# Few questions

1. Jun 9, 2004

### BH20

The first question deals with

Last edited: Jul 1, 2004
2. Jun 9, 2004

### Gokul43201

Staff Emeritus
MW (TiO2) = 48+32 = 80 g/mol
So, no. of moles of TiO2 = 1.2/80 = 0.015 = no. of moles of Ti

MW (AgCl) = 108+35.5 = 143.5 g/mol
So, no. of moles of AgCl = 6.45/143.5 = 0.045 = no. of moles of Cl

$$\frac {moles(Cl)} {moles(Ti)} = 3$$

So, the empirical formula is TiCl3.

Last edited: Jun 9, 2004
3. Jun 9, 2004

### Gokul43201

Staff Emeritus
Will get to Q2 after lunch...

4. Jun 9, 2004

### Gokul43201

Staff Emeritus
Q2)
First use the Ideal Gas Law, PV=nRT to find n, the number of moles.

Then m/n gives you the molecular weight (MW), where m is the mass(4.97g)

Now calculate 80% of MW, 8.2% of MW and 11.8% of MW.

This gives you the weights of C, H and O, per mole.

Divide these weights by the atomic weights of C, H and O to get the molar ratio.

The nearest whole number ratio tells you the molecular formula.

5. Jun 9, 2004

### BH20

Thanks for replying to the questions..its really appreciated.

Have some questions for you in regards to the answers, I want to really understand this...

In the 1st question, I dont see where you got the 1.2 from, and then 6.45.

I understand the rest though. I understand the steps, and I get why its 3 and why the empirical (simple formula) is TiCl3.

Second question..I did it..is it like this?

PV=nRT
n=PV/RT
n=(0.90atm)(0.793L)/(0.0821)(473K)
n= 0.714/38.83
n=0.01838moles

MW=m/n
MW=4.97g/0.018moles
MW=276.11g/mol

80% of 276.11 = 220.888 (Carbon)
8.2% of 276.11 = 22.6410 (Hydrogen)
11.8% of 276.11 = 32.58098 (Oxygen)

C= 220.888/12.011 = 19.390
H = 22.6410/1.0079 = 22.4635
O = 32.58098/15.9984 = 2.0365

is this correct? or is it supposed to be divided the other way around?

and if so, how would the molecular formula look if these were the answers. So if it is this way, the 5.82, would be a 6? correct?

Thanks again.

Last edited: Jun 10, 2004
6. Jun 10, 2004

### Gokul43201

Staff Emeritus
They are the weights of TiO2 and AgCl that you stated in the problem. All I was doing there was writing : no. of moles = weight/MW

Calculation error here. I get n=0.01838 moles

No. of moles is always weight/MW or weight/AW, so it's the inverse.

Yes, but you need to recalculate correctly to get the right numbers.

Last edited: Jun 10, 2004
7. Jun 10, 2004

### BH20

Question 1: duh...I need more sleep ;)

Question 2:

PV=nRT
n=PV/RT
n=(0.90atm)(0.793L)/(0.0821)(473K)
n= 0.714/38.83
n=0.01838moles

MW=m/n
MW=4.79g/0.01838moles
MW=260.60g/mol

80% of 260.60 = 208.48 (Carbon)
8.2% of 260.60 = 21.3692 (Hydrogen)
11.8% of 260.60 = 30.7508 (Oxygen)

C= 208.48/12.011 = 17.3574
H = 21.3692/1.0079 = 21.2017
O = 30.7508/15.9984 = 1.9221

Hopefully there are no calculation errors again.

So, the C would stay at 17? the H would stay 21, and the O would go to 2.

And then, would it be C17H21O2 as the molecular formula?

Thanks again...might have some more questions later!

8. Jun 10, 2004

### Gokul43201

Staff Emeritus
That seems about right, though typically the numbers you end up with will be closer to whole numbers (17.36 is not especially close to 17).

