Report, edit, etc...Posted by n2o-SiMpSoNs on 2006-11-08 at 20:13:42
lol Could someone help me with this problem. If you want ill give you some mineralz 
ok here is the problem
A confiscated white substance, suspected of being cocaine, was purified by a forensic chemist and subjected to chemical analysis. Combustion of a 50.0 mg sample yielded 150.0 mg of CO2 and 46.05 mg H2O. Analysis for nitrogen showed that is contained 9.39% N by mass. The formula for cocaine is C17H21NO4. Can the forensic chemist conclude that the suspected compound is cocaine? Please show all work.
Thank you if you do this.
ok here is the problem
A confiscated white substance, suspected of being cocaine, was purified by a forensic chemist and subjected to chemical analysis. Combustion of a 50.0 mg sample yielded 150.0 mg of CO2 and 46.05 mg H2O. Analysis for nitrogen showed that is contained 9.39% N by mass. The formula for cocaine is C17H21NO4. Can the forensic chemist conclude that the suspected compound is cocaine? Please show all work.
Thank you if you do this.
Report, edit, etc...Posted by Syphon on 2006-11-08 at 20:27:12
If it was Cocaine it would've had 21.5% Nitrogen/mass. 
CO[sub]2[/sub] + H[sub]2[/sub]O + N ↔ H[sub]2[/sub]CO[sub]3[/sub]N?
CO[sub]2[/sub] + H[sub]2[/sub]O + N ↔ H[sub]2[/sub]CO[sub]3[/sub]N?
Report, edit, etc...Posted by Mune'R0x on 2006-11-08 at 20:39:27
lol y u stab me?
Report, edit, etc...Posted by BeeR_KeG on 2006-11-08 at 20:57:31
Oh damn , I'm good at Chemical theories but I hate nomenclature.
Report, edit, etc...Posted by Syphon on 2006-11-08 at 21:16:39
QUOTE(BeeR_KeG @ Nov 8 2006, 08:57 PM)
Oh damn , I'm good at Chemical theories but I hate nomenclature.
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...Carbonic Nitride?
Report, edit, etc...Posted by Tango on 2006-11-08 at 21:24:35
Sorry I am absolutely horrible in Chemistry and dont know how to solve that. At my high school, after I took Chemistry, Physics was way easier. Just some encouragement. Physics literaly retaught chemistry at my school.
Report, edit, etc...Posted by DT_Battlekruser on 2006-11-08 at 23:15:57
QUOTE(Syphon(MM) @ Nov 8 2006, 06:16 PM)
...Carbonic Nitride?
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Isn't that better arranged CH[sub]2[/sub]NO[sub]3[/sub]?
Report, edit, etc...Posted by Syphon on 2006-11-08 at 23:40:16
QUOTE(DT_Battlekruser @ Nov 8 2006, 11:15 PM)
Isn't that better arranged CH[sub]2[/sub]NO[sub]3[/sub]?
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Yes, but maybe not. I have no idea what it is.
Report, edit, etc...Posted by ClearWater on 2006-11-09 at 00:04:22
QUOTE(n2o-SiMpSoNs @ Nov 8 2006, 08:13 PM)
lol Could someone help me with this problem. If you want ill give you some mineralz
ok here is the problem
A confiscated white substance, suspected of being cocaine, was purified by a forensic chemist and subjected to chemical analysis. Combustion of a 50.0 mg sample yielded 150.0 mg of CO2 and 46.05 mg H2O. Analysis for nitrogen showed that is contained 9.39% N by mass. The formula for cocaine is C17H21NO4. Can the forensic chemist conclude that the suspected compound is cocaine? Please show all work.
Thank you if you do this.
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ok here is the problem
A confiscated white substance, suspected of being cocaine, was purified by a forensic chemist and subjected to chemical analysis. Combustion of a 50.0 mg sample yielded 150.0 mg of CO2 and 46.05 mg H2O. Analysis for nitrogen showed that is contained 9.39% N by mass. The formula for cocaine is C17H21NO4. Can the forensic chemist conclude that the suspected compound is cocaine? Please show all work.
Thank you if you do this.
