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Design a model to calculate relative atomic mass and sketch the mass spectrum of copper given its isotopes and abundances.

Chemistry

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2. Design a model to demonstrate how to calculate the relative atomic mass of different elements.

To calculate the relative atomic mass of an element, a model would involve the following steps:

  • Identify all naturally occurring isotopes of the element.
  • Determine the atomic mass (or mass number) of each isotope.
  • Find the natural abundance (percentage or fractional) of each isotope.
  • Calculate the weighted average of these isotopic masses.

The formula for relative atomic mass is:

Relative Atomic Mass=∑(Isotope Mass×Fractional Abundance)\text{Relative Atomic Mass} = \sum (\text{Isotope Mass} \times \text{Fractional Abundance})

For example, if an element has two isotopes, Isotope A with mass MAM_A and abundance PAP_A, and Isotope B with mass MBM_B and abundance PBP_B, the relative atomic mass would be:

Relative Atomic Mass=(MA×PA)+(MB×PB)\text{Relative Atomic Mass} = (M_A \times P_A) + (M_B \times P_B)

This model demonstrates that the relative atomic mass is not a simple average but a weighted average based on how common each isotope is.

3. Copper has two isotopes, copper-63 and copper-65. The percentage abundance of copper-63 is 69% and that of copper-65 is 31%. Sketch the mass spectrum of copper.

First, let's calculate the relative atomic mass of copper, as this is often implied when isotope abundances are given.

  1. Convert percentage abundances to fractional abundances.

    Fractional abundance of copper-63 = 69%=0.6969\% = 0.69
    Fractional abundance of copper-65 = 31%=0.3131\% = 0.31

  2. Calculate the relative atomic mass.

    Relative Atomic Mass=(63×0.69)+(65×0.31)\text{Relative Atomic Mass} = (63 \times 0.69) + (65 \times 0.31) Relative Atomic Mass=43.47+20.15\text{Relative Atomic Mass} = 43.47 + 20.15 Relative Atomic Mass=63.62\text{Relative Atomic Mass} = 63.62

    Now, let's sketch the mass spectrum of copper. A mass spectrum shows the relative abundance of ions versus their mass-to-charge ratio (m/z). For single-charged ions, the m/z value corresponds to the isotopic mass.

    The mass spectrum would show two peaks:

    • A peak at m/z = 63 with a relative intensity proportional to 69%.
    • A peak at m/z = 65 with a relative intensity proportional to 31%.

    Here is a conceptual sketch of the mass spectrum:

    Relative Abundance (Intensity)
    ^
    |      * (69%)
    |      |
    |      |
    |      |        * (31%)
    |      |        |
    |      |        |
    |______|________|________> m/z
          63       65
    

    4. In a sample containing 100 atoms of boron, 80 were boron-11 atoms while 20 were the boron-10 isotope. Determine the relative atomic mass of boron.

  3. Determine the fractional abundance of each isotope.

    Total atoms = 100
    Number of boron-10 atoms = 20
    Number of boron-11 atoms = 80

    Fractional abundance of boron-10 = 20100=0.20\frac{20}{100} = 0.20
    Fractional abundance of boron-11 = 80100=0.80\frac{80}{100} = 0.80

  4. Calculate the relative atomic mass using the formula.

    Relative Atomic Mass=(Mass of Boron-10×Fractional Abundance of Boron-10)+(Mass of Boron-11×Fractional Abundance of Boron-11)\text{Relative Atomic Mass} = (\text{Mass of Boron-10} \times \text{Fractional Abundance of Boron-10}) + (\text{Mass of Boron-11} \times \text{Fractional Abundance of Boron-11}) Relative Atomic Mass=(10×0.20)+(11×0.80)\text{Relative Atomic Mass} = (10 \times 0.20) + (11 \times 0.80) Relative Atomic Mass=2.00+8.80\text{Relative Atomic Mass} = 2.00 + 8.80 Relative Atomic Mass=10.80\text{Relative Atomic Mass} = 10.80

    The relative atomic mass of boron is 10.80\boxed{\text{10.80}}.

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