Question

QUESTION 1

We will be using spectrophotometry to determine the concentration of iron in solutions. First, you will generate a calibration curve. The ----- variable is -----, and will be graphed on the x axis of the calibration curve. The -------- variable is ----------, and will be graphed on the --------- axis.

Word Bank:

-independent

-y

-absorbance

-dependent

-concentration

QUESTION 2

How many mL of a 100 mg/dL Ferrozine solution are needed to complex 0.00000048 moles of iron?

HINT: Look at the 1st equation in the lab manual (under the heading Background)

PUT YOUR ANSWER IN DECIMALS (not scientific notation)

QUESTION 3

How many moles of Fe3+ can be reduced by 1.1 mL of 0.527 M hydroxylamine solution?

Look at the 2nd equation in your lab manual under the heading Background to see the relationship between Iron and hydroxylamine.

Put your answers in DECIMALS, not scientific notation.

QUESTION 4

The trendline equation calculated from a calibration curve of absorbance vs iron concentration in μg dL-1 is y = 0.000812x + 0.00709. If a sample analyzed using the same assay gives an absorbance of 0.099, what is the iron concentration of the unknown sample in μg dL-1?

This is a simple "equation of a line" formula.

QUESTION 5

As part of this laboratory, you may have to remove iron from a solution. If you have 50 mL of a solution that contains 239 μg dL-1, how many μg of iron must be removed to reach a concentration of 150 μg dL-1?

HINT: 1. Convert 50mL to dl;

2. Figure out how much μg of solution you have in that 50 mL

3. Figure out how much μg of solution you want in 50 mL

4. Subtract the answers from 2 & 3.

QUESTION 6

As part of this laboratory, you may have to add iron to a solution. If you have 100 mL of a solution that contains 45 μg dL-1, how many grams of iron must be added to reach a concentration of 150 μg dL-1?

Put your answer in scientific notation

QUESTION 7

The concentration of the iron stock solution for this lab is 500 μg dL-1. What volume of this solution (in mL) must you add to the "patient" sample to give your patient 0.00000003 moles of iron?

Answer 1

1. We will be using spectrophotometry to determine the concentration of iron in solutions. First, you will generate a calibration curve. The independent variable is concentration, and will be graphed on the x axis of the calibration curve. The dependent variable is absorbance, and will be graphed on the y- axis

Absorbance is dependent on the concentration of the absorbing species and the relationship is given by the Lambert-Beer's law; A= where is molar absorptivity, b is the path length and c is concentration. Thus a plot of Abs vs Concentration will give a straight line.

2. Iron and Ferrozine form a 1:3 complex. i.e 1 mole of Fe2+ reacts with 3 moles of ferrozine.

thus number of moles of ferrozin required to react with 0.00000048 moles of iron = 0.00000048 x 3 = 0.00000144

Strength of the given ferrozin solution = 100 mg/ dL = 1000 mg/L or 1g/L (since 1dL= 0.1 L and 1000 mg = 1g)

Number of moles of ferrozin in 1L solution = (mass/molar mass)= 1/514.4 = 0.00194401 moles (Molecular weight of ferrozin = 514.4)

0.00194401 moles of ferrozin in 1000 ml of solution

0.00000144 moles of ferrozin will be present in (1000/0.00194401) x 0.00000144 mL = 0.74074074 mL

3. 0.527 M hydroxylamine = 0.527 moles in 1000 mL of solution

number of moles in 1.1 mL of this solution = (0.527/1000)x 1.1 = 0.0005797 moles

Since each mole of hydroxylamine can reduce 2 moles of Fe3+.

Thus number of moles of Fe3+ that can be reduced by 1.1 mL of giveb hydroxylamine solution = 2 x 0.0005797 = 0.0011594 moles.

4. equation of calibration curve is given as y = 0.000812x + 0.00709.

since absorbance is on the y axis, absorbance = 0.099 corresponds to y= 0.099

in order to find the corresponding concentration, we need to find the corresponding value of x

from the above equation, x = (y-0.00709)/0.000812 = (0.099-0.00709)/0.000812 = 113.19 μg dL^-1

5 Since 1 dL = 100 mL

50 mL = 0.5 dL

239 μg dL^-1= 119.5 μg in 0.5 dL

150 μg/ dL = 75 μg in 0.5 dL

thus the amount of iron to be removed = 119.5 - 75= 44.5 μg

6. 1 dL = 100 mL

Thus we need to go from 45 μg in 100 mL to 150μg.

so we need to add (150-45) 105 μg = 105 x 10^-6 g = 1.05 x 10^-4 g

7. 0.00000003 moles of Fe = 0.00000003 x 55.84 g = 167.52 x 10-8 g =1.67 x 10^-6 g = 1.67 μg

500 μg in 100 mL

1.67 μg in (100/500) x 1.67= 0.334 mL