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# Someone help me with this chemistry question watch

1. 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
2. (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?
3. (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???
4. (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
5. (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?
6. (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
7. (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?
8. (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
9. [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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