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  1. #1
    Super Member Ivor Joedy's Avatar
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    Lightbulb CDF - COMMON DILUTION FORMULA for some common cases

    CASE 1 : CDF-1
    Increasing lower concentration with material of higher concentration

    CASE 2 : CDF-2
    Increasing concentration with undiluted material

    CASE 3 : CDF-3
    Decreasing higher concentration with material of lower concentration

    CASE 4 : CDF-4
    Decreasing concentration with pure solvent

    CASE 5 : CDF-5
    Decreasing pure material with pure solvent

    CASE 6 : CDF-6
    Mixing several dilutions with different weights and concentrations

    Let us call

    $D target bottle
    $S source bottle, solvent

    D [g] initial weight of material in target bottle
    d [%] initial concentration in target bottle

    S [g] initial weight of material in source bottle (we are not interested in this)
    s [%] concentration in bottle

    S’[g] weight of the material to be transfered from bottle $S to bottle $D

    r [%] desired final concentration in target bottle $D

    k [ ] coefficient of dilutions

    CDF-Formula.png
    Common Dilution Formula

    S’ = D * { (r - d) / (s - r) }

    or

    S’ = D * k

    where

    k = { (r - d) / (s - r) }

    Principle

    Suppose we have a 10% dilution of some material in our target bottle and a 50% dilution in our source bottle,
    and we want to increase the concentration in the target bottle to 20%

    k = (20 - 10) / (50 - 20) = 1/3 or 0.3333

    S' = D * 0.3333

    This means: Whatever the weight in the target bottle is, in order to increase the concentration to 20%,
    we have to add material from the source bottle in a quantity equal to a third of this weight.

  2. #2
    Super Member Ivor Joedy's Avatar
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    Post CDF-1 Increasing lower concentration with material of higher concentration

    Example: You have oakmoss prediluted to 60% by volume from your supplier (bottle $S).
    Using the specific weight of oakmoss and alcohol you know, this is about 66% by weight.

    You have decided to dilute part of it to 10% (bottle $D), but accidentaly you have added to much alcohol to it,
    so you have now 6.35 gram of oakmoss diluted at 8.54 %

    You wish to know, how much of the 66% oakmoss you have to add to get the desired 10% dilution.


    D = 6.35 g initial content of target bottle
    d = 8.54 % initial concentration in target bottle

    S’= ? material to transfer
    s = 66 % concentration in source bottle

    r = 10 % desired final concentration in target bottle
    CDF-Img_and_Formula.jpg
    CDF formula

    k = (r - d) / (s - r)

    S’ = D * k

    So in this case

    k = (10 - 8.54) / (66 - 10) = 0.02607

    S’ = 6.35 * 0.02607 = 0.166


    Result

    You have to add 0.166 gram material from source bottle $S to target bottle $D to get a 10% dilution.
    Last edited by Ivor Joedy; 14th September 2021 at 11:50 PM.

  3. #3
    Super Member Ivor Joedy's Avatar
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    Post CDF-2 Increasing concentration with undiluted material

    Example: You have diluted rose absolute to 10%, thus having 8 gram.
    But now you realise that, in opposite to what you read, it is disappointingly weak.
    You have used up some material, now having 7.54 gram left. Now you want to increase the dilution to 20%

    D = 7.54 g inizial content of target bottle
    d = 10 % initial concentration of target bottle

    S = ? material to transfer
    s = 100 % pure absolute in source bottle

    r = 20 % desired final concentration in target bottle

    CDF-Img_and_Formula.jpg
    CDF formula

    k = (r - d) / (s - r)

    S = D * k

    So in this case

    k = (20 - 10) / (100 - 20) = 0.12500

    S = 7.54 * 0.12500 = 0.943 g


    Result

    You have to add 0.943 gram pure absolute from source bottle to target bottle to get a 20% dilution.

  4. #4
    Super Member Ivor Joedy's Avatar
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    Post CDF-3 Decreasing higher concentration with material of lower concentration

    Example: Finally you bought civet tincture.
    Its concentration is of 3.5% and the seller claims it is by weight. Because you read so, you have diluted
    the greater part of it to 0.1% But now, you would like to increase the dilution to 1%
    But there is only enough material to decrease the remaining 3.5% dilution of 2.50 gram with the plenty of 0.1% dilution.

