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    Moles of CHI3 = Mass/MR
    = 12.0/(106.5 + 12 + 1)
    = 0.100

    3:1 ratio between I2 and CHI3

    Moles of I2 = 0.100 x 3
    = 0.300

    Mass of I2 = Moles x MR
    = 0.300 x 71
    = 21.3g

    I think this is right.
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    (Original post by Natasha18))
    Thank you for the reply

    Can I ask why did you use the value of 106.5 when working out the number of moles of CHI3 I thought it'd be 126.9 (the mr of iodine) ?

    and when working out the mass of iodine at the end why do you use the value of 71 and not the Mr of iodine?

    Sorry for all the questions, just trying to understand how to work it all out
    I used values for Chlorine by mistake. I'm overtired, haha!

    Sorry!! Your values are right.
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    (Original post by Natasha18))
    Ahaha okay no worries thank you very much
    You're welcome. I've worked it out again and this is what I have now.

    Moles of CHI3 = Mass/MR
    = 12.0/(380.7 + 12 + 1) 380.7 comes from 126.9 x 3 (due to 3 I atoms)
    = 0.0305

    3:1 ratio between I2 and CHI3

    Moles of I2 = 0.0305 x 3
    = 0.0915

    Mass of I2 = Moles x MR
    = 0.0915 x 253.8
    = 23.2g
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    This is how I would do it:

    We know that the percentage yield is 85.1%

    Therefore to obtain 12 g of triiodomethane (CHI3) we need to obtain a theoretical yield of 12/0.851=14.1.

    So we need a theoretical yield of 14.1 g of triiodomethane (CHI3)

    Amount of moles of CHI3 = Mass/Molar Mass
    = 14.1/(3*126.9 + 12 + 1)
    = 0.0358

    3:1 molar ratio between I2 and CHI3

    Amount of Moles of I2 = 0.0358 x 3
    = 0.1074 moles

    Mass of I2 =Amount of Moles of I2 x Molar Mass of I2
    = 0.1074 x 253.8
    = 27.3g

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    (Original post by Natasha18))
    Hi I'd really appreciate any help on the step by step method and working out help for this moles calculation I'm stuck on :/

    The yield of triiodomethane (CHI3) obtained by a student was 82.1% Find the maximum mass of iodine that the student would have used to form 11.0g of triiodomethane.

    CH3CHO+ 3I2 + 4NaOH --> CHI3 + HCOONa + 3NaI + 3H2O

    Thank you,

    Natasha
    xxx
    Hi Natasha, this question is easier than it looks.

    OK, first you have to find the mols of CHI3. Youre given the grams, so you divide 11 by whatever the Mr of CHI3 is.

    Then, looking at the equation we see that the ratio of I2 to CHI3 is 3:1. So, if for part 1 you got, say, 6 mols of CHI3, youd have 18 mols of I2.

    Now for the sake of argument let's use the same numbers. We have 18 mols of I2. To find the mass needed we simply use the equation mass= mols x Mr. So just look up the Mr of I2 and multiply it by the mols you calculated.

    Hope this helped
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    (Original post by Natasha18))
    Hi I'd really appreciate any help on the step by step method and working out help for this moles calculation I'm stuck on :/

    The yield of triiodomethane (CHI3) obtained by a student was 82.1% Find the maximum mass of iodine that the student would have used to form 11.0g of triiodomethane.

    CH3CHO+ 3I2 + 4NaOH --> CHI3 + HCOONa + 3NaI + 3H2O

    Thank you,

    Natasha
    xxx
    First you need to find the amount in mole of CHI3 which is:

    n= m/M = 11/393.7 = 0.02794

    The formula for percentage yield is: % yield = actual amount in mole/ theoretical amount in mole. You are given the percentage yield and the mole I've worked out above is the actual amount in mole. You need to rearrange the equation to get the actual amount in mole by itself.

    theoretical amount in mole = actual amount in mole / % yield = 0.02794 / (82.1/100) = 0.03403

    To find the mass of Iodine rearrange the formula n=m/M to get mass on its own i.e. m= n x M and substitute in the values m= 0.03403 x 253.8 = 8.64 (<--- and that's your mass)

    Hope this helped
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    (Original post by Natasha18))
    .
    Hi, I'm not entirely sure on the answer but here's what I did.

    1) the number of moles of CHI3 is (m/Mr) * 0.821

    2) multiply this number by 3 as the ratio of the number of moles of CHI3 is 3 less than the iodine

    3) use m = nMr to find the mass which will be part 2) * Mr of I2


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    (Original post by Natasha18))
    Hi thanks for the reply, do I need to use the %yield in my calculations then? cause I'm confused half the people I've asked say I do the other half say I don't?


    xx
    Yes you need to use it.

