Chemistry 201
Lecture 4
Quantitative Analysis
NC State University
Focus on energy
The work done in the internal combustion
engine is called pressure volume work.
For a simple irreversible
stroke the work is:
work = P DV
In a 3.0 L 6 cylinder
engine DV = 0.5 L.
Assuming the initial
volume is 80 microliters, what is P?
We can estimate the
pressure from the
amount of octane is injected and combusted
From combustion to useful work
A typical fuel injector will inject 12 microliters of
octane fuel per stroke. What is the pressure of
the gas created by combustion in a volume of 80
microliters?
In our first examination of this problem we will
ignore the heating and simply calculate the
conversion from the volume of liquid fuel to the
volume of H2O and CO2 according to the reaction
stoichiometry.
Calculating pressure
First convert from volume to moles:
The change in the number of moles is:
Therefore the pressure is:
The pressure is ~100 atm.
Mass analysis of a hydrocarbon
A 10 gram sample of a hydrocarbon unknown
composition is combusted to yield 14.21 grams of
H2O and 30.87 grams CO2. What is the
hydrocarbon?
Mass analysis
A 10 gram sample of a hydrocarbon unknown
composition is combusted to yield 14.21 grams of
H2O and 30.87 grams CO2. What is the
hydrocarbon?
Solution: since it is a hydrocarbon, we know that
the stoichiometry follows:
We need to convert the grams of the products to
moles and then use the molar ratio to determine
n.
Mass analysis
The moles of products are:
and the ratio is:
From the known stoichiometry we know that:
Mass analysis
A more common solution would be to find the
molar ratios of the atoms:
We know that the ratio of H to C in a hydrocarbon
is:
With some algebra we find:
The unknown is CnH2n+2 is C8H18 (octane).
Goals
• Quantify compounds by mass/mol/number
• Quantify compounds by percent
• Find chemical formulas from mass data
• Apply Beer’s law to determine concentration
• Introduce titrations (acid-base)
Substance Stoichiometry
• How are materials quantified?
• How are chemical formulas determined?
Counting atoms and molecules
How many individual H atoms are contained in 10.0 g
of CH3OH?
Counting atoms and molecules
How many individual CH3OH molecules are contained
in 10.0 g of CH3OH?
Solution: 1. Determine the molar mass of CH3OH.
Mm = 12 + 16 + 4 = 32 grams/mole.
2. Convert to moles:
n = m/ Mm = (10 grams)/(32 grams/mole) = 0.3125 moles
This corresponds to 1.88 x 1023 molecules.
Counting atoms and molecules
What mass of CH3OH contains 1.75 mol of H atoms?
Solution: The mole ratio of H: CH3OH is 4:1 so the
number of moles of CH3OH is
The mass is
m = nMm = (0.4375 moles)(32 grams/mole) = 14 grams
Quantitative analysis
Quantitative analysis is the determination of the amount by
weight of each element or compound present.
Gravimetry, where the sample is dissolved and then the
element of interest is precipitated and its mass measured
or the element of interest is volatilized and the mass loss
is measured.
Optical atomic spectroscopy, such as flame atomic
absorption, graphite furnace atomic absorption, and
inductively coupled plasma atomic emission, which
probe the outer electronic structure of atoms.
Inductively coupled plasma (ICP)
Inductively coupled plasma (ICP)
Diagram of the principle of detection in an
ICP emission spectrometer.
ICP can also be coupled to a mass
Spectrometer.
The resulting signals give rise to a
Percentage for each element detected.
1. Take the percentage of each element found and divide by the element's mass.
2. Do this for all the elements for which you have results
3. Find the smallest value from step 1 and divide every value obtained in step 1
by this smallest value
4. Multiply the results in step 3 by a factor to obtain reasonable values for either
carbon or nitrogen and then compare to what was expected from a pure
sample of the compound.
Mass spectrometry
Requires charged molecules. Various methods used to ionize a sample.
Accelerate using electric field and deflect using a magnetic field. Analyze
based on amount of deflection (shown above) or time of flight.
Fragmentation pattern
Information on the molecular structure can be obtained from the fragmentation
pattern of a molecule. Once ionized the molecule is not stable and tends
further decompose into smaller ions. The study of this fragmentation provides
Information on molecules structure.