This is one compound with that formula : 1-phenyl-5-cyclohexyl-pentane-1,3-dionate

I'd expected something simpler. Are you sure the numbers in the question are correct ?

9. Jun 10, 2004

### BH20

Question 2: Estrone is known to contain 80% Carbon, 8.2% Hydrogen and the rest Oxygen. If 793ml of estrone is found to have a mass of 4.97g at 0.90atm (pressure) and 200degreesCelcius, find the molecular formula for estrone ?

This is exactly the question.

So since the pressure units were (atm), I knew the R would be 0.0821, I of course also changed the ml to L, and the temp to K. So everything is constant I believe in the first calculation.

BTW, surprise surprise, I did make a mistake because I punched in 4.79, instead of 4.97, so I did recalculate, and this is what I get.

PV=nRT
n=PV/RT
n=(0.90atm)(0.793L)/(0.0821)(473K)
n= 0.714/38.83
n=0.01838moles

MW=m/n
MW=4.97g/0.018moles
MW=276.11g/mol

80% of 276.11 = 220.888 (Carbon)
8.2% of 276.11 = 22.6410 (Hydrogen)
11.8% of 276.11 = 32.58098 (Oxygen)

C= 220.888/12.011 = 18.3904
H = 22.6410/1.0079 = 22.4635
O = 32.58098/15.9984 = 2.0365

so, what is the molecular formula for Estrone here? I know what you mean though, when I got these numbers, I thought the same thing.

I actually typed it into google, I typed in molecular formula for estrone and it said:

Molecular formula: C 18 H 22 O 2

So, I think you are right! :) I think maybe why its not even closer to the whole number is possibly due to rounding?

I might have some more questions later in regards to writing equations and stuff like that, still weak on that.

Thanks a lot for all the help so far..I'm learning a lot..really useful forums.

Last edited: Jun 10, 2004
10. Jun 10, 2004

### Gokul43201

Staff Emeritus
All the numbers in the question are given to 3 significant figures (eg : 4.97g, 793 mL, etc.). So all your calculations should use at least 3 significant figures. So, where you took n=0.018, it should really be 0.0184.

If you do this you get MW = 270.1 g/mol. And from this you end up getting :
C = 17.99
H = 21.97
O = 1.99

So that's really a lot better.

11. Jun 10, 2004

### BH20

thats what I thought :) Thanks

I did a few more of these types of questions from a workbook and got them right..thanks to you.

Have some more questions, I'll ask abit later..on writing equations and something else I dont get, which I've tried for awhile, but I just cant get.

12. Jun 10, 2004

### BH20

Well, been finishing some other stuff up, but cant continue with some other questions if I dont get this:
Now, I have 5 different kind of questions here, and some are difficult, so for question 2 and 3 for example, the steps are enough for me, or should be anyways, no need to do it all. Question 4 and 5 though are really difficult though and I might need to see how it is done, also question 1.

Question 1: Writing and Balancing Equations. I can balance them, but not write them. There is about 20 of these questions, but I picked some out that are abit different and then I can get them all.

s

I'd like to know what I have to look at when writing equations, what are the steps.

These next two are extremeley difficult, I dont even know where to begin.

Equation is KClO3 produces KCl + 3/2O2

Question 5: Consider the following reaction, which takes places in an autoclave at 250degreescelssius and 800atm.

NH3(g) + 7/4O2(g) produces NO2(g) + 3/2H2O(g)

Into the reaction vessel has been placed 200L of NH3(g) and 120L of O2(g). The reaction is allowed to go to completion. Determine the quantity, in moles, of the gas that remains unreacted.

Thanks to anyone who answers. Fairly new to chemistry in the sense that those are not courses that I usually took, but trying to learn.

Last edited: Jul 1, 2004
13. Jun 10, 2004

### Gokul43201

Staff Emeritus
Will get to Q1 later, but that is perhaps the most important and elementary question to be addressed before going to more advanced problems.

Q3 appears to be incomplete....unless it is part of Q2.

