Question

1. read a discussion of spectrophotometry in the literature.
2. explain why it is important to exercise extreme care when heating the asa sample with 1m naoh solution.
3. the determination of glucose concentration in blood serum is often based on the formation of a blue - green complex of glucose and o - toluidine in glacial acetic acid. this reaction is shown at the bottom of this page.

the analytical wavelength for the complex is 635 nm. a set of standard solutions was prepared by taking a known volume of a stock glucose solution, containing 1.000 g glucose dissolved in 100 ml of distilled water, and diluting to 100 ml with distilled water. these solutions and the blood samples were treated with the o - toluidine reagent, and the percent transmittance of each sample was read against a reference solution. the data obtained are

stock glucose, ml	concentration of glucose complex, m	% t	a
10.00		59.16
20.00		35.08
25.00		27.10
30.00		20.65
unk. solution		46.17

1. calculate the molar concentration of glucose in the stock solution.
2. for each standard solution and the unknown, convert the percent transmittance to absorbance, using equation 4. enter these absorbances in the table.

Explanation:

Step1: Calculate molar mass of glucose

The molar - mass of glucose ($C_6H_{12}O_6$) is $M=(6\times12.01 + 12\times1.01+6\times16.00)\ g/mol = 180.18\ g/mol$.

Step2: Calculate molar concentration of stock glucose solution

The stock glucose solution has $1.000\ g$ of glucose in $100\ mL$ ($0.1\ L$) of distilled water. Using the formula $c=\frac{n}{V}$, where $n=\frac{m}{M}$, we have $n=\frac{1.000\ g}{180.18\ g/mol}=0.00555\ mol$. Then $c=\frac{0.00555\ mol}{0.1\ L}=0.0555\ M$.

Step3: Convert percent transmittance ($\%T$) to absorbance ($A$)

The formula to convert $\%T$ to $A$ is $A = 2-\log(\%T)$.

• For $\%T = 76.91$: $A = 2-\log(76.91)=2 - 1.886 = 0.114$.
• For $\%T = 59.16$: $A = 2-\log(59.16)=2 - 1.772 = 0.228$.
• For $\%T = 35.08$: $A = 2-\log(35.08)=2 - 1.545 = 0.455$.
• For $\%T = 27.10$: $A = 2-\log(27.10)=2 - 1.433 = 0.567$.
• For $\%T = 20.65$: $A = 2-\log(20.65)=2 - 1.315 = 0.685$.
• For $\%T = 46.17$: $A = 2-\log(46.17)=2 - 1.664 = 0.336$.

Answer:

1. The molar concentration of glucose in the stock solution is $0.0555\ M$.

2.

stock glucose, mL	concentration of glucose complex, M	% T	A
10.00	-	59.16	0.228
20.00	-	35.08	0.455
25.00	-	27.10	0.567
30.00	-	20.65	0.685
unk. solution	-	46.17	0.336