We read your photo as:

The spectrometer in AAS can be either single-beam or double-beam. Give one major difference between them. State the major function of the transducer in a mass spectrometer. Use the stick diagram to il

Chemistry

ScanSolve AI Answer

Step-by-step solution

2 steps

Okay, let's continue with question 4.

4 (a) "The spectrometer in AAS can be either single-beam or double-beam." Give one major difference between them.
A major difference is that a double-beam spectrometer splits the light from the source into two paths: one through the sample and one through a reference. This allows for continuous correction for variations in source intensity, leading to greater stability and accuracy compared to a single-beam spectrometer, which passes all light through the sample.

4 (b) State the major function of the transducer in a mass spectrometer.
The major function of the transducer (detector) in a mass spectrometer is to detect the separated ions, convert their kinetic energy into an electrical signal, and amplify this signal so it can be recorded and processed.

4 (c) (i) Use the stick diagram to illustrate the fragmentation patterns of 2-methylbutane and plot a graph of relative intensity on the y-axis against mass-to-charge values on the x-axis showing the simplified mass spectrum using the following data:

  • Relative intensity (R.I): 100, 75, 50, 22
  • Mass-to-charge (m/e) values: 15, 14, 13, 12
  1. Fragmentation of 2-methylbutane.

    2-methylbutane has the formula C5H12C_5H_{12}. Its structure is:

    CH3−CH(CH3)−CH2−CH3CH_3-CH(CH_3)-CH_2-CH_3

    When ionized, it forms a molecular ion C5H12+C_5H_{12}^+. Fragmentation occurs at various C-C bonds.

    • Loss of CH3CH_3 (methyl group, m/e=15m/e=15): C4H9+C_4H_9^+ (m/e=57m/e=57)
    • Loss of C2H5C_2H_5 (ethyl group, m/e=29m/e=29): C3H7+C_3H_7^+ (m/e=43m/e=43)
    • Loss of C3H7C_3H_7 (propyl group, m/e=43m/e=43): C2H5+C_2H_5^+ (m/e=29m/e=29)
    • Loss of C4H9C_4H_9 (butyl group, m/e=57m/e=57): CH3+CH_3^+ (m/e=15m/e=15)

    The provided data (15, 14, 13, 12) are for very small fragments, likely from further breakdown of the initial fragments.

    • m/e=15m/e = 15: CH3+CH_3^+
    • m/e=14m/e = 14: CH2+CH_2^+
    • m/e=13m/e = 13: CH+CH^+
    • m/e=12m/e = 12: C+C^+
  2. Plot the simplified mass spectrum.

    A stick diagram plots relative intensity against mass-to-charge ratio (m/em/e).

    • At m/e=15m/e = 15, Relative Intensity = 100
    • At m/e=14m/e = 14, Relative Intensity = 75
    • At m/e=13m/e = 13, Relative Intensity = 50
    • At m/e=12m/e = 12, Relative Intensity = 22

    The graph would have:

    • X-axis: Mass-to-charge (m/e)
    • Y-axis: Relative Intensity (%)
    • Vertical lines (sticks) at m/em/e values 12, 13, 14, 15, with heights corresponding to their relative intensities (22, 50, 75, 100 respectively).
    Relative Intensity (%)
    ^
    |   100 |       |       |   X
    |    75 |       |   X   |
    |    50 |   X   |       |
    |    25 | X     |       |
    |     0 +---------------------> m/e
            12  13  14  15
    

    (Note: This is a textual representation of the stick diagram. In a proper diagram, the 'X' marks would be the top of vertical lines extending from the x-axis.)

    4 (c) (ii) At what m/e value is the base peak?
    The base peak is the most intense peak in the mass spectrum, assigned a relative intensity of 100%.
    From the given data:

    • Relative intensity 100 corresponds to m/e=15m/e = 15.

    The base peak is at m/e=15\boxed{m/e = 15}.

Was this helpful?

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question