Q2, Q4 and Q5 are essentially the same...the only difference being that Q2 talks about weights while Q4 &Q5 refer to volumes of gases (which are related throught the Ideas Gas Law)

Let's do Q2 ...

Step 1 : Write down the equation and balance it (see future discussion for Q1)
$$H_2SO_4 + 2NaOH -->Na_2SO_4 + H_2O$$

Step 2 : Say this in words : "1 mole of sulphuric acid reacts with 2 moles of sodium hydroxide to give 1 mole of sodium sulphate and 1 mole of water". The important part is that 1mole of sulphuric acid neutralizes 2 moles of sodium hydroxide.

Step 3 : Find the weights of
a)1mole of sulphuric acid = 1* MW(H2SO4) = 98 g
b)and 2 moles of sodium hydroxide = 2* MW(NaOH) = 2*40=80 g

Step 4 : Now it's just ratio-proportion.
98g of H2SO4 for 80g of NaOH, so how many g of H2SO4 for 10g of NaOH ?
x=10*98/80 = 12.25g

Now let's continue from here into Q3 (since it appears to belong with Q2) ...

Step 1 : completed above.

Step 2 : also done. Now the important part is that 1 mole of H2SO4 gives rise to 1 mole of H2O.

Step 3 : we are not dealing with weight for water...so we use moles and proceed.
No. of moles of H2SO4 = wt./MW = 12.25/98 = .125 moles

Step 4 : So 0.125 moles of water will be produced (since it's 1:1, from Step 2)

Step 5 : Use the Ideal Gas Law to convert moles to liters, PV = nRT. Remember to convert pressure to the right units.

That's it !!

For Q4, Q5, Step 1 is already done for you. After Step 2 you can then work backwards, converting volume of gas into moles (using Gas Law).

Do it and if you have problems, post them.

14. Jun 10, 2004

### BH20

yeah, I will get to them and come back with problems and questions for you.

but yes, question 2 and 3, it is actually 21a and b in the book, so they are indeed together.

15. Jun 10, 2004

### BH20

Alright..I understand this. Took me so long to finally get it, but I finally got it.

PV=nRT
V=nRT/P
V=(0.125moles)(8.31R)(383K)/(110kPa)
V=3.61L

Therefore, 3.61L would be produced?

Now, I have never done this kind of question, but here is my try:

First, we need to find moles, because if we get it, then we just need the MV of the compund and can find the mass..correct?

So...Question 4:

PV=nRT
n=PV/RT
n=(0.950atm)(5.00L)/(0.0821)(297K)
n=4.75/24.3837
n=0.1948moles

So, now, can we just find the total MV of pottasium chlorate:
K=39.0983
Cl=35.453
O=15.9994*3=47.9982
MV=121.546

Then, m=n*MV
m=0.1948 * 121.546
m=23.667g

Therefore, its 23.667g?

Question 5:

Now Question 5 says to find the one thats unreacted..so I'm thinking thats NH3?..which would mean that we use the 200L when we find the moles?

So, since the pressure is atm, we change the units accordingly.

but then I tried putting in the numbers, but its too high, so I'm guessing I am doing it wrong, there must be another step in there.

(and sorry about the rounding, and significant digits, I have to pay attention to that)

16. Jun 11, 2004

### Gokul43201

Staff Emeritus
Doesn't the question say 100C ? So that should be 373K, not 383K. That makes the volume,

$$V = \frac {.125*8.31*373} {110,000} = 3.52 *10^{-3} m^3 = 3.52 L$$

Perfectly true !

No, you're missing a step in the process - and this IS the most important step. n=0.1948 moles is the no. of moles of O2. You forgot the part where "1 mole of KClO3 gives 1.5 moles of O2" - without this step, the balanced equation becomes useless. Use this to get n(KClO3), which you multiply by its MW to get the mass.

Do this...then we'll get to Q5.

