Counting Particles By Mass: The Mole

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수업 노트

The Mole and Avogadro's Constant

  • The mole is the amount of substance that contains the same number of fundamental units as there are atoms in exactly 12.00 g of carbon-12.
  • The Avogadro constant (NA or L) is the number of particles equivalent to the relative atomic mass (Ar) or relative molecular mass (Mr) of a substance in grams.
  • The value of the Avogadro constant is 6.02 × 1023 mol-1.
  • The Avogadro constant applies to atoms, molecules and ions.
  • One mole of any substance contains 6.02 × 1023 fundamental units.
  • The mass of a substance with this number of particles is called the molar mass.

Relative Atomic and Molecular Mass

  • The relative atomic mass (Ar) of an element is the weighted average mass of one atom compared to one twelfth the mass of a carbon-12 atom.
  • Ar is determined using the weighted average mass of the isotopes of the element.
  • Ar has no units because it is a ratio and the units cancel.
  • Relative molecular mass (Mr) is the sum of the relative atomic masses of the atoms in a molecule.
  • For compounds containing ions, the term relative formula mass (Mr) is used and is calculated in the same way.
  • Example: Mr of H2O = (2 × 1.01) + 16.00 = 18.02.

Moles and Mass Calculations

  • The number of moles can be calculated using: moles = mass (g) ÷ molar mass (g mol-1).
  • The molar mass of a substance is its relative atomic mass (Ar) or relative formula mass (Mr) expressed in grams.
  • Molar mass is given in units of g mol-1.
  • To find the mass of a substance: mass = moles × molar mass.
  • Always show your workings in calculations to check for errors and possibly gain credit.

Moles, mass and molar mass

Moles, mass and molar mass

Moles and Particles

  • The number of particles can be calculated using: number of particles = moles × Avogadro constant.
  • To find the number of moles from particles: moles = number of particles ÷ Avogadro constant.
  • When calculating the number of atoms in a molecule, multiply the number of moles of molecules by the number of atoms in one molecule.
  • Example: 0.010 moles of CH3CHO contains 0.040 moles of H atoms, which is 2.4 × 1022 H atoms.

Empirical Formula

  • The empirical formula gives the simplest whole-number ratio of atoms of each element in a compound.
  • The molecular formula shows the actual number and type of atoms in a molecule.
  • The empirical formula can be determined from percentage composition by mass data.
  • To find the empirical formula: divide the mass or percentage of each element by its atomic mass, then divide by the smallest value to get the simplest ratio.
  • The formula of an ionic compound is always written as an empirical formula.
  • Organic compounds often have different empirical and molecular formulae.

Molecular Formula

  • To determine the molecular formula from the empirical formula: divide the relative molecular mass (Mr) by the relative mass of the empirical formula.
  • This gives a whole number multiplier.
  • Multiply the empirical formula by this number to get the molecular formula.
  • Example: If empirical formula is C4H10S and Mr = 180.42, the multiplier is 2, so molecular formula is C8H20S2.

Molar Concentration

  • Concentration tells us how much solute is dissolved in a solvent to make a solution.
  • It is usually measured as the amount of solute in 1 dm3 of solution.
  • The solute is the substance being dissolved; the solvent is the liquid that does the dissolving.
  • Concentration can be expressed as moles per unit volume (mol dm-3), mass per unit volume (g dm-3), or parts per million (ppm).
  • Formula: concentration (mol dm-3) = moles of solute (mol) ÷ volume of solution (dm3).
  • To convert cm3 to dm3, divide by 1000.
  • To convert from mol dm-3 to g dm-3, multiply the molar concentration by the molar mass.
  • To convert from g dm-3 to mol dm-3, divide the mass concentration by the molar mass.

Concentration, moles and volume

Concentration, moles and volume

Parts Per Million (ppm)

  • ppm is used for very low concentrations, such as pollutants in water or air.
  • 1 ppm is defined as 1 mg of solute dissolved in 1 dm3 of water.
  • Since 1 dm3 of water weighs 1 kg, 1 ppm is also 1 mg dissolved in 1 kg of water.
  • This is equivalent to 10-3 g in 103 g, or a concentration of 1 in 106.

Avogadro's Law and Gas Volumes

  • Avogadro's Law states that equal volumes of gases under the same conditions of temperature and pressure contain the same number of molecules.
  • At standard temperature and pressure (STP), one mole of any gas occupies 22.7 dm3.
  • STP conditions are defined as 0 °C (273 K) and 100 kPa pressure.
  • The volume ratio of gaseous reactants and products in a balanced equation is the same as the mole ratio.
  • Example: In C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l), 50 cm3 of propane requires 250 cm3 of O2 and produces 150 cm3 of CO2.
  • Gas volume relationships are only valid if all gases are measured under the same conditions of temperature and pressure.
  • If gases are not in the same ratio as the balanced equation, use limiting reactant principles to determine the actual amount of product formed.

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연습 문제

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  1. 1.Which statement defines the mole?

    Easy
    • AThe amount of substance that contains 6.02 × 10²³ elementary entities.
    • BThe mass of one atom of carbon-12.
    • CThe volume occupied by one gram of any gas at room temperature.
    • DThe number of particles in one gram of any substance.

  2. 2.What is the value of the Avogadro constant?

    Easy
    • A6.02 × 10²³ mol⁻¹
    • B6.02 × 10⁻²³ mol⁻¹
    • C12.00 g mol⁻¹
    • D22.7 dm³ mol⁻¹

  3. 3.Which statement about relative atomic mass (Ar) is correct?

    Easy
    • AIt is the weighted average mass of an atom compared to one-twelfth the mass of a carbon-12 atom.
    • BIt is the mass of one mole of an element in grams.
    • CIt is the number of atoms in one mole of an element.
    • DIt is the mass of one atom of an element in kilograms.

  4. 4.Which of the following is the relative formula mass (Mr) of ammonium sulfate, (NH₄)₂SO₄? (Ar: N=14.01, H=1.01, S=32.07, O=16.00)

    Easy
    • A132.17
    • B114.11
    • C68.14
    • D100.17

  5. 5.The empirical formula of a compound gives the simplest whole-number ratio of atoms of each element in the compound.

    Easy

    True or false?

  6. 6.How many moles are present in 2.64 g of sucrose, C₁₂H₂₂O₁₁ (Mr = 342.3)?

    Medium
    • A7.71 × 10⁻³ mol
    • B7.71 × 10⁻² mol
    • C0.771 mol
    • D1.30 × 10² mol

  7. 7.Which of the following statements about concentration are correct? (select all that apply)

    Medium
    • AConcentration in mol dm⁻³ is calculated as moles of solute divided by volume of solution in dm³.
    • B1 ppm is equivalent to 1 mg of solute dissolved in 1 dm³ of water.
    • CTo convert cm³ to dm³, divide by 1000.
    • DA concentrated solution contains a small amount of solute per unit volume.
    • EThe units of concentration can be mol dm⁻³, g dm⁻³ or ppm.

  8. 8.Match each term with its correct description.

    Medium
    • Mole
    • Molar mass
    • Empirical formula
    • Avogadro's Law
    • The amount of substance containing 6.02 × 10²³ particles
    • The mass of one mole of a substance in grams
    • The simplest whole-number ratio of atoms in a compound
    • Equal volumes of gases at the same temperature and pressure contain the same number of molecules

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