This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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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 Step 1: Fragmentation of 2-methylbutane. 2-methylbutane has the formula C_5H_12. Its structure is: CH_3-CH(CH_3)-CH_2-CH_3 When ionized, it forms a molecular ion C_5H_12^+. Fragmentation occurs at various C-C bonds. Loss of CH_3 (methyl group, m/e=15): C_4H_9^+ (m/e=57) Loss of C_2H_5 (ethyl group, m/e=29): C_3H_7^+ (m/e=43) Loss of C_3H_7 (propyl group, m/e=43): C_2H_5^+ (m/e=29) Loss of C_4H_9 (butyl group, m/e=57): CH_3^+ (m/e=15) The provided data (15, 14, 13, 12) are for very small fragments, likely from further breakdown of the initial fragments. m/e = 15: CH_3^+ m/e = 14: CH_2^+ m/e = 13: CH^+ m/e = 12: C^+ Step 2: Plot the simplified mass spectrum. A stick diagram plots relative intensity against mass-to-charge ratio (m/e). At m/e = 15, Relative Intensity = 100 At m/e = 14, Relative Intensity = 75 At m/e = 13, Relative Intensity = 50 At m/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/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 = 15. The base peak is at m/e = 15. That's 2 down. 3 left today — send the next one.