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j_sum1
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You have the masses of each element. You can then find the number of moles of each element. From that you get the molar ratio of the elements.
Present this ratio in its simplest form and you have the empirical formula.
Step two is to use your gas laws: PV=nRT to find the number of moles of the compound in your vaporised sample.
Step three is to find the molecular mass of the substance. In your vaporised sample you have a known mass and know the number of moles. It should be
easy.
Step four is to compare the molecular mass calculated in step three with the mass of the empirical formula calculated in step one. You should find
that the actual compound is a whole number multiple of your empirical formula. From there, the answer pretty much falls out.
As an aside, stacey0987, you have started a number of threads of a similar nature. It is pretty clear that you are studying for some exams or such.
I don't think anyone minds giving you a bit of assistance -- that is what this place is for in part. I think there is a general reluctance to doing
your homework for you: which I think you understand. Given the similar nature of your requests, might I suggest you keep them all in one thread. I
will see if a passing mod can merge your threads into one thread entitled "study assistance".
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Yttrium2
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If anyone could upload this problem on a paper worked clearly I'd greatly appreciate it. I have forgotten how to do this. I like Staceys questions,
they are good review for me. It seems like I've completely forgotten how to do a lot of this  It is easier when you have the book to look at (reference?), it is a hell of a lot harder when you are tested
I'm going to attempt it, a lot of the times I feel like I'll stumble and do poorly until I suck it up, and get the gumption to go ahead and take a
stab at it.
0.951g of a vaporised sample of the liquid occupied 300cm3 at 200 degrees Celsius and a pressue of 103kPa.
P = 103kPa V= .3 Liters =N is unkown R is also unkown... and T is 200 degrees celsius
does pressure need to be in ATM and does T need to be in kelvin?
P = 1.01653 atmospheres V= .3 liters N= unkown R is 0.08206 T is 473.15 Lord Kelvins...
*you have to rearrange the equation to get N (moles) by itself
PV=NRT would be rearranged to PV/RT=N*
1.01653 atm (.3L)/.08206(473.15K) = N (moles)
1.01653atm(.3L)/.08206(473.15K) = .304959/38.826689
that last division problem equals .007854.... this is the moles
[Edited on 11/25/2015 by Yttrium2]
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Yttrium2
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Can you use that unit of volume for the equation, or does it have to be in liters?
first you have to convert into Atmosphers, Kelvins, and Liters
you then solve for moles with PV=nRT, you rearrange to solve for n by dividing PV by RT to equal moles
When you have the moles of nitrogen, you find the weight of those number of moles by using nitrogens atomic mass
[Edited on 11/26/2015 by Yttrium2]
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Yttrium2
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Can someone help me work the rest of the problem?
" 20g of the liquid contains 6.28g of fluorine, 11.74g of chlorine and 1.98g of carbon. In a separate investigation 0.951g of a vaporised sample of
the liquid occupied 300cm3 at 200 degrees Celsius and a pressue of 103kPa. "
how do we get the percentages if its out of 20g when it needs to be out of 100? do we do a proportion?
so we've got the moles, then we find the empirical formula, then again, how do we go about finding the molecular mass?
I'm guessing 100grams would contain 5 times as much (how would we do this if it wasnt such a nice number like 20 but like 17.5grams? find how many
times that goes into 100 and multiply this number by each number to find out how much of each is in 100 grams? EEEEEEEk)
so 6.28g F x 5 = 31.4g F in a 100gram sample,
11.74g Cl x 5 = 58.7g Cl in a 100 gram sample,
1.98g C x 5 = 9.9g C in a 100 gram sample
if we add up the grams this equals 100.
so there is
58.7% Cl
31.4% F
9.9% C
so there are
58.7g Cl
31.4g F
9.9g C
then we use the atomic masses to find the amount of moles in each sample
58.7 g Cl x 1/35.45 = 1.65 moles Cl
31.4g F x 1/19 = 1.65 moles F
9.9g C x 1/12.01 = .824 moles C
we divide these by the least amount of moles, which is carbon, at .824
1.65 moles Cl/8.24 =.2
1.65 moles F/8.24 =.2
.824 moles C/8.24 =1
multiply these by 5 to make the two's a whole number
C5F5Cl5 ( I forget which letters go first and why)
now we have the compound, we can find the formula mass of the compound, and we have the number of moles of the compound, so we should be able to find
the overall amount of mass that was asked for. Granted, I increased the numbers by 5 to make it out of 100grams, so perhaps minusing 80% somewhere is
necessary.
Can someone please answer my questions? I'm getting all confused??????????????????????????????????????????
[Edited on 11/25/2015 by Yttrium2]
[Edited on 11/25/2015 by Yttrium2]
[Edited on 11/25/2015 by Yttrium2]
[Edited on 11/25/2015 by Yttrium2]
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gdflp
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You found the # of moles in the gaseous sample above, it gives the weight of the sample that was vaporized, thus weight/# of moles = 0.951g/.00785 =
121.15 g/mol is the molecular weight. Then for fluorine we can say (6.28g / 20g) * 121.15g / (19g / mol) = 2 moles of fluorine. For chlorine we can
say (11.74g / 20g) * 121.15g / (35.45g / mol) = 2 moles of chlorine. For carbon we can say (1.98g / 20g) * 121.15g / (12.01g / mol) = 1 mole of
carbon. This gives us a formula of CF<sub>2</sub>Cl<sub>2</sub>, or dichlorodifluoromethane.
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Yttrium2
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Quote: Originally posted by Magpie  |
But from Dalton's Law we know the ratio of the g-atoms has to be in simple ratios of integer numbers. So, we divide each number by one of the numbers
of g-atoms to get integers. The choice is arbitrary. Let's use the number for O as this will get us numbers equal to or larger than 1.
