Chemistry 201. Quantitative Analysis


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1 Chemistry 201 Lecture 4 Quantitative Analysis NC State University
2 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
3 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 H 2 O and CO 2 according to the reaction stoichiometry.
4 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.
5 Mass analysis of a hydrocarbon A 10 gram sample of a hydrocarbon unknown composition is combusted to yield grams of H 2 O and grams CO 2. What is the hydrocarbon?
6 Mass analysis A 10 gram sample of a hydrocarbon unknown composition is combusted to yield grams of H 2 O and grams CO 2. 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.
7 Mass analysis The moles of products are: and the ratio is: From the known stoichiometry we know that:
8 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 C n H 2n+2 is C 8 H 18 (octane).
9 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 (acidbase)
10 Substance Stoichiometry How are materials quantified? How are chemical formulas determined?
11 Counting atoms and molecules How many individual H atoms are contained in 10.0 g of CH 3 OH?
12 Counting atoms and molecules How many individual CH 3 OH molecules are contained in 10.0 g of CH 3 OH? Solution: 1. Determine the molar mass of CH 3 OH. M m = = 32 grams/mole. 2. Convert to moles: n = m/ M m = (10 grams)/(32 grams/mole) = moles This corresponds to 1.88 x molecules.
13 Counting atoms and molecules What mass of CH 3 OH contains 1.75 mol of H atoms? Solution: The mole ratio of H: CH 3 OH is 4:1 so the number of moles of CH 3 OH is The mass is m = nm m = ( moles)(32 grams/mole) = 14 grams
14 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.
15 Inductively coupled plasma (ICP)
16 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.
17 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.
18 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.
19 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 C 2 H 6 O. One mole of ethanol has a mass of g. The elemental formula indicates that one mole of ethanol contains two moles of carbon, six moles of hydrogen, and one mole of oxygen.
20 Composition by mass Thus the composition of the compound by mass is 2 moles C (12.01 g/mole C) % C = g ethanol 100 % = % Similarly the weight percents of hydrogen and oxygen in ethanol are 6 moles H (1.008 g/mole H) % H = g ethanol 100 % = % 1 mole O (16.00 g/mole O) % O = 100 % = % g ethanol Notice that the sum of the weight percents of all the elements in a compound must equal 100 %.
21 Determining molecular formula Acetylene and benzene have the simplest formula CH. Their molar masses are 26 and 78 respectively. What are their molecular formulae?
22 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 Divide the molar masses by the formula mass. Acetylene: 26/13 = 2 The chemical formula C 2 H 2 Benzene: 78/13 = 6 The chemical formula C 6 H 6
23 Determining the molecular formula Nicotine is 74.03% C, 8.70% H, and 17.27% N. Its M m is 162. What is the molecular formula?
24 Determining the molecular formula Nicotine is 74.03% C, 8.70% H, and 17.27% N. Its M m is 162. What is the molecular formula? Solution: 1. Multiply the molar mass by each percentage. C: x 162 = H: x 162 = 14 N: x 162 = 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 C 10 H 14 N 2
25 Beer s law We can use the absorption of light by chemical compounds to determine their concentration in solution. I 0 I
26 Beer s law Light is attenuated exponentially in the solution: The attenuation factor is A the absorbance. We can also write
27 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.
28 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 1 cm 1 at 560 nm. What concentration of Rhodamine is required?
29 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 1 cm 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.
30 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.
31 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.
32 Fluorescently labeled cell
33 Green fluorescent protein Originally derived from a jellyfish
34 Applications of GFP
35 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.
36 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.
37 Solubility rules One can use the solubility rules: 1. Compounds of NH 4 + and group 1A metal ions are soluble. 2. Compounds of NO 3, ClO 4, ClO 3  and C 2 H 3 O 2  are soluble. 3. Compounds of Cl , Br  and I  are soluble, except those of Ag +, Cu +, Tl +, Hg 2+ and Pb Compounds of SO 4 2 are soluble, except those of Ca 2+, Sr 2+, Ba 2+ and Pb Most other ionic compounds are insoluble.
38 Example What is the silver ion concentration in a solution of AgNO 3 if the addition of an excess of K 3 PO 4 to 50.0 ml of the AgNO 3 solution produces g of precipitate?
39 Example What is the silver ion concentration in a solution of AgNO 3 if the addition of an excess of K 3 PO 4 to 50.0 ml of the AgNO 3 solution produces 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.
40 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.
41 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.
42 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.
43 Solubility One can measure the spectrum and calculate the concentration of Pd 2 (dba) 3 in various solvents to determine the limit of solubility. This organometallic Compound is soluble in THF and insoluble in water.
44 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.
45 Titrations Stoichiometric amount of reactant (titrant) is added to a known volume of an analyte. One standard type is an acidbase 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.
46 Endpoint indicator In an acidbase 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.
47 Definitions ph is minus log to base 10 of [H + ]. ph = log 10 ([H + ]) Recall that K w = [H + ][OH  ] = 1014, so pk w =? We can also define poh = log 10 ([OH  ])
48 Definitions K w = [H + ][OH  ] = and pk w = 14 Therefore, pk w = 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.
49 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 + ].
50 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 104 M x 0.02 L = 2 x 106 moles. At the equivalence point the volume is 120 ml and: [H + ] = [OH  ] = 2 x 106 moles/0.12 L = 1.67 x 105 M ph = log([h + ]) = log(1.67 x 105 ) = 4.78
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