Experimental Analytical Chemistry 227 Spring 2011

Chapter 0 THE
ANALYTICAL PROCESS
Lectures 1 & 2

SAMPLING:
How
does one get a true
representative sample of the material he/she wishes to study? How
do we get the "analyte" in a form that we can analyze
quantitavely? How do we make sure that the sampel is
uniform? Seawater at the surface of the sea is very different in
composition than that even 200 ft down, and also different again at
1000 ft down.. It is heterogeneous in composition.

SAMPLE PREPARATION:
Sometimes,
dissolution
of a solid material is all that is necessary. In other cases,
liquids may be analyzed directly. Gases may be anaylyzed
directly, or bubbled into a liquid in order to cause a reaction to
occur. At times, many steps are required to be able to analyze
certain components: dissolution, extraction with various solvents,
drying, etc.

THE CHEMICAL
ANALYSIS: Difference between qualitative analysis (what it is)
and quantitative analysis (how
much of a specific analyte is present). There are many techniques
that are available for these determinations: HPLC, IR, UV-Vis
spectrometry, Mass spectrometry, titrations, Atomic Absorption
spectroscopy, electroanalytical techniques, column chromatography,
nmr. The analyst must decide on which technique will give him/her
the "best" and most reliable analysis.

CALIBRATION CURVES:
Standard
materials are
usually run to prepare calibration curves. Then the "unknown" is
run in exactly the same way, and from the calibration curve, the
quantity of the unknown can be determined with some degree of accurace
and precision.

INTERPRETING THE
RESULTS: Once the analysis
is done, then it is time to interpret the results: usually, one
does not make a single measurement and is not satisfied that he/she has
done such a magnificent job that this is the "true" answer!
Usually, analyses are done at least in triplicate, and the average
gives a much better feeling than would only a single measurement.
Also, we can calculate the standard deviation of the results; the
standard deviation is a measure of the reproducibility of the
measurements. When the standard deviation is 0, then we have the
same result over and over again; when the standard deviation is
very large, then we do not have much confidence that the results are
very reproducible. Now, we have to ask if we have done everything
that needs be done to answer some of the questions that were initially
asked: is this the only sample, does the answer we got make any sense,
etc.???

QUALITY ASSURANCE:
We
have to make sure that the
analyses are correct. One way is to use "quality assurance"
techniques. One of these methods involves "spiking" the unknown
with the specific analyte of interest, re-analyzing the sample, and
then finding the original amount of analyte has been increased by
exactly the amount of added spike. That means that you should
have great confidence in the whole method and that your measurements
were done accurately.

Chapter 1 CHEMICAL MEASUREMENTS

1-1
SI Units and
Prefixes: SI stands for "Systčme International d"Unités"
=
Fundamental
Units from which all others are derived. Length
(m), mass (kg), time (s), electric current (A), temperature (K), amount
of substance (mole). There are other derived units:
frequency (Hz) = 1/s, force (N) = m^{.}kg/s^{2};
pressure (Pa) = N/m^{2}; energy (J) = kgm^{2}/s^{2};
power
(W)
= J/s; qunatity of electricity (C) = A^{.}s; electric
potential (V) = W/A; electric resistance (ohm) = V/A

1-2
Conversion
Between
Units: ALWAYS use
units!!!!! 1 L = 1000 mL; 1 mL = 1 cc = 1 cm^{3};
1
in.
= 2.54 cm; 1 mm = 1 x 10^{-3} m; 1 µm = 1 x 10^{-6}
m;1 nm = 1 x 10^{-9} m; 1 pm = 1 x 10^{-12} m

Ex.: Convert 0.0123 m into
nm: 0.0123 m (1 nm / 1 x 10^{-9}
m) = 12300000 nm. However, we need to look at the correct number
of digits to be reported in the answer. The correct answer = 1.23
x 10^{7} nm

1-3 Chemical Concentrations

Molarity (M) and Molality (m): M = moles solute / L solution

m = moles solute / kg of solvent

These are very
different!! Will be the same
numerical value when
the density of the solvent is 1.00 g/mL = H_{2}O (because 1000
g = 1 kg = 1000 mL).
BUT, when the density is not 1.00 (all the otehr solvents) ,
then
these values are very different!

