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afchick7689

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percentage of
« on: September 04, 2005, 01:49:14 PM »
Ok heres the problem:

A 0.20g sample of a compound is decomposed.  The nitrogen, collected over KOH at 27 degrees C and 760 mm measured 20.0 ml. Calculate the percentage of nitrogen in the compound (The KOH had a water vapor pressure of 20 mm).

So im a little confused with the wording, and im not sure what calculations need to be done in order to find the percentages,

would it be N2KOH--> 2N + KOH

and then would you use PV = nRT so,
760 mm * 20 ml = n * .0821 * (27+273.15) ? But then im not sure what value it would be to know the number of moles of nitrogen?

i just need a little help figuring out the process...

Offline xiankai

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Re:percentage of
« Reply #1 on: September 04, 2005, 09:37:34 PM »
there is no such thing as N2KOH

u do not need to find the formula of the compound, nor the reaction in this case.

the KOH as mentioned earlier, is used to "collect" the N2.

for PV=nRT, P must be in atm(i think, in which case it would be 1 because i belieev u're most certainly talking about 760mm Hg), T must be in kelvin.

all u need to do is find out the mass of N2 based on the given data, because % composition is based on mass ratio.

so mass of N2/mass of compound = % composition of N2 in compound. ;)
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afchick7689

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Re:percentage of
« Reply #2 on: September 05, 2005, 10:25:32 AM »
ooo ok great, it makes sense now... can you just check and make sure my work here is right?

A 0.20 g sample of a compound is decomposed.  The nitrogen, collected over KOH at 27 C and 760 mm measured 20.0 ml.  Calculate the percentage of nitrogen in the compound.  (The KOH solution had a water vapor pressure of 20 mm.)

PV=nRT

760 mmHg = 1 atm
27 °C +273.15= 300.15 K
20.0 mL N |    1 liter N   = .02 L N
               | 1000 mL N
So...
1 atm * 0.02 L N = n * 0.0821 (L*atm/K*mol) * 300.15 K

0.02 = n * 24.642315

8.1161 x 10-4 mol N= n

8.1161 x 10^-4 mol N | 14.01 g N = 0.011371 g N
                                |  1 mol N
So...
Mass of N2/ Mass of Compound = % composition of N2 in compound
0.011371 g N2 / 0.20 g = .057= 5.7% N2 in compound

--I have 2 questions though:
1- does it matter that its N2? Should I double the molar mass when converting from moles to grams?
and
2- why is the water vapor pressure of KOH given? should i use that somewhere, because I didnt?

thanks so much for the *delete me*

Offline xiankai

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Re:percentage of
« Reply #3 on: September 05, 2005, 11:27:05 AM »
1. it should matter, Mr refers to the molecular formula, whereares Ar only refers to the elemental formula

2. im also not sure :-\, but i think its to do with the fact that there is more than just N2 acting on the KOH, like water vapour. something about partial pressure maybe?
« Last Edit: September 05, 2005, 08:46:21 PM by xiankai »
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