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### Topic: Empircal Moleuclar formula using Gas Laws  (Read 4188 times)

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#### ozzlomo

• Regular Member
•   • Posts: 22
• Mole Snacks: +0/-1 ##### Empircal Moleuclar formula using Gas Laws
« on: April 28, 2009, 02:33:14 AM »
A gaseous compound is 30.4% nitrogen and 69.6% oxygen by mass. A 5.25g sample of the gas at 4.0 degrees Celsius occupies a volume of 1.00 L and exerts a pressure of 1.26 atm. Find the molecular formula.

I'm not sure where to start. Any help would be greatly appreciated.

#### ozzlomo

• Regular Member
•   • Posts: 22
• Mole Snacks: +0/-1 ##### Re: Empircal Moleuclar formula using Gas Laws
« Reply #1 on: April 28, 2009, 02:42:05 AM »

"Hi, I found the right answer: N2O4.

Here is my works:
1. I found the mass of each gas in the compound using the given % by mass.
mass of N= .304*5.25= 1.596g N
mass of O= 5.25 - 1.596= 3.654g O

2. Mass-Mole conversion:
1.596g N* 1mol of N/ 14.01g of N = 0.1139 mol N
3.654g O* 1mol of O/ 16g of O = 0.2284 mol of O

3. Writing the Emperical Formula:
0.1139/0.1139= 1
0.1139/0.2284= 2
=> NO2

Emp. Formulas mass= 46.01g

4. Find the Molar Mass:
M= m*RT/PV
=> M= {5.25g*0.0821(L*atm/K*mol)*269K}/(1.26atm*1L)
M= 92.02g

5. Find the Multiplier (x)
x= Molar mass/Emp.Formula mass= 92.02g/46.01g
x= 2

Molecular Formula:
(NO2)*2 => N2O4" - iconwin

#### Borek ##### Re: Empircal Moleuclar formula using Gas Laws
« Reply #2 on: April 28, 2009, 03:25:10 AM »
Wasn't that hard, eh?
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