mass spectrometry - wou

Chapter 13

Mass Spectrometry

Problem 13.1: Consider the two molecules below (molecule X and molecule Y) as analyzed by MS. Both molecules can be described as a system with three possible fragments: fragment A has a molecular mass of 10 amu, fragment B has a mass of 15 amu, and fragment C has a mass of 20 amu. Assume that (1) the charges generated on the fragments are either +1 or +2 and (2) only one bond is cleaved for each fragmentation. List all possible ions that could be generated, and determine which could be distinguished explicitly and which would have m/z similar to other ions. Molecule X: A–B–B–C Molecule Y: C–A–C–B

Molecule X:

Molecular ion: (A-B-B-C)+, m/z = 60 (A-B-B-C)2+, m/z = 30

Fragments: (A-B-B)+ , m/z = 40 (A-B-B)2+ , m/z = 20

C+, m/z = 20 C2+, m/z = 10

(A-B)+, m/z = 25 (A-B)2+, m/z = 12.5

(B-C)+, m/z = 35 (B-C)2+, m/z = 17.5

(B-B-C)+, m/z = 50 (B-B-C)2+, m/z = 25

A+, m/z = 10 A2+, m/z = 5

Molecule Y:

Molecular ion: (C-A-C-B)+, m/z = 65 (C-A-C-B)2+, m/z = 32.5

Fragments: (C-A-C)+ , m/z = 50 (C-A-C)2+ , m/z = 25

B+, m/z = 15 B2+, m/z = 7.5

(C-A)+, m/z = 30 (C-A)2+, m/z = 15

(C-B)+, m/z = 35 (C-B)2+, m/z = 17.5

(A-C-B)+, m/z = 45 (A-C-B)2+, m/z = 22.5

C+, m/z = 20 C2+, m/z = 10

Explicitly unique: both molecular monovalent ions; (A-B-B)+

All the others will be ambiguous because more than one ion with the same m/z is possible.

There are some divalent ions that have unique masses, but divalent ions are very rare, and so

the peaks would be small and would likely be ignored in the analysis

Problem 13.2: The analytical chemist is often faced with having to make either a qualitative or quantitative determination. Given only what we have discussed so far in this chapter, which type of ion source (hard or soft) would you select for qualitative and which would you select for quantitative analysis? Explain your reasoning.

A hard ion source would be best for qualitative analysis. While we might not be able to see a molecular ion peak, we would observe many peaks for different product ions – that is, we would get a picture of the molecular fragments that make up the molecule as a whole, which gives us insight into the identity of the substance. Given only what we have covered so far, a soft ion source would be a logical selection for quantitative analysis. To determine how much of the substance is present, we would generally desire an intense signal, and we can expect to have a strong signal for the molecular ion using a soft ion source.

Problem 13.3: What would be the velocity of a methyl ion (CH3+) if it were accelerated across a

voltage of 5.0 x 102 volts?

Equation 13.3 is relevant here, but we need the mass of the methyl ion in units of kg.

m =15.034 g

mol x

1 kg

1000 g x

mol

6.022 x 1023 ions = 2.4965 x 10-26 kg

Also, recall that the fundamental electronic charge is 1.6022 x 10-19 C, that a coulomb is equivalent to a J/V, and a Joule is equivalent to a kg·m2/s2, so we can represent the fundamental electronic charge as 1.6022 x 10-19 kg·m2/V·s2

v = √2 z e V

m = √

2 (1)(1.6022 x 10−19 kg∙m2

V∙s2⁄ )(5.0 x 102 V)

2.4965 x 10−26 kg = 8.1 x 104 m/s

Problem 13.4: What voltage would be needed to accelerate a tert-butyl dication (C4H92+) to a

velocity of 1.0 × 106 m/s?

v = √2 z e V

m so V =

m v2

2 z e=

(9.484 x 10−26 kg)(1 x 106m

s)2

(2)(2)(1.602 x 10−19 C) = 1.48 x 105 V = 150 kV

Problem 13.5: Demonstrate that the units in Equation 13.4 cancel out properly.

