Question

Determine the mass of butane gas that was released from the lighter and collected in the gas collection tube.
I have the graphs and I will post them if your willing to help (:

Answer 1

use the formula:
PV=nRT
you should know the volume of the gas, its pressure, and its temperature. Just adjust the formula and solve for n.

Answer 2

mass of lighter before (g)
32.105 g
volume of butane gas collected (mL)
90. mL
mass of Lighter after (g)
31.844 g
room temperature (K)
297 K
air pressure (atm)
1.2 atm
so (90)(1.2) = (297) ?would be how its set up?

Answer 3

yes, it is (90)(1.2)=n(0.0821)(297)
0.0821 is the gas constant.
once you calculate the moles of butane, divide the number of grams (32.105 - 31.844) by that number of moles to get the molar mass.

Answer 4

I got 2.75? is that right..

Answer 5

Actually, it needs to be 0.090 L in the equation instead of 90 mL, since the units get messed up w/ R. I got 0.00443 moles, and a molar mass of 58.93.
The actual molar mass of butane (C4H10) is 58.12, so your experiment had about a 1.4% error. great job!

Answer 6

Okay so my final answer would be 58.12? (:

Answer 7

No, your answer is 58.93 .
If you had a flawless experiment, then it would be closer to 58.12. But a 1.5% error is very good. If you put 58.12, your teacher will probably get mad and suspect that you changed the numbers.

Answer 8

okay thanks ben (: your a live savor! lol

Answer 9

Any time! (not too literally, though)