EXAM 1 Name



EXAM 1 Name ______________________________________

1. Signal-to-noise ratio of some measurement is 50. What is the relative standard deviation of this measurement? Is it suitable for quantitative determination?

RSD = 1/(SN_ratio)=0.02 or 2%

2. Using Fig.1, estimate S/N ratio of the peak.

Peak height is around 20 arbitrary units (21 to be more precise)

Peak to peak noise is around 2 units (pessimistic estimate) and is equal to 3 standard deviations. So, 20-30 is a good estimate of S/N ratio

3. If the peak in Fig.1 corresponds to 10 ppm of the compound A, what is the detection limit for A?

From the previous problem, S/N_ ratio is 20-30. At detection limit, it is 3. Threfore, the detection limit somewhere is between 1 and 2 ppm (remember, it is a single significant digit number!).

4. How many scans are necessary to receive a S/N ratio of 100 (assume that the peak in Fig.1 is a result of one scan) ?

You need to increse S/N_ ratio 3-5 fold> To do that, you need x2=9-25 experiments. I would ask for 16 or for 32 in the instruments menu.

4. Convert transmittance of 0.100 into absorbance.

A=-logT=1.0

5. In laboratory experiment No.4, the Na and K emission peaks were narrow while Ca peak was visibly wide. Why?

Na and K – atomic lines, Ca band is a molecular band ( of an unstable CaOH molecule)

6. Transmittance of some solution at 500 nm is 0.9 in a 1 cm cuvette. What is the transmittance of the same solution in a 4 cm cuvette?

0.9 ( 0.9 ( 0.9 ( 0.9 = 0.65 or

A(1)=-log(0.9) A(2)=4*A(1) T=10-A(2) = 0.65

Both approaches result in the same answer

7. Molar absorptivity for a cobalt complex with 8-quinolinol is 3529 at 365 nm. Calculate the absorbance of a 2 (10-4 M solution of the complex in a 1 cm cell.

A=((C(l=3529 ( 2 (10-4 ( 1 =0. 706

8. Molar absorptivities are often reported for chemical compounds in solution but practically never for atomic absorption procedures. Explain why.

It is relatively easy to calculate the absorbances for solutions (see above). In atomic absorption, the apparent absorbances are related to a particular instrument + today’s weather + bandwidth + numerous parameters. It has no practical use.

9.. To determine the amount of potassium in a water sample, the emission measurements of the following measurements were made:

#1 pure deionized water I = 0

#2 10 mL of water sample was diluted to 50 mL I = 50

#3 10 mL of water sample + 2.0 mL of K+ solution (1.0 (g/mL) was diluted to 50 mL I = 100

Calculate the concentration of potassium in the water sample.

Total amount of K = 2.0 mL ( 1.0 (g/mL ( (50)/(100-50)= 2.0 (g

Total volume is 10 mL

Thus, concentration is 0.2 (g/mL - or corresponding molarity, but I did not ask about it!

10. Some coating reflects around 40% of light at any wavelength between 400 and 800 nm. What color do you expect to see?

Gray ( or dirty-gray-brownish – something like that)

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Figure 1

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