Determining Riboflavin Content in a Multivitamin: Things to do before the lab Calculate everything you can that you can in advance Step 1: Calculate the dilution of a 4.0 M solution to 1 L of a 0.02 M solution (to make your solvent) Step 2: Calculate how to make 500 ml of a 50.0 ppm solution of riboflavin Half the class will do step 2, and half will do step 3. Everyone needs the calculations for step 2, even if you aren t doing that step. ppm = parts per million In our case, this means! "#$%&'()#* +, -!.%'/0#%*. Conveniently, our solvent is water, which has a density of 1 g/ml. Therefore, 1 ppm = 1 mg/l Step 4: Calculate dilutions of the 50.0 ppm solution to concentrations listed in the lab manual don t be intimidated by the unusual unit. 1 + 2 + = 1 4 2 4 still works. We will discuss how to do standard addition calculations on Wednesday during lecture
Absorbance Review Using your phone or a laptop, go to kahoot.it
Ferrozine absorbs and transmits at!" #$ + 3!' #(!"(!'), -( Ferrozine (Fz 2- ) is a metal ligand transparent purple Ferrozine is a green crystal that is transparent in solution until it binds to iron! Only as an iron complex does it absorb around 590 nm and transmits about 700 nm and 420 nm
Absorbance and emission spectroscopy are dependent upon All of the above! Light interacting with a sample Unique energy transitions between electrons A monochrometer
Spectrometers measure P 0 P Only light transmitted (P) makes it to the detector! Transmittance: T = P/P 0 Absorbance: A = log(p 0 /P)
Beer s Law relates
Absorbance spectroscopy is Handy but limited!
Absorption, Emission and Fluorescence Spectroscopies Riboflavin R. Corn and K. Kartub - Chem M3LC
Three Spectroscopic Processes: Absorption:! + h$! Emission:!! + h$ Fluorescence:! + h$!! + h$ ' h$ > h$ '
Fluorescence Spectroscopy Fluorescence is the process that first consumes a photon and puts the atom or molecule in an excited state... M + hν M*
Fluorescence Spectroscopy And then emits a photon of lower energy which takes the the atom or molecule back to the ground state. M* M + hν hν > hν
Fluorescence Spectroscopy Net reaction: M + hν M* M + hν hν > hν The emitted photon has less energy than the absorbed photon because the molecule loses some energy (by vibrating and rotating) in the excited state: M* hν M
The Fluorescence Spectrum plots the amount of light emitted by a sample as a function of photon wavelength. P0 P PF P0 = power of incident light beam (units: W) P = power of transmitted light beam. PF = power of emitted fluorescence.
Fluorescence Spectroscopy The absorption and fluorescence spectra of riboflavin. riboflavin absorption fluorescence
Discuss with your Partner: Why would we want to take both an absorbance and a fluorescence spectrum of a fluorescent molecule like riboflavin? When analyzing a complex system like a multivitamin, why might fluorescence be useful? Why is the detector in an fluorescence spectrometer in a different location compared to an absorbance spectrometer?
Let s Play a Guessing Game How many MnM s are in this bag? What if we weighed it? What if we added one MnM? How much does it weigh now? Ok add another, and another. How does the weight change?
Let s be smart about our guessing Let s say our original bag weighs 303.3 grams After each addition of a single M&M s here is what the new mass was: Number of M&M s added Mass (g) 0 303.3 1 304.2 2 305.4 3 306.3 4 307.2 5 308.2 Using this information, calculate the original number of M&M s
Answer time! What kind of math did you use? Probably some combination of finding the mass of a single M&M through the change in weight, and then seeing how many units fit into the original mass If you used a graph to perform this math, it is the same thing as using y = mx + b Where y is the weight, m (slope) represents the change in weight per M&M, and x is the number of M&M s Since we don t know how many M&M s are in here, we assume x =0 when y = 303.3 g and back calculate to see how many M&M s must have been in the bag We get about 330 M&M s
Method of Standard Addition You can determine the concentration of an unknown solution C by fluorescence using the method of standard addition. 1. Make five solutions C + n where n= 0 to 4 2. Measure the fluorescence from these five solutions and record the values F0 to F4. This leads to five (x,y) data points: (C, F0), (C +, F1), (C + 2, F2), (C + 3, F3), (C + 4, F4). 3. Graph the following five (x,y) data points: Point # 1 2 3 4 5 x 0 1 2 3 4 y F0 F1 F2 F3 F4 (0, F0), (, F1), (2, F2), (3, F3), (4, F4). We will get a straight line that can be fit with the linear equation y = mx + b.
Method of Standard Addition This line is the same line as a standard calibration curve, but shifted to the left by an amount equal to C. To get the value of C, we set y=0 and calculate the value for the x intercept x0 = x at y=0: x0 = b/m = -C therefore: C = b/m in units of. From Wikipedia.
Questions?