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Signal characteristics - wavenumber

Using equations to predict the wavenumber for different types of chemical bonds. Created by Jay.

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  • blobby green style avatar for user Christopher Truszkowski
    At , it is mentioned that the k value for the single bond between C - O will be 5x10^5, which was the same k value for the bond C - H. Could anyone tell me where this number was taken from? I understand that a double bond would be double the constant but where does the original approximation come from?
    (20 votes)
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    • aqualine ultimate style avatar for user JLD
      The approximation comes from modelling the bonds as a spring: https://www.khanacademy.org/science/organic-chemistry/spectroscopy-jay/infrared-spectroscopy-theory/v/bonds-as-springs

      The amount of energy that a spring can hold varies depending on the type of spring. To account for this, springs are multiplied by a constant value.

      The potential energy stored in a spring is given by the equation:

      U=½kx^2

      where U is equal to potential energy, x is displacement, and k is the spring constant.

      k = "stiffness" of the spring.

      Stiffness describes the rigidity of an object. A double bond is stiffer than a single bond, thus k increases with increasing bond order.

      References
      single bond between carbon and oxygen, k = 5 x 10^5 dyne/cm
      double bond between carbon and oxygen, k =10 x 10^5 dyne/cm
      triple bond between carbon and oxygen, k = 15 x 10^5 dyne/cm.
      (15 votes)
  • blobby green style avatar for user Tina S
    at , why do we divide by avagadro's number to turn grams to amu? If we have grams written as k/grams (spring constant/reduced mass) then shouldn't we multiply by 1gram/6.022 X 10^23 amu to cancel out the grams and be left with only amu?
    (11 votes)
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    • piceratops seedling style avatar for user fguo
      I had the same question before, so I watched this and the last video several times. Then I noticed he did make a mistake, but it is not about dividing or multipling the Avogadro number. On the last video https://www.khanacademy.org/test-prep/mcat/physical-processes/infrared-and-ultraviolet-visible-spectroscopy/v/bonds-as-springs at , he said that the unit of the reduced mass is amu, not gram. However, he said it is gram in this video by mistake. Actually what he intended to do is to convert amu to gram, which requires dividing the Avogadro number. That makes sense because the unit of K is N/cm, N is g*cm/s^2, so the actual unit of K is g/s^2, and if we convert the unit of M from amu to gram, the only thing left inside the square root is 1/s^2. finally, the unit of 1/c is s/cm, so we could get the correct unit for the wave-number which is 1/cm.
      (7 votes)
  • blobby green style avatar for user Rose Haft
    why do we use the reduced mass? Perhaps I missed it, but how is this derived? Thank you!
    (5 votes)
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  • blobby green style avatar for user Hanan Abuzeineh
    How can we know the value of K the force constant?
    (6 votes)
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  • blobby green style avatar for user Rose Haft
    Where can we get more ks for various different bonds?
    (6 votes)
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  • piceratops seed style avatar for user Zikesha  Bellamy
    The speed of light in this video was recorded as 3x10^10 instead of 3x10^8.
    (3 votes)
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  • male robot hal style avatar for user the ULTIMATE GEEK
    where is the region for a double bond to an atom that's not carbon?
    (3 votes)
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  • piceratops sapling style avatar for user anuj.prasad27
    why use wavenumber instead of wavelength ?
    (3 votes)
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    • primosaur seedling style avatar for user dipakpatil888
      As we look for the electromagnetic radiation spectrum and compared different regions of it with respect to energy, frequency, and wavelength.
      Then, the energy, frequency both are directly proportional to each other whereas energy, frequency both are inversely proportional to wavelength.
      e.g. In the case of UV-Visible spectroscopy, at 200 nm (shorter wavelength) the energy and frequency of UV-Visible radiation are high as compared to 800 nm (Longer wavelength).
      Now, wavenumber expressed as reciprocal to wavelength is expressed in centimeters. When wavelength converted to wavenumber, then, Wavenumber is directly proportional to energy and frequency (In case of wavelength, it is inversely proportional to energy and frequency).
      So, to use direction correlation, between wavenumber, energy, and frequency, wavenumber to be used in IR
      eg. In the case of the IR region, the range is 2.5 to 25 micron. At 2.5 micron (wavenumber 4000 cm-1) the energy and frequency of IR radiation are high as compared to 25 microns (wavenumber 400 cm-1).
      (1 vote)
  • blobby green style avatar for user ltbd78
    The units of K is in dynes/cm because the reduced mass is in grams, correct?
    (2 votes)
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  • blobby green style avatar for user lisa rossi
    Is the wavenumber related to IR absorption? I'm a bit confused in what it is meant by one term or the other, do they refer to the same thing?
    (1 vote)
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Video transcript