Composition by mass
• The composition of a compound is often expressed
in terms of the weight percent of each element in
the compound.
• For example, ethanol has the formula C2H6O. One
mole of ethanol has a mass of 46.07 g. The elemental
formula indicates that one mole of ethanol contains
two moles of carbon, six moles of hydrogen, and one
mole of oxygen.
Composition by mass
• Thus the composition of the compound by mass is
2 moles C (12.01 g/mole C)
%C=
100 % = 52.14 %
46.07 g ethanol
Similarly the weight percents of hydrogen and
oxygen in ethanol are
6 moles H (1.008 g/mole H)
%H=
100 % = 13.13 %
46.07 g ethanol
1 mole O (16.00 g/mole O)
%O=
100 % = 34.73 %
46.07 g ethanol
Notice that the sum of the weight percents of all the
elements in a compound must equal 100 %.
Determining molecular formula
Acetylene and benzene have the simplest formula CH.
Their molar masses are 26 and 78 respectively. What
are their molecular formulae?
Determining molecular formula
Acetylene and benzene have the simplest formula CH.
Their molar masses are 26 and 78 respectively. What
are their molecular formulae?
Solution:
1. Calculate the formula mass of the simplest formula.
CH has a mass of 13
2. Divide the molar masses by the formula mass.
Acetylene: 26/13 = 2
The chemical formula C2H2
Benzene: 78/13 = 6
The chemical formula C6H6
Determining the molecular formula
Nicotine is 74.03% C, 8.70% H, and 17.27% N.
Its Mm is 162. What is the molecular formula?
Determining the molecular formula
Nicotine is 74.03% C, 8.70% H, and 17.27% N.
Its Mm is 162. What is the molecular formula?
Solution:
1. Multiply the molar mass by each percentage.
C: 0.7403 x 162 = 119.9
H: 0.087 x 162 = 14
N: 0.1727 x 162 = 27.9
2. Divide each resulting fractional mass by the atomic mass of each element.
C: 119.9/12 = 10
H: 14/1 = 14
N: 27.9/14 = 2
Note that these values must be rounded to the nearest integer value.
The molecular formula is C10H14N2
Beer’s law
We can use the absorption of light by chemical
compounds to determine their concentration in
solution.
I0
I
Beer’s law
Light is attenuated exponentially in the solution:
The attenuation factor is A the absorbance. We can
also write
Absorbance
Absorbance depends linearly on concentration, c, and
on path length, d:
The quantity is the extinction coefficient. It is a
measure of the ability of a particular molecule to
absorb light.
Rhodamine concentration
Rhodamine is a laser dye. To get the laser to run you
need an absorbance of 0.5 in a jet that is 100 microns
thick. For Rhodamine = 55,000 M-1cm-1 at 560 nm.
What concentration of Rhodamine is required?
Rhodamine concentration
Rhodamine is a laser dye. To get the laser to run you
need an absorbance of 0.5 in a jet that is 100 microns
thick. For Rhodamine = 55,000 M-1cm-1 at 560 nm.
What concentration of Rhodamine is required?
Solution:
1. Use Beer’s law,
Determine which quantity is the unknown and solve for it. Here we know
everything except the concentration, c.
Measuring the
extinction coefficient
The extinction coefficient can be measured by making
solutions of various masses of a compound dissolved in
a solvent. Then the absorption spectra of each are
determined. A plot of absorbance vs. concentration
should be a straight line with a slope equal to .
Note that the units of are M-1 cm-1.
Fluorescence
Fluorescence is the process of emission of light from an
excited state created by absorption. Fluorescence is
very sensitive, and can even be measured from single
molecules. A key parameter is the fluorescence
quantum yield, which gives the efficiency of emission
following excitation.
Fluorescently labeled cell
Green fluorescent protein
Originally derived from a jellyfish
Applications of GFP
Gravimetric analysis
One can determine the mass of an ion in solution
by causing a precipitation. The precipitant can then be
weighed to provide a measurement of the mass.
Gravimetric analysis
One can determine the mass of an ion in solution
by causing a precipitation. The precipitant can then be
weighed to provide a measurement of the mass.