17. Jun 11, 2004

### Gokul43201

Staff Emeritus
Actually, you don't have to change units, just use R=0.0821 L.atm/K.mol.
Remember, when you use SI units, and R=8.314 J/K.mol, the corresponding units for P and V must also be SI...ie. P in Pa (or N/m^2) and V in m^3 (NOT Liters !). Did you use V in Liters ? If you did not convert it to m^3 you will get an answer that it 1000 times too high. So, if this is what you did, divide by 1000 to get the correct n. Also, check this is right using R = 0.0831 and the given units.

Secondly, it is not correct to assume that NH3 is going to be left unreacted. You have to figure this out using the ratio of moles of NH3 and O2 in the balanced equation and compare that with the ratio of moles put into the reaction vessel.

18. Jun 11, 2004

### BH20

Question 3: Yep, you are right. It is 100degreescelsius. I was looking at a lot of things at once, the book, the screen, and some webpages, going back and forth to understand the steps, and looked at the pressure instead which was 110.

Question 4: I cant say I quiet understand the ratio proportion stuff...I understand why you have to do it, but I dont quiet get it. I understood what you did in the other question, but not here.

Do I have to get all the MW of all the compounds in the equation ? They are as follows...
KClO3 is 121.55g, KCL is 74g and O2 is 48g

I get that since 3/2 is infront of the O2, when finding the MW, you have to times it by 1.5..but then, I dont get it after that.. because after you did this proportion stuff wouldn't you already have a mass and be done the question after 3 steps?

19. Jun 11, 2004

### Gokul43201

Staff Emeritus
The equation says that for every mole of KClO3 used, you get 1.5 moles of water.

Think of it like say, a cooking recipe : "1 cup of sugar goes into making 4 cups of cake. How much sugar is contained in half a cup of cake, if a cup of sugar weighs 160gm ?"

4 cups cake => 1 cup sugar, so 0.5 cups cake => 0.5*1/4 = 0.125 cups sugar => 0.125*160 = 20 gm.

It's the same kind of thing with the equations. Replace 'cup' with 'mole' and 'weight of a cup' with the 'MW'.

Now using this in Q4 :

n(O2) = 0.195 moles from PV = nRT for oxygen.
But 1.5 moles(O2)=>1 mole(KClO3) So, 0.195 mol(O2)=>0.195*1/1.5 = 0.13 mol(KClO3) =>0.13*121.55g=15.8g of KClO3

NOTE : The MW is simply the weight of 1 mole of the compound (like the weight of 1 cup of sugar)...not the weight of 'so many moles'. So the MW of O2 is always 32g. Then you can multiply this number by the number of moles of O2 to fing the weight of O2 - though we don't need that for this problem.

Hope this helps.

Last edited: Jun 11, 2004
20. Jun 11, 2004

### BH20

Ok, let me try to explain this out loud to see if it makes any sense...

- We are trying to find the mass of pottassium chlorate (KClO3).
- We are given the info for the oxygen gas (O2), in which we will be able to find the number of moles. So we use the Ideal Gas Law(PV=nRT). We found it to be 0.195moles. And we want to find moles because along with the MW, we can then find the mass of KClO3.

- We then have to look at the ratio proportion..O2 is 1.5moles, and KCL03 is 1 mole.
To find it we use the moles we found O2(0.195*1(KClO3)/1.5...we get 0.13moles....

- Now we can use the moles of KClO3 (0.13moles) and times it by its MW (121.55) to get its mass of 15.8g. (using formula n=m/MW)

- Ok, now I realize all the steps, and how we came to the answer. I'm pretty sure though it will take another couple of questions to really get it going, because its much easier when you actually see it done.

One thing that confused me was the we found O2 to be 0.195moles, but I also saw 1.5moles infront of it, so I was getting confused. BUT, what I gather it means is that O2 is 0.195moles, but so its propertionate and balanced, it will take 1.5moles for the whole equation to be balanced..is that correct?

I'll get to question 5 and give it a try, but for it...it says "Determine the quantity in moles of the gas that remains unreacted". so how to we know which one that is?

Last edited: Jun 11, 2004