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A forensic chemist wants to test the accuracy of a gas chromatograph that is to be used for the analysis of blood alcohol content - VCE - SSCE Chemistry - Question 5 - 2010 - Paper 1
Question 5
A forensic chemist wants to test the accuracy of a gas chromatograph that is to be used for the analysis of blood alcohol content. A blood sample may contain a numbe... show full transcript
Worked Solution & Example Answer:A forensic chemist wants to test the accuracy of a gas chromatograph that is to be used for the analysis of blood alcohol content - VCE - SSCE Chemistry - Question 5 - 2010 - Paper 1
Step 1
What evidence is presented in the chromatogram to support this claim?
Answer
The chromatogram shows that the peak corresponding to ethanol has a distinct retention time of 0.9 minutes, which is well-separated from other volatile chemicals. The presence of other peaks, representing different compounds, does not overlap with the ethanol peak, indicating that the identification of ethanol remains unaffected despite the presence of these other chemicals.
Step 2
To determine the percentage of alcohol in a blood sample only the peak at a retention time of 0.9 minutes is measured. Explain why.
Answer
The measurement focuses exclusively on the peak at 0.9 minutes because this peak is specifically associated with ethanol, the alcohol of interest. Other volatile compounds may interfere with the measurement if included, leading to inaccuracies. Therefore, to ensure an accurate determination of alcohol content, only the relevant peak should be considered.
Step 3
Determine the percentage (m/v) of alcohol in the driver's blood if the peak area at a retention time of 0.9 minutes was found to be 110000.
Answer
To determine the percentage (m/v) of alcohol, the formula:
ext{Percentage (m/v)} = rac{ ext{mass of alcohol (g)}}{ ext{volume of solution (mL)}} \times 100
is used. Assuming the peak area correlates to the concentration of ethanol, further calibration is needed to convert the area into mass. If, for example, it is established that a peak area of 110000 corresponds to 11.0 g of ethanol in 100 mL of blood, the calculation would be:
ext{Percentage (m/v)} = rac{11.0 ext{ g}}{100 ext{ mL}} \times 100 = 11 ext{ \\%}.