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from ppm to % w/v, v/v and w/w

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  • from ppm to % w/v, v/v and w/w

    For example:

    juice sample has a density 1.025g/ml.
    ethanol density is 0.789 g/ml.
    final concentration of ethanol in juice is 50ppm w/v.

    I need help converting the 50ppm to % w/v , % v/v and %w/w.

    Thank you!

  • #2
    Originally posted by 001con View Post
    For example:

    juice sample has a density 1.025g/ml.
    ethanol density is 0.789 g/ml.
    final concentration of ethanol in juice is 50ppm w/v.

    I need help converting the 50ppm to % w/v , % v/v and %w/w.

    Thank you!
    Percent is parts per hunded, so for ppm of the same "flavor," just divide by 104. 50 ppm w/v = 0.005% w/v.

    For the other conversions, it is easier to express 50 ppm w/v with units 50 mg/L

    For v/v, divide by density of the numerator substance (ethanol) or multiply by reciprocal

    50 mg/L x 1 mL/0.789 g = 63.4 L/L = 0.00634 %

    For w/w, divide by density of the mixture (multiply by reciprocal)

    50 mg/L x 1 L/1.025 kg = 48.8 mg/kg = 0.00488 %

    Comment


    • #3
      Originally posted by JohnS View Post

      Percent is parts per hunded, so for ppm of the same "flavor," just divide by 104. 50 ppm w/v = 0.005% w/v.

      For the other conversions, it is easier to express 50 ppm w/v with units 50 mg/L

      For v/v, divide by density of the numerator substance (ethanol) or multiply by reciprocal

      50 mg/L x 1 mL/0.789 g = 63.4 L/L = 0.00634 %

      For w/w, divide by density of the mixture (multiply by reciprocal)

      50 mg/L x 1 L/1.025 kg = 48.8 mg/kg = 0.00488 %
      thanks for John's fast reply.

      i am wondering when i do this, is it wrong and no meaning.

      50 mg/L x 1 mL/0.789 g x 1.025 kg / 1L = 0.0065%(?)

      thanks again

      Comment


      • #4
        Originally posted by 001con View Post

        thanks for John's fast reply.

        i am wondering when i do this, is it wrong and no meaning.

        50 mg/L x 1 mL/0.789 g x 1.025 kg / 1L = 0.0065%(?)

        thanks again
        Depends on what you have and what you are trying to get to. If starting with w/v, it has no meaning. If you were starting with w/w, it would convert to v/v.

        Comment


        • #5
          Originally posted by JohnS View Post

          Depends on what you have and what you are trying to get to. If starting with w/v, it has no meaning. If you were starting with w/w, it would convert to v/v.

          So, it means mg/kg x ml/g x g/ml= ethanol ml/juice ml =(v/v)

          (1)how about the following calculation? Can it be equal to v/v or what am i missing to make it equal to v/v if this is from w/v?

          50ppm
          ---------------------------------------
          0.789g/ml x 1.025 g/ml


          (2) The standard is prepared from 0.1g of ethanol to 10 ml volmetric flask in water solution. So, it means
          0.1g/10ml x 1000000ug/1g
          =0.1ug/10 ml x 1000000
          =10000ug/ml=10000ppm w/v

          making a100ppm ethanol solution in 10 ml volumetric flask needs 0.100ml of 10000ppm standard solution, so the this standard is still w/v .

          about the juice preparation is 1g in 10ml v flask with water and the ethanol result would be 5 ug/ml (from the calibration curve)x 10ml/1g=50ug/ml.

          So, the 50ppm result is w/v.

          if 1g of juice and the total weigh of sample and water solution is 10g, the ethanol result would be
          5ug/ml (from the calibration curve) x 10g/1g = 50w/w?

          Thanks

          Comment


          • #6
            Originally posted by 001con View Post


            So, it means mg/kg x ml/g x g/ml= ethanol ml/juice ml =(v/v)

            (1)how about the following calculation? Can it be equal to v/v or what am i missing to make it equal to v/v if this is from w/v?

            50ppm
            ---------------------------------------
            0.789g/ml x 1.025 g/ml
            If you are starting with w/v, and want v/v, you are only converting weight of solute to volume of solute and you only divide by the density of the solute (ethanol in this case).

            (2) The standard is prepared from 0.1g of ethanol to 10 ml volmetric flask in water solution. So, it means
            0.1g/10ml x 1000000ug/1g
            =0.1ug/10 ml x 1000000
            =10000ug/ml=10000ppm w/v

            making a100ppm ethanol solution in 10 ml volumetric flask needs 0.100ml of 10000ppm standard solution, so the this standard is still w/v.
            I agree to this point. I don't understand the section below or have any idea where the 5 g/mL came from.

            about the juice preparation is 1g in 10ml v flask with water and the ethanol result would be 5 ug/ml (from the calibration curve)x 10ml/1g=50ug/ml.

            So, the 50ppm result is w/v.

            if 1g of juice and the total weigh of sample and water solution is 10g, the ethanol result would be
            5ug/ml (from the calibration curve) x 10g/1g = 50w/w?

            Thanks
            Last edited by JohnS; 08-02-2017, 09:48 AM.

            Comment


            • #7

              Originally posted by JohnS View Post
              I agree to this point. I don't understand the section below or have any idea where the 5 g/mL came from.
              I was wondering when i prepare sample in w/v and in w/w and using the same calibration curve w/v to calculate the ethanol concentration , is my final concentration and the concentration unit correct.

              after gc running all the standards (unit is w/v) and plotting a calibration curve, then the GC software calculates the sample concentration, which is 5ug/ml. next step is correcting the concentration with the dilution factor.

              Now, the juice is weighed 1g to 10ml volumetric flask and is diluted with water. the ethanol concentration get from the calibration curve calculation is 5 ug/ml.
              The final ethanol concentration in juice would be

              5ug/ml x 10ml/1g=50ug/ml w/v

              if 1g of juice is diluted with water and made the final weigh of the sample solution is 10g, the ethanol concentration get from the same calibration curve is 5ug/ml.

              So, the final ethanol concentration in juice would be

              5ug/ml x 10g/1g x 1ml/0.789g = 63.4 ug/g w/w?

              Thanks for all your help.



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