- [Voiceover] In the last video we saw that the frequency of bond vibration can be thought about like a spring oscillation and so it's dependent on two things. It's dependent upon K, which is the spring constant, or the force constant. And the reduced mass, where the reduced mass is equal to the mass m1, at times m2, over m1 plus m2 and so here's your bond as a spring, with m1 and m2 on either end. And so, if you increase the force constant, increase the spring constant, so that's like increasing the strength of the bond, obviously you would increase the frequency just looking at the map here. So you increase the frequency of bond vibration. So a stronger bond vibrates faster than a weaker bond. And also, if you decrease the reduced mass, right? So if you decrease the reduced mass, mathematically that's also going to increase the frequency of vibration here, like that. In the first video, we talked about relating the frequency to wave numbers. So a wave number, you take the frequency divided by the speed of light, in centimeters per second. So if you take this over here and just divide by the speed of light, which is C, you're going to get wave numbers. So this is good to use to approximate the wave number, where you would find your signals in IR spectroscopy. All right, there's one more thing we have to do though, to get the proper units for wave number and that's because here, the reduced mass is in grams and so we need to make that atomic mass units, or AMU. So we can divide by Avogadro's number to get that. So I'm not too concerned with units here. I'm just going to show you, real quickly, where this equation comes from. Wave number is equal to one over two pi C, times the square root of K, over the reduced mass and to get our proper units, we have to divide by Avogodro's number which is 6.02 times 10 to the 23. All right, so let's continue here, that'll be one over two pi C times the square root. We'll make a little bit of a long square root sign here because we still have the force constant over the reduced mass. And now we can just move Avogadro's number up to the numerator here, so that's 6.02 times 10 to the 23. Let's go ahead and find these square roots of Avogadro's number. So the square root of Avogadro's number is equal to... So we have square root of 6.02 times 10 to the 23 and we get, that number is 7.76 times 10 to the 11. So let's go ahead and write that down here. So now we wave numbers equal to 7.76 times 10 to the 11, divided by 2 pi, C is the speed of light and centimeters per second. That's 3 times 10 to the tenth. And then we still have the square root of the force constant over the reduced mass. So let's go ahead and solve it even further here. We need to divide that number by two. We need to divide that number by pi. And we need to divide that number by the speed of light in centimeters per second. So divide by 3 times 10 to the tenth and we get 4.12 and so we arrive at this wave number as equal to 4.12 times the square root of K over the reduced mass. So here we have a nice little equation where we can approximate the wave number for different bonds. So where we would expect to see the signal for different bonds stretching. And let me use a different color to point these things out here so, K, call this the spring constant or the force constant. The units are dynes per order, dynes per centimeter and then we have the reduced mass here in AMUs. So let's do some calculations and let's see if we can approximate where we would find the signals for some bonds. And so let's get some more room down here. And let's start with a carbon hydrogen bond. So we have a carbon hydrogen bond here. First thing we could do is calculate the reduced mass, so reduced mass is equal to... Atomic mass of carbon is 12, mass of hydrogen is one, so 12 times 1, over 12 plus 1. And we did this calculation in the previous video and we got .923. So we're going to plug this into our equation for wave number. So wave number is equal to 4.12 times the square root of the force constant, for a single bond. We're talking about a carbon hydrogen bond here. It's a single bond. You can use a force constant of 5 times 10 to the 5, dynes per centimeter. And we're going to divide that by 0.923, the reduced mass and so let's get out the calculator and do that calculation. All right, so we have 5 times 10 to the 5. We're going to divide that number by 5.923. We're going to take the square root of our answer here and then we're going to multiply that by 4.12. And this gives us 3032, so 3032 wave numbers. 3032, units would be 1 over centimeters. So this is what we calculated, so this is the approximate wave number. So this is approximately where you would find the signal, for a carbon hydrogen bond stretching. So if you're looking on an IR spectrum and the actual signals is pretty close to this number, so this is a good approximation. And that's why it's kind of useful to use this equation here. Let's do it again for carbon oxygen. So it's still a single bond but this is carbon oxygen this time. So the reduced mass is going to change. But the force constant is going to stay the same because in both cases we're talking about a single bond here. We're going to pretend like they're exactly the same strength. Which, just to help us for our calculations to approximate things. All right, so the reduced mass would be equal to, carbon is 12, oxygen is 16. So we have 12 times 16 over 12 plus 16 and let's do that math really quickly. So we have 12 times 16, which is 192, divided by 28, gives us about 6.9. So we're going to say the reduced mass is approximately 6.9. And let's calculate the approximate wave number. So where would we expect to find this signal? So 4.12 times the square root-- We're going to use the same number. We're going to use 5 times 10 to the 5. So we're saying we have a single bond here, so again, these are just approximations. So this is 5 times 10 to the 5. We're going to divide that by the reduced mass which is 6.9 and let's see what we get for the wave number. So let's do that math. So we have 5 times 10 to the 5, going to divide that by 6.9. We're going to take the square root of our answer and then we're going to multiply by 4.12 to get the approximate wave number, which is 1109. So we have 1109, 1 over centimeters, for our wave number here. And let's compare these two wave numbers. So what did we do in our calculations? Let me use blue here. So what did we do? We switched out a hydrogen for an oxygen. So