Quantitative Analysis

Chemistry 201

NC State University

Lecture 4

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 micro-

liters, what is P?

We can estimate the

pressure from the

amount of octane is injected and combusted

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.

From combustion to useful work

First convert from volume to moles:

The change in the number of moles is:

Therefore the pressure is:

The pressure is ~100 atm.

Calculating pressure

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?

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).

Mass analysis

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 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.

Quantitative analysis

Inductively coupled plasma (ICP)

Inductively coupled plasma (ICP)

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.

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.

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) 46.07 g ethanol Similarly the weight percents of hydrogen and oxygen in ethanol are

6 moles H (1.008 g/mole H) 46.07 g ethanol 1 mole O (16.00 g/mole O) 46.07 g ethanol Notice that the sum of the weight percents of all the elements in a compound must equal 100 %.

100 % = 52.14 % % C =

% H = 100 % = 13.13 %

100 % = 34.73 % % O =

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