Sciencemadness Discussion Board - Stacey0987's study assistance thread!

Stacey0987's study assistance thread!

stacey0987 - 23-11-2015 at 14:02

A compound containing hydrogen, carbon and oxygen only contains 74.2% carbon and 7.9% hydrogen. Its Mr is found to be 178 by mass spectroscopy. Find its empirical and molecular formula.

can someone help me with this question and how they got the answer.

[Edited on 23-11-2015 by stacey0987]

[Edited on 26-11-2015 by Bert]

chemistry

stacey0987 - 23-11-2015 at 14:06

balance these equations
3NH₃+H₃PO₄------>3(NH4)₃PO₄

Li₂CO₃----->2CO₂ + 2Li₂O

P₄ + 5OH- + H₂O ----> PH₃ + 5H₂PO₂_-

[Edited on 23-11-2015 by stacey0987]

[Edited on 23-11-2015 by stacey0987]

[Edited on 23-11-2015 by stacey0987]

[Edited on 23-11-2015 by stacey0987]

[Edited on 23-11-2015 by stacey0987]

Fulmen - 23-11-2015 at 14:20

Few will do your homework for you. You should try to solve it yourself or at least describe what you understand about solving it, this way we can help you where you get stuck.
Also the 3. equation makes no sense, both sides must have the same net charge.

ziqquratu - 23-11-2015 at 14:26

Hi stacey,

Start with the elemental composition. You know the proportions - by mass - of C and H, and can therefore calculate the percentage of O (since those are the only elements present).

From there, you can calculate what percent of the molecular mass is contributed by each element (multiply molecular mass by mass percent of each element - so, for example, (178 * 0.742) = mass of carbon present). Finally, divide those numbers by the atomic mass of the corresponding element to get the number of atoms of each element (178 * 0.742) / 12, for example, gives the number of carbon atoms in the formula - you might need to round off, of course!).

So, in short:
1) calculate percentage of oxygen
2) calculate mass of each element (m = Mr * percentage)
3) calculate number of atoms of each element (n = m / atomic mass)

aga - 23-11-2015 at 14:44

I seem to recall posting a simple guide to balancing equations using simple algebra in response to one of your earlier questions.

Not being a trained teacher, i appologise that it was not clear enough for you to understand.

Nobody here will do your homework for you.

If you do NOT understand the questions, your 100% best course of action is to talk directly to your Teacher.

They are professional Teachers : if you say you do not understand, they will attempt to help you.

Involves at least some desire on Your part to understand, and some Effort.

Yttrium2 - 23-11-2015 at 18:00

I am a little confused here, it has been a while since i've done problems that are similar to this. I'd have to check my notes. However, I've got a lot of math to do so chemistry questions have resorted to something to just satisfy my curiousity. Eventually, I'll be studying chemistry again, but for now it is math.

I guess this is kind of a math problem. It seems like it is an empirical formula problem / a percent composition problem, is been a while and this type of stuff sticks better on a flash card or USB stick where I can more easily find it versus committing it to memory. Atleast that was in the past, perhaps it will stick better now. I'm going to try to answer it...

edit empircal formula, and moleculr fomula

[Edited on 11/24/2015 by Yttrium2]

Yttrium2 - 23-11-2015 at 18:18

74.2 % Carbon
7.9 % Hydrogen

the sum of these is 82.1, this taken away from 100 is 17.9%

so there is 17.9 % Oxygen

the next step for findng the empirical formula would be to change these numbers to grams

so 74.2 g of C
7.9 g of H
17.9g of O

then we multiply these by 1 mole over the AMU(I think?)

carbons amu is 12.011 gram 74.2 x 1/12.011 = 6.177
hydrogens amu is 1.00794 7.9 x 1/1.00794 = 7.837 <(these numbers)
oxygens amu is 15.9994 17.9 x 1/15.9994 = 1.118

next we make sig figs, we use least amount there are 2 so C is 6.1 H is 7.9 and O is 1.1
Then we divide these numbers by the smallest molar amount, which is oxygen, to find the whole number ratio of each compound.