    D = 2.50 g initial content of target bottle
    d = 3.5 % initial concentration in target bottle

    S’= ? material to transfer
    s = 0.1 % concentration in source bottle

    r = 1 % final concentration in target bottle

    CDF-Img_and_Formula.jpg
    CDF formula

    k = (r - d) / (s - r)

    S’ = D * k

    So in this case

    k = (1 - 3.5) / (0.1 - 1) = 2.77778

    S’ = 2.5 * 2.77778 = 6.945


    Result

    You have to add 6.945 gram material from the 0.1% source bottle to the 3.5% target bottle
    to get (2.5 + 6.945) = 9.445 gram of an 1% civet dilution.
    Last edited by Ivor Joedy; 14th September 2021 at 11:57 PM.

  5. #5
    Super Member Ivor Joedy's Avatar
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    Post CDF-4 Decreasing concentration with pure solvent

    Example: To test & wear it as perfume on itself, you bought sandalwood.
    To catch up with the low cost oriental perfumes, you have increased the concentration to 30% but still could not
    achieve their projection. Being pretty expensive you decide to lower the concentration of your remaining
    42 gram natural sandalwood Mysore perfume, and because you read from Luca Turin that 17% is the best
    concentration for a perfume, you would like to try out this. Obviously you must add some pure alcohol.

    But you suspect, Mr. Turin was referring to 17% by volume, so using the average middle specific weight
    of Santalum Album (0.9825), this is about the same as 21 % by weight.

    D = 42.00 g inizial content of target bottle
    d = 30.00 % initial concentration in target bottle

    S’= ? material to transfer
    s = 0 % pure alcohol (grade is irrelevant)

    r = 21 % final concentration in target bottle
    CDF-Img_and_Formula.jpg
    CDF formula

    k = (r - d) / (s - r)

    S’ = D * k

    So in this case

    k = (21 - 30) / (0 - 21) = 0.42857

    S’= 42.00 * 0.42857 = 17.99994


    Result

    You have to add 18 gram ethanol to get a sandalwood dilution of 21% by weight, i.e. ca. 17% by volume.

  6. #6
    Super Member Ivor Joedy's Avatar
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    Post CDF-5 Decreasing pure material with pure solvent

    Example: You bought 1 ml (ca. 0.8942 gram) of expensive tonka bean absolute
    and decide to dissolve it to 10 % strait in the bottle (target bottle) not to lose a bit of the solid stuff.
    How much alcohol to add into suppliers bottle ?

    D = 0.8942 g inizial content of target bottle
    d = 100 % pure tonka absolute

    S'= ? pure ethanol to add
    s = 0 % pure ethanols concentration of oil

    r = 10 % final concentration in target bottle
    CDF-Img_and_Formula.jpg
    CDF formula

    k = (r - d) / (s - r)

    S' = D * k

    So in this case

    k = (10 - 100) / (0 - 10) = 9

    S' = 0.8942 * 9 = 8.0478


    Result

    You have to add 8.0478 gram of alcohol to the supplier's bottle to get a 10% tonka dilution.

  7. #7
    Super Member Ivor Joedy's Avatar
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    Post CDF-6 Mixing several dilutions with different weights and concentrations

    Example: You have got new bottle with 10 gram of oakmoss, prediluted by the supplier
    to 66% by weight. But there are still 2.68 gramm of oakmoss in the old bottle which you have prediluted down to 50%
    And of course there ist the 10% dilution with remaining 4.37 gram

    You wish to put the content of all three bottles together before diluting it further to exact numbers.

    A' = 10.00 g material from first bottle, if whole content used then A'= A
    a = 66 % concentration of the dilution in first bottle

    B' = 2.68 g material from second bottle, if whole content used then B'= B
    b = 50 % concentration of dilution in the second bottle

    C' = 4.37 g material from third bottle, if whole content used then C'= C
    c = 10 % concentration of dilution in the third bottle

    r = ? resulting concentration of the mix

    CDF-Img_and_multiple_Formula.jpg
    CDF' formula rearranged and extended

    r = (a*A' + b*B' + c*C' ...) / (A' + B' + C' ...)

