If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Worked example: Calculating concentration using the Beer–Lambert law

AP.Chem:
SAP‑8 (EU)
,
SAP‑8.C (LO)
,
SAP‑8.C.1 (EK)
,
SAP‑8.C.2 (EK)
The Beer–Lambert law relates the absorption of light by a solution to the properties of the solution according to the following equation: A = εbc, where ε is the molar absorptivity of the absorbing species, b is the path length, and c is the concentration of the absorbing species. In this video, we'll use the Beer–Lambert law to calculate the concentration of KMnO₄ in an unknown solution. Created by Sal Khan.

Want to join the conversation?

  • mr pants teal style avatar for user Markus Hjorth
    When using the other numbers that are given in the table for calculating epsilon I don´t get the same value for epsilon. Why?

    For example: 0,840/0,150 = 5,6 (and not 5,4).
    (28 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user WilssonLowe
      This is because they are (supposed to simulate) real world measurements, which are never perfect, so each pair of values will give you a slightly different epsilon value. This is also the reason why they give us more than just one set of measurements - if you want to be as accurate as possible you should calculate epsilon for each value set and then take the average of all these epsilons and use that. Sal doesn't do it in the video, probably mostly because it takes more time, but that's kinda okay anyway if you consider that these kinds of spectrometric measurements usually have a pretty high level of precision and the measurement of the cell width (1.0) only has two significant figures.

      Hope that helps!
      (63 votes)
  • leaf blue style avatar for user Mr. Cavin
    At he said 'spectrometer', we he soon corrected. I wonder, what is the difference between spectrometer and a spectrophotometer?
    (16 votes)
    Default Khan Academy avatar avatar for user
    • piceratops ultimate style avatar for user anderson.o.chen
      A spectrometer is 'An apparatus used for recording and measuring spectra, esp. as a method of analysis.'. However, a spectrophotometer is ;An apparatus for measuring the intensity of light in a part of the spectrum, esp. as transmitted or emitted by particular substances.;,
      Hope that helped!
      (43 votes)
  • blobby green style avatar for user sethduban
    What is the purpose of knowing that the solution was measured at 540nm? and was it just coincidence that epsilon = 5.40?
    (15 votes)
    Default Khan Academy avatar avatar for user
  • piceratops ultimate style avatar for user ben
    Is mole spelled mole or mol? Sal spells it both ways.
    (4 votes)
    Default Khan Academy avatar avatar for user
    • spunky sam blue style avatar for user Ernest Zinck
      mole is the word used to describe Avogadro's number of particles. mol is the SI symbol for the unit. It is not an abbreviation.
      In SI, you use words with words but symbols with numbers. Thus, you would write "two moles" but "2 mol". Note that symbols are not pluralized : it is incorrect to write "2 mols".
      (19 votes)
  • blobby green style avatar for user Jannie Khang
    what if the length was not given? how do i find the molar concentration?
    (6 votes)
    Default Khan Academy avatar avatar for user
    • purple pi purple style avatar for user ScienceMon
      As long as the length is constant, there will be a linear relationship between concentration and absorbance.
      The biggest difference is that without knowing length, you could not solve for epsilon (the E symbol). Therefore, all unknowns would have to have the same length as your original sample.
      (7 votes)
  • blobby green style avatar for user Michael
    How did Sal get liter per cm times mole? also how can you have a liter per mole? And why did Sal do mole per liter at the end instead of liter per mole? I'm really confused.
    (6 votes)
    Default Khan Academy avatar avatar for user
    • leaf blue style avatar for user Paolo Miguel Bartolo
      You're probably referring to the unit of the epsilon constant. Usually, constants have complicated units in order to make sure that the answer has the correct unit and that the other units are cancelled out.

      So according to the Beer-Lambert law, absorbance equals epsilon times length of container (or the length that the light has to travel through to pass through the solution) times concentration. The unit of concentration is molarity, which is moles over liter. The unit of length can be centimeters. Absorbance has no unit. If you would try to multiply the units of epsilon, length, and concentration, you should get the unit for absorbance which has no unit. Multiplying liter per cm times mole, cm, and mole per liter would result to all units cancelling out, resulting in no unit.

      Sal wanted to find concentration, so he used moles per liter.
      (6 votes)
  • winston default style avatar for user Jared Desai
    I just realized something. At , Sal says that the length of the vial is 1 centimeter, while the problem says that it is 1.0 centimeter. When he gets to his final answer at , he reports the answer to three significant digits. Since the length of the vial only had 2 significant digits, shouldn't the answer be 0.010, instead of 0.0998? Thanks for the input.
    (3 votes)
    Default Khan Academy avatar avatar for user
  • primosaur ultimate style avatar for user Oliver Worley
    How do you measure the absorbency of a solution without knowing the concentration?
    (4 votes)
    Default Khan Academy avatar avatar for user
    • leaf blue style avatar for user Paolo Miguel Bartolo
      You just need to know the intensities of the light before and after it passes through the solution.

