Last exam questions Flashcards
Describe a fluorimeter, what parts the instrument is made out of and their function
Light source: The light source provides the excitation energy necessary to induce fluorescence in the sample. Common light sources include high-intensity LEDs, xenon lamps, or lasers.
Monochromator: The monochromator is used to filter the light source, selecting the excitation wavelength. This allows the user to selectively excite the sample at a specific wavelength.
Sample holder: The sample holder is used to hold the sample in place during measurement. It may be a cuvette or a microplate, depending on the type of sample being analyzed.
Detector: The detector is used to measure the intensity of the emitted fluorescence. Typically, a photomultiplier tube or a CCD camera is used for detection.
Filters: Fluorimeters may use filters to remove any scattered light or stray excitation light, which can interfere with the fluorescence signal.
Data analysis software: The data analysis software is used to analyze the fluorescence data and to calculate the concentration of the fluorescent molecule in the sample.
For a fluorimeter, what substances can be analyzed and what advantages there is to florescent spectroscopy compared to absorption spectroscopy
Sensitivity: Fluorescence spectroscopy is a more sensitive technique than absorption spectroscopy, as it is capable of detecting much lower concentrations of molecules. This is because fluorescence measurements are made on the basis of the emission of photons, rather than the absorption of photons, and fluorescence signals can be amplified through signal integration over time.
Selectivity: Fluorescence spectroscopy offers greater selectivity than absorption spectroscopy, as it is possible to selectively excite specific molecules or groups of molecules using a specific excitation wavelength. This selectivity can be used to detect and quantify specific molecules in complex mixtures, such as biological fluids.
Structural information: Fluorescence spectroscopy can provide valuable information on the structure and conformation of molecules. For example, changes in the fluorescence emission spectrum of a protein can provide insights into changes in its structure or conformation.
Non-destructive: Fluorescence spectroscopy is a non-destructive technique, which means that the sample is not altered or destroyed during analysis. This makes it possible to analyze the same sample repeatedly, or to recover the sample for further analysis using other techniques.
A wide variety of samples can be analyzed such as nucleotides and proteins, chemicals and drug detection in samples, pollutions, certain metals etc
Based on the chemical structure of paracetamol, explain why varying pH during UV absorbance spectroscopy may have an effect of the calibration curve (it has a NH group, O, CH3, HO and a carboxygen ring
At acidic pH values, the phenol group may become protonated, resulting in the formation of a positively charged species. This can lead to a shift in the absorption peak to a lower wavelength, which can affect the calibration curve. Additionally, at high acidic pH values, the amide group may undergo hydrolysis, leading to the formation of a degradation product that may also have a different UV absorbance spectrum compared to paracetamol, leading to further distortions in the calibration curve.
At alkaline pH values, the amide group in paracetamol may become deprotonated, resulting in the formation of a negatively charged species. This can lead to a shift in the absorption peak to a higher wavelength, which can again affect the calibration curve.
Furthermore, variations in pH can also affect the solubility and stability of paracetamol, which can further complicate the UV absorbance measurements. For example, at very high or very low pH values, the solubility of paracetamol may be reduced, which can affect the amount of sample that can be analyzed and the accuracy of the measurements.
If the absorption ϵ, is 11,329 at 243 nm and the cuvette is 1cm and the result is 0.75 what are the concentration of paracetamol?
The Beer-Lambert law states that the absorbance (A) of a sample is directly proportional to the concentration (c) of the absorbing species and the path length (l) through which the light passes through the sample, according to the formula:
A = εcl
where ε is the molar absorptivity or molar extinction coefficient of the absorbing species.
Given the following parameters:
ε = 11,329 M^-1cm^-1 at 243 nm
l = 1 cm
A = 0.75
We can rearrange the Beer-Lambert law to solve for the concentration (c) of the absorbing species:
c = A / (εl)
Substituting the values given above, we get:
c = 0.75 / (11,329 M^-1cm^-1 x 1 cm) = 6.62 x 10^-5 M
Therefore, the concentration of the absorbing species is 6.62 x 10^-5 M.
How do you calculate retention factor? Do question 3 A.
k = (tr - t0) / t0
where tr is the retention time of the compound and t0 is the retention time of an unretained compound or the dead time of the column.