we increased the mass of the second atom, or the nucleus, if you want to think about it that way. So you increase the mass of the second atom and what happened to the wave number? The wave number decreased. We went from 3032 to 1109 and noticed what we did, right? The reduced mass was .923 and it went up to 6.9. So we increased the reduced mass and we decreased the frequency, or we decreased the wave number. And so again, this is approximately where you would find the signal for a carbon oxygen single bond. Not exact, because obviously we're just using really simple numbers here but it's pretty close. All right, let's do a carbon carbon double bond now. So carbon carbon double bond. Let's calculate the reduced mass. The reduced mass would be 12 times 12, over 12 plus 12. And when you do that math, you're going to get six. So I think we did that in the previous video too. Let's plug this into our equation. So the approximate wave number would be equal to 4.12 times the square root. All right, so what are we going to plug in now for the force constant? Well, we have a double bond here. We have a double bond, so a carbon carbon double bond. And if we just approximate and say that a double bond is twice as strong as a single bond, we can just take this number and multiply it by two. So that would be 10 times 10 to the 5. So we're just saying that a double bond is close to being twice as strong and again, not a perfect number, but it's a good approximation and it just gives us an idea about where do these wave numbers come from. Now, we solve for the wave number, the approximate wave number here. So we have 10 times 10 to the 5. We're going to divide that by six. We're going to take the square root of our answer and then we're going to multiply by 4.12 and this is going to give us somewhere around 1682. So our calculation gives us 1682, where we'd find the signal. Again, not perfect, but a decent approximation. Not a perfect number but pretty close to where you would find the signal for a carbon carbon double bond on an IR spectrum. Let's do one more. One more, let's get some more room down here. Let's see if we can squeeze in one more calculation. Let's do a carbon triple bond. So a carbon carbon triple bond. The reduced mass obviously would be the same as above, at six. So we're going to be changing the force constant here. So the wave number would be equal to 4.12 at two times. So what would we plug in for the force constant now, right? This would be... This was double a single bond, so we need to go for triple a single bond. We have a triple bond here. So 3 times 5 times 10 to the 5. So that's 15 times 10 to the 5, so this would be three, we're approximating this, saying it's three times as strong as a single bond, 15 times 10 to the 5. We're going to divide that by six. And let's see what happens now. So we have 15 times 10 to the 5. We're going to divide that by six. We're going to take the square root of our answer once again. So the square root of our answer and we get 500, when you multiply that by 4.12 and we get 2060. So we get 2060 here for the wave number, so however you want to write this. So what happened? What happened between these last two example? We increased K. That's the only thing that changed between these two calculations. We increased K, we increased the force constant. We're saying a triple bond is stronger than a double bond. And what happened to the frequency? Or the wave number? The wave number increased. So increase K, you increase the frequency, you increase the wave number here. Finally, let's plot these on an IR spectrum. So let's first start with the carbon hydrogen single bond. So somewhere in the 3032, so let's go down here for our spectrum. So 3032, so that would be somewhere around in here. So this would be approximately where we would find the signal for a carbon hydrogen bond. And this just allows us to think about regions of our IR spectrum. So somewhere in this region, in this region right in here, is where you would find a bond to hydrogen. So we just said this was carbon to hydrogen here but you could generalize this by saying it's any bond to hydrogen because it's the hydrogen, it's the smaller mass, that causes the increased frequency. The increased value for the wave number. Let's look at our next one here. Let's look at carbon oxygen. So for carbon oxygen, we have 1109 so that's a big difference. It's a single bond. 1109, that's way down here. So 1109 would be somewhere around in here for this carbon oxygen bond. And so, it turns out that this region right in here is where youwould find, it's a single bond region, so this is the single bond region, and when you're not talking about hydrogen. So this would be the single bond region. This over here would be the bond to hydrogen region. And then we have two more regions to talk about, right? We have our double bond region, let's use green for that. So carbon carbon double bond, we calculated it's approximately 1682 wave numbers. So let's find that on our spectrum here. So 1682, so that would be approximately in here. This would be like six, somewhere in there, in that range. And so we could say that it's approximately where we would find the signal for a carbon carbon double bond. And your double bond region... All right, so let me write, "double bond region." Double bond region is right around in that area. So double bond region. Expect to see a signal for a double bond in this area here. So somewhere in here. So this is approximately your double bond region on IR spectrum. And then finally, our last thing that we're going to think about is our triple bond. So our triple bond, we calculated an approximate value for a carbon carbon triple bond. We got 2060, so 2060 would be somewhere in here. So that's approximately where we would find our triple bond region. And that usually does go about 2100 to 2300, so in here somewhere so pretty small but just to give you an idea of these different regions. So here's the triple bond region, triple bond. So hopefully that allows you to think about why you get these different wave numbers. It has to do with two factors. It has to do with the force constant and it has to do with the reduced mass. If you think about those two factors you can think about where the signal should appear for these bonds, for these bonds stretching. So you can figure out the approximate wave number. And you should know these wave numbers because it's going to help you when you read your IR spectrum.