λ = kT

Pd2π√2

m = (J ∙ K−1)(K)

(Pa)(m2)π√2

m = (kg ∙ m2/s2)

(kg/m ∙ s2)(m2)

m = m

Problem 13.6: Rewrite Equation 13.4 to relate λ to concentration of particles in molarity.

We can use the ideal gas law to derive molarity, or

M =n

V=

P

RT then isolate the quantity T/P as seen in Eq. 13.4:

T

P=

1

MR

λ = kT

Pd2π√2 = (

T

P) (

k

d2π√2 ) = (

1

MR) (

k

d2π√2 ) = (

k

MRd2π√2 )

Problem 13.7: Estimate the mean free path of a water molecule (2.8 Å diameter) at standard temperature and pressure.

λ = kT

Pd2π√2 =

(1.38 x 10−23)(273.15)

(101325)(2.8 x 10−10 m)2(π√2)= 1.068 x 10-7 m = 1.1 x 10-7 m

Problem 13.8: Assuming the water molecule in Problem 13.7 is traveling at about 610 m/s, how many collisions per second is it undergoing? In Problem 13.7 we calculated the mean free path of the water to be

λ = kT

Pd2π√2 =

(1.38 x 10−23)(273.15)

(101325)(2.8 x 10−10 m)2(π√2)= 1.068 x 10-7 m

So it will collide, on average, every 1.068x10-7 m it travels.

Also, we are told that in 1 second, it travels 610 m.

1 collision

1.068 x 10−7 m x

610 m

s = 5.71 x 109 collisions per second = 5.7 x 109 collisions/sec

Problem 13.9: If we want the mean free path of a water molecule (2.8 Å diameter) in a CI system to be 0.10 mm, at what pressure would we need to introduce reagent gas? Not given a specific temperature, let’s assume room temperature of 25oC.

λ = kT

Pd2π√2 so P =

kT

Ld2π√2 =

(1.38 x 10−23J/K)(298.15K)

(0.10 x 10−3 m)(2.8 x 10−10 m)2(π√2)= 118.12 Pa = 0.001166 atm

Problem 13.10: Using the pressure you found in Exercise 13.9, estimate the mean free path length of the theoretical organic molecule described in Example 13.2.

λ = kT

Pd2π√2 =

(1.38 𝑥 10−23 𝐽/𝐾)(298.15 𝐾)

(118.12 𝑃𝑎)(5.0 𝑥 10−10 𝑚)2 (𝜋√2) = 3.136 x 10-5 m = 3.1 x 10-5 m

Problem 13.11: The mass spectrometric analysis of the common pesticide DDT (C14H9Cl5) resulted in a base peak of m/z = 352, with the only other major peaks being 317, 235, and 165. Postulate about the type of ionization (EI or CI), the most logical mode of ionization (negative or positive mode); the ionization reaction responsible for the base peak; and the m/z= 317 peak.

Because the base peak is the molecular ion (with 5 x 35Cl) and we only see a few other major peaks for a moderately sized molecule, we can propose that chemical ionization was used. Either positive or negative mode would work for this molecule, but with such a large proportion of its mass and volume made up of chlorine atoms, the most logical choice would be negative mode if it were available. If we were then using negative mode CI, then the reagent gas served as a buffering medium to

moderate the velocity of the electrons from the filament, allowing the molecules to directly capture electrons (via the chlorines): C14H9Cl5 + e- (slowed) C14H9Cl5- The ion at m/z=317 is likely the molecular ion minus one chlorine atom:

C14H9Cl5 + e- C14H9Cl4- + Cl.

Problem 13.12: What is the final velocity of the m/z = 235 ion in Problem 13.11 if an accelerating voltage of 1.3 kV is used in the mass analyzer?

m =235 g

mol x

1 kg

1000 g x

mol

6.022 x 1023 ions = 3.902 x 10-25 kg

v = √2 z e V

m = √

2 (1)(1.6022 x 10−19 kg∙m2

V∙s2⁄ )(1.3 x 103 V)