    So in this case

    r = (66 * 10 + 50 * 2.68 + 10 * 4.37) / (10 + 2.68 + 4.37) = 837.7 / 17.05 = 49.132


    Result

    The beaker contains 17.05 gram oakmoss dilution in a concentration of 49.132 %

  8. #8

    Default Re: CDF - COMMON DILUTION FORMULA for some common cases

    This seems like creating enormous complication around something that couldn't be simpler: the mass fraction of a component in a mixture is the mass of that component in the mixture divided by the total mass of the mixture. If you want this in percent, multiply by 100; if you want parts per thousand, multiply by 1000. That's it!

  9. #9

    Default Re: CDF - COMMON DILUTION FORMULA for some common cases

    Thanks for taking the time to share this! Not to be ungrateful but I agree with Mike with it being overcomplicated (to me anyways) but people interpret maths differently to others so it might still help people!

    In most cases though I use these:

    You want 50g of perfume at 15%? Multiply 50 by 15%, you get 7.5, so add 7.5g of fragrance on your scale, then fill it up with alcohol to 50g, and done! (This is assuming your fragrance formula only uses pure (undiluted) materials, if not see the last formula)
    ______________________________

    You want 10g of oakmoss abs at 5%? Same as for the perfume; 10*5%=0.5, so add 0.5g of oakmoss abs then fill up with dpg or whichever until you get 10g total
    _______________________________

    You only have 3.4g of cetalox left and you want it at 13% (for whichever reason, using random numbers just to show the formula)
    Use this: (1/desired percentage) x the amount of material you have:

    (1/13%) x 3.4 = 26.146. Take your 3.4g of cetalox and add dpg etc until you get a total of 26.146g, and voila you have cetalox at 13%
    ______________________________

    You have 5g of Aurantiol already at 25% but you want it at 3%? Then do this: existing percentage x current total weight, you get the actual amount of pure material in your solution. Then after that use the same as above, (1/desired percent) x the amount of pure material
    5 x 25% = 1.25
    (1/3%) x 1.25 = 41.666

    Take the 5g you already have, fill it until you get 41.66g and voila it's at 3% (if you want to be sure, 41.66 x 3% = 1.25 (rounded up), and there is indeed 1.25 of pure material in your mix.
    ______________________________

    And finally, the most annoying but most needed method, if your finished fragrance already has some solvents.
    You want to make 50g of perfume at 12%?
    Then first, you'll have to find how much pure materials are in your fragrance (I suggest using/making a spreadsheet for this). Lets say your finished formula consists of 86% of pure fragrance (because you used a lot of 50% galaxolide as an example)
    First, use the first method I wrote here; 50 x 12% = 6

    But if you put 6g of your fragrance, you'll only actually have 6 x 86% = 5.16g of actual desired materials, so your finished perfume will NOT be at 12%. So the extra step is to put the right amout of your fragrance to arrive at 6g of pure materials:
    Needed amount of pure materials, in this case 6 / your fragrance's dillution percentage, in this case 86%

    6 / 86% = 6.976

    So you need to add 6.976g of your fragrance (which contains 6g of pure materials), then fill up to 50g, and you'll have 12% Perfume! It's not easy but the more you do it the easier it becomes!

    And finally, for ease you can condense into one formula: (Total amount desired of final perfume x Desired perfume concentration) / Your fragrance's dillution percentage

    (50g x 12%) / 86% = 6.976g
    ______________________________

    There are much simpler methods if you only work with values like 10g at 10%, 100g at 1% etc but the above "formulas" work in any percentage. These are of course methods I like to use but I do NOT proclaim these are the correct or best methods, they just work for me.

    Hope this can help someone!
    Last edited by Stefan.E; 15th September 2021 at 10:16 AM.

  10. #10
    Super Member Ivor Joedy's Avatar
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    Smile Re: CDF - COMMON DILUTION FORMULA for some common cases

    Quote Originally Posted by Stefan.E View Post
    ... it being overcomplicated ...
    Of course I know what you mean, but I don't think the formula is complicated. It is the constant need to rethink
    each case, as you demonstrate it, that is prone to failure.

    The formula is so easy, especially the factor "k", that I can do most cases in my head within seconds.
    Or I use it for control.

    I made this formula for the simple reason that most real world cases I meet
    are not of the simple type you cite.

    They are closer to the following type:

    Once I had 56.7 gram of a perfume, containing oakmoss, patchouli and castoreum.
    Now it is only 36.47 gram. I know the concentration of oakmoss in the whole was (and is) after many dilutions 4.73%
    I decide to increase it to 7.5% My oakmoss is prediluted to 66% by weight.

