How to calculate molecular formula

[Gokul43201]

1a) Okay (b,c) If you have the # of moles, you don't need to find the mass. Just multiply the # of moles by N = 6.02*10^23 atoms/mole, to get the # of atoms.

2) Depends on what you mean by V1 and V2. What did you use for P1, T1, P2, T2 ?

3a) Correct.
(d) "Hydrogen is hotter because it is a non-metal." Heavens, NO. What about chlorine ? It is also a gas. They are put together into a cylinder, so they mix with each other and get to be the same temperature. When 2 things are in contact for enough time they reach a common temperature.
(b) Temperature is just a measure of the Average Kinetic Energy of the molecules. Since their temperatures are equal, so must be their KEs. KE = (1/2)mv^2
(c) Since their KEs are equal and the Cl2 molecule has a greater mass, it must have a lower speed.

PS : Read Ch 6 (Kinetic Th. of Gases) in the link I gave you before.

 

[BH20]

1a) Okay (b,c) If you have the # of moles, you don't need to find the mass. Just multiply the # of moles by N = 6.02*10^23 atoms/mole, to get the # of atoms.

2) Depends on what you mean by V1 and V2. What did you use for P1, T1, P2, T2 ?

3a) Correct.
(d) "Hydrogen is hotter because it is a non-metal." Heavens, NO. What about chlorine ? It is also a gas. They are put together into a cylinder, so they mix with each other and get to be the same temperature. When 2 things are in contact for enough time they reach a common temperature.
(b) Temperature is just a measure of the Average Kinetic Energy of the molecules. Since their temperatures are equal, so must be their KEs. KE = (1/2)mv^2
(c) Since their KEs are equal and the Cl2 molecule has a greater mass, it must have a lower speed.

PS : Read Ch 6 (Kinetic Th. of Gases) in the link I gave you before.

1 b) gave me 1.5mol of carbon, and wanted me to find the number of atoms.
c) gave me 1.5mol of Carbon Dioxide to find the number of atoms.

the mm should be in this though..no? because we just have the moles of carbon..or am I getting mixed up?

2. Original Question: Find the volume of a gas at -30degreescelsius and 0.75atm if it occupies a volume of 255L at 40degreescelcius and 1.25atm... I used the General Gas Equation.

V1=?
T1=-30dc=243K
P1=0.75atm
V2=255L
T2=40.0dc=313K
P2=1.25atm

P1V1T2=P2V2T1

??

3d) oops..yeah, you are right, that made no sense.

 

[Gokul43201]

1) There is no place for the molar mass in (b,c). A mole is simply a quantity equal to a fixed number (the Avogadro Number, N(av) = 6.02*10^23) of atoms/molecules. So if you have 1 mole of anything, there are N(av) number of atoms/molecules in it.

In (1a) you used the mass and molar mass to find the # of moles. In (b,c) you are already given the # of moles, so why would you want to use the molar mass ?

2) Correct.

 

[BH20]

In (b,c) you are already given the # of moles, so why would you want to use the molar mass ?

I guess its a trick question then?

because b) had 1.5 mol and c) had 1.5mol..so would we not get the same answer for b and c?

but b) had 1.5 mol of carbon and c) had 1.5 mol of carbon dioxide

so that's where I got mixed up I guess.

 

[Gokul43201]

Yes, it's the same answer. The idea of the question is to reinforce the concept of a mole.

 

[BH20]

doing more questions here, stuck on one...

What experimental evidence requries that the emission of energy by an atom be quantized?

 

[BH20]

and on another question, I answered that the atomic number has more to do with the chemical behaviour of an element than atomic mass, because the atomic number is also the number of electrons an element has, and electrons dictate the type of bond and behaviour of the element.

 

[Gokul43201]

Yes, that's right.

and look up the Photoelectric Effect.

 

[BH20]

What could you tell me about Neutralization reactions, and ionic equations?

I'll jist give one question from each:

Neutralization:

Write equations for the neutralizations reactions that result in the formation of the following salts:

a) lithium carbonate

Ionic equations:

Identify those for which a reaction is likley to occur. For those that do occur, write a net ionic equation.

a) Chromium dipped into silver nitrate.

 

[Gokul43201]

Neutralization : acid + base --> salt + water

eg : H2SO4 + 2NaOH --> Na2SO4 + 2H2O

Recall my previous description of acids and bases. The reaction is basically a double-displacement reaction. Notice how the salt directly tells you what acid and base were used.

Ionic equation likely to occur : Look up the (Pauling) electronegativity series. There are 2 basic (equivalent rules). They are :

1. A more electropositive (or less electronegative) cation will displace a less electropositive (or more electronegative) cation. eg : Na(En=0.9) will displace Cu(En=1.9). The reverse will not happen.

