How to calculate molecular formula

[BH20]

The first question deals with

 

[Gokul43201]

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.

 

[Gokul43201]

Will get to Q2 after lunch...

 

[Gokul43201]

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.

 

[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 don't 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.

 

[Gokul43201]

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 don't see where you got the 1.2 from, and then 6.45.

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

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=1.86moles

Calculation error here. I get n=0.01838 moles

MW=m/n
MW=4.79g/1.86moles
MW=2.58g/mol

80% of 2.58 = 2.064g (Carbon)
8.2% of 2.58 = 0.211g (Hydrogen)
11.8% of 2.58 = 0.3044g (Oxygen)

C= 12.011/2.064 = 5.82
H = 1.0079/0.211 = 4.78
O = 15.9984/0.3044 = 52.6

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

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

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.

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

 

[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!

 

[Gokul43201]

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 ?

 

[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.

 

[Gokul43201]

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.

 

[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 don't get, which I've tried for awhile, but I just can't get.

 

[BH20]

Well, been finishing some other stuff up, but can't continue with some other questions if I don't 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 don't 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.

 

[Gokul43201]

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.

 

[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.

 

[BH20]

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

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

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.

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

Therefore, 3.61L would be produced?

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).

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 that's unreacted..so I'm thinking that's 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)

 

[Gokul43201]

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

Therefore, 3.61L would be produced?

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

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?

Perfectly true !

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?

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.

 

[Gokul43201]

Question 5:

Now Question 5 says to find the one that's unreacted..so I'm thinking that's 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.

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.

 

[BH20]

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.

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 can't say I quiet understand the ratio proportion stuff...I understand why you have to do it, but I don't 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 don't 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?

 

[Gokul43201]

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.

 

[BH20]

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.

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?

 

[Gokul43201]

The equation for the reaction only tells you the ratio of how many mole react with how many moles to give how many moles etc. If we multiply the equation throughout by some contant number, it is still correct. In a way, that is exactly what we are doing.

KClO3 --> KCl + 1.5 O2
can also be written as

5KClO3--> 5KCl + 7.5 O2

or 100KClO3--> 100KCl + 150 O2

For simplicity, it is written in the first form or sometimes multiplied by 2 so there are no fractions. What we are doing is multiplying this equation by some number which then makes the coefficient of O2 to become 0.195. Clearly that number is 0.195/1.5 = 0.13

So then you can write :

\frac {0.195} {1.5} KClO_3 --> \frac {0.195} {1.5} KCl + \frac {0.195} {1.5} *1.5 O_2

OR...


0.13 KClO_3 --> 0.13 KCl + 0.195 O_2

So this tells you how many moles of KClO3 were used up.

Like you said, familiarity leads to comprehension. Do more problems and you'll get the hang of it.

 

[Gokul43201]

For Q5, convert the volumes into moles using the Gas Law. Now find the constant that you need to multiply the reaction equation by to give you this number of moles - there will be 2 numbers, one that works for NH3 and another that works for O2 - the smaller of these 2 numbers is the right one. Think about why it is. So this will tell you how many moles get used up (of NH3 and O2 - one of them, completely) in the reaction. But you know how many moles were put into the container. So the difference is the number of unreacted moles.

 

[BH20]

For Q5, convert the volumes into moles using the Gas Law. Now find the constant that you need to multiply the reaction equation by to give you this number of moles - there will be 2 numbers, one that works for NH3 and another that works for O2 - the smaller of these 2 numbers is the right one. Think about why it is. So this will tell you how many moles get used up (of NH3 and O2 - one of them, completely) in the reaction. But you know how many moles were put into the container. So the difference is the number of unreacted moles.

sadly, I'm really confused about this WHOLE question, even though it is abit like the other one. Not sure what you even mean by some of the things. Never converted volume into moles before...I don't actually know what to do here at all.

 

[Gokul43201]

sadly, I'm really confused about this WHOLE question, even though it is abit like the other one. Not sure what you even mean by some of the things.

You'll understand once you start.

Never converted volume into moles before...I don't actually know what to do here at all.

You're kidding ?! How did you find the no. of moles of O2 in Q4 from the volume of O2 ?

 

[BH20]

Question: 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.

So we will use the Ideal Gas Law, and put in both volumes and covert them into moles..and then the right answer will be the lower moles one? or I should say the one we will use?

