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*Hints:
Know what the Law of Conservation of Mass
and predict it'sits products
Use stoichiometry:
aA + bB ---> cC + dD

Chemical Properties of C21H34O2:
Solubility = Since this is an oil, the substance is highly soluble in water, due to density differences, but readily soluble in ethers (organic compounds that is two alkyl or aryl groups connected to an Oxygen atom) and alcohols (organic compounds where a carbon atom is connected to a hydroxyl group [-OH]).
those rainbow colorcolored oil pools
{http://upload.wikimedia.org/wikipedia/commons/thumb/5/51/Ether-%28general%29.png/800px-Ether-%28general%29.png} This is the general denoted formula of an ether R - O - R' {http://upload.wikimedia.org/wikipedia/commons/thumb/d/d4/Alcohol_general.svg/185px-Alcohol_general.svg.png} This is a picture of alcohol
Elemental Composition: A cathecol ring with a C15 side chain
It has not been synthesized in a lab yet, I believe.
This is legal since all you have to do is extract it from a very common plant.
The hopes isare that certain removed while the the miscible
Combustion of urushiol:
2C21H34O2(L)+57O2(g) ---> 34H2O(g)+ 42CO2(g)

his visiting mother;mother, collapsed and
{2932490445_f56f5424d7_o.jpg} A representation of how Ryan was found
The motive was unclear, but it seemed to be suicide.

Disclaimer's Note: This wiki page is a HIGH SCHOOL chemistry project that contains factual information, but a FICTIONAL murder mystery. All circumstances, persons and events portrayed are purely imaginary and fabricated. Any similarities between actual circumstances, persons and events are completely coincidental.
barely know it'sits name. However, species of theToxicodendronthe Toxicodendron genus, the
Urushiol is an incredibly powerful irritant and allergen. A single nanogram (ng, which is 10-9, one BILLIONTH of a gram) of the oil can cause a rash, but in general it requires 100 ng to induce a rash. Urushiol samples are still potent enough to cause after A CENTURY. At minimum, a specimen of urushiol oil can stay active for 1~5 years.
A common phrase uttered by people to avoid urushiol is "leaves of three, let them be."

2) Then multiply the subscripts by that number (42 / 2 = 21, 68 / 2 = 34, 4 / 2 = 2)
The True Ending:
{http://upload.wikimedia.org/wikipedia/commons/9/92/US_Army_CID_crime_scene_investigator.jpg} Investigator The The lab results gotten away.
{http://l.yimg.com/l/tv/us/img/site/97/68/0000039768_20070516140920.jpg?y=626&sig=FnxD9b7IEkRNLaDrHIVqhg--} Samantha Lily Fine, you got me. I was just so mad at Ryan for dumping me! I didn't want to escalate to killing him, just getting back at him. I saw on T.V. once that two people smoked poison sumac and died because they had an allergic reaction. I used a lot less than the guys on the television did, but I honestly didn't know that it would kill Ryan! Plus, using a plant would frame Melisa, I hated her as much as I was mad at Ryan.
{http://upload.wikimedia.org/wikipedia/commons/9/92/US_Army_CID_crime_scene_investigator.jpg} Investigator Well, you're going away for a long time. Maybe you can find another boyfriend once your out of jail in a couple of decades.

