Q.Calculate the mass of $0.25\ \text{mol}$ of carbon dioxide gas, $\text{CO}_2$ (molar mass $44.01\ \text{g mol}^{-1}$).

Atoms and molecules are far too small and too numerous to count one by one, so chemists count them in a fixed-size bundle called the mole. This concept is the arithmetic of that bundle — converting freely between mass, moles, number of particles, and volume of a gas.

1 — The mole and Avogadro's number. One mole is the amount of a substance that contains exactly Nₐ = 6.022×10²³ elementary entities (atoms, molecules, ions, electrons — whatever is specified). So number of particles = moles × Nₐ, and moles = number of particles / Nₐ. One mole of water contains 6.022×10²³ molecules; one mole of Na⁺ contains 6.022×10²³ ions.

2 — Molar mass. The molar mass M (in g/mol) is the mass of one mole. Numerically it equals the relative atomic mass (for an element) or the sum of the atomic masses in the formula (for a compound): H₂O has M = 2(1) + 16 = 18 g/mol; glucose C₆H₁₂O₆ has M = 6(12) + 12(1) + 6(16) = 180 g/mol. The central bridge is moles = mass / molar mass, i.e. n = m / M.

3 — The interconversion web. Everything hangs off moles as the hub:

moles = mass / M = number of particles / Nₐ = gas volume at STP / 22.4 L.

Pick the box you are given, convert to moles, then convert out to the box asked for. For example, mass → number of molecules is a two-step chain: m → n = m/M → N = n·Nₐ.

4 — Gas volume at STP. One mole of any ideal gas occupies a fixed molar volume: 22.4 L at the old STP (0 °C, 1 atm), or 22.7 L at the current IUPAC STP (0 °C, 1 bar). So moles of gas = volume at STP / 22.4 L. Use the molar volume the question specifies; JEE problems usually still take 22.4 L unless "1 bar" is stated. This only applies to gases — never to a solid or liquid.

5 — Counting a specific sub-species. To count the atoms of one particular element in a sample, first get the moles of the compound, then multiply by the number of that atom in one formula unit, then by Nₐ. In 1 mole of H₂SO₄ there are 4 moles of O atoms, i.e. 4 × 6.022×10²³ = 2.409×10²⁴ oxygen atoms. The same logic gives the total atoms (multiply by all atoms per formula unit), or the number of electrons/protons (moles × electrons-per-formula-unit × Nₐ).

6 — Average (relative) atomic mass. An element's tabulated atomic mass is the abundance-weighted average of its isotopes: M̄ = Σ (fractional abundance × isotope mass). For chlorine, 75% of mass-35 and 25% of mass-37 give 0.75(35) + 0.25(37) = 35.5. The reverse question — find an isotope's abundance from the known average — is solved by letting the fraction be x, writing the weighted-average equation, and solving the single linear equation. …