Percent Composition: % composition = [# g of analyte / sample wt (g)] x 100

Parts per Million and Parts per Billion:

1 ppm = 1 mg of solute /1000 g solvent = 1 mg of solute / 1 L solvent (usually water)

1 ppm = 1 μg of solute / 1 g solvent = 1 μg of solute / 1 mL of water

1 ppb = 1 μg of solute / 1000 g solvent = 1 μg of solute / 1 L solvent (usually water)

1 ppb = 1 ng of solute / 1 g solvent = 1
ng of solute / 1 mL of water

Ex.: Find the M
of concentrated H_{2}SO_{4} (MW = 98.02 g/mol).
The density of this H_{2}SO_{4} is 1.800 g/mL and it is
98.00% pure.

The easiest way to do this problem is:
Take 1L of H_{2}SO_{4}. = This is 1000
mL. 1000 mL H_{2}SO_{4}
(1.800 g H_{2}SO_{4} / mL H_{2}SO_{4})
(0.9800)(1 mol_{ }H_{2}SO_{4 } / 98.02 g H_{2}SO_{4}_{})
= 18.00 M H_{2}SO_{4
}

Ex.: Find the M
of 42.5 g H_{2}SO_{4} (MW = 98.02 g/mol) dissolved in
enough water to make 500.0 mL of solution. The density of H_{2}SO_{4}
is 1.800 g/mL and it is 98.00% pure.

M = 42.5 g H_{2}SO_{4
}(1 mol_{ }H_{2}SO_{4 } / 98.02 g H_{2}SO_{4})
/ 0.500 L = 0.8672 M H_{2}SO_{4 }

1-4
Preparing
Solutions: NEVER dissolve any solid by putting the solid in a
volumetric flask and then adding solvent to dissolve it in the
flask. You should dissolve the solid in the solvent in a
BEAKER. After it is dissolved, then trnasfer thiw solution
quantitatively into th eflask and bring it up to the mark (BEFORE
putting on the cap and shaking it to make it uniform!!)

1-5 The Equilibrium Constant:

Manipulating Equilibrium Constants

Le Chatelier’s Principle

Chapter 2 TOOLS OF THE TRADE

2-1
**SAFETY, waste Disposal, and Green Chemistry**: eye protection, gloves, ALWAYS use the waste bottles to
get rid of chemicals; do not pour used chemicals down the sink (see TAs
for
help).

2-2
**Your Lab Notebook**:
IMPORTANT!!!!

2-3
**The Analytical Balance**:
Make sure the balance is level!
Reset the “zero” on the balance.
Do NOT use weighing paper unless you are directed to do so. DO NOT ADD ANY CHEMICALS INTO THE WIEGHING
CHAMBER, **EVER!!!!** Take the empty
weighed flask or beaker out of the balance, put sample into it, and
then
re-weigh.

2-4
**Burets**: Make
sure that the buret is CLEAN! Rinse the
buret with ~ 5 mL of the solution, making sure to run it through the
stopcock. Do this twice.
Then fill the buret BEYOND the 0 mark, open
the stopcock full-bore open to push out any entrained air in the
stopcock, then
bring the level of the solution BELOW the 0.00 mL mark and take the
initial
reading.

2-5
**Volumetric Flasks**:
Never try to dissolve a solid in a volumetric flask!! First, dissolve the solid in a small
amount
of distilled water and then quantitatively transfer this to the
volumetric
flask. After the quantitative transfer,
then fill the flask to the mark with distilled water and finally, put
the
ground glass stopper on the flask and invert it and shake it.

2-6
**Pipets and Syringes**:
Always use a bulb for pipetting solutions. Used
syringes
CANNOT
be
thrown
in the basket; they MUST be
handled as though they were medical waste.

2-7 **Filtration**

**
**2-8

have to be handled accordingly. Also, once a sample is dry, then it becomes necessary to keep the sample dry by storing it in a desiccator over calcium chloride (why this material?), only taking the material out of the desiccator to weigh a sample and then

immediately putting it back in to keep it dry.

2-9**
** **Calibration
of
Volumetric
Glassware**

2-10
**Methods of Sample Preparation**:
gases, liquids, solids (ex.: how
sample an oil spill in the ground?),
biological fluids (ex.: when blood was collected, how was it kept? = on
ice in
a cooler?)

