Selenium from 0.108 g of Brazil nuts was converted into a fluorescent product, which was extracted into 10.0 mL of cyclohexane. Then 2.00 mL of the cyclohexane solution were placed in a cuvet for fluorescence measurements.
A.) Standard additions of fluorescent product containing 1.4ug Se/mL are given in a table:
Volume of standard added (uL):
0, 10, 20 , 30 40
Fluorescence intensity:
41,4, 49.2, 56.4, 63.8, 70.3
What is the weight % of Se in the nuts?
I made a graph from the values to get a line y = 0.724x + 41.74. Setting y = 0, x = 57.65
I'm not sure what's the next step in solving this.
Well, it seems like you've done a great job making the graph and finding the value of x. Now, let's figure out the next step together, shall we?
To find the weight % of Se in the nuts, we need to use the equation:
% Se = (weight of Se / weight of nuts) x 100
We know that the weight of Se in the nuts is 0.108 g, so now we just need to find the weight of the nuts.
Since you have the volume of the cyclohexane solution (2.00 mL) and the concentration of the fluorescent product (1.4 ug Se/mL) as given in the table, we can calculate the weight of the nuts.
Weight of nuts = (Volume of cyclohexane x Concentration of fluorescent product) / 1000
Weight of nuts = (2.00 mL x 1.4 ug Se/mL) / 1000
Weight of nuts = 0.0028 g
Now, we can plug in these values into our equation:
% Se = (0.108 g / 0.0028 g) x 100
% Se ≈ 3857.14
Wow, that's a lot of selenium! It seems like the nuts are doing an excellent job of providing you with this essential mineral. Just remember not to eat too many Brazil nuts at once, or else you might end up glowing like the fluorescent product! Keep up the good work, and let me know if you have any more questions!
To determine the weight % of Se in the nuts, you can use the standard addition method and the calibration curve you obtained from the fluorescence measurements.
The weight % of Se in the nuts can be calculated using the equation:
% Se = (μg/mL Se in standard added / μg/mL Se in nuts) × (volume of standard added / weight of nuts) × (100 / dilution factor)
First, let's calculate the μg/mL Se in the nuts using the calibration curve you obtained. From the equation of the line (y = 0.724x + 41.74), where x represents the volume of standard added, and y represents the fluorescence intensity:
For x = 57.65 (where y = 0):
0 = 0.724(57.65) + 41.74
Solving for the μg/mL Se in nuts, we can find:
μg/mL Se in nuts = 57.65 (y-intercept) / 0.724
Next, calculate the weight % of Se in the nuts using the given values:
Volume of standard added (μL): 0, 10, 20, 30, 40
Fluorescence intensity: 41.4, 49.2, 56.4, 63.8, 70.3
Substitute the values into the equation:
% Se = (1.4 μg/mL Se in standard added / μg/mL Se in nuts) × (volume of standard added / weight of nuts) × (100 / dilution factor)
Let's assume the dilution factor is 1 (since the standard addition is done in the same cyclohexane solution).
For each data pair (volume of standard added and corresponding fluorescence intensity), calculate the weight % of Se in the nuts using the equation above.
For example:
For 10 μL standard added (volume of standard added = 10; fluorescence intensity = 49.2):
% Se = (1.4 μg/mL Se in standard added / μg/mL Se in nuts) × (10 μL / weight of nuts) × (100 / 1)
Solve for the weight % of Se in the nuts for each data point, repeating the calculation for the other volumes of standard added and corresponding fluorescence intensities.
Finally, calculate the average weight % of Se in the nuts by taking the average of all the weight % values obtained.
This average weight % of Se in the nuts will give you the answer to your question.
To determine the weight percentage of Selenium (Se) in Brazil nuts, we can use the information provided about the fluorescence intensity and the standard additions. The fluorescence intensity of the cyclohexane solution can be directly correlated to the concentration of Se in the solution.
Firstly, let's understand the concept of standard additions. In this technique, known amounts of a standard solution (fluorescent product with a known concentration of Se) are added to create a series of samples. By measuring the fluorescence intensity of each sample, we can establish a relationship between the concentration of Se and the fluorescence intensity.
In this case, the standard additions are given in the table:
Volume of standard added (uL):
0, 10, 20, 30, 40
Fluorescence intensity:
41, 4, 49.2, 56.4, 63.8, 70.3
From the graph you created, where the line equation is y = 0.724x + 41.74, you have correctly determined that when the fluorescence intensity (y) is 0, the corresponding x-value (volume of standard added) is approximately 57.65 uL.
To calculate the weight percentage of Se in the nuts, we need to utilize the relationship between the concentration of Se and the volume of the standard added. Here's the next step:
1. Calculate the concentration of Se in the cyclohexane solution for the corresponding volume of the standard added.
- Using the equation obtained from the graph (y = 0.724x + 41.74), substitute x = 57.65 and solve for y.
y = 0.724 * 57.65 + 41.74
y ≈ 82.4644
2. Convert the concentration of Se to weight concentration (μg/mL to mg/mL).
- The concentration of Se obtained from step 1 is 82.4644 μg/mL.
- To convert μg to mg, divide by 1000.
Concentration of Se (mg/mL) = 82.4644 μg/mL / 1000
Concentration of Se (mg/mL) ≈ 0.0825 mg/mL
3. Calculate the amount of Se in the 2.00 mL cyclohexane solution used for fluorescence measurements.
- Multiply the concentration from step 2 by the volume used for measurement (2.00 mL).
Amount of Se (mg) = 0.0825 mg/mL * 2.00 mL
Amount of Se (mg) = 0.165 mg
4. Calculate the amount of Se in the 0.108 g of Brazil nuts.
- Based on the given information, the 0.108 g of Brazil nuts was converted into the fluorescent product.
- Since the fluorescent product contains Se, we need to calculate the amount of Se in the nuts.
Amount of Se in nuts (mg) = 0.108 g * (% Se / 100)
Amount of Se in nuts (mg) = 0.108 g * (weight % of Se / 100)
5. Solve for the weight % of Se in the nuts.
- Set the amount of Se in nuts from step 4 equal to the amount of Se in the cyclohexane solution from step 3 and solve for the weight % of Se.
0.165 mg = 0.108 g * (weight % of Se / 100)
weight % of Se = (0.165 mg / 0.108 g) * 100
Therefore, by calculating the weight % of Se using the given information and following these steps, you will find the weight percentage of Selenium in the Brazil nuts.