[right][snapback]585919[/snapback][/right]
Combustion products: 150.0mg CO[sub]2[/sub] and 46.05mg H[sub]2[/sub]O + Heat
Convert to moles:
QUOTE
Molar mass of CO[sub]2[/sub]:
12.01g(Carbon)+2*16.00g(O)=44.01g
44.01g * 1000mg / 1g = 44010mg
150.0mg CO[sub]2[/sub] * 1mole / 44010mg = .003408moles
Mass of C
.003408mol(CO[sub]2[/sub]) * 1mol(Carbon) / 1mol(CO[sub]2[/sub]) * 12.01g(Carbon) / 1mol = .04093g(Carbon)
Mass of O
.003408mol(CO[sub]2[/sub]) * 2mol(Oxygen) / 1mol(CO[sub]2[/sub]) * 16.00g(Oxygen) / 1mol = .1091g(Oxygen)
Molar mass of H[sub]2[/sub]O:
2*1.01g(H)+16.00g(O)=18.02g
18.02g * 1000mg / 1g = 18020mg
46.05mg H[sub]2[/sub]O * 1mole / 18020mg = .002555moles
Mass of H
.002555mol(H[sub]2[/sub]O) * 2mol(Hydrogen) / 1mol(H[sub]2[/sub]O) * 1.01(Hydrogen) / 1mol = .005161g(Hydrogen)
Mass of O
.002555mol(H[sub]2[/sub]O) * 1mol(Oxygen) / 1mol(H[sub]2[/sub]O) * 16.00(Oxygen) / 1mol = .005161g(Oxygen)
Mass of N
50.0g * 9.39% = 4.695g
12.01g(Carbon)+2*16.00g(O)=44.01g
44.01g * 1000mg / 1g = 44010mg
150.0mg CO[sub]2[/sub] * 1mole / 44010mg = .003408moles
Mass of C
.003408mol(CO[sub]2[/sub]) * 1mol(Carbon) / 1mol(CO[sub]2[/sub]) * 12.01g(Carbon) / 1mol = .04093g(Carbon)
Mass of O
.003408mol(CO[sub]2[/sub]) * 2mol(Oxygen) / 1mol(CO[sub]2[/sub]) * 16.00g(Oxygen) / 1mol = .1091g(Oxygen)
Molar mass of H[sub]2[/sub]O:
2*1.01g(H)+16.00g(O)=18.02g
18.02g * 1000mg / 1g = 18020mg
46.05mg H[sub]2[/sub]O * 1mole / 18020mg = .002555moles
Mass of H
.002555mol(H[sub]2[/sub]O) * 2mol(Hydrogen) / 1mol(H[sub]2[/sub]O) * 1.01(Hydrogen) / 1mol = .005161g(Hydrogen)
Mass of O
.002555mol(H[sub]2[/sub]O) * 1mol(Oxygen) / 1mol(H[sub]2[/sub]O) * 16.00(Oxygen) / 1mol = .005161g(Oxygen)
Mass of N
50.0g * 9.39% = 4.695g
QUOTE
Molar mass of C[sub]17[/sub]H[sub]21[/sub]NO[sub]4[/sub]:
17*12.01g(Carbon)+21*1.01g(H)+14.01g(N)+4*16.00g(O)=303.39g
303.39g * 1000mg / 1g = 303390mg
50.0mg C[sub]17[/sub]H[sub]21[/sub]NO[sub]4[/sub] * 1mole / 303390mg = .000165moles
Mass of C
.000165mol(C[sub]17[/sub]H[sub]21[/sub]NO[sub]4[/sub]) * 17mol(Carbon) / 1mol(C[sub]17[/sub]H[sub]21[/sub]NO[sub]4[/sub]) * 12.01g(Carbon) / 1mol = .0337g(Carbon)
Mass of H
.000165mol(C[sub]17[/sub]H[sub]21[/sub]NO[sub]4[/sub]) * 21mol(Hydrogen) / 1mol(C[sub]17[/sub]H[sub]21[/sub]NO[sub]4[/sub]) * 1.01g(Hydrogen) / 1mol = .00350g(Hydrogen)
Mass of N
.000165mol(C[sub]17[/sub]H[sub]21[/sub]NO[sub]4[/sub]) * 1mol(Nitrogen) / 1mol(C[sub]17[/sub]H[sub]21[/sub]NO[sub]4[/sub]) * 14.01g(Nitrogen) / 1mol = .00231g(Nitrogen)
17*12.01g(Carbon)+21*1.01g(H)+14.01g(N)+4*16.00g(O)=303.39g
303.39g * 1000mg / 1g = 303390mg
50.0mg C[sub]17[/sub]H[sub]21[/sub]NO[sub]4[/sub] * 1mole / 303390mg = .000165moles
Mass of C
.000165mol(C[sub]17[/sub]H[sub]21[/sub]NO[sub]4[/sub]) * 17mol(Carbon) / 1mol(C[sub]17[/sub]H[sub]21[/sub]NO[sub]4[/sub]) * 12.01g(Carbon) / 1mol = .0337g(Carbon)
Mass of H
.000165mol(C[sub]17[/sub]H[sub]21[/sub]NO[sub]4[/sub]) * 21mol(Hydrogen) / 1mol(C[sub]17[/sub]H[sub]21[/sub]NO[sub]4[/sub]) * 1.01g(Hydrogen) / 1mol = .00350g(Hydrogen)
Mass of N
.000165mol(C[sub]17[/sub]H[sub]21[/sub]NO[sub]4[/sub]) * 1mol(Nitrogen) / 1mol(C[sub]17[/sub]H[sub]21[/sub]NO[sub]4[/sub]) * 14.01g(Nitrogen) / 1mol = .00231g(Nitrogen)
.165moles C[sub]17[/sub]H[sub]21[/sub]NO[sub]4[/sub] + "unlimited" O[sub]2[/sub] -> (Yields) 3.408moles CO[sub]2[/sub] + 2.555moles H[sub]2[/sub]O
Well I don't feel like showing Oxygen but this should be sufficient proof to show that there is no constant ratio for Carbon, Hydrogen, and Nitrogen to equate the masses.
