Homework Help: Chemistry

Posted by Anonymous on Monday, January 17, 2011 at 6:54pm.

"The analyzer said that at 20degreesC the unknown gas was 82.8% carbon by weight and 17.2% hydrogen. The analyzer also diffused the gas through a special apparatus and measured the rate of its diffusion compared to how fast oxygen gas diffused under the same conditions. The rates of diffusion were 5.0cm(cubed)/sec for oxygen gas and 3.7cm(cubed)/for unknown gas. With this information figure out the molecular formula for the unknown gas

Chemistry - DrBob222, Tuesday, January 18, 2011 at 12:06am
You go through and redo the math for more accurate numbers but here is how you do with some estimates along the way.
Take a 100 g sample. This will give you
82.8 g C and 17.2 g H. Convert to moles.
grams/atomic mass = moles.
moles C = 82.8/about 12 = 7
moles H = 17.2/about 1 = about 17
Now find the ratio of the two elements to each other. The easy way to do this is to divide the smaller number by itself, which assures you of getting 1.00 for that. Then divide the other number by that same small number. I get
7/7 = 1.00
17/7 = about 2.4
I expect when you do the numbers more accurately you will obtain 1 to 2.5 which obviously is a whole number ratio of 2 to 6 so the empirical formula is C2H5.

From the diffusion data:
(5/3.7) = sqrt(Munk/MO2)
Solve for Munk (molar mass unknown). I think you will get something like 58 (but not exactly). Then
So the formula for the compound must be (C2H5)n and you want to determine n.
The empirical formula mass is about 29 (2*12 + 5*1) = about 29.
n = 58/29 = 2 and the formula will be
(C2H5)2 or C4H10.

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