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So it's really late and I'm studying for this test I have and I ran into a problem I cant figure out. This is basic chemistry, so anyone with basic knowledge can help me, the noob business major. Problem is as follows:
Given this unbalanced chemical equation:
CxHyNz(l) + O2(g) --> CO2(g) + H2O(l) + N2(g)
Question: when 0.312g of dimethylhydrazine (CxHxNz) is combusted, 0.458g of carbon dioxide, 0.374g of water and 0.145g nitrogen are collected. What is the empirical formula of dimethylhydrazine?
What I know: usually we know either the percentage of elements for the empirical formula, or the amount of each in grams, so that I can determine the mole to mole ratio. But now I only know the weight of the products, so I'm guessing that I need to work backwards somehow but I'm lost as to where to start @_@
TY for help!!!
GOT IT I LOVE EVERYONE IN THIS THREAD
molCO2 = .0104
molH2O = .0208
molN2 = .0052
then
molCO2 to mol C = .0104
molH2O to mol H = .0416
molN2 to mol N = .0104
which simplifies to CH4N
 
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iNcontroL
USA29055 Posts
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On February 22 2010 18:02 {88}iNcontroL wrote:
COMMON MAN?
CHEMISTRY'S NOT MY THING >_< >_< >_<
i do calculations, accounting, business law, managerial and financial shit. i dont play this stupid equation game ~~~~~~
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wiki would be great in the middle of the test. but im pretty sure they dont allow that. ALMOST positive.
i just need a quick build order for how to tackle this beast.
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First you have to calculate the amount n (in mole) of each product using the molar mass M (M =n/m, m is given).
If you have that you can figure out the C/H/N ratio pretty easily.
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CA10824 Posts
On February 22 2010 18:02 {88}iNcontroL wrote:
COMMON MAN?
he is indeed a commoner.
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Follow Yizuo.
Actually its easier if you just calculate the mass for each element in the left side. Say 1g of CO2 equals x g of C and y g of O. Same with water and nitrogen then procede to calculate the total mass and youll find the numbers.
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On February 22 2010 18:08 Yizuo wrote:
First you have to calculate the amount n (in mole) of each product using the molar mass M (M =n/m, m is given).
If you have that you can figure out the C/H/N ratio pretty easily.
okay so i found
molCO2 = .0104
molH2O = .0208
molN2 = .0052
...now what do i do 
EDIT: do i use CO2 to find C, H2O to find H, and N2 to find N? fml this is frustrating...
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ok, appearantly you have 4 H-Atoms for every C-Atom (CO2/H2O ratio) and the same number of C- and N-atoms (N2/CO2 ratio)
so you have CxNxH4x
I'm not sure if you can calculate the exact formula
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hm so I think I figured it out...
but if I'm going from
.0052mol N2 ----> mol N
would the ratio be
.0052(2/1) or .0052(1/1) ?
its like, for every mol N2 there is 2 mol N? its making my head spin
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On February 22 2010 18:15 mOnion wrote:Show nested quote +On February 22 2010 18:08 Yizuo wrote:
First you have to calculate the amount n (in mole) of each product using the molar mass M (M =n/m, m is given).
If you have that you can figure out the C/H/N ratio pretty easily.
okay so i found molCO2 = .0104 molH2O = .0208 molN2 = .0052 ...now what do i do EDIT: do i use CO2 to find C, H2O to find H, and N2 to find N? fml this is frustrating...
Divide each of the mole values you found by the smallest value which would be N2.
The reason you do this is to have an integer value for the number of moles.
In this case, the values you found are all easily divisible by .0052.
If it wasn't you'd multiply until integers are attainable.
So, molCO2 = 2.
molH20 = 4.
molN2 = 1.