Report, edit, etc...Posted by Rantent on 2006-11-09 at 02:29:15
A confiscated white substance, suspected of being cocaine, was purified by a forensic chemist and subjected to chemical analysis. Combustion of a 50.0 mg sample yielded 150.0 mg of CO2 and 46.05 mg H2O. Analysis for nitrogen showed that is contained 9.39% N by mass. The formula for cocaine is C17H21NO4. Can the forensic chemist conclude that the suspected compound is cocaine? Please show all work.
12.01 + 15.99*2 = 43.99 g/mol CO2
% C in CO2 = 27.3%
1.01*2+15.99 = 18.01 g/mol H2O
%H in H2O = 11.2%
150.0*27.3% = 40.95 mg C
46.05*11.2% = 5.157 mg H
12.01*17+1.01*21+14.00+15.99*4 = 303.34 g/mol Cocaine
12.01*17/303.34 = 67.3% C in Cocaine
1.01*21/303.34 = 6.99% H in Cocaine
14.00/303.34 = 4.61% N in Cocaine
15.99*4/303.34 = 21.0% O in Cocaine
Results:
9.39% N in Substance
81.9% C in Substance
10.3% H in Substance
Oops I forgot oxygen.
Screw the whole thing.
It's not cocaine btw.
12.01 + 15.99*2 = 43.99 g/mol CO2
% C in CO2 = 27.3%
1.01*2+15.99 = 18.01 g/mol H2O
%H in H2O = 11.2%
150.0*27.3% = 40.95 mg C
46.05*11.2% = 5.157 mg H
12.01*17+1.01*21+14.00+15.99*4 = 303.34 g/mol Cocaine
12.01*17/303.34 = 67.3% C in Cocaine
1.01*21/303.34 = 6.99% H in Cocaine
14.00/303.34 = 4.61% N in Cocaine
15.99*4/303.34 = 21.0% O in Cocaine
Results:
9.39% N in Substance
81.9% C in Substance
10.3% H in Substance
Oops I forgot oxygen.
Screw the whole thing.
It's not cocaine btw.
Report, edit, etc...Posted by n2o-SiMpSoNs on 2006-11-09 at 16:47:03
QUOTE(Omega @ Nov 8 2006, 09:24 PM)
Sorry I am absolutely horrible in Chemistry and dont know how to solve that. At my high school, after I took Chemistry, Physics was way easier. Just some encouragement. Physics literaly retaught chemistry at my school.
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hahaha I've never had so much trouble with a class in my life. I don't get it at all. In any other class I've taken I can pay attention in class and get it easily but I do that plus tryo to do the work but I still don't get it. not to mention my teacher isn't good at all.
Also, thanks ClearWater and Rentent so much and Syphon some too
Report, edit, etc...Posted by Syphon on 2006-11-09 at 19:38:01
QUOTE(n2o-SiMpSoNs @ Nov 9 2006, 04:47 PM)
hahaha I've never had so much trouble with a class in my life. I don't get it at all. In any other class I've taken I can pay attention in class and get it easily but I do that plus tryo to do the work but I still don't get it. not to mention my teacher isn't good at all.
Also, thanks ClearWater and Rentent so much and Syphon some too
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Also, thanks ClearWater and Rentent so much and Syphon some too
[right][snapback]586291[/snapback][/right]
I have it the other way around, I have a 35 in french, 36 in math, and 70 in science.
In science I don't have a textbook, notebook, binder or pay any attention in class, and I haven't done one shred of homework in the class all year, nor have a I taken a note.