6.1/1.1 = C5.54
7.9/1.1 = H7.18
1.1/1.1 = O1

then to get the lowest mole ratio we have to find a mutiple to make these each whole numbers

so we find that multiplying each by 6 makes whole numbers, or very close (not sure how close it needs to be)
C33H43O43

[Edited on 11/24/2015 by Yttrium2]

Yttrium2 - 23-11-2015 at 18:32

Quote: Originally posted by ziqquratu

Hi stacey,

Start with the elemental composition. You know the proportions - by mass - of C and H, and can therefore calculate the percentage of O (since those are the only elements present).

From there, you can calculate what percent of the molecular mass is contributed by each element (multiply molecular mass by mass percent of each element - so, for example, (178 * 0.742) = mass of carbon present). Finally, divide those numbers by the atomic mass of the corresponding element to get the number of atoms of each element (178 * 0.742) / 12, for example, gives the number of carbon atoms in the formula - you might need to round off, of course!).

So, in short:
1) calculate percentage of oxygen
2) calculate mass of each element (m = Mr * percentage)
3) calculate number of atoms of each element (n = m / atomic mass)

isn't this incorrect? Don't ya divide the percentages by the atomic mass of each respective element? What is this question saying? What is it saying about Mr being found by spectroscopy? What is Mr, and what is being found by spectroscopy here? Dont you just need the different percentages, then divide this by the amu of each respective elements, then divide the lowest mole ratio into each amount of mole present, then take these numbers and multiple by 1, 2, 3.... untill you find the smallest whole number molar ratio? And once this is found isn't this the empirical formula?

[Edited on 11/24/2015 by Yttrium2]

CharlieA - 23-11-2015 at 18:46

But the formula weight from MS is 178. The empirical formula that you have, C5.5H7.0O1 (my figures are just slightly different than yours) can be multiplied by 2 to give C11H14O2, which has a formula weight of 178.

I hate to post the solution of this problem in detail, but then again, if you have to ask....

Yttrium2 - 23-11-2015 at 19:06

3NH3 + H3PO4 -> (NH4)3PO4

on the left on the right

3Nitrogens 3 Nitrogen
12 hydrogens 12 hydrogens
1 Phosphorus 1 Phosphorus
4 oxygens 4 Oxygen

I thought I was going to really have to dig deep into everything I saved from general chemistry. After playing with the equation in the right way, I got it easy.

How do you make the subscripts symbols?

for the second equation, somehow I fumbled and thought it'd be a difficult equation involving fractional coefficients or something which is totally something that I forgot how to do. The equation is simple

Li2CO3 -> Co2 +Li2O
2 Li on each side
1 C on each side
3 O on each side

[Edited on 11/24/2015 by Yttrium2]

Yttrium2 - 23-11-2015 at 19:10

Quote: Originally posted by CharlieA

But the formula weight from MS is 178. The empirical formula that you have, C5.5H7.0O1 (my figures are

how did you get 7.0 for your hydrogen???

Yttrium2 - 23-11-2015 at 19:17

how did you get 7.0 for your hydrogen???

Yttrium2 - 23-11-2015 at 19:30

damn typo! Keyboard was lagging and I must have pressured 9 instead of 8! Agh!

what all happened in this problem? I'm getting wayyyyy confused

[Edited on 11/24/2015 by Yttrium2]

woelen - 24-11-2015 at 00:17

The intermediate rounding is not OK, only round in the last step of calculations, or when presenting answers. I use 6 digits in intermediate calculations.

So, the ratios are

C: 6.17767/1.11879 = 5.52174
H: 7.83777/1.11879 = 7.00558
O: 1

Now, the estimates are much easier to convert to integer values. I would say that the empirical formula is C11H14O2. The molecular weight must be an integer multiple of C11H14O2.

The molecular weight is measured as 178, which perfectly matches the molecular weight of C11H14O2. So, the molecular formula also is C11H14O2.

woelen - 24-11-2015 at 00:23

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]

aga - 24-11-2015 at 03:40

Quote: Originally posted by Yttrium2

How do you make the subscripts symbols?

Like this :
SO(sub)4(/sub)

but use Square brackets :
SO₄

TheAlchemistPirate - 24-11-2015 at 09:05

*is surprised people are actually helping with these copy-and-pasted questions*

Nicodem - 24-11-2015 at 10:01

Is this truly a demand to solve a homework problem without even a "please" word? Must be some kind of a joke.

Magpie - 24-11-2015 at 21:21

This problem has already been solved by others but I wonder if Stacey understands the basis for how the answer was derived. My derivation below is a more long-winded approach in an effort to provide the basis for the solution.

