Question

Section 1
Tube	[FeSCN^+] M	Absorbance
1	0.0005	0.016
2	0.0011	0.0422
3	0.00367	0.0917
4	0.00727	0.224
5	0.00965	0.267
6	0.0137	0.398
Part B	unknown sol.	0.175

SCENARIO: Five standard solutions containing different known concentrations of an iron (II) thiocyanate (FeSCN+ ) complex are analyzed using spectrophotometry, a technique which measures the quantity of light absorbed by the solution as a function of the concentration of the analyte in solution (in this case, FeSCN+ ). The results of this analysis are included in your assigned data set labeled “Section I”.

1. Look at your data set. • Which variable is the independent variable? What axis will this be in the graph? • Which variable is the dependent variable? What axis will this be in the graph?

2. Launch Excel by clicking on the icon.

3. Enter your data into the spreadsheet. • Note that you have three columns of information. Which two columns contain experimental data to be graphed? Put the x values into the spreadsheet to the left of the y values. • Type labels for each column in the first row. Be concise but descriptive.

Highlight the set of data which you wish to graph and choose “Scatter” from the “Insert” tab at the top of the window. Choose the scatter option that shows only points, no lines. Your graph will appear in your spreadsheet.

5. Resize the chart. Click on the chart to activate it. Move the mouse cursor over one corner of the chart until it becomes a two-headed arrow. Then click and drag the chart to the desired size. Make sure the chart is large enough to see all of the data clearly and the axes are readable.

6. From the “Layout” tab, enter the chart title and titles for both of the the axes (including units, if appropriate).

7. Then right click on one of the data points and select "Add Trendline". Under the “Trendline Options” tab, select "Display equation on chart" and "Display R-squared value on chart". The equation that appears represents the equation of the trendline that appears on the graph. EXAMPLE: y = 0.275x - 0.003 R 2 = 0.9987 NOTE: The program will fit a straight line to the data no matter how good or how awful the data. You must judge the quality of the fit and the suitability of this type of fit to your data set.

8. Record the equation for the trendline and the R2 value from your graph on the report form. Does your trendline have a good fit to the data?

9. Select one of your original values for FeSCN+ concentration that does not appear to sit on the best fit line. Put that concentration into your equation to calculate the “best-fit” absorbance and demonstrate that the equation returns a value close to but not the same as the measured absorbance for that data point. (It will not be the exact value unless the points fall exactly on the line.) Show this calculation on your report form.

Part B: Using graphs to predict an unknown value for a variable By graphing the standard solutions, you have established the relationship between absorbance and concentration for that particular experiment. Your graph contains a visual representation of that relationship (the plot) as well as a mathematical expression of it (the equation). Using that information it is possible to predict the concentration of an unknown sample using either or both of these tools. SCENARIO: You are handed a new sample containing an unknown concentration of iron thiocyanate ions. The measured absorbance of the sample is listed in the Section I data set labeled “Unknown Solution”.

1. Use your graph to interpolate the concentration of the FeSCN+ complex in solution from the given absorbance. • Estimate the value by drawing a straight line from your given absorbance to the trendline, then down from the trendline to the x-axis.

2. Use the equation for the trendline to calculate the concentration of the FeSCN+ complex in solution from the given absorbance.

3. Which method is more accurate? The calculation or the interpolation?

I am confused on how to finish part A and begin part B.

Answer 1

Ans:

Part A is graphing the data as given in the table.

We can see the trendline on the graph is the best fit line as it touches maximum points on the graph.

And we get the equation for the line by selecting the options mentioned in the steps here.

Part B

We can find the unknown concentration by using the equation on the graph more accurately.(We can use the graph also by interpolation by drawing the y point as 0.0175 and find the corresponding x point,that too is correct,it also gives a value close to 0.00608)

y = 28.63x + 0.001

0.175 = 28.63x + 0.001

0.174=28.63x

0.174/28.63 = x

x = 0.006077

unknown concentration is 0.00608