The combustion of benzene (C6H6), which is a hydrocarbon hexagonal ring, is 2C6H6 9(l)+ 15O2(g) ---> 12CO2 (g)+ 6H2O(g): two moles of benzene when reacted with fifteen moles of oxygen GAS will produce twelve moles of carbon dioxide and six moles of water vapor. {http://upload.wikimedia.org/wikipedia/commons/thumb/2/23/Benzene-2D-flat.png/522px-Benzene-2D-flat.png} Benzene ring
Note the discrepancies: 54 moles of CO2 were created and 1.824 kg of EXCESS oxygen gas in respects to the benzene combustion was consumed! Using those two bits of information you could find out the formula of the substance.
Find out the combustion equation of the substance.
a) 54 moles of CO2 were produced when only twelve should of been. That means the substance combusted to produce 42 moles of CO2.
b) 1.824 kg O2 * 1000 grams O2 = 1824 grams O2
.......................................................................................................................1.. kg O2
i) 1824 grams O2 * 1 mole O2 = 57 moles O2
................................................. 32 g O2
ii) Deduce that since 57 moles of excess O2 were used, that means it was consumed by the combustion reaction of the poison.
At this point the equation is aCaHbOc + 57 O2(g) --> 42CO2(g) + dH2O(g)
------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
Solve for the coefficients of the poison; this is the tricky, tricky, tricky, part.
Since the law of conservation of mass dictates that matter can not be created nor destroyed, that means 42 moles of Carbon were reactants.
aC42HbOc + 57 O2(g) --> 42CO2(g) + dH2O(g)
Since there are 42 moles of C in the poison, 42 moles C * 12.01 grams C (the molar mass of Carbon) = 504.42 grams C.
................................................................................................................................................................................................................................................................................................................................................. 1 moles C
Since 504.42 grams C is 79% of the chemical formula, you can find out the other subscripts by setting up ratios.
504.42 g = b (1.008 [molar mass of H]) grams
..........0.79.........................................................................................................0.17..
Cross multiply: 85.7514 = 0.79(1.008b)
Isolate b: 85.7514/(0.79*1.008)
b = 67.78 => 68
aC42H68Oc + 57 O2(g) --> 42CO2(g) + dH2O(g)
504.42 g = c (16.00 [molar mass of O]) grams
..........0.79.........................................................................................................0.10..
Cross multiply: 50.442 g = 0.79(16c)
Isolate c: 50.442 / (.79*16)
c = 3.99 => 4
aC42H68O4 + 57 O2(g) --> 42CO2(g) + dH2O(g)
------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
............3.Solve for d and a
Since there are 68 moles of H in the reactant side, there has to be 68 moles of H in the product side. To achieve that, d must equal 34, since 34 moles of H2 equals 68 moles of H.
Finally a = 1 since everything balances out.
That means the equation is C42H68O4 + 57 O2(g) --> 42CO2(g) + 34H2O(g)
..........4. Using the equation from problem 1 and the fact that 6 moles of water are created in the combustion of benzene, the creation 40 moles of water will validate ... .......... .......... .......... .............your .answer.
..........5.Find the molecular formula to find out the substance is actually C21H34O2
To find molecular formula, 1)you must divide the actual mass by the empirical mass: (318.482 / 636.964 => .5)
2) Then multiply the subscripts by that number (42 / 2 = 21, 68 / 2 = 34, 4 / 2 = 2)
The True Ending:
{http://upload.wikimedia.org/wikipedia/commons/9/92/US_Army_CID_crime_scene_investigator.jpg} Investigator The lab results came back in; the lab that processed the ivy sample from your house returned an urushiol sample with 21 carbons, 34 hydrogens, and 2 oxygens thus matching the sample solved by our ace chemist. Question now remains, why and where did you come up with such an idea? If we didn't have such an apt chemist, you would of certainly gotten away.

Given Information #3: When the lab burned a cigarette from the packet they recovered, there was an unusual result: 54 moles of CO2 (g) were produced and an excess of 1.824 kg of oxygen gas disappeared!
After breaking apart the tobacco mixture, scientists and forensic experts found substances that did not belong. Among them was a substance with a massive molecular composition. About 79.00% was carbon, about 10.7% hydrogen and about 10.00% oxygen. No other tests could be conducted since the lab used the only clean sample it had; when Ryan's body was moved, the samples were contaminated as it fell into water. What substance is it?
you identified?
If If not, what is the correct formula? Why?
If Dr. Melisa 318.482 grams, validate her answer using chemical formula you found in question 1.find the molecular formula.
*Hints:
Know what the Law of Conservation of Mass

Given Information #3: When the lab burned a cigarette from the packet they recovered, there was an unusual result: 54 moles of CO2 (g) were produced and an excess of 1.824 kg of oxygen gas disappeared!
After breaking apart the tobacco mixture, scientists and forensic experts found substances that did not belong. Among them was a substance with a massive molecular composition. About 79.00% was carbon, about 10.7% hydrogen and about 10.00% oxygen. No other tests could be conducted since the lab used the only clean sample it had; when Ryan's body was moved, the samples were contaminated as it fell into water. What substance is it?
If 40 moles of H2O (g) were created, would that confirm the substance you identified?
If Dr. Melisa told you the molar mass of the substance was 318.482 grams, validate her answer using chemical formula you found in question 1.
*Hints:
Know what the Law of Conservation of Mass
cC + dD ---> aA + bB
The benzene ring is a hexagon.
Remember how to do empirical formula.

Given Information #2: Benzene is a C6 RING.
Given Information #3: When the lab burned a cigarette from the packet they recovered, there was an unusual result: 54 moles of CO2 (g) were produced and an excess of 1.824 kg of oxygen gas disappeared!
was carbon, 10.50%about 10.7% hydrogen and
*Hints:
Know what the Law of Conservation of Mass