Chemistry question from BMAT

Practice Question Solving
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Hello! This question is from the BMAT 2014 and I was wondering how to work with the ratios. It seems simple but it confuses me :sadhug:
What I did was finding the number of moles of the compound in 6g. That would be 0.1 mol.
Then I inferred that 1 mol would be 60gr, so the Mr of the compound is 60. Do I just have to guess the molecular formula with the ratios or there’s any calculation I could do?
Thank you so much in advance! :sunflower:

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Hey!

[Album] imgur.com

Hope my answer’s correct!

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hello! :sunflower: why did you multiply by 0,5 ? :pleading_face:

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:heart:
I considered each part as a gram.
Like 6 part is 6 grams…
So 6 part is 0.5 mole in carbon atom.

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thank you so much :two_hearts::ok_woman:

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You’re welcome :smiling_face_with_three_hearts::heart::pray:t2:

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hi everyone!

i’m really sorry i don’t understand the worked solution, could someone try explaining it differently? How did you get 0,5 from 6 parts?

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Hi Juliette!
This is what Dorsa did:
firstly she calculated the moles of C,H and O from the given parts in the question.(considering it given mass)
C= 6/12=0.5
H=1/1=1
O= 8/16=0.5.
After that we calculated/found the lowest whole no. ratio of the different components(by dividing each mole value with the lowest no. which here is 0.5)
C=0.5/0.5=1
H=1/0.5=2
O=0.5/0.5=1
from here we have the empirical formula which is CH2O.
from here we will find the molar mass of the gas.
6g of the gas has a volume of 2.4dm^3.
considering molar volume(24dm^3) we have 60g of the gas.
empirical formula is 30g.
using Molecular mass=n x empirical formula mass.
n=60/30=2.
Thus molecular formula=2(CH2O)=C2H4O2.
Hope it helps:)

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hey! now that I re read the exercise I dont get where it says that the compound weights 60grams :pleading_face:

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Thank you so much Ujjwal!!!
it’s much more clear now!

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yes! but how did you know that we have 1 mole of the compound? I mean, the question mentions that in the case we have 6grams, we have 0.1 moles.
in other words, when should I infer that the exercise asks for 1 mole of the compound?

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yes🙆‍♀️ I understood that 1 mole occupies 24dm3. I was confused because I thought this was a general rule for gases at room temperature. So the problem is telling us, in that sentence, that not only that rules apply here but that we also have 1 mole of the compound? (in this exercise) What I mean is that the problem itself does not states specifically that we have 1 mol of the compound, so we just infer it from the general rule? Thank you soo much x your patience

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Hi!
Ya you’re right that’s a general rule that 1 mol of any gas occupies 24dm^3 at room temperature and pressure.
And here we have 0.1 moles you can infer it from the fact that we have 2.4dm^3 of gas present and molar volume is 24dm^3.
Thus, n=2.4/24=0.1moles.
Now, you can use this mole to calculate the molar mass of the given gas.
n=m/Mr
0.1=6/Mr
Mr = 6/0.1=60g.
I think my above explanation was bit sketchy this should make things clear for you :slight_smile:

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OMG i cant believe I missed that. Im sincerely grateful to you, thank you sooo much for your kidness and patience

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