How do you calculate plate number? Do question 3b
N = L / (5.54 * (tR / w0.5)^2)
where L is the length of the column, tR is the retention time of a peak, and w0.5 is the width of the peak at half-height.
To calculate the plate number, the retention time and peak width at half-height must be measured for a peak of interest. The length of the column is a known value and must be entered into the formula.
The factor 5.54 is a constant that depends on the flow rate of the carrier gas and the temperature of the column. It is commonly used for columns operated at flow rates of 1-100 mL/min and temperatures of 20-30°C.
How do you calculate resolution? How can we see if we have a baseline peak based on the resolution of two peaks? Do question 3c and d
The resolution (R) is a measure of the separation between two adjacent peaks in a chromatogram and is calculated using the following formula:
R = 2 (t2 - t1) / (w1 + w2)
where t1 and t2 are the retention times of the two peaks, and w1 and w2 are their respective peak widths at half height.
To determine if there is a baseline peak based on the resolution of two peaks, one can calculate the resolution using the formula above. A resolution value of 1.5 or higher indicates a good separation between the two peaks, with little or no overlap. A resolution value of less than 1.0 indicates that the two peaks are not well separated and may be overlapping, indicating the presence of a baseline peak.
How can you tell based on 2 different chromatogram looking at the peaks which chromatogram had the thickest film the one with no baseline separation and lower peak heights or the one with baseline separation and higher peak heights? Why?
The one with high narrow peaks and seperation has thinner film since the analyte will interact less with the stationary phase and pass faster through the column which geaves more narrow and sharp peaks
What three column parameters can be changed to increase plate number N?
Column Length: By increasing the length of the column, the number of theoretical plates increases, leading to an increase in the plate number N. However, increasing the column length also increases the retention time and the analysis time.
Particle Size: Smaller particle size packing materials in the column lead to a higher number of theoretical plates and thus an increase in the plate number N. However, decreasing the particle size can also increase the pressure drop in the column, requiring higher operating pressure.
Column Diameter: By decreasing the column diameter, the band broadening and dispersion effects are reduced, leading to an increase in the plate number N. However, decreasing the column diameter can also increase the backpressure and reduce the sample loading capacity.
Draw the chemical structure of a common phase in gas crhomatographt and give trhe chemical namwe of this chromatography. Both the polar and non polar one
SiO2 and Si-O-Si-C18H37
How do you achieve gradient solution in gas chromatography? Give an example of such gradient
One example of a gradient solution in GC is a temperature program, where the temperature of the column is increased over time to elute different compounds from the sample. For example, if analyzing a mixture of compounds with different boiling points, a temperature gradient can be used to selectively elute each compound as the temperature of the column increases.
Give an example of a suitable stationary phase and mobile phase in isocratic elution with reversed HPLC
HPLC is C18-bonded silica as stationary phase and water and organic solvent such as methanol for example usually 50/50
Give an example of the mobile phase composition can be changed in gradient elution reversed phase HPLC. How does the eluent strength change during the gradient? Explain
One can change the concentration of water and methanol to detemrine how fast the analyte shoudl move throught he column. The more it interacts with the mobile phase the faster it will go so increaisng or decreasing methanol to slow down or speed up.
In reversed phase HPLC what will be the elution order for oxazepam and diazepam? Explain based on the chemical structure of compounds and their interactions with the stationary phase
Diazepam is more hydrophobic then oxazepam and will therefore elute after oxazepam. This is because oxazepam has an NH group and an OH group which diazepam doesn’t which makes the group less hydropohibic
In what three ways can separation factor be affected for HPLC
Changing the mobile phase composition: The separation factor is affected by the composition of the mobile phase, which can be adjusted by changing the ratio of solvents used in the mobile phase or the pH of the buffer solution. By changing the mobile phase composition, the separation factor can be adjusted to improve the selectivity of the separation.
Changing the column temperature: The separation factor is also affected by the column temperature. Increasing the column temperature can decrease the retention time and change the separation factor for some compounds, which can be useful for improving selectivity and/or speeding up the separation.
Changing the stationary phase: The separation factor is also influenced by the chemical and physical properties of the stationary phase. Different types of stationary phases, such as C18, C8, or phenyl columns, can have different selectivity for different types of solutes. Changing the stationary phase can therefore be a useful approach for optimizing the separation factor and improving the selectivity of the chromatographic separation.