Handling the sample: minimize contamination, loss, decomposition, matrix change.

Have to be wary of contamination by:
container, atmosphere, light.
Ex.: SO_{2} often reacts
with container materials.

How measure the sample?

1) Weighing

2) Volume (aliquot = definition)

Analyte = definition

Chapter 3 MATH TOOLKIT

3-1
**Significant Figures**:
All of the known numbers in a measured quantity + the last one
that has
the error. Ex.: the
number
12.3456
has
6
significant digits
= the 1^{st} five of them are absolutely known quantities, and
the last
digit (the 6) is the one with the error.
Usually, the error is taken as ±1;
if it is not that value, it will be given.

3-2**
****Significant Figures in Arithmetic**

**Addition and
Subtraction**: The answer CANNOT have
more decimal places
than the number in the problem with the LEAST number of decimal places.

Ex.: 12.1234 + 1.089 – 0.12 = 13.0924 is the INCORRECT answer => 13.09 is the CORRECT answer, because it has 2 decimal places just like 0.12.

Another ex.: 13 + 1.2345 + 123.21 + 4.23 = 141.6745 is the INCORRECT answer => 142 is the CORRECT answer, because it has NO decimal places just like the 13.

When do you round up or down? Always round up if the digit being dropped is larger than 5. If the digit being dropped is 5, look at the next digit to see if you should round the 5 up or down.

**Multiplication and
Division**: The answer CANNOT have more
significant
figures than the number in the problem with the LEAST number of
significant
figures.

Ex.: 12.123 x 17.23 / 134.6521 = 1.551251633 is the INCORRECT answer => 1.551 is the CORRECT answer, because it has 4 significant figures just like 17.23 has 4 significant figures.

Another ex.: 315.2 mg x 0.9995 / 42.11mL x 74.55 mg/mmol = 0.100354331 M is the INCORRECT answer => 0.1004 M is the CORRECT answer, because it has 4 significant figures just like all of the numbers in the problem.

**Logarithms and
Antilogarithms**: Only use the ** mantissa**
part of
the log (the decimal part) when you consider sig. figs.
The

Ex.: log 3.462 = 0.539327064 is the INCORRECT answer => 0.539 is the CORRECT answer, because it has 3 decimal places just like 0.462.

Ex.:
The **anti**log of 0.1064 = 1.277614884 is the INCORRECT
answer => 1.2776
is
the
CORRECT
answer,
because it has 4 decimal places
just like 0.1064.

3-3**
****Types of Error**

**Gross Error**: negligence = spill material on the counter,
pick it up and put it back in the flask; spill material outside of the
flask
when you are weighing, splash titrant outside of the Erlenmeyer flask
while
titrating.

**Systematic Error**
(determinate error): repeatable if you do
everything the same way over and over again.
Can be discovered and corrected, usually by someone watching
what you
are doing and finding out that you are doing something incorrectly over
and
over again.

0.4436** **±0.0003
g or
0.4436(3) g. This latter is read
as is the previous value where the number in parentheses is known to be
in the
last digit.

**Absolute and
Relative
Uncertainty**: or Absolute and
Relative Error:

ABSOLUTE ERROR: difference between the true value and a measured value

Ex.: Known % Cu in a sample = 6.48%. We measure 6.63%.

Therefore, absolute error = 6.48% - 6.63% = 0.15%

RELATIVE ERROR: the absolute error divided by the true value, expressed in %, ppt, ppm, ppb

Relative
error
in
the
above
measurement = [(6.48% – 6.63%) / 6.48%] x 100 = 2.47% = 2.47 pph

OR = 24.7 ppt = [(6.48% – 6.63%) / 6.48%] x 1000

Ex.: Measure the N of NaOH solution: 0.1023, 0.1026, 0.1034, 0.1027

Average = Xbar = (0.1023 + 0.1026 + 0.1034 + 0.1027) / 4 = 0.1027 N

Relative error (in ppt) = [(high – low) / Xbar] x 1000 = [(0.1034 – 0.1023) / 0.1027] x 1000 = (0.0011 / 0.1027) x 1000 = 11 ppt

3-4
**Propagation of Uncertainty**:
after all systematic errors have been eliminated, there still
are
random errors to be considered.