3.902 x 10−25 kg = 3.267 x 104 m/s = 3.3 x 104 m/s

Problem 13.13: Given that the half-life of the 238U-to-206Pb decay series is 4.47 billion years, how long ago was a zircon crystal formed if the current 238U concentration is 3.97 ppb and the current 206Pb concentration is 224 ppm? (Hint: Recall that radioactive decays follow first-order kinetics.) ** There is an error concentration – the 206Pb concentration should be 2.24 ppb, not 224 ppm ** t1/2 = 4.47 x 109 years 0.693 = kt1/2 = (k)(4.47 x 109 yr) k = 1.5503 x 10-10 yr-1 Concentration of 238U from concentration of Pb:

2.24 𝑛𝑔 𝑃𝑏

𝑔 𝑐𝑟𝑦𝑠𝑡𝑎𝑙 𝑥

𝑛𝑚𝑜𝑙 𝑃𝑏

206 𝑛𝑔 𝑃𝑏 𝑥

1 𝑛𝑚𝑜𝑙 𝑈

1 𝑛𝑚𝑜𝑙 𝑃𝑏 𝑥

238 𝑛𝑔 𝑈

𝑛𝑚𝑜𝑙 𝑈 = 2.488 ppb 238U

Original concentration of 238U: 2.488 ppb + 3.97 ppb = 6.458 ppm 238U ln(A)t = -kt + ln(A)0 ln(3.97 ppb) = -(1.5503 x 10-10 yr-1)·t + ln(6.458 ppb) t = 3.138 x 109 yr = 3.14 billion years Problem 13.14: A common ion formed in the ICP source is Ar2

+. This argon ion dimer might interfere with the analysis of what atomic ion? Ar2

+ has a m/z = 2(39.95) = 79.9 g/mol. An initial glance at the periodic table would seem to indicate that this would interfere with Br+; however, ICPMS detects primarily individual atoms, and no single Br atom has a mass of 79.9 amu. The mass on the periodic table is a weighted average of two isotopes, 79Br and 81Br, each of which has approximately 50% abundance. However, the most abundant isotope of selenium (80Se, with about 50% abundance) has a mass of 79.9 amu. The dimer Ar2

+ might interfere with analysis of selenium.

Problem 13.15: In wastewater analysis, the molecular ions ArCl+ and ClOH+ are not uncommon. These molecular ions might interfere with the analysis of what atomic ions? ArCl+ = 75 (75%) and 77 (25%). This might interfere with 75As (a common pollutant) or 77Se, although 77Se has a very low abundance (0.09%) ClOH+ = 52 (75%) and 54 (25%). This might interfere with 52Cr (another industrial pollutant) or 54Fe (low abundance, 6%) Problem 13.16: What resolution would be necessary in order to resolve peaks from univalent ions having molar masses of 1,058.0 g/mol and 1,058.5 g/mol?

Res = m̅

∆m=

1058.25

0.5 = 2116.5

Problem 13.17: Assuming the resolution of the mass analyzer considered in Problem 13.16 is fixed across the mass spectrum, what two m/z values could be resolved with respect to a mean m/z of 111?

2116.5 = 111

∆m Δm = 0.0524 So we could resolve 111.026 from 110.974 m/z

Problem 13.18: A certain MS uses a magnetic sector mass analyzer having a fixed accelerating voltage of 2.95 kV and a radius of 0.233 m. What magnetic field (in T) must be used to focus a carbon dioxide ion [CO2

+] on the detector?

m

z=

B2 ∙e ∙R2

2 V

(7.308 x 10−26 kg)

(1)=

B2 ∙(1.602 x 10−19C) ∙(0.233m)2

2 (2950 J

C)

B2 = (7.308 x 10−26 kg)(2)(2950(kg ∙

m2

s2 ))

(1.602 x 10−19)C2(0.05429 m2)

B = 0.222 kg

C∙s = 0.22 T

Problem 13.19: If the same mass analyzer as described in Problem 13.18 were used and the accelerating voltage was kept constant, what magnetic field would be required to measure a bis-imidazolium dication, (C7H8N4)2+?

m = 148 g

mol x

kg

1000 g x

mol

6.022 x 1023 ions = 2.458 x 10-25 kg

m

z=

B2 ∙ e ∙ R2

2 V

B2 = (2.458 x 10−25 kg)(2)(2950(kg ∙

m2

s2 ))