2. A more electronegative anion will displace a less electronegative anion. eg : F(En=4) will displace Br(En=2.8).

In short, electropositive cations and electronegative anions are better at forming compounds.

 

[BH20]

thanks..got it.

Few more easier ones but to make sure its right.

rounding to 3 significant figures...
i) 3.1747=3.17
ii) 0.0024652=0.00247
iii) 101.5=102
iv) 1.3463=1.35
v) 7.0021=7.0021

??

and another question asks if using a more expensive balance mattered in meausuring a mass of a sample..with the more expensive one, the mass was 75.357...using the less expensive it was 75.4...how do they differ? does it matter?

 

[BH20]

and just to check an answer...

if you are trying to find ATOMS..and it gives you a mass of 52.5g of calcium metal, and we know the molarmass is 40.1gm/mol, we can find the moles..and then we times that by 6.02*10tothe23? what do you get as the answer? second time I did it, got a different answer than first time, so just wondering.

 

[Gokul43201]

thanks..got it.

Few more easier ones but to make sure its right.

rounding to 3 significant figures...
i) 3.1747=3.17
ii) 0.0024652=0.00247
iii) 101.5=102
iv) 1.3463=1.35
v) 7.0021=7.0021

??

and another question asks if using a more expensive balance mattered in meausuring a mass of a sample..with the more expensive one, the mass was 75.357...using the less expensive it was 75.4...how do they differ? does it matter?

v) 7.0021 ~ 7.00 (to 3 sig. figs)

The balance question : This depends on other data in the problem. If you are given other numbers to only 2 or 3 sig figs, it doesn't help to buy the expensive balance because the error will be dominated by the accuracy of those numbers, not by the balance. If all other numbers are quoted to 5 sig figs or better, then the hot-shot balance would be handy.

The ATOMS question : (doing this in my head) The answer should be a little less than 8*10^23. The method you described is correct.

 

[BH20]

v) 7.0021 ~ 7.00 (to 3 sig. figs)

The balance question : This depends on other data in the problem. If you are given other numbers to only 2 or 3 sig figs, it doesn't help to buy the expensive balance because the error will be dominated by the accuracy of those numbers, not by the balance. If all other numbers are quoted to 5 sig figs or better, then the hot-shot balance would be handy.

The ATOMS question : (doing this in my head) The answer should be a little less than 8*10^23. The method you described is correct.

I got, 7.826 * 10tothe24? sound about right?

and the other question..(the balances) its in reagards to how much heat was lost or gained. Q=c*m*T...it says that in the example, and likley key part, "for both meauseraments, the last digit is uncertain"..so that woud likley mean it would not matter which one is used?

 

[Gokul43201]

Depends on how many sig figs before that "last digit"...

 

[BH20]

Depends on how many sig figs before that "last digit"...

ah ok..got it.

 

[BH20]

they have this in the book...

2.0g divided by 71.0g/mol * 6.02 *10tothe23

somehow when I calculate the EXACT same thing, I get 1.7*10tothe23..they get to the 22.

any idea?

 

[BH20]

they have this in the book...

2.0g divided by 71.0g/mol * 6.02 *10tothe23

somehow when I calculate the EXACT same thing, I get 1.7*10tothe23..they get to the 22.

any idea why?

 

[Gokul43201]

1.7 * 10^22 is correct. You must be making a mistake. Multiply numerators, 6*2=12
Now 12/71 =0.17. So, it 0.17*10^23 = 1.7*10^22

 

[BH20]

hmm..still don't get that..weird.

what would the correct formula for Manganese(II) iodite be?

 

[Gokul43201]

Mn(IO_2)2

Go to Ch. 4 : http://www.egusd.k12.ca.us/elkgrovehigh/Summer/APchemSUMMERWORK.htm

 

[BH20]

Some more questions:

1. What were the limitations of Dalton's atomic theory?

2. What do you get when you calculate 1.30mol * 6.02*10tothe23...since I keep getting the wrong answer when I type in the books example, want to see if I'm getting this wrong too.

 

[Gokul43201]

1. try googling
2. about 7.8*10^23

 

[BH20]

hmm, yeah, tried that right of the bat, I can find obviously the main points about the theory, but can't find the limitations anywhere.

 

[FZ+]

http://dl.clackamas.cc.or.us/ch104-04/dalton's.htm

Reading from this, a number of limitations leap to mind.

1. He characterises elements as atoms with the same mass. We now know that mass isn't important, chemistry wise. What is important is the electron configuration. An element is defined by proton number, not mass number.

2. Atoms are not indivisible, and indestructible, as we now know.

3. Compounds are not defined by the ratio of elements, but by their relationship to each other. Molecular formula, as opposed to empirical formula.