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  • Help with this calculation exam question please!

    20 ferrous gluconate tablets (mass = 6.9417g) were powdered and 1.2174g of the powder was dissolved in 20ml water and 20ml 1M sulphuric acid. On titration with 0.0973M ammonium cerium (IV) sulphate solution using ferroin as indicator 22.41ml was required to reach the end point. Given that each ml of 0.1M ammonium cerium(IV) sulphate is equivalent to 5.585mg of Fe(ii), calculate the ferrous iron content (mg) for an average tablet.

    How do I go about working this out??? I am so confused!
    Thanks for your help!

  • #2
    Re: Help with this calculation exam question please!

    Well a bit late for my tired old brain, but looks like simple proportion to me.
    johnep

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    • #3
      Re: Help with this calculation exam question please!

      Originally posted by johnep View Post
      Well a bit late for my tired old brain, but looks like simple proportion to me.
      johnep
      Do you think you could help me out by giving me a starting position please? Do I use the formula:
      mass = conversion factor*volume*concentration/ concentration of conversion factor

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      • #4
        Re: Help with this calculation exam question please!

        Originally posted by shenzys View Post
        20 ferrous gluconate tablets (mass = 6.9417g) were powdered and 1.2174g of the powder was dissolved in 20ml water and 20ml 1M sulphuric acid. On titration with 0.0973M ammonium cerium (IV) sulphate solution using ferroin as indicator 22.41ml was required to reach the end point. Given that each ml of 0.1M ammonium cerium(IV) sulphate is equivalent to 5.585mg of Fe(ii), calculate the ferrous iron content (mg) for an average tablet.
        We don't use formulae, but proportions. So we may not be able to help you. But here goes...
        22.41ml of 0.0973M was required, so 22.41*.0973/.1ml of 0.1M would be required. A bit less, see?
        That can be converted to mg Fe present in the sample dissolved.
        This can then be proportioned up to the amount in 20 tablets (by *6.9417/1.2174, answer has to be more, see) and then divided by 20 to get the amount per tablet.

        The same idea can be used for most of the calculations you ask about.
        ....just my opinion

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        • #5
          Re: Help with this calculation exam question please!

          22.41ml x (0.0973M/0.1M) = 21.81ml

          5.585mg x 21.81ml = 121.78mg of iron

          121.78mg x (6.9417g/1.2174g) = 694.40mg of iron

          694.40mg/20 = 34.72mg per tablet

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