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Baa?21242 Posts
On February 22 2010 18:21 Yizuo wrote:
ok, appearantly you have 4 H-Atoms for every C-Atom (CO2/H2O ratio) and the same number of C- and N-atoms (N2/CO2 ratio)
so you have CxNxH4x
I'm not sure if you can calculate the exact formula
edit: ignore me i fail
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to OP:
maybe it's easier to not think of moles, but of atoms and molecules (a mole is only a number after all)
for every molecule N2 we have 2 atoms of N etc
molCO2 = .0104
molH2O = .0208
molN2 = .0052
-for every molecule of CO2 we have one atom of C (.0104 C atoms)
-for every molecule H2O we have two atoms of H (0.0208 *2 H atoms)
-for every molecule of N2 we have two atoms of N (0.0052 * 2 N atoms)
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On February 22 2010 18:32 Carnivorous Sheep wrote:Show nested quote +On February 22 2010 18:21 Yizuo wrote:
ok, appearantly you have 4 H-Atoms for every C-Atom (CO2/H2O ratio) and the same number of C- and N-atoms (N2/CO2 ratio)
so you have CxNxH4x
I'm not sure if you can calculate the exact formula So wrong :3 Yes you have 2 mols of Nitrogen for each mol of N2. You can calculate how many grams of C, H, and N are on the right side of the equation. You have the starting mass of the unknown compound. To find the specific mass of each element in the compound, take the starting mass and subtract it with the masses of two of the elemnts for the mass of the third element in the compound. Then convert that mass to mols. Repeating this for each element should give you x, y, and z, and then its a simple matter of converting to integers.
i claim steal.
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Separate the C's, O's, H's, and N's. Scrap the Oxygens because it doesn't come from the dimethylhydrazine.
molCO2 = mol C = .0104
molH2O = (2)mol H = .0416
molN2 = (2)mol N = .0104
Divide each of those numbers by the lowest (.0104) to get the ratio between the Carbon/Hydrogen/Nitrogen. So it would be + Show Spoiler +
Editted: Mistyped the mol H
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On February 22 2010 18:32 Carnivorous Sheep wrote:Show nested quote +On February 22 2010 18:21 Yizuo wrote:
ok, appearantly you have 4 H-Atoms for every C-Atom (CO2/H2O ratio) and the same number of C- and N-atoms (N2/CO2 ratio)
so you have CxNxH4x
I'm not sure if you can calculate the exact formula So wrong :3 Yes you have 2 mols of Nitrogen for each mol of N2. You can calculate how many grams of C, H, and N are on the right side of the equation. You have the starting mass of the unknown compound. To find the specific mass of each element in the compound, take the starting mass and subtract it with the masses of two of the elemnts for the mass of the third element in the compound. Then convert that mass to mols. Repeating this for each element should give you x, y, and z, and then its a simple matter of converting to integers.
The way I described definatly works and is easier than yours imo...
gtg now
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Baa?21242 Posts
On February 22 2010 17:58 mOnion wrote:
I THINK I GOT IT:
molCO2 = .0104
molH2O = .0208
molN2 = .0052
then
molCO2 to mol C = .0104 followed by mol C to grams C = .1248g
molH2O to mol H = .0416 followed by mol H to grams H = .0416g
molN2 to mol N = .0104 followed by mol N to grams N = .1456g
so the molecular formula would be
C.1248 H.0416 N.1456
which simplifies to
C6H2N7
amirite?
