirical formula CH2O has mass 30. If the compound's molar mass is 180, then n = 180 / 30 = 6, so molecular formula = C6H12O6. If they do not give you the molar mass, you can only find the empirical formula, not the molecular one.
NEET often gives you only some percentages and says 'the remaining is element X'. Just subtract the given percentages from 100 to get the missing one. Example: if a compound is 32% A and 20% B, then C = 100 - 32 - 20 = 48%. Then continue with the normal steps. Missing this subtraction is the most common careless mistake.
Yes. The ratio 1 : 1 : 3 means one atom of A, one atom of B, and three atoms of C, which is written as ABC3 (the '1' is not written). So a ratio like 1 : 1 : 3 gives ABC3, and 1 : 3 gives CH3. Always convert the final ratio directly into subscripts.
An organic compound contains 78% (by wt.) carbon and the remaining percentage of hydrogen. The empirical formula of the compound is [at. wt.: C = 12, H = 1]
A compound X contains 32% of A, 20% of B and the remaining percentage of C. The empirical formula of X is (atomic masses: A = 64, B = 40, C = 32 u)
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Yes. When the atom ratio is already the smallest whole-number ratio, both are the same. Examples: H2O, CO2, NH3, and CH4. Here n = 1, so molecular formula = empirical formula.
No. To find only the empirical formula you just need the percentages and atomic masses. You need the molar mass only when the question asks for the molecular formula, because you need it to calculate n.
Glucose (C6H12O6) has empirical formula CH2O. Benzene (C6H6) has empirical formula CH. Both molecular formulas are simple multiples of their empirical formulas (n = 6 for both).
Empirical and molecular formula questions appear almost every year in Some Basic Concepts of Chemistry. They are quick, scoring, and only need the mole idea plus simple division. Practising the 4-step method saves time in the exam.
