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The "Worked Example 2.2" illustrates how to determine the relative mass of an unknown substance using a beam balance, comparing it to a carbon-12 sample.

Worked Example 2.2 Explanation:
The example uses an analogy with beads.

  • a) A carbon-12 sample is represented by 12 beads. By definition, carbon-12 has a relative atomic mass of 12. This implies that each carbon-12 bead in this analogy has a relative mass of 1 unit.
  • b) An unknown sample is represented by xx beads.
  • c) To find the relative mass of the unknown sample, the example assumes that xx beads of the unknown sample have the same total mass as the 12 carbon-12 beads when balanced. Therefore, the total relative mass of the xx unknown beads is 12 units. The formula A=12Number of beads in unknown sample=12x=yA = \frac{12}{\text{Number of beads in unknown sample}} = \frac{12}{x} = y calculates the relative mass per bead of the unknown sample. So, yy represents the relative mass of one bead of the unknown substance.

Activity 2.2: Understanding relative molecular mass
Relative molecular mass (MrM_r) is the sum of the relative atomic masses (ArA_r) of all the atoms in a molecule. We will use the approximate relative atomic masses: Hydrogen (H\text{H}) ≈1.0\approx 1.0, Carbon (C\text{C}) ≈12.0\approx 12.0, and Oxygen (O\text{O}) ≈16.0\approx 16.0.

1) For Water (H2O\text{H}_2\text{O}):

  1. Identify the atoms and their quantities.

    The molecule H2O\text{H}_2\text{O} contains 2 hydrogen atoms and 1 oxygen atom.

  2. List the relative atomic masses.

    Ar(H)=1.0A_r(\text{H}) = 1.0
    Ar(O)=16.0A_r(\text{O}) = 16.0

  3. Calculate the relative molecular mass.

    Mr(H2O)=(2×Ar(H))+(1×Ar(O))M_r(\text{H}_2\text{O}) = (2 \times A_r(\text{H})) + (1 \times A_r(\text{O})) Mr(H2O)=(2×1.0)+(1×16.0)M_r(\text{H}_2\text{O}) = (2 \times 1.0) + (1 \times 16.0) Mr(H2O)=2.0+16.0M_r(\text{H}_2\text{O}) = 2.0 + 16.0 Mr(H2O)=18.0M_r(\text{H}_2\text{O}) = 18.0

    The relative molecular mass of water is 18.0\boxed{\text{18.0}}.

    2) For Carbon Dioxide (CO2\text{CO}_2):

  4. Identify the atoms and their quantities.

    The molecule CO2\text{CO}_2 contains 1 carbon atom and 2 oxygen atoms.

  5. List the relative atomic masses.

    Ar(C)=12.0A_r(\text{C}) = 12.0
    Ar(O)=16.0A_r(\text{O}) = 16.0

  6. Calculate the relative molecular mass.

    Mr(CO2)=(1×Ar(C))+(2×Ar(O))M_r(\text{CO}_2) = (1 \times A_r(\text{C})) + (2 \times A_r(\text{O})) Mr(CO2)=(1×12.0)+(2×16.0)M_r(\text{CO}_2) = (1 \times 12.0) + (2 \times 16.0) Mr(CO2)=12.0+32.0M_r(\text{CO}_2) = 12.0 + 32.0 Mr(CO2)=44.0M_r(\text{CO}_2) = 44.0

    The relative molecular mass of carbon dioxide is 44.0\boxed{\text{44.0}}.

    3) For Methane (CH4\text{CH}_4):

  7. Identify the atoms and their quantities.

    The molecule CH4\text{CH}_4 contains 1 carbon atom and 4 hydrogen atoms.

  8. List the relative atomic masses.

    Ar(C)=12.0A_r(\text{C}) = 12.0
    Ar(H)=1.0A_r(\text{H}) = 1.0

  9. Calculate the relative molecular mass.

    Mr(CH4)=(1×Ar(C))+(4×Ar(H))M_r(\text{CH}_4) = (1 \times A_r(\text{C})) + (4 \times A_r(\text{H})) Mr(CH4)=(1×12.0)+(4×1.0)M_r(\text{CH}_4) = (1 \times 12.0) + (4 \times 1.0) Mr(CH4)=12.0+4.0M_r(\text{CH}_4) = 12.0 + 4.0 Mr(CH4)=16.0M_r(\text{CH}_4) = 16.0

    The relative molecular mass of methane is 16.0\boxed{\text{16.0}}.

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