Chemistry
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OpenStudy (anonymous):
How many grams of ethane gas (C2H6) are in a 12.7 liter sample at 1.6 atmospheres and 24°C? Show all work used to solve this problem.
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OpenStudy (joannablackwelder):
Any ideas on how to do this one?
OpenStudy (anonymous):
Use the formula and plug in the atmosphere and temp
OpenStudy (anonymous):
and volume
OpenStudy (joannablackwelder):
To find what?
OpenStudy (anonymous):
N of C2H6 and use the formula again
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OpenStudy (joannablackwelder):
Close! When you get n of C2H6, what formula will you use to get g of C2H6?
OpenStudy (anonymous):
i dont know
OpenStudy (joannablackwelder):
From moles of C2H6, you can use the molar mass to get to mass of C2H6.
OpenStudy (anonymous):
would that be what we did last time of multiplying?
OpenStudy (joannablackwelder):
Yup, something like that.
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OpenStudy (anonymous):
oh ok
OpenStudy (anonymous):
n=1.03
OpenStudy (joannablackwelder):
Hm, I get it a little different.
OpenStudy (anonymous):
What did you get?
OpenStudy (joannablackwelder):
0.83 mol
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OpenStudy (anonymous):
ok so i did 1.6*12.7=n*.0821*24 == 20.32=n*19.704
OpenStudy (joannablackwelder):
You have to change your temp to K.
OpenStudy (anonymous):
ohh thats why
OpenStudy (anonymous):
297.15k?
OpenStudy (joannablackwelder):
Yup, remember the units of R. They all have to match the units you are using in the rest of the equation.
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OpenStudy (anonymous):
ok got it n=.833
OpenStudy (anonymous):
.833 mol C2H6
OpenStudy (joannablackwelder):
Yay!
OpenStudy (anonymous):
now do we just plug it in again?
OpenStudy (joannablackwelder):
Plug it into what?
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OpenStudy (anonymous):
PV=nRT
OpenStudy (anonymous):
no right
OpenStudy (joannablackwelder):
We have moles of C2H6, so we just need to use molar mass to get mass of C2H6.
OpenStudy (anonymous):
what is the formula for that again?
OpenStudy (joannablackwelder):
12.01*2+1*6=30.02g/mol
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OpenStudy (anonymous):
where did you get all those numbers from
OpenStudy (joannablackwelder):
atomic mass of each element*number of that element all summed up
OpenStudy (anonymous):
ohh ok
OpenStudy (anonymous):
so would 30.02 be the Volume
OpenStudy (anonymous):
NVM thats our new N
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OpenStudy (joannablackwelder):
No, it's the molar mass of C2H6.
OpenStudy (anonymous):
ok and now we get the mass with that
OpenStudy (joannablackwelder):
You got it :)
OpenStudy (anonymous):
wait
OpenStudy (anonymous):
whats the difference between the molar mass and mass
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OpenStudy (joannablackwelder):
molar mass is the mass per mole
OpenStudy (anonymous):
and mass would be the whole thing?
OpenStudy (joannablackwelder):
Mass is the mass given in the problem.
OpenStudy (anonymous):
which is 12.7L right
OpenStudy (anonymous):
Im lost
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OpenStudy (joannablackwelder):
L is a unit of volume
OpenStudy (anonymous):
yeh so what would be the mass that was given in the problem
OpenStudy (joannablackwelder):
In this case, mass is what we are looking for.
OpenStudy (anonymous):
ok how do we find it using molar mass
OpenStudy (joannablackwelder):
But it is dependent on the problem. The molar mass is the same for each compound, no matter the problem.
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OpenStudy (joannablackwelder):
0.833 moles C2H6 *(30.02 g C2H6/1 mol C2H6) = ? g C2H6
OpenStudy (anonymous):
Thats what had me confused because in the last problems we would use two different compounds i did not know you could use the same for the formula
OpenStudy (joannablackwelder):
Hm, I'm not sure what you mean.
OpenStudy (anonymous):
For example .885 mol Ca*(1 mol O2/2 mol Ca) we used O2 and Ca
OpenStudy (joannablackwelder):
That is a molar ratio based on the balanced reaction.
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OpenStudy (joannablackwelder):
What we are using now is the molar mass.
OpenStudy (anonymous):
ohhh ok
OpenStudy (anonymous):
so what would be the answer to 0.833 moles C2H6 *(30.02 g C2H6/1 mol C2H6) = ? g C2H6
OpenStudy (joannablackwelder):
Multiply by the numerator and divide by the denominator.
OpenStudy (anonymous):
Yeh i know that part but it dosent matter if we have different unit like Mol and Grams
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OpenStudy (joannablackwelder):
You can use a ratio as long as the ratio is an equality.
OpenStudy (joannablackwelder):
30.02g C2H6 = 1 mol C2H6
OpenStudy (anonymous):
so would the answer be 36.038
OpenStudy (joannablackwelder):
I get 25 g
OpenStudy (joannablackwelder):
.833*30.02
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OpenStudy (anonymous):
ahh got it
OpenStudy (anonymous):
ok so 25.006
OpenStudy (joannablackwelder):
:)
OpenStudy (anonymous):
and now we have our n
OpenStudy (joannablackwelder):
That is mass.
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OpenStudy (joannablackwelder):
25 g
OpenStudy (anonymous):
now we transfer the mass to moles
OpenStudy (joannablackwelder):
We wanted mass, so we are done! :)
OpenStudy (anonymous):
ohh right
OpenStudy (anonymous):
so the answer is 25g
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OpenStudy (anonymous):
but if we wanted to convert it to mole we would do 25/30?
OpenStudy (anonymous):
ok i got it
OpenStudy (anonymous):
thats the part that always gets me
OpenStudy (joannablackwelder):
You got it!
OpenStudy (joannablackwelder):
Just look at the units.