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    http://pmt.physicsandmathstutor.com/...s%201%20QP.pdf


    Question 1d.ii

    Someone please help me I have been stuck on it for hours.

    Answer:
    FIRST CHECK THE ANSWER ON THE ANSWER LINE
    IF answer = 1.35 (g) award 3 marks
    IF answer = 0.54 (g) award 2 marks (no scale-up)
    IF answer = 0.216 (g) award 2 marks (incorrect scale-up)
    n(compound D) = 1.73/346 = 0.00500 mol
    n(1,3-diaminobenzene) required = 100/40 x 0.005
    = 0.0125 mol
    Molar mass of 1,3-diaminobenzene = 108 (g mol–1)
    AND
    Mass of 1,3-diaminobenzene = (108)(0.0125) = 1.35 g
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    (Original post by chemquestion)
    http://pmt.physicsandmathstutor.com/...s%201%20QP.pdf


    Question 1d.ii

    Someone please help me I have been stuck on it for hours.

    Answer:
    FIRST CHECK THE ANSWER ON THE ANSWER LINE
    IF answer = 1.35 (g) award 3 marks
    IF answer = 0.54 (g) award 2 marks (no scale-up)
    IF answer = 0.216 (g) award 2 marks (incorrect scale-up)
    n(compound D) = 1.73/346 = 0.00500 mol
    n(1,3-diaminobenzene) required = 100/40 x 0.005
    = 0.0125 mol
    Molar mass of 1,3-diaminobenzene = 108 (g mol–1)
    AND
    Mass of 1,3-diaminobenzene = (108)(0.0125) = 1.35 g
    moles of D = mass/Mr = 1.73/346 = 0.005 mol

    This represents 40%, therefore 100% would be 0.005 x 100/40 = 0.0125 mol

    Mr of reactant, 1,3-diaminobenzene, C6H8N2 = 108

    Hence mass of reactant needed = 0.0125 x 108 = 1.35g

    So what's the problem?
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    (Original post by charco)
    moles of D = mass/Mr = 1.73/346 = 0.005 mol

    This represents 40%, therefore 100% would be 0.005 x 100/40 = 0.0125 mol

    Mr of reactant, 1,3-diaminobenzene, C6H8N2 = 108

    Hence mass of reactant needed = 0.0125 x 108 = 1.35g

    So what's the problem?
    do you know anyone else who is good at chemistry as well or is doing A level chemistry?

    1. I don't get why you have to do 100/40?
    also how is it related to percentage yield???
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    (Original post by chemquestion)
    do you know anyone else who is good at chemistry as well or is doing A level chemistry?

    1. I don't get why you have to do 100/40?
    also how is it related to percentage yield???
    That's a strange question! I know lots of people who are "good" at chemistry.

    You are told in the question that the actual percentage yield is 40%, i.e. 40/100

    and

    percentage yield = actual yield/theoretical yield

    So rearranging:

    theoretical yield = actual yield/percentage yield = 1.73/0.4 g
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    (Original post by charco)
    That's a strange question! I know lots of people who are "good" at chemistry.

    You are told in the question that the actual percentage yield is 40%, i.e. 40/100

    and

    percentage yield = actual yield/theoretical yield

    So rearranging:

    theoretical yield = actual yield/percentage yield = 1.73/0.4 g

    Because I want to tag them
    this question I asked before: Calculate the mass of phenylamine that was produced from the 3.69 g of
    nitrobenzene in this experiment. Give your answer to three significant figures.
    Mr
    : nitrobenzene,123; phenylamine, 93.1

    IN QUESTION: Purification gave a 72.1% yield of phenylamine.


    ''purification gave 72.1% yield of phenylamine''

    Does this 72.1 also mean the actual yield that is formed? same as the question I asked at the beginning of this thread?
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    (Original post by chemquestion)
    Because I want to tag them
    this question I asked before: Calculate the mass of phenylamine that was produced from the 3.69 g of
    nitrobenzene in this experiment. Give your answer to three significant figures.
    Mr
    : nitrobenzene,123; phenylamine, 93.1

    IN QUESTION: Purification gave a 72.1% yield of phenylamine.


    ''purification gave 72.1% yield of phenylamine''

    Does this 72.1 also mean the actual yield that is formed? same as the question I asked at the beginning of this thread?
    Yes, if the yield is 72.1% it means that you get 72.1/100 of the theoretically possible amount produced.

    In a perfect world we would always get 100% of the amount expected, but in the real world with side-reactions and losses due to work-up we often get much less.

    Tag "Pigster" - he's a grumpy old bugger, but he'll get the job done
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    (Original post by charco)
    Yes, if the yield is 72.1% it means that you get 72.1/100 of the theoretically possible amount produced.

    In a perfect world we would always get 100% of the amount expected, but in the real world with side-reactions and losses due to work-up we often get much less.

    Tag "Pigster" - he's a grumpy old bugger, but he'll get the job done
    Oh I get that now,
    but why does it say in the markscheme 100/40?
    I get 40/100 but not 100/40?
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    (Original post by chemquestion)
    Oh I get that now,
    but why does it say in the markscheme 100/40?
    I get 40/100 but not 100/40?
    It's basic maths.

    If you divide by 40/100 it's the same as multiplying by 100/40
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    [QUOTE=chemquestion;76560192]
    (Original post by chemquestion)

    so 40/100 is same as 100/40

    so if we decide we want to use 40/100 we have divide 0.005 by 40/100? I just checked and it gives the same answer.
    No, 40/100 is NOT the same as 100/40.

    I said that if you DIVIDE a number by 40/100 it is the same as MULTIPLYING the same number by 100/40

    num/40/100 = num x 100/40

    And yes, you can tag me. I may not reply, but you can tag me - I love a good game of tag.
 
 
 
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