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Chemistry 7 Online
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.

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

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.

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

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.

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?

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

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

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

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

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.

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.

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

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

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.

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

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.

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