The Mole And The Avogadro Constant

The Mole and Avogadro Constant

• The mole (mol) is the SI unit for amount of substance.
• One mole contains exactly 6.02 \times 10^{2}^{3} particles (Avogadro constant).
• Particles can be atoms, molecules, ions, or formula units.

How many atoms are in one mole of sodium (Na)?

• 6.02 \times 10^{2}^{3} atoms
• 12.04 \times 10^{2}^{3} atoms
• 6.02 \times 10^{2}^{2} atoms
• 1 mole of atoms

Molar Mass

• The mass of one mole of a substance is its molar mass.
• For an element, molar mass = relative atomic mass (Ar) in grams.
• For a compound, molar mass = relative formula mass (Mr) in grams.

The molar mass of carbon $(Ar = 12)$ is ____ g/mol.

Moles, Mass and Molar Mass

• The relationship is: moles = mass / molar mass.
• Use the formula triangle: cover the quantity you want.
• Mass = moles × molar mass; molar mass = mass / moles.

What is the mass of 0.250 moles of zinc $(Ar = 65)$?

• 16.25 g
• 260 g
• 0.00385 g
• 65 g

Molar Volume of Gas

• At room temperature and pressure (RTP: 20°C, 1 atm), one mole of any gas occupies $24 dm^{3} (24,000 cm^{3})$.
• This is called the molar gas volume.
• Volume $(dm^{3}) =$ moles × 24; Volume $(cm^{3}) =$ moles $\times 24000$.

3 moles of hydrogen gas occupy ____ dm^{3} at RTP.

Reacting Masses

• Use balanced equations to find mole ratios.
• Convert given masses to moles, then use ratio to find moles of other substances.
• Finally convert moles back to mass.

In 2Mg + O₂ → 2MgO, how many moles of MgO are produced from 6.0 g of Mg? (Ar $Mg = 24)$

• 0.25 mol
• 0.50 mol
• 1.0 mol
• 0.125 mol

Limiting Reactants

• The limiting reactant is used up first and determines the amount of product.
• The other reactant is in excess.
• To find the limiting reactant: calculate moles of each, compare with the balanced equation ratio.

Which statements about limiting reactants are true?

• The limiting reactant is the one with the smallest mass.
• The reaction stops when the limiting reactant is used up.
• The amount of product depends on the limiting reactant.
• The excess reactant is completely consumed.

Worked Example: Limiting Reactant

• 9.2 g Na reacts with 8.0 g S to form Na₂S.
• Moles $Na = \frac{9.2}{23} =$ 0.40; moles $S = \frac{8.0}{32} = 0.25$.
• Equation: $2Na + S$ → Na₂S. Ratio $Na:S = 2:1$. 0.40 mol Na needs 0.20 mol S; S is in excess, Na is limiting.

In the reaction $2Na + S$ → Na₂S, which is the limiting reactant if 0.40 mol Na and 0.25 mol S are used?

• Na
• S
• Both are limiting
• Neither

Calculating Number of Particles

• Number of particles = moles × Avogadro constant (6.02 \times 10^{2}^{3}).
• For compounds, consider the number of atoms or ions per formula unit.
• Example: 1 mole MgCl₂ contains 6.02 \times 10^{2}^{3} molecules, 3 \times 6.02 \times 10^{2}^{3} atoms, 2 \times 6.02 \times 10^{2}^{3} Cl⁻ ions.

One mole of water (H₂O) contains ____ molecules.

Summary of Key Equations

• moles = mass / molar mass
• mass = moles × molar mass
• molar mass = mass / moles
• volume of gas $(dm^{3}) =$ moles $\times 24$
• moles of $gas =$ volume $(dm^{3}) / 24$
• number of particles = moles \times 6.02 \times 10^{2}^{3}

What is the volume of 0.5 mol of CO₂ at RTP?

• $12 dm^{3}$
• $24 dm^{3}$
• $48 dm^{3}$
• $6 dm^{3}$