PV=nRT
n(0.20m3)=PV/RT
n=(800atm)(0.20m3)/(0.0821)(523K)
n=160/42.9393
n=3.726moles

n(0.12m3)=2.235moles

hmm..this doesn't look right..I've tried some other similar questions, and I didn't get them either.

I really apoligize, but I seriously don't get this one...

 

[Gokul43201]

So we will use the Ideal Gas Law, and put in both volumes and covert them into moles..and then the right answer will be the lower moles one? or I should say the one we will use?

PV=nRT
n(0.20m3)=PV/RT
n=(800atm)(0.20m3)/(0.0821)(523K)
n=160/42.9393
n=3.726moles

n(0.12m3)=2.235moles

hmm..this doesn't look right..I've tried some other similar questions, and I didn't get them either.

I really apoligize, but I seriously don't get this one...

First make sure you understand what units to use when. If you use R=0.0821 L-atm/K-mol, then
V : liters,
P : atm
T : K

If you use R = 8.315 J/K-mol
V : m^3
P : Pa or N/m^2
T : K

Now what values of n(O2) and n(NH3) do you get ?

 

[BH20]

Yeah, I used the 0.821 (R) so I knew to use L, atm and K as well, but the number was too high.

because it was 800atm*200L? or is that right?

 

[Gokul43201]

It will be high @ 800 atm, so that's okay.

 

[BH20]

Now what values of n(O2) and n(NH3) do you get ?

PV=nRT
n(02)=PV/RT
n=(800atm)(200L)/(0.0821)(523K)
n=160,000/42.9383
n(O2)=3,726.63moles

n(NH3)=2,235.77moles

ok, so I get N(O2) 3,726.63 moles. n(NH3) is 2,235.77moles.

 

[Gokul43201]

I think it's the other way round.

 

[BH20]

I think it's the other way round.

yep, sorry.

n(NH3) 3,726.63 moles. n(O2) is 2,235.77moles.

 

[Gokul43201]

Yes, now if you multiply the equation

NH3(g) + 1.75 O2(g) -> NO2(g) + 1.5 H2O(g) (just changed fractions to decimals)
throughout by the number 2235.77/1.75=1277.6 , you get

1277.6 NH3(g) + 2235.77 O2(g) -> 1277.6 NO2(g) + (1.5*1277.6) H2O(g)

This converts the equation to a form that involves the actual number of moles in the reaction vessel. So, this says that all the O2 will get used up but only 1277.6 moles of NH3 will react. So the unreacted number is 3726.6 - 1277.6 = 2449 moles of NH3.

 

[BH20]

Yes, now if you multiply the equation

NH3(g) + 1.75 O2(g) -> NO2(g) + 1.5 H2O(g) (just changed fractions to decimals)
throughout by the number 2235.77/1.75=1277.6 , you get

1277.6 NH3(g) + 2235.77 O2(g) -> 1277.6 NO2(g) + (1.5*1277.6) H2O(g)

This converts the equation to a form that involves the actual number of moles in the reaction vessel. So, this says that all the O2 will get used up but only 1277.6 moles of NH3 will react. So the unreacted number is 3726.6 - 1277.6 = 2449 moles of NH3.

alright thanks..can we get to the the writing equations and balancing? Maybe that is the reason I did not pick this up abit better.

sinc sulphate + barium nitrate produces ...
gold (III) sulphate + barium chloride produces...
tin + antimony(V) chloride produces...
ferrous bromide + phosporic acid produces...
the complete combustion of ethene, C2H4 produces...

when writing equations, do you basically look at the ions? I do know though how the compounds come together.

 

[Gokul43201]

You need to read up the different types of reactions...I cannot describe them all here. And you must understand valency, oxidation numbers, etc. Look up your elementary chemistry textbook.

Also see : http://members.tripod.com/~EppE/reaction.htm

Start at : http://members.tripod.com/~EppE/index.htm

Start from the intro and go through stoichiometry (and Gas Laws, if necessary)

PS : these pages take a while to load...thanks to a slow ad-server.

 

[BH20]

You need to read up the different types of reactions...I cannot describe them all here. And you must understand valency, oxidation numbers, etc. Look up your elementary chemistry textbook.

Also see : http://members.tripod.com/~EppE/reaction.htm

Start at : http://members.tripod.com/~EppE/index.htm

Start from the intro and go through stoichiometry (and Gas Laws, if necessary)

PS : these pages take a while to load...thanks to a slow ad-server.