(1.602 x 10−19)C2(0.05429 m2)

B = 0.1167 kg

C∙s = 0.12 T

Problem 13.20: Assuming the same parameters described in Problem 13.18, what magnetic field range would you need to scan in order to sweep a m/z range of 40 to 4,000 Da? 40 amu = 40 g/mol

m1 =40 g

mol x

1kg

1000 g x

mol

6.022 x 1023 ions = 6.64 x 10-26 kg

m2 =4000 g

mol x

1kg

1000 g x

mol

6.022 x 1023 ions = 6.64 x 10-24 kg

B12 =

(6.64 x 10−26 kg)(2)(2950(kg ∙m2

s2 ))

(1.602 x 10−19)C2(0.05429 m2)

B1 = 0.212 T = 0.21 T

B22 =

(6.64 x 10−24 kg)(2)(2950(kg ∙m2

s2 ))

(1.602 x 10−19)C2(0.05429 m2)

B2 = 2.12 = 2.1 T

The magnetic field would need to range from 0.21 to 2.1 T. Problem 13.21: What is the magnitude of electrostatic field (E, in V/m) that would be needed in order to impart a curvature (r) of 0.233 m with a fixed accelerating voltage of 2.95 kV?

We use Eq. 13.18 here.

r = 2V

E but given r, we need to calculate E:

E = 2V

E=

2 (2.95 x 103 V)

0.233 m = 25320 V/m = 25.3 kV/m

Problem 13.22: Calculate the kinetic energy of a univalent ion in the spectrometer described in Problem 13.21. We use Eq. 13.19 to relate to kinetic energy:

r = 2(KE)

zeE but solve for KE: KE =

rzeE

2=

(0.233 m)(1)(1.602 x 10−19J/V)(25320V

m)

2

= 4.7 x 10-16 J Problem 13.23: Determine the time of flight for an 800 m/z ion accelerated across 6.5 kV and allowed to drift through a length of 2 m.

tof = L ∙ √m

z ∙

1

2V= (2 m) ∙ √(800

kg

C) ∙

1

2(6500 V)= (2 m)√0.06154

kg

C∙

C

kg ∙m2

s2

tof = 0.496 sec = 0.50 s

Problem 13.24: If the instrument used to measure the 800 m/z ion, as described in Problem 13.23, has a resolution of 15,000, what is the expected peak width for that ion?

R = m̅

∆m ∆m =

m̅

R=

800

15000 = 0.0533 m/z

tof = L ∙ √m

z ∙

1

2V= (2 m) ∙ √(0.0533

kg

C) ∙

1

2(6500 V)= 0.00405 s = 0.0040 s

Problem 13.25: Demonstrate that the units in Equations 13.23 and 13.24 cancel each other out. Equation 13.23:

rc = m

z∙

vxy

B

m = kg

1∙

m/s

T =

𝑘𝑔

𝐶∙

𝑚/𝑠

𝑘𝑔/𝐶∙𝑠 m = m

Equation 13.24:

fc = z

m∙

B

2π

s−1 = C

kg∙

kg/C∙s

1 s-1 = s-1

Problem 13.26: What is the resonance orbital radius of the CH3+ with a vxy of 3.0 x 104 m/s in a

7.035 T magnetic field? What is the precession frequency?

m1 =15 g

mol x

1kg

1000 g x

mol

6.022 x 1023 ions = 2.49 x 10-26 kg

rc = m

z∙

vxy

B=

2.49 𝑥 10−26𝑘𝑔

1 𝐶∙

3.0 𝑥 104𝑚/𝑠

7.035 𝑇 = 1.062 x 10-22 m = 1.1 x 10-22 m

fc = z

m∙

B

2π=

1 𝐶

2.49 𝑥 10−26 𝑘𝑔 ∙

7.035 𝑇

2𝜋 = 4.497 x 1025 s-1 = 4.5 x 1025 s-1

Problem 13.27: To avoid collision with the cell walls, what is the maximum pulse width for a 2 V excitation signal that could be used for a 2 cm cell in a 7.035 T magnetic field? We can use Eq. 13.25 to find the excitation pulse time required to achieve a resonance radius of 2 cm.