edit: ignore me i fail
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Baa?21242 Posts
On February 22 2010 18:37 Milestone wrote:Separate the C's, O's, H's, and N's. Scrap the Oxygens because it doesn't come from the dimethylhydrazine. molCO2 = mol C = .0104 molH2O = (2)mol H = .416 molN2 = (2)mol N = .0104 Divide each of those numbers by the lowest (.0104) to get the ratio between the Carbon/Hydrogen/Nitrogen. So it would be + Show Spoiler +
edit: Ignore me I fail D:
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On February 22 2010 18:32 Carnivorous Sheep wrote:Show nested quote +On February 22 2010 18:21 Yizuo wrote:
ok, appearantly you have 4 H-Atoms for every C-Atom (CO2/H2O ratio) and the same number of C- and N-atoms (N2/CO2 ratio)
so you have CxNxH4x
I'm not sure if you can calculate the exact formula So wrong :3 Yes you have 2 mols of Nitrogen for each mol of N2. You can calculate how many grams of C, H, and N are on the right side of the equation. gives me
molCO2 to mol C = .0104 followed by mol C to grams C = .1248g
molH2O to mol H = .0416 followed by mol H to grams H = .0416g
molN2 to mol N = .0104 followed by mol N to grams N = .1456g You have the starting mass of the unknown compound. To find the specific mass of each element in the compound, take the starting mass and subtract it with the masses of two of the elemnts for the mass of the third element in the compound. Then convert that mass to mols. the total for the 3 masses i calculated is .312, which is the mass of the unknown compound, so subtracting the total for 2 of the elements from the total mass of the unknown compound seems reduntant? Repeating this for each element should give you x, y, and z, and then its a simple matter of converting to integers.
my questions are in the quote ^^^
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Baa?21242 Posts
On February 22 2010 18:48 mOnion wrote:
my questions are in the quote ^^^
Bah I'm basically stupid, the methodology I gave you was for the combustion of an organic compound (CxHyOz), which necessitated figuring out exactly how much mass of the element, specifically O, is in the compound and not in the air.
Because the three components of this compound are separated out into three separate compounds on the right side you can basically ignore everything I said and just go with what the other people said, my bad if I confused you T_T
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0.458g CO2 => 0.125g C+ 0.333g O
0.374g H2O => 0.0416g H + 0.332g O
0.145g N2 => 0.145g N
====> 0.665g O + 0.125g C + 0.0416g H + 0.145g N in the left side.
====> 0.312g of dimethylhydrazine = x* 12*0.312/(12*x+1*y+14*z) g C= 0.125g C
====> 0.312g of dimethylhydrazine = y* 1*0.312/(12*x+1*y+14*z) g H = 0.0416g H
====> 0.312g of dimethylhydrazine = x* 14*0.312/(12*x+1*y+14*z) g N = 0.145g N
Then solve the 3x3 system.
am i right¿
IM right¿
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He's looking for the empirical formula, not the molecular one.
The other people are right.
I suggest learning to find the molecular formula since it'll be more helpful later on.
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On February 22 2010 18:54 Carnivorous Sheep wrote:Bah I'm basically stupid, the methodology I gave you was for the combustion of an organic compound (CxHyOz), which necessitated figuring out exactly how much mass of the element, specifically O, is in the compound and not in the air. Because the three components of this compound are separated out into three separate compounds on the right side you can basically ignore everything I said and just go with what the other people said, my bad if I confused you T_T
lol its fine dude, you can do no wrong while trying to help, ty so much for the help! everyone in thread was amazing
figured it out, should blow this test outta the water with TL power behind me ^_^
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On February 22 2010 18:56 TheBeardedWonder wrote:
He's looking for the empirical formula, not the molecular one.
The other people are right.
I suggest learning to find the molecular formula since it'll be more helpful later on.
yeh yeh we find the molecular formula in the subsequent question
using the empirical formula find the molecular formula if the weight of dimethylhydrazine is 60g/mol.
which is simple obv once you have empirical ^_^
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Its C2H8N2
1 mole of CxHyNz makes =
x = moles of CO2 = 0.0104 moles
y/2 = moles of H2O = 0.0208 moles
z/2 = moles of N2 = 0.00521 moles
divide by smallest number of moles (because it is the limiting product of the reaction)
so x/z, y/z, z/z
you get ratios of C= 2 H = 8, and N = 2
So its C2H8N2
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Valhalla18444 Posts
On February 22 2010 18:02 {88}iNcontroL wrote:
COMMON MAN?
of course hes a common man
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you can PM me if you have more questions 
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EPT