A balance.
Solubility rules
One can use the solubility rules:
1. Compounds of NH4+ and group 1A metal ions are
soluble.
2. Compounds of NO3-, ClO4-, ClO3- and C2H3O2- are
soluble.
3. Compounds of Cl-, Br- and I- are soluble, except
those of Ag+, Cu+, Tl+, Hg2+ and Pb2+.
4. Compounds of SO42- are soluble, except those of
Ca2+, Sr2+, Ba2+ and Pb2+.
5. Most other ionic compounds are insoluble.
Example
What is the silver ion concentration in a solution of
AgNO3 if the addition of an excess of K3PO4 to 50.0 mL
of the AgNO3 solution produces 0.3634 g of
precipitate?
Example
What is the silver ion concentration in a solution of
AgNO3 if the addition of an excess of K3PO4 to 50.0 mL
of the AgNO3 solution produces 0.3634 g of
precipitate? (Ag = 107.8; P = 30.9; K = 39.0)
Solution: Step 1 Determine the identity of the precipitant.
Step 2. Write a balanced equation.
Step 3. Calculate the number of moles of the precipitant.
Step 4. Determine the initial concentration.
Volumetric flasks
Use volumetric flasks when accurate solution volumes are
desired. Weighed components added to the precise volume
provide a means to obtain accurate concentrations.
Solubility
Compounds that are soluble in water are often
insoluble in organic solvents and vice versa.
To measure the solubility you first create a saturated
solution. Then centrifuge that solution and measure
the amount of the analyte in the supernatant.
Solubility
Compounds that are soluble in water are often
insoluble in organic solvents and vice versa.
to measure the solubility you first create a saturated
solution. Then centrifuge that solution and measure
the amount of the analyte in the supernatant.
Solubility
One can measure the spectrum and calculate the
concentration of Pd2(dba)3 in various solvents to
determine the limit of solubility. This organometallic
Compound is soluble in THF and insoluble in water.
Solubility
Analysis can use absorption spectrometry. However,
to determine the calculation using the absorption
spectrum one needs the extinction coefficient.
From an undergraduate laboratory at Dartmouth College.
Titrations
Stoichiometric amount of reactant (titrant) is added to
a known volume of an analyte. One standard type is an
acid-base titration. In this case, the titrant is often
dispensed using a buret.
The stopcock of the buret
can be opened to permit a
known volume to flow into
the analyte solution.
Endpoint indicator
In an acid-base titration one uses an indicator dye to
give a visual signal that the titration end point has
been reached. For example, a frequent application is
the neutralization of an acid or base. In that case, the
endpoint occurs when the solution has reached pH 7.
Colorless solution.
Equivalence (pink) Gone too far.
Definitions
pH is minus log to base 10 of [H+].
pH = -log10([H+])
Recall that Kw = [H+][OH-] = 10-14, so pKw = ?
We can also define
pOH = -log10([OH-])
Definitions
Kw = [H+][OH-] = 10-14 and pKw = 14
Therefore, pKw = pH + pOH
This is useful if we are given pH and we wish to
calculate [OH-].
Of course, when a solution is neutral [H+] = [OH-]
or pH = pOH = 7.
Detecting acidity in lakes
Lakes in Sweden that are on granite have little buffering
capacity, and are therefore often acidic. An water
quality chemist will titrate the solution to determine
the [H+] concentration. If 20 mL of a pH 10 solution is
added to 100 mL of lake water before the equivalence
point is reached calculate [H+].
Detecting acidity in lakes
Lakes in Sweden that are on granite have little buffering
capacity, and are therefore often acidic. An water
quality chemist will titrate the solution to determine
the [H+] concentration. If 20 mL of a pH 10 solution is
added to 100 mL of lake water before the equivalence
point is reached calculate [H+].
Solution: The number of moles of [OH-] is 10-4 M x 0.02 L = 2 x 10-6 moles.
At the equivalence point the volume is 120 mL and:
[H+] = [OH-] = 2 x 10-6 moles/0.12 L = 1.67 x 10-5 M
pH = -log([H+]) = -log(1.67 x 10-5) = 4.78