    If the student forms 12g of triiodomethane (Mr = 394) this is equivalent to 12/394 mol = 0.00305 mol

    ... and he only succeeded in extracting 85.1% then the theoretical yield = 0.00305 * 100/85.1 mol = 0.0358 mol

    From the equation 3 mol of iodine make 1 mol of triiodomethane

    hence mol of iodine used = 3 * 0.0358 = 0.107 mol

    Mr of iodine = 254

    Mass of iodine used = 0.107 * 254 = 27.27 g
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    (Original post by Natasha18))
    Right okay so you work out the number of moles of triiodomethane then multiply it by the yield
    no, you find out the moles of triiodomethane which you are told accounts for only 85% of the total possible product and then scale it up to 100% to get the theoretical (100%) yield

    and not find out the theoretical yield first then use that to work out the number of moles? I ask because doesn't the question ask for the maximum mass of iodine that would have been used so the theoretical yield should be used to work out the theoretical yield of triiodomethane first then the rest, instead of multiplying the number of moles of triiodomethane by the % yield?
    No, you are confused ...
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    (Original post by charco)
    Yes you need to use it.

    If the student forms 12g of triiodomethane (Mr = 394) this is equivalent to 12/394 mol = 0.00305 mol

    ... and he only succeeded in extracting 85.1% then the theoretical yield = 0.00305 * 100/85.1 mol = 0.0358 mol

    From the equation 3 mol of iodine make 1 mol of triiodomethane

    hence mol of iodine used = 3 * 0.0358 = 0.107 mol

    Mr of iodine = 254

    Mass of iodine used = 0.107 * 254 = 27.27 g
    Why is it 12? Doesn't the student form 11g of triiodomethane? And why do you multiply the moles by 3?
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    (Original post by Natasha18))
    I mean do you work out the theoretical yield of moles or theoretical yield in terms of mass? I don't get why you work out the theoretical yield of moles and not theoretical mass yield?
    You cannot get the mass without working out the moles first ...
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    (Original post by Natasha18))
    Hi thanks for the reply, do I need to use the %yield in my calculations then? cause I'm confused half the people I've asked say I do the other half say I don't?


    xx
    Yes you need to calculate the theoretical yield.

    The desired experimental yield is 12 g and you know that the percetage yield is 85.1%.

    Percetage yield = (Experimental Yield)/(Theoretical Yield)

    0.851= 12g/(Theoretical Yield)

    Therefore the Theoretical Yield = 12g/0.851 = 14.1g

    I.e. You theoretically need to make 14.1g of triiodomethane, since your percentage yield is 0.851 giving you the desired experimental yield of 12 g of triiodomethane.

    So we need a theoretical yield of 14.1 g of triiodomethane (CHI3)

    Amount of moles of CHI3 = Mass/Molar Mass
    = 14.1/(3*126.9 + 12 + 1)
    = 0.0358 moles

    3:1 molar ratio between I2 and CHI3

    Amount of Moles of I2 = 0.0358 x 3
    = 0.1074 moles

    Mass of I2 =Amount of Moles of I2 x Molar Mass of I2
    = 0.1074 x 253.8
    = 27.3g

    Hopefully this helped

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    I'm rather confused as to why people keep using 12g CHI3 when it clearly says in the question that 11.0g is produced, with a yield of 82.1%? :confused:
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    (Original post by thegodofgod)
    I'm rather confused as to why people keep using 12g CHI3 when it clearly says in the question that 11.0g is produced, with a yield of 82.1%? :confused:
    hmmmm....

    This thread is the result of merging two threads together and I see that the very first post has been edited to nothing.

    I suspect that the same question was posted twice with different values. The original 12g was deleted, I suspect...

    :holmes:
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    (Original post by charco)
    Yes you need to use it.

    If the student forms 12g of triiodomethane (Mr = 394) this is equivalent to 12/394 mol = 0.00305 mol

    ... and he only succeeded in extracting 85.1% then the theoretical yield = 0.00305 * 100/85.1 mol = 0.0358 mol

    From the equation 3 mol of iodine make 1 mol of triiodomethane

    hence mol of iodine used = 3 * 0.0358 = 0.107 mol

    Mr of iodine = 254

    Mass of iodine used = 0.107 * 254 = 27.27 g
    hi charco,

    Can you please explain to me why you multiplied by 3?
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    (Original post by Natasha18))
    You multiply by 3 because the ratio of CHI3 to I2 is 1:3
    Oh I get it, thanks
 
 
 
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