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  • Convert back to per gram dried weight

    Hi guys please help

    I had 1g dried sample extract with 10 ml of water then I took 10 uL of water reacting with 300 uL of DPPH the reading that I got was equal to 0.12 mM Trolox then I added another 10 ml MeOH to extract dried sample ( total volume was 20 mL) and took 10 uL of water /MeOH reacting with 300 uL of DPPH and the reading that I got was equal to 0.07mM Trolox. How should I convert back to how many trolox per gram dried weight of each steps ?

    my calculation is 10 uL = 0.12 mM and 10mL = 10000 uL therefore 10mL equal to =0.12 mM x 10000 =1200 mM.

    and 10 uL = 0.07 mM and 20mL =20000 uL therefore 20 equal to = 0.07mM x 20000=1400 mM

    According to above, first step will be 1200 mM of Trolox per 1 g of dried sample and secod step will be 1400mM per 1 g of dried sample.

    Is the calculation right ?

    Thank you for the help.

  • #2
    Re: Convert back to per gram dried weight

    Originally posted by Unregistered View Post
    Hi guys please help

    I had 1g dried sample extract with 10 ml of water then I took 10 uL of water reacting with 300 uL of DPPH the reading that I got was equal to 0.12 mM Trolox then I added another 10 ml MeOH to extract dried sample ( total volume was 20 mL) and took 10 uL of water /MeOH reacting with 300 uL of DPPH and the reading that I got was equal to 0.07mM Trolox. How should I convert back to how many trolox per gram dried weight of each steps ?

    my calculation is 10 uL = 0.12 mM and 10mL = 10000 uL therefore 10mL equal to =0.12 mM x 10000 =1200 mM.

    and 10 uL = 0.07 mM and 20mL =20000 uL therefore 20 equal to = 0.07mM x 20000=1400 mM

    According to above, first step will be 1200 mM of Trolox per 1 g of dried sample and secod step will be 1400mM per 1 g of dried sample.

    Is the calculation right ?

    Thank you for the help.
    Mrs X is a chemist and I'm not. Until she comes along, I'll have a go at it. I think there may be some confusion betweens moles (amount of subtance) and molarity, a concentration in moles per liter.

    On the initial 10 mL sample. you mixed 10 無 of it with 300 無 of something else, so the measured result is diluted 31:1 from the initial sample. 31 * 0.12 mM is 3.72 mM (or 3.72 mmol/L) in the 10 mL sample. Molar mass of Trolox is 250.29 g/mol and you have 10 mL of it

    3.72 mmol/L x 250.29 g/mol x 10 mL = 9.3 mg. This is all attributable to the original 1 g sample dissolved in the 10 mL of water, hence 9.3 mg/g.
    Alternatively 3.72 mmol/L x 10 mL = 37.2 痠ol, also attributable to the 1 g, hence, 37.2 痠ol/g

    In the 2nd case, the concentration of the 20 mL sample (I'm ignoring the 10 無 already removed) is 2.17 mmol/L. Multiplying it by 20 mL, and attributing it all to the 1 g sample, you have 43.4 痠ol/g.

    I don't know whether the two results represent reasonable experimental error, better extraction with methanol, or something else.

    Comment


    • #3
      Re: Convert back to per gram dried weight

      Originally posted by Unregistered View Post
      Hi guys please help

      I had 1g dried sample extract with 10 ml of water then I took 10 uL of water reacting with 300 uL of DPPH the reading that I got was equal to 0.12 mM Trolox then I added another 10 ml MeOH to extract dried sample ( total volume was 20 mL) and took 10 uL of water /MeOH reacting with 300 uL of DPPH and the reading that I got was equal to 0.07mM Trolox. How should I convert back to how many trolox per gram dried weight of each steps ?

      my calculation is 10 uL = 0.12 mM and 10mL = 10000 uL therefore 10mL equal to =0.12 mM x 10000 =1200 mM.

      and 10 uL = 0.07 mM and 20mL =20000 uL therefore 20 equal to = 0.07mM x 20000=1400 mM

      According to above, first step will be 1200 mM of Trolox per 1 g of dried sample and secod step will be 1400mM per 1 g of dried sample.

      Is the calculation right ?

      Thank you for the help.
      Hi, I'm not sure i understand what you are doing correctly?

      Like JohnS says, mM means millimoles per litre, so it is a concentration. Milli moles is just written millimoles so it doesn't get confused with millimeters (mm).

      Solution A: 1g dried sample extract with 10 ml of water

      Solution B: 10 uL of water reacting with 300 uL of DPPH the reading that I got was equal to 0.12 mM Trolox then I added another 10 ml MeOH to extract dried sample ( total volume was 20 mL)

      Solution C: 10 uL of water /MeOH reacting with 300 uL of DPPH and the reading that I got was equal to 0.07mM Trolox.

