Stuck on a few Chemistry Questions?

Hey guys, I%26#039;m having some amount of difficulty with the following questions..





1) To determine the percentage of arsenic in a particular brand of pesticide , a 2.15 g sample is treated to precipitate all the arsenic as its sulfide, As2S3 .


If 0.353g of precipitate is obtained, calculate the percentage of arsenic in the pesticide.





2)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 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 0.564g.


a) Calculate the mass of sulfur in the precipitate and hence the percentage of sulfur in the sample as analyzed.


b) What is the effect on the final result if the precipitate is not washed with water before drying?





Hope you can help

Stuck on a few Chemistry Questions?
1. You have 0.353g of precipitate, but it%26#039;s not pure arsenic, so you have to find the mass of the arsenic within the precipitate, then compare that to the original sample.





A quick perusal of a periodic table shows that the atomic mass of arsenic is 74.92 and the atomic mass of sulfur is 32.06. As2S3 would give a mass of 74.92 x 2 + 32.06 x 3 = 246.02 g / mol





Arsenic makes up 74.92 x 2 / 246.02 or 0.609 of the total mass. That means arsenic makes up 0.609 x 0.353 g = 0.215 g of the pesticide.





To find the percentage of arsenic in the pesticide, divide the two, then multiply by 100. ( 0.215 / 2.15) x 100 = 10%





2. a) BaSO4 (barium sulfate) has a molar mass of 233.43 grams/mol. Sulfur has a molar mass of 32.06 g, so the fraction of sulfur in the precipitate will be 32.06 / 233.43 = 0.1373





The precipitate had a mass of 0.564, so the mass of sulfur in it was 0.1373 x 0.564 g = 0.0774





The percent of sulfur is the mass of sulfur divided by the sample mass times 100


% S = 0.0774 / 2.322 x 100 = 3.2 %





b) If you don%26#039;t thoroughly wash the precipitate, there may be some other chemicals present in the powder that would increase the measured mass of the precipitate. By rinsing with distilled water, you remove all chemicals other than the precipitate and water, which will evaporate when drying to leave only the precipitate.





Hope that helps!
Reply:This type of gravimetric q%26#039;s aren%26#039;t much fun.





1) pest (2.15 g) ====%26gt; As2S3 (0.353g)





moles As2O3 = 0.353 / 74.92x2 + 3x16 = 0.00178 mol





gives: 0.000353 mol As


mass As = 0.00357 x 74.92 = 0.267g





% As = (0.267g / 2.15g) x 100 = 12.4%





2) Fert (2.322g) + water ====%26gt;SO4(2-) + Ba2+ =====%26gt;


BaSO4(s), 0.564





NOTE: best way to do these is by a crude flow chart.





There is one mole of S for every mole BaSO4





Just follow (1)





b) duh, increase % of S since other compounds (undissolved fert) may be present.



beauty