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# Signal characteristics - wavenumber

## Video transcript

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 is dependent upon K which is the spring constant or the force constant and the reduced mass with the reduced mass is equal to the mass m1 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 right increase the spring constants that's like increasing the strength of the bond obviously you would increase the frequency just looking at the math here so you increase the frequency of bond vibration and so a stronger bond a stronger bond vibrates faster than a weaker bond and also if you if you decrease the reduced mass right so if you decrease the reduced mass mathematically that's also to increase the frequency of vibration here like that in the first video we talked about relating the frequency to wave numbers right so the wave a wave number you take the frequency divided by the speed of light in centimeters per second and so if you take this and take this over here and just divide by the speed of light which is C you're going to get wave numbers and so this is a this is good to use to approximate the wave number where you would find your signals in an 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 the reduced mass is in grams and so we need to make that atomic mass units or amu and so we can divide by Avogadro's number to get that so not not too concerned with with units here I'm just going to show you real quickly where this equation comes from so 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 Avogadro's number which is 6.02 times 10 to the 23rd all right so let's let's continue here that'd be 1 over 2 pi C times the square root we 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 could just move God rose number up to the numerator here so that's 6.02 times 10 to the 23rd let's go ahead and find the square root of Avogadro's number right so the square root of avocados number is equal to so we have square root of 6.02 times 10 to the 23rd and we get that number is seven point seven six times ten to the eleventh so let's go ahead and write that down here so now we have wave numbers equal to seven point seven six times 10 to the 11 divided by two pi c is the speed of light and centimeters per second that's three times ten to the tenth and then we still have we still have this 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 divided by three times ten to the tenth and we get four point one two and so we arrive at this this this wave number is equal to four point one two times the square root of K over the reduced mass and 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 vines per centimeter and then we have the reduced mass here in am use and so let's let's do some calculations and so 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 let's start with a carbon hydrogen bonds we have a carbon hydrogen bond here first thing we could do is calculate the reduced mass so that the reduced mass is equal to atomic mass of carbon is 12 mass of hydrogen is 1 so 12 times 1 over 12 plus 1 and we did this calculation in the previous video and we got point nine two three so we're going to plug this into our equation for wave number so wave number is equal to four point one two times the square root of the force constant for a single bond right we're talking about a carbon hydrogen bond here it's a single bond you can use a force constant of five times 10 to the fifth Dynes per centimeter we're going to divide that by zero point nine two three at the reduced mass and so let's get out the calculator and do that calculation all right so we have five times 10 to the fifth we're going to divide that number by by 0.92 three we're going to take the square root of our answer here and then we're going to multiply that by four point one two and this gives us three thousand and thirty-two so 3032 wave numbers so three thousand and thirty two units would be one over centimeters and so this is this is what we calculated right 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 at 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 all right so still a single bond but this is carbon oxygen this time and so the reduced mass is going to change but the force constant is going to stay the same because in both cases right in both cases we're talking about a single bond here so we're going to pretend like they're exactly the same strength which which just to help us for our calculations to approximate things all right so the reduced mass will 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 4.1 2 times the square root we're going to use the same number right we're used 5 times 10 to the fist so we're saying we have a single bond here so again these are just approximation so this is 5 times 10 to the 5th I'm going to divide that by the reduced mass which is 6.9 now let's see what we get for the wave number so let's do that math alright so we have 5 times 10 to the 5th I'm 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 4.1 - to get the approximate wave number which is 1 1:09 all right so we have 1 1 0 9 1 over centimeters for our wave number here and let's compare these two wave numbers so what did we do in our calculations so let me use all amuse let me use blue here so what do we do we we switched out a hydrogen right for an oxygen so we increased the mass of the second atom or the nor the nucleus if you want to think about that way so you increase the mass of the second atom and what happened to the wave number at the wave number decrease 1 from 3032 to 1109 and notice what we did right the reduced mass was 0.9 to 3 and it went up to 6.9 so we increased we increased the reduced mass and we decreased we decrease the frequency or we decrease 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 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 of B 12 times 12 over 12 plus 12 and when you do that math you're going you get six so I think we did that in the previous video too so let's let's plug this into our equation so the approximate wave number would be equal to four point one two times the square root alright so what are we going to plug in now for the force constant well we have a double bond here alright 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 all right we could just take this number and multiply it by two right so that would be 10 times 10 to the fifth so we're just saying that a double bond is is close to being twice as strong and again not a perfect not a perfect number but it's a good approximation and it just gives us an idea about where these wave numbers come from so now we solve for the wave number the approximate wave number here so we have 10 times 10 to the fifth we're going to divide that by 6 we're going to take the square root of our answer and then we're going to multiply by 4 0.12 and this is going to give us somewhere around 1682 all right so our calculation gives us 1682 where we'd find we would find this signal again not perfect but a decent approximation all right not a perfect not a perfect number but pretty close to where you would find the signal for a carbon-carbon double bond and an IR spectrum let's do one more one more let's get some more room down here so let's see if we can squeeze in one more calculation let's do a carbon-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 be equal to four point one two times so what would we plug in for the force constant now right well it's this would be alright 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 fifth all right so that's 15 times 10 to the fifth so this would be 3 we're approximating in this thing it's 3 times it's wrong as a single bond 15 times 10 to the fifth we're going to vide that by sick all right and let's let's see what happens now so we have 15 times 10 to the fifth we're going to divide that by 6 we're going to take the square root of our answer once again so square root of our answer and we get 500 we multiply that by 4 point one two and we get 2060 so we get two thousand 2060 here for the wave number all right so however you want to write this so what happened what happened between these this last two examples here what we increased K all right that's the only thing that changed between these two calculations we increased K we increase the force constant we're saying a triple bond is stronger than a double bond and what happened to the frequency right or the wave number the wave number increased right so increase K you increase the frequency you increase the wave number here finally let's 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 for our spectrum so three thousand three thousand and thirty two so that would be somewhere around in here all right 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 spectra so somewhere in this region right in this region right in here is where you would find a bond to hydrogen so we just we've 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 a smaller mass that causes the increased frequency the increased value for the wave number let's look at our next one here and let's look at let's look at carbon oxygen right so for carbon oxygen we have 1109 and so that's a big difference it's a single bond alright 1109 that's that's way down here so 1109 would be somewhere somewhere around in here for this carbon oxygen bond and so it turns out that this region right in here is where you find it's a single bond region right so this is the single bond region when you're not talking about hydrogen so this would be the single single bond region this over here would be the the bond to hydrogen region and then we have a we have two more two more regions to talk about alright we have our double bond region let's use green for that so carbon-carbon double bond be calculated it's approximately one thousand six hundred and eighty two wave numbers so let's find that on our spectrum here so 1682 so that would be that would be approximately in here so this would be this would be like six somewhere in there in that range and so we could say that's approximately where we find a signal for a carbon-carbon double bond and your double bond region alright so let me write double bond region double r on region is right around in that in that area so double bond region expect to see a signal for a double bond in this area here so somewhere somewhere in here alright so this is approximately your double bond region on an ir spectrum and then finally our last thing that we're going to think about is our triple bonds right 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 alright so that's approximately where we'd find our triple bond region right and that usually does go about about a 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 so triple bond and so hopefully hopefully that allows you to think about why you get these different you get these different wave numbers right it has to do with two factors has to do with the force constant and it has to do with the reduced mass and if you think about those two factors right 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 spectra