Post by: kaanonball on December 31, 2008, 03:57:12 pm
The Question
Sulfur in azalea fertiliser is present in the form of soluble sulfates. To check the percentage of sulfur in such a product, a 2.322 g sample was weighed
out and dissolved in water. Barium chloride was then added under carefully controlled conditions in order to precipitate out all the sulfate present as
barium sulfate. After filtering, washing with water and drying, the mass of precipitate obtained was 0.564 g.
Calculate the mass of sulfur in the precipitate and hence the percentage of sulfur in the sample as analysed.
Post by: polky on December 31, 2008, 04:25:11 pm
As you added Barium Chloride into a solution which has all these SO4 2- ions floating around, the Barium (which is present as ion Ba 2+ floating around in the solution) bonds with the Sulfate to form BaSO4, which is a solid and precipitated out.
Calculations are:
- You have the mass of the precipitate, so you have the mass of BaSO4. You can find out the n (mol) of BaSO4, thus the n (mol) of Sulfur (mol ratios) and from there the mass of sulfur in the precipitate.
- Once you get the mass of Sulfur, just use that to figure out the % by mass of sulfur in the sample by taking (mass of sulfur)/(mass of sample) x100.
(Sorry for the crappy latex-less reply, I suck at latex!) Hope this helps.
Post by: kaanonball on January 01, 2009, 12:13:50 pm
but is the equation
Post by: Mao on January 01, 2009, 06:13:13 pm
Post by: kaanonball on January 01, 2009, 07:41:23 pm