Then the number of moles are:
6.1111/1.1188 = 5.4621 for C
7.8373/1.1188 = 7.0051 for H
1.1188/1.1188 = 1.0000 for O
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How did Dalton come up with dividing everything by one of the numbers to come up with full numbers? I know there cant be half an atom, but how did he
come up with dividing everything by one of the numbers and then finding a multiple of it to find the emperical formula?
And what do you mean the numbers are arbitrary? Don't you have to divide by the smallest one?
[Edited on 11/26/2015 by Yttrium2]
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Yttrium2
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even though it only said 20 grams of the sample, and I multiplied everything by 5 to make it out of 100 grams? How could this be the amount of moles
in the sample if I had scaled everything in the sample up by x5?
I'm a little confused here, what is this particular type of problem called, where could I find it in a textbook?
[Edited on 11/26/2015 by Yttrium2]
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Yttrium2
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| Quote: Originally posted by ziqquratu | | but when you know Mr you can simplify the math by starting with a mole of substance (and thus integer moles of each element). |
Care to elaborate?
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diddi
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this looks like the blind leading the blind ?
Beginning construction of periodic table display
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Magpie
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| Quote: Originally posted by Yttrium2 |
How did Dalton come up with dividing everything by one of the numbers to come up with full numbers? I know there cant be half an atom, but how did he
come up with dividing everything by one of the numbers and then finding a multiple of it to find the emperical formula?
And what do you mean the numbers are arbitrary? Don't you have to divide by the smallest one?
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All of these numbers just represent the ratios of the number of g-atoms. The numbers are arbitrary it's just the ratios that count. 100g basis is
arbitrary as explained by zigaratu.
Dalton only said that the number of atoms in the molecular formula can be expressed by small whole numbers (integers). The mathematical manipulations
that I chose to use are again arbitrary.
I could have divided the g-atoms by the number of g-atoms for H instead of O. Then the numbers would be:
0.7797 for C
1.0000 for H
0.1428 for O
So then it's not so obvious what multiplier I would need to get the small whole numbers. As a guess let's try the number 14.  This then get's us the smallest whole numbers, ie
14 x 0.7797 = 10.9158 for C
14 x 1 = 14 for H
14 x 0.1428 = 1.9992 for O
which gives C11H14O2
The discrepancies are due to measurement errors in the analytical work, same as before.
The single most important condition for a successful synthesis is good mixing - Nicodem
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Yttrium2
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Can someone please do the problem on paper, make each step clear, and upload it?
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gdflp
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Here is a written solution to the problem. Let me know if you have any questions.
Attachment: scan0177.pdf (55kB)
This file has been downloaded 358 times
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Magpie
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Yes, that is a better approach than what I suggested. This gives the answer directly. 
The single most important condition for a successful synthesis is good mixing - Nicodem
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CharlieA
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| Quote: Originally posted by woelen | An important lesson you can learn from this example:
Never ever perform heavy rounding during calculations in physics or chemistry calculations. If your input values are given at N digit precision, then
use much more than N digits for all intermediate results and only when the final values to be presented are determined, round to N digits again.
The above is a nearly general thing, it applies to nearly all scientific calculations with input data of limited precision. Only in specialized
situations, the accuracy of the final answers is much lower or much higher than the accuracy of the input data.
[Edited on 24-11-15 by woelen] |
How can the accuracy of a final answer be higher than the accuracy of the input date? ...except by chance, that is...
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j_sum1
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The Arrhenius equation is the one that springs to mind. https://en.wikipedia.org/wiki/Arrhenius_equation
In a certain range, the temperature is highly critical. But when the temperature is sufficiently high, a moderate change in temperature might have a
very minimal effect on the value calculated. For example, a 10% change in T might only affect the calculated value to at the fourth or fifth
significant figure.
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Bert
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Threads Merged
25-11-2015 at 20:39 |
Bert
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Threads Merged
25-11-2015 at 20:41 |
Bert
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Threads Merged
25-11-2015 at 20:42 |
Etaoin Shrdlu
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| Quote: Originally posted by CharlieA | | Quote: Originally posted by woelen | An important lesson you can learn from this example:
Never ever perform heavy rounding during calculations in physics or chemistry calculations. If your input values are given at N digit precision, then
use much more than N digits for all intermediate results and only when the final values to be presented are determined, round to N digits again.
The above is a nearly general thing, it applies to nearly all scientific calculations with input data of limited precision. Only in specialized
situations, the accuracy of the final answers is much lower or much higher than the accuracy of the input data.
[Edited on 24-11-15 by woelen] |
How can the accuracy of a final answer be higher than the accuracy of the input date? ...except by chance, that is... |
In context, I assume this question is about precision. This can happen with multiple trials where you can determine precision from the variance in
your data. Assume reading absorbance of a solution:
0.796
0.796
0.797
0.796
0.795
0.797
0.796
0.796
0.796
0.797
Average is 0.7962, using normal significant figure rules that would be rounded to 0.796, but the standard deviation is 0.0006 so this can be reported
to 4 decimal places as 0.7962.
Similarly:
0.796
0.780
0.755
0.802
0.769
Average is 0.7804, using normal significant figure rules this would be rounded to 0.780, but the standard deviation is 0.02 so this should only be
reported to 2 decimal places as 0.78.
EDIT: There are...arguments about whether you should report a result that wound up "too accurate" to the precision of the original measurements, but
the uncertainty itself can still be narrower than your original reading uncertainty.
[Edited on 11-26-2015 by Etaoin Shrdlu]
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