A hypothetical rock was collected during space exploration. Trapped inside was a no- ble gas. A volume of 1.05 mL of the gas was extracted from the rock at STP. The mass of the gas sample was determined to be 0.188 mg. Identify the gas.

Question

ipwnbunnies · Answer

Use the ideal gas equation.
$$PV = nRT$$

ipwnbunnies · Answer

We want to solve for 'n', the amount of moles of the noble gas. We are given volume, pressume (1 atm), and temperature (273 K).

anonymous · Answer

Hmm I see.  So the n moles I get from that I would have to match it to a gas?

anonymous · Answer

why do they give me the mass?

ipwnbunnies · Answer

Can you figure it out? If we have the mass of the gas, and we know how much gas there is, can't we find the mass per mole of gas? What will that be?

anonymous · Answer

ok so first,  I have PV/RT = n  

[(1atm) (0.00105L)]/[(0.082057)(273.15) = 3.4952  mol

anonymous · Answer

so..   
0.000188g * 3.4952moles = 0.0006570976 g per moles?

anonymous · Answer

looks kind of wrong

ipwnbunnies · Answer

You got the amount of moles right. Look again at your last step.

ipwnbunnies · Answer

Why would you multiply mass (in grams) and moles to get g/mol? Hmm...

ipwnbunnies · Answer

Ooo, check your amount of moles actually lol. That can't be right.

anonymous · Answer

miss understood "grams per mole" sorry.

0.188mg -> 0.000188grams 

0.000188grams/0.34952moles = 0.00053788052186 g/moles=  5.38 * 10^-4

ipwnbunnies · Answer

Yeah, that's the problem. The molar mass cannot be that low. Again, check to make sure you have the right amount of moles.

anonymous · Answer

0.000188grams / 0.00004684 moles = 4.0132967 g/moles which correspond to Hellium??

ipwnbunnies · Answer

There ya go.

anonymous · Answer

I had the wrong number of moles as you said ^ :) thanks a lot!!! keep pwning bunnies

ipwnbunnies · Answer

That hypothetical rock is floating, I guess.