Moles And Molar Mass

Individual atoms and molecules are far too small and numerous to count directly during laboratory work. The mole connects a measurable macroscopic quantity — mass in grams — to the number of particles through Avogadro's number, allowing chemists to relate balance readings to the particle-level changes described in chemical equations.

Equation used: $$\text{Number of molecules}=n\times {N}_{A}$$ Where: n = number of moles (mol), Nᴀ = Avogadro’s number (6.022 × 10²³ mol⁻¹). Given: $$n=2.50\text{ mol}$$ $${N}_{A}=6.022\times {10}^{23}{\text{ mol}}^{-1}$$ Working: $$\text{Number of molecules}=2.50\times 6.022\times {10}^{23}$$ $$\text{Number of molecules}=1.51\times {10}^{24}$$ The sample contains 1.51 × 10²⁴ molecules of carbon dioxide.

Both the amu and the mole are defined relative to the same carbon-12 reference. One amu is 1/12 the mass of a carbon-12 atom, and one mole is defined so that 12.00 g of carbon-12 contains exactly 6.022 × 10²³ atoms. This shared reference ensures that the numerical value of a single particle's mass in amu equals the mass of one mole of those particles in grams.

Step 1 — Convert mass to moles. Equation used: $$n=\frac{m}{M}$$ Given: $$m=25.0\text{ g}$$ $$M=58.44{\text{ g mol}}^{-1}$$ Working: $$n=\frac{25.0}{58.44}$$ $$n=0.4278\text{ mol}$$ Step 2 — Convert moles to formula units. Equation used: $$\text{Number of formula units}=n\times {N}_{A}$$ Given: $$n=0.4278\text{ mol}$$ $${N}_{A}=6.022\times {10}^{23}{\text{ mol}}^{-1}$$ Working: $$\text{Number of formula units}=0.4278\times 6.022\times {10}^{23}$$ $$\text{Number of formula units}=2.58\times {10}^{23}$$ The sample contains 2.58 × 10²³ formula units of sodium chloride.

10.0 g of methane contains more molecules. The molar mass of H₂O is 18.02 g mol⁻¹, giving $$\frac{10.0}{18.02}=0.555\text{ mol}$$ The molar mass of CH₄ is 16.04 g mol⁻¹, giving $$\frac{10.0}{16.04}=0.623\text{ mol}$$ Because methane has a smaller molar mass, 10.0 g represents a greater number of moles and therefore a greater number of molecules.