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Semi OT: Calculating dilution

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  • Semi OT: Calculating dilution

    I used to know how to do this, but having not needed to do so for a long time, I have forgotten the process.

    I have 1355 grams of a solution containing 20% 2-Butoxyethanol. I need to raise the percentage of 2-Butoxyethanol to 25%. How much 100% 2-Butoxyethanol do I need to add? Total weight after addition is irrelevant.
    Kevin

    More tools than sense.

  • #2
    Assuming all percentages are by weight, you should add 90.33g of the 2-BE.

    Ed
    For just a little more, you can do it yourself!

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    • #3
      As near as dammit 90 grams.

      Comment


      • #4
        Originally posted by MichaelP
        You need to add 5% of the stuff. 1355/100*5=67.75(g)
        Hopefully you are not a pharmacist
        (271+61.75)/(1355+67.75)*100% = 23.3%
        Location: Helsinki, Finland, Europe

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        • #5
          Originally posted by MattiJ View Post
          Hopefully you are not a pharmacist
          (271+61.75)/(1355+67.75)*100% = 23.3%
          I think you have a typo in your numerator. (=> 23.8%)

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          • #6
            Originally posted by ed_h View Post
            Assuming all percentages are by weight, you should add 90.33g of the 2-BE.

            Ed
            Thanks,

            Your assumption is correct, everything is by weight. Adding the 90 g will be easy (shop scale reads to grams). So what formula did you use to get there, just in case I ever have to do it again? The solution is for an airbrush cleaner for acrylic paint. I used all the 2-BE I had on hand, but now that I have resupplied, I want to bring the solution to the "proper" percentage.
            Kevin

            More tools than sense.

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            • #7
              Originally posted by tomato coupe View Post
              I think you have a typo in your numerator. (=> 23.8%)
              It's 23.387% unless you suck at math.

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              • #8
                Originally posted by MattiJ View Post
                Hopefully you are not a pharmacist
                (271+61.75)/(1355+67.75)*100% = 23.3%
                You lost me there: where did the 61.75 come from? I get the 271 (20% of 1355) and the 67.75 (suggested addition).

                Thinking further: should the 61.75 actually be 67.75 (the addition)?
                Last edited by KJ1I; 03-19-2018, 06:52 PM.
                Kevin

                More tools than sense.

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                • #9
                  Originally posted by KJ1I View Post
                  You lost me there: where did the 61.75 come from? I get the 271 (20% of 1355) and the 67.75 (suggested addition).
                  Typo.
                  And luckily I'm not pharmacist either
                  Location: Helsinki, Finland, Europe

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                  • #10
                    Originally posted by 3 Phase Lightbulb View Post
                    It's 23.387% unless you suck at math.
                    See posts #9 and #10. MattiJ just had a typo, that's all, and it gave him the wrong number.

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                    • #11
                      Originally posted by KJ1I View Post
                      Thanks,

                      Your assumption is correct, everything is by weight. Adding the 90 g will be easy (shop scale reads to grams). So what formula did you use to get there, just in case I ever have to do it again? The solution is for an airbrush cleaner for acrylic paint. I used all the 2-BE I had on hand, but now that I have resupplied, I want to bring the solution to the "proper" percentage.
                      Your 20% solution has 0.2 x 1355 g = 271 g of BE

                      You want to add an unknown weight, X, of BE to bring the solution to 25%.

                      The new weight of BE will be 271 + X.

                      The new weight of the entire solution will be 1355 + X.

                      So

                      271 +X = 0.25(1355 + X)

                      Solve for X = 90.33 g

                      Ed
                      For just a little more, you can do it yourself!

                      Comment


                      • #12
                        Once more, thank you.

                        Kj
                        Kevin

                        More tools than sense.

                        Comment


                        • #13
                          Kevin,

                          Sorry, I deleted my post by mistake. To recoup, I stated that if, by your request, we ignore the final solution weight, you'd need to add 67.75g of BE. And I noted that, in reality, the final concentration will become 23.8% instead of the needed 25%.

                          This difference is because we ignored the new total solution weight.

                          To reach 25% solution in real life, you need to follow Ed's calculations. That's how it's done properly. As you can see, the difference is stunning.
                          Last edited by MichaelP; 03-19-2018, 07:17 PM.
                          Mike
                          WI/IL border, USA

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                          • #14
                            complicated thinking here. 5% of 1355 = 67.75. thats it.

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                            • #15
                              Originally posted by dian View Post
                              complicated thinking here. 5% of 1355 = 67.75. thats it.
                              And what is "it", other than the incorrect answer to the question?

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