      You first need to find the transmittance of the solution, which is the intensity of the light after it passes through the solution divided by the intensity of the incident light, which is the light before it enters the solution. Then you just get the negative log of the transmittance (use base 10) in order to get the absorbency.

      Unless the solution has zero transmittance, then the absorbency should be a positive number. I'm not sure what will be the absorbency of a solution with zero transmittance (probably infinite?)
      (3 votes)
  • aqualine ultimate style avatar for user Nandagopal M
    Will the absorbance be zero when Molarity is zero? Does pure water absorb some light?
    (4 votes)
    Default Khan Academy avatar avatar for user
    • piceratops ultimate style avatar for user Just Keith
      Yes, water will absorb and scatter some light. So, what we do with a spectrophotometer is use what is called a "blank". The blank will NOT contain the substances whose absorbance we're interested in (most of the time the blank is water plus the indicator). And we just treat the absorbance of this blank as if it were 0. This is known as "zeroing out" or sometimes as "blanking out" the spectrophotometer. The way that you do this depends on how sophisticated the method you're using is.

      So, the answer to the question is that we have to declare something to be 0 absorbance. So, normally we would treat a cuvette containing 0 M of the substance we're interested in as having 0 absorbance -- we just tell the spectrophotometer to treat that as 0. But, strictly speaking, no, it is not actually 0 absorbance.
      (3 votes)
  • male robot hal style avatar for user James Knight
    At , Sal explains that we've proven the Beer-Lambert law to be true because we can see that the relationship of the points is linear. The points in the table, though, do not form a straight line - calculating the slope by using different pairs of the coordinates yields slightly different slopes for each pair of coordinates, and if the relationship was linear, based on the first set of coordinates, the last coordinate would be (.15, .81) rather than .(.15,.84). So is it the case that the Beer-Lambert relationship is "almost' linear or is the discrepancy due to imperfection in measuring instruments. Thanks.

    James Knight
    (2 votes)
    Default Khan Academy avatar avatar for user
    • piceratops ultimate style avatar for user Just Keith
      Beer-Lambert is only approximately true. I wouldn't trust it for any absorbance greater than 0.400 myself. (My research required much better accuracy and precision than I student would need, so you might get away with a little higher.)

      Depending on many circumstances, the curve of absorbance vs. concentration begins to flatten out above 0.3 absorbance.

      The way that we accommodate for this is to analyze standards of known concentrations, and only use those that are within what is called "the linear range" -- in other words where Beer-Lambert applies.
      (3 votes)

Video transcript

- [Instructor] So I have a question here from the Kotz, Treichel, and Townsend Chemistry and Chemical Reactivity book, and I got their permission to do this. And it says a solution of potassium permanganate has an absorbance of 0.539 when measured at 540 nanometers in a one centimeter cell. What is the concentration of the potassium permanganate? Prior to determining the absorbance for the unknown solution the following calibration data were collected for the spectrophotometer. So the way that we would tackle this is we know that there is a linear relationship between absorbance and concentration. We could describe it something like this, that absorbance is going to be equal to sum slope times are concentration. And you could say sum y-intercept, if we're a purist about it, then the y intercept should be zero because at a zero concentration, you should have a zero absorbance. But the way that chemists would typically do it, is that they would put these points into a computer and then a computer do a linear aggression. You could also do that by hand but that's a little bit out of the scope of this video. And I did that, I went to Desmos and I typed in the numbers that they gave. And this is what I got, so I just typed in these numbers and then it fit a linear regression line to it and it got these parameters, m is equal to this and b is equal to this. Now we could say significant figures it seems like the small significant figures here we have have our three, but we could just view the m and the b as intermediate numbers in our calculations. What I'm going to do is I'm gonna use m and b, and then my final I'll answer I'm going to round to three significant figures. So what this tells us, is that absorbance is going to be 5.65333 times our concentration minus 0.0086. And now they've given us what A is. Let me get rid of all of this stuff here. They told us that our absorbance is 0.539, so we know that 0.539 is equal to 5.65333C minus 0.0086. And then if you wanna solve for C, let's see, we could add this to both sides first. So you get 0.539 plus 0.0086 is equal to 5.65333C, and then divide both sides by this, and you would get C is equal to, is going to be approximately equal to, be a little careful all of these would really be approximate. C is gonna be equal to 0.539 plus 0.0086 divided by 5.65333. And of course we want to round to three significant figures. All right, 0.539 plus 0.0086 is equal to that, divided by 5.65333 is equal to this, so if we go three significant figures this is going to be 0.0969. So I would write the concentration is approximately 0.0969 Molar.