How to calculate molecular formula

  • Thread starter BH20
  • Start date
In summary, the first question deals with calculating the number of moles of TiO2 and AgCl using their respective molecular weights and given masses. The empirical formula is then determined to be TiCl3. In the second question, the Ideal Gas Law is used to find the number of moles of a compound, which is then used to calculate its molecular weight. By finding the weights of carbon, hydrogen, and oxygen in the compound, the molar ratio can be determined to find the molecular formula. In question 2, a calculation error was made in determining the number of moles, resulting in an incorrect molecular weight and molar ratio. In question 3, the correct calculations are shown, resulting in a
  • #71
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?
 
Chemistry news on Phys.org
  • #72
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.
 
  • #73
BH20 said:
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.
 
  • #74
Gokul43201 said:
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?
 
Last edited:
  • #75
Depends on how many sig figs before that "last digit"...
 
  • #76
Gokul43201 said:
Depends on how many sig figs before that "last digit"...

ah ok..got it.
 
  • #77
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?
 
Last edited:
  • #78
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?
 
  • #79
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
 
  • #80
hmm..still don't get that..weird.

what would the correct formula for Manganese(II) iodite be?
 
  • #81
Mn(IO_2)2

Go to Ch. 4 : http://www.egusd.k12.ca.us/elkgrovehigh/Summer/APchemSUMMERWORK.htm
 
Last edited by a moderator:
  • #82
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.
 
  • #83
1. try googling
2. about 7.8*10^23
 
  • #84
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.
 
  • #85
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.
 
Last edited by a moderator:
Back
Top