Yes, I'm aware of the different types of reactions. The issue is writing the equation.

For example, the first example:

zinc sulphate + barium nitrate produces

This is a double displacement reaction. obviously, and I know how to interchange them (come up with the product).

ZnSo4 + BaNO_3(2) produces BaNO3 + BaSo4

The issue is, when I produce the new compounds, what is the deciding factor in if its Ba_2NO3, etc... I thought it was the valency? is it?

Is there some sort of special valency for elements like nitrate? I know what it is for Nitrogen, but not nitrate, etc.

And if so, what if a an element has more than one valency states, which one do you use?

I understand that based on the reactants the products will vary based on the type of reaction it is.

Thanks

 

[Gokul43201]

Yes it's the valency. If element A has valency x and element B has valency y. Then the compond formed by them will be A_yB_x
ie. interchange the valencies. Now if x,y have some common factor, then you reduce this to the simplest ratio.

eg : Zn = 2; O = 2 => Compound = Zn2O2 = ZnO

Some elements exhibit more than one valency, though one of these is more common than the other. If not specified, you should assume the more common valency. But most sensible people specify the valency in such cases. eg: Fe(III), Cr(VI), Zn(IV)

Valencies of common radicals : NO3 = 1, SO4 = 2, PO4 = 3, CO3 = 2, OH = 1, SO3 = 2, NH4 = 1

PS : Check your double-decomposition reaction, and correct it.

 

[BH20]

Yes it's the valency. If element A has valency x and element B has valency y. Then the compond formed by them will be A_yB_x
ie. interchange the valencies. Now if x,y have some common factor, then you reduce this to the simplest ratio.

eg : Zn = 2; O = 2 => Compound = Zn2O2 = ZnO

Some elements exhibit more than one valency, though one of these is more common than the other. If not specified, you should assume the more common valency. But most sensible people specify the valency in such cases. eg: Fe(III), Cr(VI), Zn(IV)

Valencies of common radicals : NO3 = 1, SO4 = 2, PO4 = 3, CO3 = 2, OH = 1, SO3 = 2, NH4 = 1

PS : Check your double-decomposition reaction, and correct it.

You do the crossover though..right?

For example in that first example:

Zinc is +2, SO4 is -2 Ba is +2 NO3 is -1

So, let's look at ZnSO4. Since Zinc has a valency of +2, it will go across and should really be ZnSO4(2) and then, since SO4 is -2, the total should be Zn2SO42. As you said though, we want to reduce to the simplest ratio, so since its both 2's, we can just leave it like that, it would reduce to 1..right?

And then if we look at BaNO3...Barium is +2, so we crossover and it should make it NO3(2), NO3 has a valency of -1, so that means we don't put anything infront of Ba..so its BaNO3(2).

ZnSO4 + BaNO3(2) produces ZnNO3(2) + BaSO4

Is this right..and it already looks balanced..the question states to balance it too..so maybe its wrong?

One small issue for the elements with more valency's...is that I'm not really sure which one is the common.

 

[Gokul43201]

So far so good ! And it is balanced, so there's nothing more to do there.

"I'm not sure which valency is more common"

Nor am I...that's why it's usually specified. But for Zn, II is more common than IV, and for Cr, II and III are more common than VI. Fe II and III are both fairly common. Cu, II more than I. It's only with the transition metals (d-block) that you see these multiple valencies.

 

[BH20]

ok..thanks

The second has:

Gold(III) Sulphate + Barium Chloride produces

Does the (III) mean we use the +3 valency, and if so, do we cross that over too?

So it would be:

Au2SO4(3) + BaCl2 produces Au2Cl3 + BaSO4

Does it look right, and if so, how would we balance it?

 

[Gokul43201]

ok..thanks

The second has:

Gold(III) Sulphate + Barium Chloride produces

Does the (III) mean we use the +3 valency, and if so, do we cross that over too?

Yes, always !

So it would be:

Au2SO4(3) + BaCl2 produces Au2Cl3 + BaSO4

Does it look right, and if so, how would we balance it?

Mistake on the right side...find it. Then we can get to balancing.

 

[Gokul43201]

For balancing an equation, you need to make sure there are an equal number of atoms of each element on the LHS and RHS. You cannot change the molecular formula to get this right. What you can do is change the coefficients (number of moles/molecules) before each compound.