This is a problem in finding the simplest ratios of the atoms (or g-atoms) of C, H, and O, as put forth in Dalton's Law. The atoms combine in ratios of small integers to form molecules.
That is, if the molecular formula is CxHyOz, then x, y, and z will be small integers.

Take as a basis 100g of this substance. Then the weights of each element would be:

74.2g C, 7.9g H, and 17.9g O

The g-atoms would be:

74.2/12.011 = 6.1111 for C
7.9/1.008 = 7.8373 for H
17.9/15.999 = 1.1188 for O

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

The g-atoms for O is an integer, and that for H is very nearly an integer, but that for C is not. But if we multiply all by 2 we will have all at very near integers, ie:

10.924 for C
14.010 for H
2.000 for O

Therefore the empirical formula is C11H14O2.

The molecular weight for this empirical formula just happens to be 178 so this is also the molecular formula.

ziqquratu - 24-11-2015 at 22:01

I started from the molecular mass, since it was provided, as you can calculate the number of moles of each element that are present in one mole of the substance - and therefore should get the right numbers straight up (after appropriate rounding, of course, because the elemental analysis might be a little off or whatever).

Apologies if I've confused anyone - I simply wished to provide a guide without providing the answer outright, using a shortcut applicable to the question as presented. If you don't know the Mr, you'd usually start with assuming 100g, as clearly described by woelen and Magpie, but when you know Mr you can simplify the math by starting with a mole of substance (and thus integer moles of each element).

Maths difficulties :(

stacey0987 - 25-11-2015 at 10:17

Hello I am having difficulty with this question
What mass of nitrogen (Mr=28) occupies 25.0dm₃ at 1.89x 105pA and 198 degrees Celsius.

Could someone please talk me through this, I do not just want an answer I want to no the stages so that I can do it independently next time.

[Edited on 25-11-2015 by stacey0987]

DutchChemistryBox - 25-11-2015 at 10:51

It is always the best to learn it by finding it out yourself.

I'll give you a hint:
PV = nRT aka the ideal gas law

aga - 25-11-2015 at 10:54

From experience, it is wise to assume that the questioner will NOT know what the P,V etc mean, so best to point them out DutchChemistryBox.

Without knowing what each term actually means, the equation is useless.

Maths help

stacey0987 - 25-11-2015 at 14:15

Hi I have a question that I find really hard I need someone to talk me through it and explain how to do it, I do not just want a answer I want the method so I will no how to do it independently.
Thank you for taking your time in helping me Your great

A liquid from an unlabeled aerosol was analysed and was found to be a single compound. 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.

I need to find the relative molecular mass of the liquid ????
Calculate the empirical and molecular formula for this compound ?????
I would do this question but I have NO IDEA HOW TO

Magpie - 25-11-2015 at 14:37

You can get the empirical formula the same way that your first problem was solved. Then as Dutch Chemistry Box said use the Ideal Gas Law, PV=nRT, to find the molecular weight, MW. n is the number of moles. Google will explain the PV=nRT.

n = weight/MW

j_sum1 - 25-11-2015 at 14:41

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".

Yttrium2 - 25-11-2015 at 15:13

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]

Yttrium2 - 25-11-2015 at 15:14

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]

Yttrium2 - 25-11-2015 at 15:40

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]

gdflp - 25-11-2015 at 16:02

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.

Yttrium2 - 25-11-2015 at 16:03

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

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]

Yttrium2 - 25-11-2015 at 16:11

Quote: Originally posted by gdflp

You found the # of moles in the gaseous sample above

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]

Yttrium2 - 25-11-2015 at 16:23

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?

diddi - 25-11-2015 at 16:43

this looks like the blind leading the blind ?

Magpie - 25-11-2015 at 16:49

Quote: Originally posted by Yttrium2

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.

Yttrium2 - 25-11-2015 at 16:55

Can someone please do the problem on paper, make each step clear, and upload it?

gdflp - 25-11-2015 at 17:23

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 359 times

Magpie - 25-11-2015 at 18:16

Yes, that is a better approach than what I suggested. This gives the answer directly.

CharlieA - 25-11-2015 at 19:12

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...

j_sum1 - 25-11-2015 at 19:48

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.

Etaoin Shrdlu - 25-11-2015 at 20:43

Quote: Originally posted by CharlieA

Quote: Originally posted 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]