Chemistry Problem!!! % composition!?

Question

29g of Ag combines completely with 4.3g of S to make Ag2S. What is the percent composition of Ag and S in this compound?

i think i found the answer to this problem but i'm really not sure. i got 43.5% of Ag and 12.9% of S.

please help!!! thanks!

Answer 1

Weight Percent:

Total weight of Ag2S = 29+4.3 =33.3 g
(Due to complete reaction an conservation of mass)
_________ __________ _________ __________ ____________
You can check the above fact:
Moles of Silver = 29/107.8682 = 0.2688
But 2 mol of silver react with 1 mol of sulphur to form 1 mol Ag2S
Mols of Silver required per mol of Sulphur = 0.2688/2 = 0.2688/2
=0.1344

Moles of Sulphur = 4.3/32.066 = 0.1342

Moles are having very close values!
___________ ___________ _____________ ______________

Weight % of Ag = 29/33.3 = 87.07%

Weight % of S = 100-87.07 = 12.93%

Answer 2

Well, you should be certain that the answer you got is NOT correct because only Ag and S are present and their total must equal 100%, not the 56.4% you came up with...right? :-))

Som's answer is correct...but assumes that the questioner is not being devious, which is often the case in problems like this. The question states that the Ag combines completely, but does not say the same about the S...what if 4.3g S is an excess amount, some of which did not combine?

The way to be certain you have the correct answer is merely to calculate the % composition of Ag2S...
AW of Ag is 107.86, so one mol of Ag2S wd contain 215.72g Ag
AW of S is 32.06 so ome mol wd contain 32.06g S...
for a total of 247.78g
%Ag=100 x 215.72 / 247.78 = 87.06%
% S =100 x 32.06 / 247.78 = 12.94%
Considering the significant figures in question, the best answer is 87%Ag/13%S

So the instructor did not 'throw you a curve', all 4.3g of S were used...but it's better to be safe than sorry, isn't it? :-))

Later edit: Sheen was on the right track, but unfortunately shifted the Ag and S coefficients in his equation, which led to an incorrect answer...easy mistake to make.

Answer 3

You might have been able to spot that your answer is wrong because the two percentages do not add up to 100.

The formula makes no difference to the percentages - just the mass of each element as a fraction of the total mass.

Answer 4

Ag + 2S = Ag2S
107.87 g + 2 x 32.06 g >> 171.99 g
(1 mole) + (2 moles) >> (1 mole)
1:2
29 g of Ag = 29 / 107.87 moles = 0.2688 moles
Sulfur required to react completely
= 0.2689 x 2 = 0.5376 moles
= 0.5376 x 32.06 g = 17.24 g
They don't react completely.

4.3 g of S = 4.3 / 32.06 moles = 0.1341 moles
Silver required to react completely
= 0.1341 / 2 moles
= 0.1341 / 2 x 107.87 g
= 7.23 g

Silver left over = 29 - 7.23 = 21.77 g

You did percentage for mixture, not a compound
>>

If you want percentage in the compound>> always same>>
Ag + 2S = Ag2S
107.87 g + 2 x 32.06 g >> 171.99 g
(1 mole) + (2 moles) >> (1 mole)
Silver : Sulfur >> 1:2 (by weight)
>>
Percentage of silver in the compound
= 107.87 / (171.99) x 100 = 62.72%

Percentage of Sulfur in the compound
= 32.06 x 2 / (171.99) x 100 = 37.28%