**Addition and
Subtraction**: the absolute error of the
answer is found
from the absolute errors of each of the values: e(answer) = (e_{1}^{2}
+ e_{2}^{2} + e_{3}^{2} + ...)^{1/2}

0.125 (±0.003) + 1.232(±0.001) – 0.5821(±0.0008) = 0.775 is the answer.

But, what is the __error__
associated with this answer? The
absolute error e = (e_{1}^{2}
+ e_{2}^{2} + e_{3}^{2} + ...)^{1/2}

Therefore: e
= [(0.003)^{2
}+ (0.001)^{2} +
(0.0008)^{2}]^{1/2 }=
(0.00001064)^{ 1/2 }=
0.003261901

e = ±0.003 Therefore, the answer with its error is: 0.775 ± 0.003

This is called __absolute
error__
or absolute uncertainty.

To find the relative error = (absolute error / value) x 100 = (0.003/0.775)x 100 = 0.4% (only 1 sig. fig. because the error only has 1 sig. fig). Therefore: 0.775 ± 0.4 %

**Multiplication and
Division**: the relative error of the
answer is found by
the relative errors of each of the values:
%e(answer) = (%e_{1}^{2}
+ %e_{2}^{2}
+ %e_{3}^{2} + ...)^{1/2}

3.26(±0.03) x 1.893(±0.003) / 0.591(±0.003) x 13.21(±0.02) = 0.790 is the answer

But, what is the __relative
error__
associated with this answer?

The relative error %e(answer)
= (%e_{1}^{2}
+ %e_{2}^{2} + %e_{3}^{2} + ...)^{1/2}

Therefore: %e
= {[(0.03/3.26)x100]^{2 }+ [(0.003/1.893)x100)]^{2} + [(0.003/0.591)x100)]^{2}] +
[(0.02/13.21)x100)]^{2}]} ^{1/2
}= (5.129639)^{ 1/2 }=
±2 %

%e = ±2 % Therefore, the answer with its error is: 0.790 ± 2 %

This is called __relative
error__
or relative uncertainty.

To find the absolute error = (relative error x value) / 100 = 0.790 x (2 / 100) = 0.0158 => 0.016 Therefore: 0.790 ± 0.016 (the error must have same number of decimals as the answer).

For the numerator: answer = 2.4087 => 2.41 with the absolute error:

e = (e_{1}^{2} + e_{2}^{2}
+ e_{3}^{2} + ...)^{1/2} = [(0.002)^{2 }+
(0.02)^{2} + (0.0008)^{2}]^{1/2 }=
(0.00040464)^{ 1/2 }= **±** 0.02012
=> **± **0.02

^{
}

For the relative
error: %e(answer)
= (%e_{1}^{2}
+ %e_{2}^{2} + %e_{3}^{2} + ...)^{1/2
}

= {[(0.02/2.41)x100]^{2 }+ [(0.002/0.678)x100)]^{2} + [(0.02/1.22)x100)]^{2}]}^{1/2
} =
(3.463159)^{ 1/2} = 1.8610
=> ±2 %

And, therefore the absolute error = 2.91 x (2 / 100) = 0.0482 => 0.05

The answer with its
absolute error
is: 2.91 **± **0.05

The answer with its
relative error
is: 2.91 **± **2 %

Ex.: 27.52 (± 0.03) / 29.01 (± 0.02) = 0.948638401 => 0.949 (± 0.1%) RELATIVE ERROR = 0.949 (± 0.001) ABSOLUTE ERROR and NOT = 0.9486 (± 0.0009)

Ex.: 27.58 (± 0.03) / 25.01 (± 0.02) = 1.102759 => 1.103 (± 0.1%) RELATIVE ERROR = 1.103 (± 0.001) ABSOLUTE ERROR

Ex.: 0.1971 (± 0.0005) / 2.007 (± 0.004) = 0.098206 => 0.0982 (± ?) OR 0.09821 (± ?)

= 0.0982 (± 0.20%) RELATIVE ERROR

= 0.0982 (± 0.0002) ABSOLUTE ERROR

3-5
Introducing Spreadsheets: Learn how to use
XL spreadsheets. This will simplify many of the routine
calculations thta you do in the lab, allow you to do plotting and doing
calculations such as finding the average of a series of numbers, the
standard deviation, etc.

3-6
Graphing in Excel: We have to learn how to
use XL for graphing.

RAL011311