 

[BH20]

dont see the mistake...Au is +3 (in this case anyways) Cl +2..so Au2Cl3
and then Ba is +2 and SO4 is -2, so it cancels out.

 

[Gokul43201]

Is Cl really 2 ? Then how did you get BaCl2 ?

 

[Gokul43201]

Cl valency is -1, not -2.

 

[BH20]

Cl valency is -1, not -2.

yep, my bad..dont know what made me think nit was -2.

Au2SO4(3) + BaCl2 produces AuCl3 + BaSO4

LS has to equal RS

LS= Au=2, SO4=3, Ba=1, Cl=2
RS= Au=1, SO4=1 Ba =1, Cl=3

Balanced:

Au_2(SO4)_3 + 3BaCl2 produces 2AuCl_3 + 3BaSO4

LS= Au=2, SO4=3, Ba=3, Cl=6
RS= Au=2, SO4=3, Ba=3, Cl=6

looks good to me.

 

[Gokul43201]

And to me.

 

[BH20]

I have some other ones to do which are different, more complex:



1. Zinc + mercurous nitrate produces - single displacement reaction.

Zn + HgNO3 produces Hg + Zn(NO3)_2

Balance:

Zn + 2HgNO3 produces 2Hg + Zn(NO3)_2

2. tin + Antimony(V) chloride produces - single displacement reaction.

Sn + SbCl4 produces Sb + SnCl4

Balance:

Looks fine. Altought Tin does have a valency of +2 and +4, but I think this one is right.

3. Calcium + water produces - synthesis reaction

Ca + H20 produces

not sure how this one comes together...pretty sure its not CaH20.

Found this on the net, but not sure why its like that:

Ca(s) + 2H2O(l) ——> Ca(OH)2(aq) + H2(g

Have a few more I will bring up.

 

[BH20]

4. Ferrous Bromide + Phosphoric Acid - not sure what type of reacion

FeBr + H3O4P produces

no idea how to do this one...

5. the complete combustion of ethene, C2H4 produces -

not sure either

6. calcium carbonate + hydrobromic acid produces - guess double displacement?

Ca2(Co3)_3 + HBr produces not sure

 

[Gokul43201]

I have some other ones to do which are different, more complex:



1. Zinc + mercurous nitrate produces - single displacement reaction.

Zn + HgNO3 produces Hg + Zn(NO3)_2

Balance:

Zn + 2HgNO3 produces 2Hg + Zn(NO3)_2

2. tin + Antimony(V) chloride produces - single displacement reaction.

Sn + SbCl4 produces Sb + SnCl4

Balance:

Looks fine. Altought Tin does have a valency of +2 and +4, but I think this one is right.

3. Calcium + water produces - synthesis reaction

Ca + H20 produces

not sure how this one comes together...pretty sure its not CaH20.

Found this on the net, but not sure why its like that:

Ca(s) + 2H2O(l) ——> Ca(OH)2(aq) + H2(g

Have a few more I will bring up.

Always treat H2O as though it were written HOH, with H (valency +1) being the cation, and OH(valency -1) being the anion.

So, Ca + HOH ---> Ca(OH)2 + H but then H exists only as H_2
So, Ca + H2O ---> Ca(OH)2 + H2, and now balance this to get the above equation.

PS : When a metal exhibits more than two valencies (eg : Fe II and III), the lower valency is indicated by an -ous ending and the higher valency by an -ic ending. So Fe(II) is called ferrous and Fe(III) is calles ferric.

That's how you should know that the mercurous in (Q1) above is Hg(I) and not Hg(II)

 

[Gokul43201]

Acids are made up of a H (cation) and any anion that is usually among Cl, Br, SO4, PO4, NO3, etc. The number of H atoms depend on the valency of the anion.

PO4 is -3. So phosphoric acid is H3PO4.
SO4 is -2. So sulphuric acid is H2SO4.
Cl, Br, NO3 are all -1. So you have HCl (hydrochloric), HBr (hydrobromic) and HNO3 (nitric acid).

Likewise, a base is usually made up of any common cation (Ca, Na, Mg, K, Cu, Zn, etc.) and OH (-1) as the anion. They are named as hydroxides. eg : Ca(OH)2 is calcium hydroxide, NaOH is sodium hydroxide.

When a hydrocarbon combusts, it is reacting with oxygen (O2) to give CO2 and H2O. So hydrocarbon + O2 ---> H2O + CO2. All you have to do is balance this. (Hint: First balance C and H, then do O)