Sec. 001
Report I, Unit C
BMB 471 – Report I – Experimental Unit C Answers
1.

a.
What quantity of solid material from the bottle labeled FMN did you
weigh on the analytical balance?
i.
0.0670 grams FMN
b.
Show how you calculated the true concentration of the stock FMN
solution, in mM, from this weight and the information on the Certificate
of Analysis (do not show how you calculated the amount of FMN to
weigh).
i.
0.06697 grams FMN=(478.33 g FMN/1 mol FMN)*x moles FMN
x=0.00014 mol FMN
purity = (100%4.8% water loss) x 74% = 70.45%=0.7045
0.00014 mol FMN (pure)*0.7045=9.86e^5 moles FMN (as is)
9.86e^5 moles FMN/0.1 L solution=0.000986M FMN
0.000986 M FMN *1000 = 0.986 mM FMN
2.

a.
What is your measured value for the molar
extinction coefficient of
FMN?
i.
Value without filter:
10972.222/M*cm
ii.
Value without filter (2
nd
trial):
10988.095/M*cm
1.
Filter position used:
0 position
iii.
Value with filter:
11174.44/M*cm
iv.
Value with filter (2
nd
trial):
11195.0/M*cm
1.
Filter position used:
1 position
3.

a.
Describe the procedure you used to obtain the values of the extinction
coefficients (without and with the filter) from the raw absorbance values
obtained from the spectrophotometer, i.e. the steps you went through
following collection of the data until the extinction coefficients were
obtained.
i.
The extinction coefficient can be obtained for each data set by
finding the slope of the line (linear portion) on a graph of
absorbance vs. concentration. To simplify, an online linear
regression tool was used (www.xuru.org), where absorbance and
concentration values were entered and an equation for the line of
the data (linear portion) was produced (y=mx+b format). The value
of “m” in this equation is also known as the slope of the line,
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 Spring '08
 Bowlby
 FMN, Absorbance Values

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