      For solution B, did you make up to 20mL with water, and did the 10無 come from solution A?

      Was the total volume of solution C 310無? again, did the 10無 come from solution A?

      For the 0.12mM trolox, and 0.07mM trolox, where the readings definitely mM, (millimoles per litre?)



      If you could just clarify the above, that would help.

      Comment


      • #4
        Re: Convert back to per gram dried weight

        Originally posted by Mrs X View Post
        Hi, I'm not sure i understand what you are doing correctly?

        Like JohnS says, mM means millimoles per litre, so it is a concentration. Milli moles is just written millimoles so it doesn't get confused with millimeters (mm).

        Solution A: 1g dried sample extract with 10 ml of water

        Solution B: 10 uL of water reacting with 300 uL of DPPH the reading that I got was equal to 0.12 mM Trolox then I added another 10 ml MeOH to extract dried sample ( total volume was 20 mL)

        Solution C: 10 uL of water /MeOH reacting with 300 uL of DPPH and the reading that I got was equal to 0.07mM Trolox.

        For solution B, did you make up to 20mL with water, and did the 10無 come from solution A?

        Was the total volume of solution C 310無? again, did the 10無 come from solution A?

        For the 0.12mM trolox, and 0.07mM trolox, where the readings definitely mM, (millimoles per litre?)



        If you could just clarify the above, that would help.
        Thank you for your quick reply

        I put 1 g sample into a test tube and added with 10 ml water shaking for 10 min then took 10 無 from the test tube reacting with 300 無 DPPH. I added another 10mL MeOH into the same tube to make up 20mL and shaking for another 10 min then took 10無 from the water/MeOH mixture and react with 300 無 DPPH.

        I had two solution: solution A was 10 無 from 10mL water extract (10 min sample) and solution B was 10 無 from 20 mL water/MeOH mixture (20 min sample).

        Total reaction volume was 310 無 (10 無 of each solution with 300 無). The concentration was millimoles per liter.

        I'm sorry that English is not my first language.

        Thank you for the time and help.

        Comment


        • #5
          Re: Convert back to per gram dried weight

          Originally posted by Unregistered View Post
          Thank you for your quick reply

          I put 1 g sample into a test tube and added with 10 ml water shaking for 10 min then took 10 無 from the test tube reacting with 300 無 DPPH. I added another 10mL MeOH into the same tube to make up 20mL and shaking for another 10 min then took 10無 from the water/MeOH mixture and react with 300 無 DPPH.

          I had two solution: solution A was 10 無 from 10mL water extract (10 min sample) and solution B was 10 無 from 20 mL water/MeOH mixture (20 min sample).

          Total reaction volume was 310 無 (10 無 of each solution with 300 無). The concentration was millimoles per liter.

          I'm sorry that English is not my first language.

          Thank you for the time and help.
          So the concentration of solution A should be twice the concentration of solution B, which is almost what you have, allowing for rounding etc.

          Solution B: Your concentration of B is 0.07mmol/L, you used 10無 from 20mL total sample. 10無 is 0.01mL, so the calculation is 0.07/1000 x 20/0.01 = 0.14mmol/g


          Solution A: Your concentration of A is 0.12mM, and you took 10無 from 10mL, so you had 0.12/1000 x 10mL/0.01mL = 0.120mmol in the original 1g.

          You can use the whole range (0.120 - 0.140millimoles/g), go for the middle range (0.130millimoles/g)

          Comment


          • #6
            Re: Convert back to per gram dried weight

            Originally posted by Mrs X View Post
            So the concentration of solution A should be twice the concentration of solution B, which is almost what you have, allowing for rounding etc.

            Solution B: Your concentration of B is 0.07mmol/L, you used 10無 from 20mL total sample. 10無 is 0.01mL, so the calculation is 0.07/1000 x 20/0.01 = 0.14mmol/g


            Solution A: Your concentration of A is 0.12mM, and you took 10無 from 10mL, so you had 0.12/1000 x 10mL/0.01mL = 0.120mmol in the original 1g.

            You can use the whole range (0.120 - 0.140millimoles/g), go for the middle range (0.130millimoles/g)
            Mrs X,
            Perhaps I am confused. Are not solutions B and C diluted in the ratio 10 無/310 無 from the original A (10 mL) and A' (20 mL) or is the dilution somehow compensated by the measurement method?

            Comment


            • #7
              Re: Convert back to per gram dried weight

              ??? When i re-read it again, it would seem so. However, that would make 4 solutions? - OP, so sorry, but could you please clarify again?

              Comment

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