April 29, 2024, 10:18:59 AM
Forum Rules: Read This Before Posting


Topic: Moles  (Read 5298 times)

0 Members and 1 Guest are viewing this topic.

Offline NTP

  • Regular Member
  • ***
  • Posts: 17
  • Mole Snacks: +0/-0
Moles
« on: March 19, 2007, 11:58:41 AM »

Hi,

Hi,

CAN SOMEONE EXPLAIN THE REASONING BEHIND THIS SUM, THANK YOU.

Write down the mass of calcium which has the same number of atoms as 12g of magnesium.



Not sure if I worked it right, I first worked out the number of atoms of 12gmagnesium and the result was 3 *10 23. Then I used this and to find the mass of calcium by doing moles of atoms= mass / ram, and my final answer was 1.2 *10 25

wOULD APPRECIATE A FEEDBACK , THANK YOU

Offline AWK

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 7979
  • Mole Snacks: +555/-93
  • Gender: Male
Re: Moles
« Reply #1 on: March 19, 2007, 12:10:09 PM »
12 g of Mg is equivalent to 1/2 mole. How many grams weights 1/2 mole of Ca?
AWK

Offline NTP

  • Regular Member
  • ***
  • Posts: 17
  • Mole Snacks: +0/-0
Re: Moles
« Reply #2 on: March 19, 2007, 12:16:12 PM »
20g is 1/2 of Ca

Offline xiankai

  • Chemist
  • Full Member
  • *
  • Posts: 785
  • Mole Snacks: +77/-37
  • Gender: Male
Re: Moles
« Reply #3 on: March 20, 2007, 04:25:58 AM »
and there is your correct answer
one learns best by teaching

Offline NTP

  • Regular Member
  • ***
  • Posts: 17
  • Mole Snacks: +0/-0
Re: Moles
« Reply #4 on: March 20, 2007, 07:28:02 AM »
But if the elements are from different groups not like it was with the previous answer both group 2 elements, it has to be worked different?
Example:

Write down the mass of silver which has the same number of atoms as 3g of aluminium.

I worked it out the same way and my answer is 7.2 * 1024g of Ag. Not sure if it correct.

Offline AWK

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 7979
  • Mole Snacks: +555/-93
  • Gender: Male
Re: Moles
« Reply #5 on: March 20, 2007, 11:50:09 AM »
3 g of Al is approx. 1/9 mole - take 1/9 mole of Ag
AWK

Offline DevaDevil

  • Chemist
  • Full Member
  • *
  • Posts: 690
  • Mole Snacks: +55/-9
  • Gender: Male
  • postdoc at ANL
Re: Moles
« Reply #6 on: March 20, 2007, 06:19:32 PM »
Write down the mass of silver which has the same number of atoms as 3g of aluminium.

I worked it out the same way and my answer is 7.2 * 1024g of Ag.

O.o, that would mean that 3g aluminium has the same amount of moles as 7,200,000,000,000,000,000 tonnes of silver? :o

Somehow I don't think that's correct ;)



You do not need Avogadro's number for these calculations. Just calculate mass to moles (mass [g] / molar mass [g/mol] = moles, do not calculate to molecules) and then back to mass of the other element.
As a mol of a compound has the same number of molecules independant of the compound, moles are easier to work with than molecules. That is precisely why the unit was introduced :)

Offline AWK

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 7979
  • Mole Snacks: +555/-93
  • Gender: Male
Re: Moles
« Reply #7 on: March 21, 2007, 02:11:09 AM »
Quote
I worked it out the same way and my answer is 7.2 * 1024g of Ag. Not sure if it correct.
It does not matter 10-24 or 1024, isn't it?
AWK

Offline NTP

  • Regular Member
  • ***
  • Posts: 17
  • Mole Snacks: +0/-0
Re: Moles
« Reply #8 on: March 21, 2007, 06:50:53 AM »

I worked it out this way using moles so it is correct now hehe  :)

Number of moles  = mass / ram    (3g of Al)

N = 3/ 27
N= 0.1111111

Therefore: number of moles =  mass /ram   (mass of Ag).
                  0.1111111  =    mass / 108
                0.1111111 * 108
 Mass of Ag = 12g

My Answer is 12g of Ag .
Still not sure but hehe

Offline xiankai

  • Chemist
  • Full Member
  • *
  • Posts: 785
  • Mole Snacks: +77/-37
  • Gender: Male
Re: Moles
« Reply #9 on: March 21, 2007, 08:02:18 AM »
right on the dot again you are. its quite easy and merely involves simple arithmetic.
one learns best by teaching

Offline NTP

  • Regular Member
  • ***
  • Posts: 17
  • Mole Snacks: +0/-0
Re: Moles
« Reply #10 on: March 21, 2007, 02:29:27 PM »

Thank you all for the help  ;)

Sponsored Links