Example vibrational frequency calculation

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Introduction to Molecular Spectroscopy

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University of Manchester

Introduction to Molecular Spectroscopy

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The course introduces the three key spectroscopic methods used by chemists and biochemists to analyse the molecular and electronic structure of atoms and molecules. These are UV/Visible , Infra-red (IR) and Nuclear Magnetic Resonance (NMR) spectroscopies. The content is presented using short focussed and interactive screencast presentations accompanied by formative quizzes to probe understanding of the key concepts presented. Numerous exercises are provided to facilitate mastery of each topic. A unique virtual spectroscopic laboratory is made available to enable students to measure and analyse spectra online. Assessment is via summative quizzes completed during the course period.

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Infrared Spectroscopy

In this module we introduce the theory underpinning infrared (IR) spectroscopy and show examples of analysis using the technique. Transitions between the vibrational energy levels of molecules occurs in the infrared region of the electromagnetic spectrum. We start with the theory underlying vibration using the simple harmonic oscillator model. Analysis of more complex molecules is introduced using group frequencies and number of vibrational modes. You will also be shown how to obtain an infrared spectrum and will have an opportunity to run your own spectrum. At the end of this module you are given a link to view how to obtain an infra red spectrum in the laboratory. Don't forget to complete the end of week laboratory quiz which contributes to your final mark for this course.

Infrared spectroscopy13:15

Vibrational frequency12:01

Example vibrational frequency calculation10:08

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Patrick J O'Malley, D.Sc
Reader
Chemistry

Right, so go ahead and just do a,

because we're talking about a lot of new things perhaps to you, and

for some reduced mass, I am going to do a calculation and

it usually helps to make it a bit clearer.

So let me, Move on to a,

right, so we have this example here.

So you have the CO group, as you all know that's part of a peptide bond.

So what you can regard that as isolated from the rest of the molecule.

And we move into this a bit later.

But some CO stretching, vibration, and

most molecules, including peptides, is quite separate.

There's no other motion going on.

So it's not a bad approximation.

And here you've been given that the force constant in a carbonated group is

908 Newtons per meters, and what you've been asked then in the question,

you want to calculate vibration frequency of a, Carbon-12, Oxygen-16.

So that's the natural isotopes,

Carbon-12 is 99% of the carbon.

You also then want to compare that with

the frequency you calculate for carbon-13.

And this is a very simple example this is a common technique in spectroscopy,

you can change the isotopes of certain groups and sometimes what you're

doing in practical in spectroscopy is you're trying to assign a different band.

Sometimes it can be difficult to assign as we'll see later on.

But if you can make the isotopic substitution like this,

what's going to happen is you 're changing the reduced mass for

that vibrator and that will change your frequency.

So it's an often technique used In biological systems

to assign infrared bands to particular groups within a complex module.

So let's just go through this.

So you have to calculate the vibration frequency.

We're going to calculate it in hertz.

I should have said hertz there, we're going to calculate it in hertz.

So we go from the fundamental equation, which you need to remember,

that mu in hertz is equal to one over two pi, square root of K over mu.

And you know now that K is the force constant and Mu is the reduced mass.

Right so let's do the first one first.

A, you have carbon 12,

and it's 16 O.

So we know pretty much everything here.

We know two pi, and then we're given the force constant minus one.

So this is just an exercise really in calculating the reduced mass which

we talked about.

So the reduce mass is m one.

It's the product of the masses divided by the sum.

That will give you grams per mole and then you need to multiply it by this u value,

which is, that's ten minus three times, Avogadro, one over Avogadro's number.

So you have the masses here, you just take the most approximate masses,

so it's 12.01 multiplied by 16.00,

and then you divide it by the sum of these.

So it's 12.01 plus 16.00, and

that'll give it to you in grams per mol, but we were asked to do this is Hertz.

So you need to then multiply that by this u value,

this unified atomic mass unit again, which you will get in tables.

And that is 1.661 x 10 to the -27.

And that could give it to you in kilograms, which is the SI unit of weight.

So you put that into your calculator,

and at least I come out with 1.14 by

10th to the minus 26 kilograms.

Okay?

So, I've calculated my mu, so

I've got everything I want from the above equation into that.

So you just plug it in, so now you say mu is equal

to one over two pi square root of first constant,

which is given in the question as 908 Mutants.

These are -1.

And it's divided by this reduced mass mu,

which is 1.14 by 10 to the -26 kilograms.

Right, so it we want the answer, the answer should be in

seconds minus one.

So you can actually break this down.

You should be able to do this.

So Newton is force.

And if you break it down into fundamental constituents, that's going to be

kilograms, meter, seconds to the minus two.

Force is equal to mass by acceleration.

Mass is in kilograms, acceleration is meters per second squared.

So, that's kilograms, meter,

seconds to minus two, meters to minus one.

So, what you'll find is you'll have,

let's just write this little change to kilograms.

Seconds to minus two.

All right?

So we'll have kilograms to minus two above the line.

We'll have kilograms below the line.

The kilograms will cancel.

We're left with seconds minus two.

We get the square root of seconds minus two is seconds to minus one, so

that's where you get your second minus one.

So as I say when you're doing calculations you always need to be able to quote

the units.

A calculation without units Is meaningless.

And you should really be able to do this type of manipulation with the units.

Okay. So we've worked out, we put that in.

And again, as you plug that into your calculator,

what I get, I get 4.49 by 10 to the 13 seconds minus 1.

So that's your bond that peptide is vibrating.

What that means it's vibrating 4.49 by 10 to so 13 times per second.

So it's pretty quick.

Let's quickly go through the second bet,

the second bet is just pretty much the same.

The second bet was just where you replaced carbon 12 with carbon 13.

So again, you had the oxygen 16 isotope.

So, again, all you had to do was work out the mu for that.

And the mu, of course,

is going to be different, because now you've got the carbon 13.

So, it will be, we'll just go through it quickly,

it's 30.0 by 16.0 mass of the oxygen atom, and

then it's the sum of these at the bottom, and then you multiply by

it by this atomic mass unit which is 1.661 by 10 to the -27.

And that'll give it to you in kilograms.

So work that out, plug it in.

You should get 1.19 x 10

to the -26 kilograms.

So that gives you reduced mass.

You plug that then into your equation,

so you get Nu = 1/2 pi,

square root of 908 Newtons meters minus one,

divided by 1.19 by 10 to the minus 26.

Kilograms this occasion, and if you work that out,

you'll get 4.39 x 10 to the 13 per second mass.

Okay, so what you can see is, it's smaller isn't it,

than the frequency for that is smaller than it is for the 12.

Because you've increased,

that's due to the fact that you've increased the reducement.

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