rexcited = Vp−p ∙texcite

2 ∙d ∙B so texcite =

2 ∙d ∙B∙r

Vp−p=

2(0.002 𝑚)(7.035 𝑘𝑔

𝐶∙𝑠)(0.002 𝑚)

2 𝐽/𝐶

= 2.814 𝑥 10−5 𝑚2∙

𝑘𝑔

𝑠𝑘𝑔∙𝑚2

𝑠2

= 2.8 x 10-5 s = 28 μs

EXERCISE 13.1: Consider the electron ionization source depicted in Figure 13.4. What do you think would be the effect of decreasing the voltage between the filament and the target? What effect would using an unheated filament have? Decreasing the voltage difference between the filament and target would decrease the energy with which electrons would impact gas state molecules that enter the source. Furthermore, the electrons would have a more random path distribution. If the filament were unheated, there would be no thermally desorbed electrons – the electron flux would diminish dramatically. EXERCISE 13.2: What would be the velocity of an imidazolium ion (C3H4N2

+) if it were accelerated across a voltage of 22.3 kV?

m = 148 g

mol x

kg

1000 g x

mol

6.022 x 1023 ions = 2.458 x 10-25 kg

v = √2 z e V

m = √

2 (1)(1.6022 x 10−19 kg∙m2

V∙s2⁄ )(22.3 x 103 V)

2.458 x 10−25 kg = 1.705 x 105 m/s = 1.71 x 105 m/s

EXERCISE 13.3: What voltage would be needed to accelerate an imidazolium ion (C3H4N2

+) to a velocity of 4.2 × 105 m/s?

m = 148 g

mol x

kg

1000 g x

mol

6.022 x 1023 ions = 2.458 x 10-25 kg

v = √2 z e V

m 𝑉 =

𝑣2𝑚

2𝑧𝑒=

(4.2 𝑥 105 𝑚/𝑠)2(2.458 𝑥 10−25𝑘𝑔)

(2)(1)(1.6022 x 10−19 kg∙m2

V∙s2⁄ ) = 135.3 V = 140 V

EXERCISE 13.4: Given that the half-life of the 235U-to-207Pb decay series is 704 million years, how long ago was a zircon crystal formed if 1.0 g of the crystal contains 4.32 ng of 235U and 1.92 ng of 207Pb? t1/2 = 704 x 106 years 0.693 = kt1/2 = (k)(704 x 106 yr) k = 9.844 x 10-10 yr-1 Concentration of 235U from concentration of Pb:







235 𝑛𝑔 𝑈

𝑛𝑚𝑜𝑙 𝑈 = 2.1797 ppb 235U

Original concentration of 235U: 2.1797 ppb + 4.32 ppb = 6.4997 ppm 235U ln(A)t = -kt + ln(A)0 ln(4.32ppb) = -(9.844 x 10-10 yr-1)·t + ln(6.4997 ppb) t = 4.1497 x 108 yr = 415 million years EXERCISE 13.5: What would the age of the zircon crystal in Exercise 13.4 be if the amounts of 235U and 207Pb were reversed? t1/2 = 704 x 106 years 0.693 = kt1/2 = (k)(704 x 106 yr) k = 9.844 x 10-10 yr-1 Concentration of 235U from concentration of Pb:







235 𝑛𝑔 𝑈

𝑛𝑚𝑜𝑙 𝑈 = 4.904 ppb 235U

Original concentration of 235U: 4.904 ppb + 1.92 ppb = 6.824 ppm 235U ln(A)t = -kt + ln(A)0 ln(1.92ppb) = -(9.844 x 10-10 yr-1)·t + ln(6.824 ppb) t = 1.288 x 109 yr = 1.29 billion years EXERCISE 13.6: Explain the function of the quadrupole ion deflector seen in Figure 13.15. The main function of the QID is to eliminate interferences. The QID has four quadrant electrodes. The ones on the right in the figure have a positive potential applied to them, while the ones on the left have a negative potential applied. Positive ions are drawn toward the left and guided between the two negative poles. Negative ions are directed to the right and so do not interfere with the detector. UV photons travel straight through the QID, exiting between the top two poles, and thus are also removed so that they cannot affect detection of positive ions.

