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  • mg/L to % (w/w)

    Could someone please explain the steps to convert mg/L to % (w/w) with the details below;

    weight of sample = 1.05 g
    concentration of sample in solution = 82.6 mg/L
    Dilution factor = 10

    I did not carry out specific gravity for this sample and I've used Agilent HPLC

    I would really appreciate for your help! :-)

  • #2
    Originally posted by Ajchemed View Post
    Could someone please explain the steps to convert mg/L to % (w/w) with the details below;

    weight of sample = 1.05 g
    concentration of sample in solution = 82.6 mg/L
    Dilution factor = 10

    I did not carry out specific gravity for this sample and I've used Agilent HPLC

    I would really appreciate for your help! :-)
    I do not know the operation of gas chromatography equipment or relevance of the dilution factor of 10.

    If I focus solely on the concentration of 82.6 mg/L, it is measured on weight/volume (w/v) basis and you want w/w. You need the density of your solution.
    I emphasize solution, not just the solvent, although for very dilution solutions, the density of the solvent is a reasonable approximation. Although not a real percent, w/v concentrations are often expressed on a basis of 100% = 1 kg/L. Your solution is 82.6 x 10-6 kg/L x 100% = 0.00826 % w/v.

    If you knew the density of your solution, you would divide by density in kg/L to convert the volume in the denominator to weight. If you solvent is water, and there are no other solutes, then the density is very close to 1 kg/L. However, if your solvent was pure ethanol (0.797 g/L), you would divide by that number. If your solution is concentrated, like a stock chemical, be sure to use its actual density.

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    • #3
      Thank you JohnS for your explanation :-) I have carried our density reading on my sample which turned out to be 1.086 g/ml (the sample was a bit viscous - not a free flow liquid)

      Would the following calculation be right for finding the final concentration in mg/kg?

      (82.6 mg/l x 10) / (1.05 g x 1.086 kg/l) = 724 mg/kg
      724 mg/kg / (1000 x 10) = 0.0724 % (w/w) (= 0.0724 g/100 g)

      Thank you

      Comment


      • #4
        Originally posted by Ajchemed View Post
        Thank you JohnS for your explanation :-) I have carried our density reading on my sample which turned out to be 1.086 g/ml (the sample was a bit viscous - not a free flow liquid)

        Would the following calculation be right for finding the final concentration in mg/kg?

        (82.6 mg/l x 10) / (1.05 g x 1.086 kg/l) = 724 mg/kg
        724 mg/kg / (1000 x 10) = 0.0724 % (w/w) (= 0.0724 g/100 g)

        Thank you
        If the "10" is the effect of your dilution factor, that is correct. You have handled the density correctly. As I said, I'm not familiar with the equipment or sure of what the dilution factor represented.

        Comment


        • #5
          Thank you JohnS again for your reply but I am still confused about the unit of weight of sample..

          In the calculation above,

          (82.6 mg/l x 10) / (1.05 g x 1.086 kg/l) = 724 mg/kg

          the litre (l) cancels out but the gram (g) from the sample unit doesn't.. so wouldn't the final concentration unit
          not be mg/kg but mg/ (g x kg)?

          I am so confused...:-(

          Thank you.

          Comment


          • #6
            Originally posted by Ajchemed View Post
            Thank you JohnS again for your reply but I am still confused about the unit of weight of sample..

            In the calculation above,

            (82.6 mg/l x 10) / (1.05 g x 1.086 kg/l) = 724 mg/kg

            the litre (l) cancels out but the gram (g) from the sample unit doesn't.. so wouldn't the final concentration unit
            not be mg/kg but mg/ (g x kg)?

            I am so confused...:-(

            Thank you.
            I'm sorry, I missed the 1.05 g. The mass of the sample is not material. I assume the factor of 10 is dilution and correct
            (82.6 mg/kg x 10) / (1.086 kg/L) = 760.6 mg/kg = 760.6 x 10/6 kg/kg x 100% = 0.07606% w/w (or 760.6 ppm)

            Comment


            • #7
              Thank you JohnS for your explanations :-)

              Yes 10 is the dilution factor. Could you please explain what you mean by 'mass of the sample is not material' as I'm still unsure
              the unit (g) for mass of the sample can;t cancel out..

              Would the density be the same as specific gravity (relative to water density) to be used in this calculation?

              Thank you :-)

              Comment


              • #8
                Originally posted by Ajchemed View Post
                Thank you JohnS for your explanations :-)

                Yes 10 is the dilution factor. Could you please explain what you mean by 'mass of the sample is not material' as I'm still unsure
                the unit (g) for mass of the sample can;t cancel out..

                Would the density be the same as specific gravity (relative to water density) to be used in this calculation?

                Thank you :-)
                The mass of the sample should not be used in the calculation if you know the density. If you don't know, then you need both the mass and volume of the sample to calculate it. Just divide the w/v concentration by density.

                Comment


                • #9
                  Thank you again for your answers :-)

                  But as it is the weight of sample I weighed out, shouldn't 1.05 g be considered in the calculation?

                  Thank you :-)

                  Comment


                  • #10
                    Originally posted by Ajchemed View Post
                    Thank you again for your answers :-)

                    But as it is the weight of sample I weighed out, shouldn't 1.05 g be considered in the calculation?

                    Thank you :-)
                    Only if you also measure the volume and use the two to cancel density. You are measuring and converting concentrations. The same conversion between w/v and w/w would exist with a larger or smaller amount of the solution.

                    Comment


                    • #11
                      Thank you so much for all of your explanations and calculations - I feel more confident with the calculation of my results :-)

                      Would products like pure paint, adhesives, for example, be considered in the 'stock chemical' group to measure their density values rather than that of the solvent or
                      solution of the dissolved products?

                      Thank you :-)

                      Comment


                      • #12
                        Originally posted by Ajchemed View Post
                        Thank you so much for all of your explanations and calculations - I feel more confident with the calculation of my results :-)

                        Would products like pure paint, adhesives, for example, be considered in the 'stock chemical' group to measure their density values rather than that of the solvent or
                        solution of the dissolved products?

                        Thank you :-)
                        Using the actual density of the mixture is always the correct answer.

                        If the mixture is very dilute, for example contaminants in drinkable water, it may be reasonable to say it is not much different from the density of the solvent; this is always an approximation and is never strictly correct. But even sea water with about 3.5% salt content has a density around 1.025 kg/L instead of 1 kg/L for pure water. A paint or adhesive to be useful will almost always be a high concentration of useful ingredients in a relatively small amount of solvent and require measured density.

                        Comment


                        • #13
                          Thank you again for all of your help and explanations :-)

                          I'm sorry I'm confused again as I was told by my lecturer that the weight (in grams) should be considered in the calculation
                          and the final concentration would be 0.0724 g/100g (as shown below)


                          (82.6 mg/l x 10) / (1.05 g x 1.086 kg/l) = 724 mg/kg
                          724 mg/kg / (1000 x 10) = 0.0724 % (w/w) (= 0.0724 g/100 g)


                          Thank you :-)

                          Comment

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