one q for stoichiometry (1 Viewer)

[s_t_a_r1234]

q.a) if 12 g of gas composed of nitrogen and oxygen occupies a volume of 4.9 L, what is its molecular mass?
b)What is the molecular formulae of the gas. Give reasons

 

[serge]

does it say what temperature / pressure...
I need to use the right constant

 

[s_t_a_r1234]

**

yea woops its, 25C and 101.3kPa

 

[serge]

n= m/mm = V/MV

12/mm=4.9/24.47

24.47x12/4.9= mm

9MV=24.47 for that temp and pressure0

so mm=59.9

knowing that Nitrogen's mm=14
and Oxygen's mm=16
(plug in a few variations that you know)
and eventually

N2O2 is the gas

 

[s_t_a_r1234]

sorry could u tell me how u worked out hte molecular formulae plz, n could you tell me what those pronumerals stand for, sorry i really suck at this

 

[Dreamerish*~]

sorry could u tell me how u worked out hte molecular formulae plz, n could you tell me what those pronumerals stand for, sorry i really suck at this

Look at this thread. I think I tried to explain everything.

Just in case you can't be bothered: Your molecular mass was 60 g. You know that nitrogen is 14 and oxygen is 16. So just do a little bit of trial and error, and you'd get N₂O₂.

n = moles
m = mass
M = molar mass (using mm is incorrect)
V = volume
MV = molar volume (not sure if we can use that one, either)

Also, At 0°C and 1.00 atm pressure, one mole of any gas occupies 22.41 L.
At 25°C and 1.00 atm pressure, one mole of any gas occupies 24.79 L.

24.47 is the old one. Don't use it.

 

[s_t_a_r1234]

so was he incorrect? could you please mark my answer, n tell me where i went wrong refer to the q i posted above.
a)4.9L=n x24.79L
N=0.19 MOL
M=12gx0.19mol
=2.28g

 

[serge]

so was he incorrect? a)4.9L=n x24.79L
N=0.19 MOL
M=12gx0.19mol
=2.28g

nope

I wasnt incorrect, just a few decimal places less accurate than Dreamerish
but that wont stop the answer from being N2O2

n= m/M =v/24.79 in other words 12/M=4.9/24.79

(you didnt rearrange the fractions properly in you answer)
(You were meant to divide 12 by 0.19)

thats why your molecular mass is 2.3
when it is actually 60

still dont understand?

 

[Dreamerish*~]

so was he incorrect? could you please mark my answer, n tell me where i went wrong refer to the q i posted above.
a)4.9L=n x24.79L
N=0.19 MOL
M=12gx0.19mol
=2.28g

The part in bold is incorrect.

Also, take care that you use "n", not "N", and "mol" instead of "MOL".

n = 0.19766 (mol is not necessary as n = moles), don't round off during calculations. Write up to at least 4 decimal places. Anyway, you rounded 0.19766 to 0.19 which was incorrect.
m = 12 g

n = m/M
.: M = m/n

M = 12/0.19766 = 60.71 g

 

[serge]

dont write molecular mass in grams
its AMU (atomic mass units)

 

[s_t_a_r1234]

oh yea woops, but you guys still didnt explain how you worked out the molecular formula

 

[serge]

trial and error

 

[nit]

Ok, just a few points - firstly, most glaring, is :

Originally posted by Dreamerish*~
(mol is not necessary as n = moles)

You always have to specify units, every step of the way here. There are plenty of units that the chemical amount can take - millimoles, micromoles, nanomoles to name a few. Hence, the use of mol is essential here.

Also, just take care when using pronumerals to represent things - don't just write down an equation such as n=m/M - explain what each pronumeral means before substituting in. I prefer using abreviations such as FW for formula mass (formula weight), which are quite clear in terms of what they represent. While I think it is more standard in the HSC to use M for molecular weight, that can become confused with other concepts such as molarity if you don't explain what it means. "mm" is not incorrect as long as you clearly provide a definition for it.

Serge: The units of molecular mass aren't that important - grams/mol or just grams suffice in this case - AMU is hardly necessary, though the formula weight may be expressed in terms of them. But it is undoubtedly necessary to specify some correct units at least.

 

[Dreamerish*~]

Thank you.

 

[serge]

The units of molecular mass aren't that important - grams/mol or just grams suffice in this case - AMU is hardly necessary, though the formula weight may be expressed in terms of them. But it is undoubtedly necessary to specify some correct units at least.

by saying grams will suffice for molecular mass
you ruin the equation, its not scientifically correct

plus if you have a weight in grams and a molecular mass in grams
it'll confuse the markers

grams will suffice for mass, not molecular mass

its all spelt out clearly on the periodic table

 

[nit]

OK, yeh my post above was slightly incorrect due to the way the question was worded. Molecular mass is, as it suggests, the mass of one molecule of the substance. Hence the units of AMU are most correct in answering this question. Grams will still work if you use the conversion factor that

1 AMU = 1.66E(-27)g or whatever it is.

So:

Originally posted by serge
by saying grams will suffice for molecular mass
you ruin the equation, its not scientifically correct

Since you want a molecular mass, grams will certainly suffice for the answer, but in this case it makes more sense to use AMU.

Originally posted by serge
plus if you have a weight in grams and a molecular mass in grams
it'll confuse the markers

hmm?

If, as questions generally do, you're asked for the molar mass of the substance, then the units required will be grams/mol or grams etc., depending on the way you phrase your answer.

 

[s_t_a_r1234]

what are nanomoles, millimoles and all those other moles

 

[s_t_a_r1234]

so there are no accurate methods of actually finding out the molecular formula besides rtial and error?

 

[Dreamerish*~]

so there are no accurate methods of actually finding out the molecular formula besides rtial and error?

milli - 1/1000 of something.
nano - even smaller.

Don't worry, HSC chemistry doesn't usually deal with these terms. Just moles is all you need.

For your question, there was no other way than trial and error, but since the identities of the elements (oxygen and nitrogen) were already given to you and the molecular mass is relatively low, it's not too difficult.

You may be thinking of the empirical formula calculations, look here.

 

[s_t_a_r1234]

yea man im fine with empiracal im just still stuck on molecular formula could you pleaase help me with that, but thanks for the info on related