EXERCISE 13.7: What is the mass of a univalent ion if it acquired a velocity of 5.5 × 105 m/s when accelerated across a potential of 1,250 V?

v = √2 z e V

m

𝑚 = 2𝑧𝑒𝑉

𝑣2 = (2)(1)(1.6022 𝑥 10−19

𝑘𝑔∙𝑚2

𝑉∙𝑠2⁄ ) (1250𝑉)

(5.5 𝑥 105𝑚

𝑠)2

= 1.324 x 10-27 kg

EXERCISE 13.8: Which method of ionization would you suggest for each of the following samples? Explain your reasoning. (a) Dioxin, a byproduct of the bleaching of paper by chlorine (b) Horseradish peroxidase (c) Steel shavings (a) While positive ion EI would work for dioxin, the fact that electronegative elements make up a significant volume of the molecule indicate that NICI would be a useful method for

quantitative purposes. (b) Horseradish peroxidase is an enzyme with a large molar mass. Either MALDI or ESI would be appropriate for ionization of this sample. (c) Steel shavings by themselves could not be introduced into a MS using methods discussed in this chapter. However, if the samples were digested into an aqueous solution, then an appropriate method for the analysis of the metals in the steel would be ICP-MS. EXERCISE 13.9: Naphthalene, a polyaromatic hydrocarbon (C10H8), was analyzed by MS using chemical ionization and a quadrupole mass analyzer. The base peak occurred at m/z = 129, and the largest ion mass in the spectrum was seen at m/z = 156. Explain these observances. The base peak has a mass one amu greater than the formula mass of the parent molecule. This is probably due to proton transfer to the gas phase molecule via a process like this: C10H8 (g) + C2H5

+ (g) C10H9+ (g) + C2H4 (g)

The ion seen at 156 is probably due to the formation of an adduct species between the parent ion and C2H4.

EXERCISE 13.10: To what excited state radius will ions in an ICR chamber be moved if a 2.5 V radio frequency signal is applied for 500 μs across a 2.3 cm cell while a 4.71 T field is applied?


2 ∙d ∙B=

(2.5 𝑘𝑔∙𝑚2

𝑠2𝐶⁄ )(500 𝑥 10−6𝑠)

(2)(0.0023 𝑚)(4.71 𝑘𝑔

𝐶∙𝑠⁄ ) = 0.00578 m

= 5.8 cm (this would exceed the size of the cell)

EXERCISE 13.11: What is the maximum time the pulse described in Exercise 13.10 could be applied to avoid having ions collide with the cell walls?


2 ∙d ∙B so texcite =

2 ∙d ∙B∙r

Vp−p=

2(0.0023 𝑚)(4.71 𝑘𝑔

𝐶∙𝑠)(0.0023 𝑚)

2.5 𝐽/𝐶

= 1.993 x 10-5 s = 19.9 μs EXERCISE 13.12: What is the maximum voltage that could be applied for 500 μs in the cell described in Exercise 13.10 to avoid having ions collide with the cell walls?


2 ∙d ∙B so Vp−p =

2 ∙d ∙B∙r

texcite=

2(0.0023 𝑚)(4.71 𝑘𝑔

𝐶∙𝑠)(0.0023 𝑚)

(500 𝑥 10−6 𝑠)

= 0.09966 V = 99.7mV EXERCISE 13.13: Your ICM-MS spectrometer has a resolution of 175 when the average mass of univalent ions is around 120 Da. Would the molecular ion Ar–Br+ interfere with the analysis of any elemental ions? Explain your answer. There are two possible masses, of approximately equal probability, that Ar-Br+ can take. 40Ar-79Br = 119 amu 40Ar-81Br = 121 amu

R = m̅

∆m 175 =

120

∆m ∆m = 0.686 Da

We could fully resolve a unique peak from the 40Ar-79Br+ signal at (119-0.686) = 118.314 and at 119.686 Da. Two ions, 118Sn (24% abundant) and 119Sn (8% abundant) could see interference from the 40Ar-79Br+ molecular ion. We could fully resolve a unique peak from the 40Ar-81Br+ signal at 120.314 and 121.686 Da. The 120Sn (32% abundant) and 121Sb (57% abundant) might see interference. EXERCISE 13.14: What resolution would be necessary in order to resolve peaks from divalent ions having molar masses of 1,000 g/mol and 1,001 g/mol? (m/z)1 = 1000/2 = 500 (m/z)2 = 1001/2 = 500.5 Average m/z = 500.25

