1 detection limits all instrumental methods have a degree of noise associated with the measurement...

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Detection Limits

All instrumental methods have a degree of noise associated with the measurement that limits the amount of analyte that can be detected.

1. Detection Limit is the lowest concentration level that can be determined to be statistically different from an analyte blank.

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When a graphical display of results is obtained, the detection limit of the instrument can be defined as the concentration of analyte resulting in a signal that is twice as the peak to peak noise (the distance between the two dashed lines in the schematic below).

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Peak-to-peak noise level as a basis for detection limit.

A “detectable” analyte signal would be 12 divisions above a line drawn through the average of the baseline fluctuations

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2. Detection Limit is the concentration that gives a signal three times the standard deviation of the background signal.

To calculate the detection limit:

a. Find the average of the blank signal

b. Find the standard deviation of the blank

c. Find the net analyte signal

analyte conc. * 3*s

analyte signalDL =

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Example A blank solution in a colorimetric analysis resulted

in absorbance readings of 0.000, 0.008, 0.006, and 0.003. A 1 ppm standard solution of the analyte resulted in a reading of 0.051. Calculate the detection limit.

The standard deviation of the four data points of the blank can be calculated to be + 0.0032 and the mean of the blank is 0.004

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The net reading of the standard = 0.051 – 0.004 = 0.047

The detection limit is the concentration which results in three times the standard deviation (3 x 0.0032 = 0.0096).

Detection limits = 1 ppm x 0.0096/0.047 = 0.2 ppm

The absorbance reading of the least detectable concentration = 0.0096 + 0.004 = 0.014

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Stoichiometric Calculations

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Review of Fundamental Concepts

Formula Weight

It is assumed that you can calculate the formula or molecular weights of compounds

from respective atomic weights of the elements forming these compounds. The

formula weight (FW) of a substance is the sum of the atomic weights of the elements from which this substance is formed from.

The formula weight of CaSO4.7H2O is

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Element Atomic weight Ca 40.08 S 32.06 11 O 11x16.00 14 H 14x1.00 FWt

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The Mole

The mole is the major word we will use throughout the course. The mole is defined as gram molecular weight which means that:

Mole Grams 1 mol H2 2.00 g 1 mol O2 32.00 g 1mol O 16.00 g 1mol NaCl 58.5 g 1 mol Na2CO3 106.00 g

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Assuming approximate atomic weights of 1.00, 16.00, 23.00, 35.5, and 12.00 atomic mass units for hydrogen, oxygen, sodium, chlorine atom, and carbon, respectively.

The number of moles contained in a specific mass of a substance can be calculated as:

mol = g substance/FW substance

The unit for the formula weight is g/mol

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In the same manner, the number of mmol of a substance contained in a specific weight of the substance can be calculated as

mmol = mol/1000

Or,

mmol = mg substance/FW substance

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The number of mmol of Na2WO4 (FW

= 293.8 mg/mmol) present in 500 mg of Na2WO4 can be calculated as

? mmol of Na2WO4 = 500 mg/293.8

(mg/mmol) = 1.70 mmol

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The number of mg contained in 0.25 mmol of Fe2O3 (FW = 159.7 mg/mmol) can be

calculated as

? mg Fe2O3 = 0.25 mmol Fe2O3 x 159.7

(mg/mmol) = 39.9 mg

Therefore, either the number of mg of a substance can be obtained from its mmols

or vice versa.

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Calculations Involving Solutions

Molarity (M)

Molarity of a solution can be defined as the number of moles of solute dissolved n 1 L of solution. This means that 1 mol of solute will be dissolved in some amount of water and the volume will be adjusted to 1 L. The amount of water may be less than 1 L as the final volume of solute and water is exactly 1 L.

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Calculations Involving Molarity Molarity = mol/L = mmol/mL This can be further formulated as Number of moles = Molarity X volume in Liters,

or

mol = M (mol/L) x VL

 Number of mmol = Molarity x volume in mL , or 

mmol = M (mmol/mL) x VmL

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Also you should know that 250 mL = 0.25 LNow one can calculate the molarity directly fromMolarity = mol/L M = 7.42x10-3 mol/0.25 L = 0.0297 mol/L

Find the molarity of a solution resulting from dissolving 1.26 g of AgNO3 (FW = 169.9 g/mol)

in a total volume of 250 mL solution.

First find mol AgNO3 = 1.26 g AgNO3 / 169.9 g/mol

= 7.42x10-3 mol

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We can find the molarity directly in one step using dimensional analysis

? mol AgNO3 / L = (1.26 g AgNO3 / 250 mL)

x ( mol AgNO3/169.9 g AgNO3) x (1000

mL/1L) =0.0297 M

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Let us find the number of mg of NaCl per mL of a 0.25 M NaCl solution

First we should be able to recognize the molarity as 0.25 mol/L or 0.25 mmol/mL. Of course, the

second term offers what we need directly

? mg NaCl in 1 mL = (0.25 mmol NaCl/mL) x (58.5 mg NaCl/mmol NaCl) = 14.6 mg NaCl/mL