    How much of the predilution have I to add? I like to know it within 1 minute, error free and without messing with maths.
    For this I consider all other ingredients as solvent.

    k = (7.5% - 4.73%) / (66% - 7.5%) = 0.0474

    So I have to add 36.47g * 0.0474 = 1.729 gram of the prediluted oakmoss to obtain ca. 38.2 g perfume with 7.5% oakmoss inside.

    I hope you can do it similar fast and error free with your method. And - that's important to me - without
    beeing doomed to only ratios of simple numbers. (BTW: I like to use Fibonacci numbers as starting point for ratios.)

    As for me, I keep the stuff as easy as possible and do not like to rethinking the same stuff every time,
    once having understood the general principle. I have done the previous calculation first within 20 seconds with my
    pocket calculator and without need to sit down and write something.

  11. #11

    Default Re: CDF - COMMON DILUTION FORMULA for some common cases

    I think this all creates alotta confusion bcs ppl think they need to "learn a method", when it is literally nothing more than grade school arithmetic.

  12. #12

    Default Re: CDF - COMMON DILUTION FORMULA for some common cases

    Quote Originally Posted by Ivor Joedy View Post
    Of course I know what you mean, but I don't think the formula is complicated. It is the constant need to rethink
    each case, as you demonstrate it, that is prone to failure.

    The formula is so easy, especially the factor "k", that I can do most cases in my head within seconds.
    Or I use it for control.

    I made this formula for the simple reason that most real world cases I meet
    are not of the simple type you cite.

    They are closer to the following type:

    Once I had 56.7 gram of a perfume, containing oakmoss, patchouli and castoreum.
    Now it is only 36.47 gram. I know the concentration of oakmoss in the whole was (and is) after many dilutions 4.73%
    I decide to increase it to 7.5% My oakmoss is prediluted to 66% by weight.

    How much of the predilution have I to add? I like to know it within 1 minute, error free and without messing with maths.
    For this I consider all other ingredients as solvent.

    k = (7.5% - 4.73%) / (66% - 7.5%) = 0.0474

    So I have to add 36.47g * 0.0474 = 1.729 gram of the prediluted oakmoss to obtain ca. 38.2 g perfume with 7.5% oakmoss inside.

    I hope you can do it similar fast and error free with your method. And - that's important to me - without
    beeing doomed to only ratios of simple numbers.

    As for me, I keep the stuff as easy as possible and do not like to rethinking the same stuff every time,
    once having understood the general principle. I have done the previous calculation first within 20 seconds with my
    pocket calculator and without need to sit down and write something.
    I absolutely agree! That's why I mentionned that everyone does maths differently, so I really appreciate that you took the time to write down your own system, which seems more robust. It was interesting to write down my methods too as, like yourself, I don't really think about too much, but maybe my method could help someone so I thought it was worth sharing as well

  13. #13
    Super Member Ivor Joedy's Avatar
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    Default Re: CDF - COMMON DILUTION FORMULA for some common cases

    Quote Originally Posted by mnitabach View Post
    I think this all creates alotta confusion ...
    I am of the opposite opinion. And also: It's just an offer to someone who likes this way of thinking.

  14. #14

    Default Re: CDF - COMMON DILUTION FORMULA for some common cases

    Quote Originally Posted by mnitabach View Post
    I think this all creates alotta confusion bcs ppl think they need to "learn a method", when it is literally nothing more than grade school arithmetic.
    I see your point, but people come from many, many different backgrounds. What might seem simple maths to people with "standard" education, might not be the case for everyone. Once you learn and are used to methods, teachings etc, it's easy to forget some people think differently or were taught differently. I think it's useful to share your method, complicated or not, because it might just "click" with specific people that will help them on the long run, which is the whole point of this post!

  15. #15

    Default Re: CDF - COMMON DILUTION FORMULA for some common cases

    Very good points!!!

  16. #16

    Default Re: CDF - COMMON DILUTION FORMULA for some common cases

    Quote Originally Posted by mnitabach View Post
    Very good points!!!
    Glad you think so! Very happy to have discussions about different thoughts and opinions

  17. #17
    Super Member Ivor Joedy's Avatar
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    Smile Re: CDF - COMMON DILUTION FORMULA for some common cases

    Good point indeed. And a matter of course.




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