May 01, 2024, 03:23:34 AM
Forum Rules: Read This Before Posting


Topic: Mass Percent Problem~  (Read 3183 times)

0 Members and 1 Guest are viewing this topic.

Offline needhelpplz

  • Very New Member
  • *
  • Posts: 1
  • Mole Snacks: +0/-0
Mass Percent Problem~
« on: September 10, 2007, 08:01:22 PM »
Douglasite is a mineral with the formula 2 KCl·FeCl2·2 H2O. Calculate the mass percent of douglasite in a 578.0 mg sample if it took 37.20 mL of a 0.1000 M AgNO3 solution to precipitate all of the Cl- as AgCl. Assume the douglasite is the only source of chloride ion.


First I got mole for AgNO3, which is 0.00372mol.  Then, what I did was I counted number of chlorids in Douglasite, which is 4, and according to that I wrote following equation:   

2 KCl·FeCl2·2H2O + 4AgNO3 --> 4AgCl + ????(we don't really need this part rite?)

then, I did 0.00372 X mole ratio X MM of Douglasite(which is 311.88416)

and I ended up with weird number...  I have no clue with this problem.. can somebody help me??^^ thx

Offline AWK

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 7979
  • Mole Snacks: +555/-93
  • Gender: Male
Re: Mass Percent Problem~
« Reply #1 on: September 11, 2007, 02:45:07 AM »
 Note:
1 mole of douglasite ==> 4 mole of chloride anions
It means mass of pure douglasite is equal to MMxmoles_of_chloride/4
(this mass is in grams!)
AWK

Sponsored Links