R = m̅

∆m=

500.25

(500.5−500) = 1000.5

EXERCISE 13.15: Assuming the resolution of the mass analyzer considered in Exercise 13.14 is constant across the mass spectrum, what two m/z values could be resolved if around a mean m/z of 2.5 × 103?

R = m̅

∆m 1000.1 =

2.5 x 103

∆m Δm = 2.499

We could resolve (2500+2.499) from (2500 – 2.499), or 2502.5 from 2497.5 EXERCISE 13.16: Estimate the resolution of the mass analyzer used to obtain the spectrum seen in Figure 13.2(a). We will use the base peak to estimate the resolution. It appears that we can see the base peak at 68 and another one at 69.

R = 68.5

(69−68) = 68.5

EXERCISE 13.17: Estimate the mean free path of an ethanol molecule (0.44 nm diameter) at 25oC and 1.0 torr.

λ = kT

Pd2π√2 =

(1.38 𝑥 10−23 𝐽/𝐾)(298.15 𝐾)

(133.3 𝑃𝑎)(0.44 𝑥 10−9 𝑚)2 (𝜋√2) = 3.588 x 10-5 m = 3.6 x 10-5 m

EXERCISE 13.18: Assuming the ethanol molecule in Exercise 13.17 is traveling at about 555 m/s, how many collisions per second is it undergoing? In Exercise 13.17 we calculated the mean free path of the ethanol to be

λ = kT

Pd2π√2 =

(1.38 𝑥 10−23 𝐽/𝐾)(298.15 𝐾)

(133.3 𝑃𝑎)(0.44 𝑥 10−9 𝑚)2 (𝜋√2) = 3.588 x 10-5 m = 3.6 x 10-5 m

So it will collide, on average, every 3.588x10-5 m it travels.

Also, we are told that in 1 second, it travels 555 m.

1 collision

3.588 x 10−5 m x

555 m

s = 1.55 x 107 collisions per second = 1.6 x 107 collisions/sec

EXERCISE 13.19: If we want the mean free path for an ethanol molecule in a CI chamber to be 0.098 mm, what pressure would be required in the chamber?

λ = kT

Pd2π√2 so

P = kT

λd2π√2 =

(1.38 𝑥 10−23 𝐽/𝐾)(298.15 𝐾)

(0.098 𝑥 10−3 𝑚)(0.44 𝑥 10−9 𝑚)2 (𝜋√2) = 48.81 Pa = 4.8 x 10-4 atm or 0.37 Torr

Advanced Exercises EXERCISE 13.20: Given the NMR and mass spectra shown, determine the molecular formula and structure of the compound.

NMR: triplet at 1.4 ppm (adjacent to CH2) doublet at 6.7, doublet at 7.8

(symmetrically di-substituted aromatic) quartet at 4.3 (adjacent to CH3), well downfield (possibly ester) possible amine or hydroxide at 4.2 MS: base peak = 120 molecular ion = 165 Peaks at ~136, 92, 65 (165-136) = 29 [poss. C2H5 or CHO] (165-120) = 45 CH3CH2-O 92 C6H5-CH2+H 65 C5H5

(commonly indicates benzene ring) Indicates benzocaine:

EXERCISE 13.21: Given the NMR and mass spectra shown, determine the molecular formula and structure of the compound.

NH2

O

O

NMR: complex multiplet at 5.8 and complex triplet at 5.1: probably R1-CH=CH-R2 multiplets at 1.4 and 2.2 possibly indicate an alkane chain

MS: Base peak: 56 (C4H8) Parent peak: 84 (30% of base peak height) Major peaks at 41 (95%) C2H3N 27 (32%) C2H3 69 (24%) C4H7N

H2N – CH = CH – CH2 CH2 – CH3

mass spectrometry - wou

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