A 29.0 g Ag combine completely with 4.30 g S to form a compound. What is the % composition of this compound?
2007-01-24 00:04:30 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
G:
29 g- mass of Ag
4.30 g- mass of S
R: % composition
Sol'n:
Ag + S= 29 + 4.30 = 33.3
% 29/33.3 x 100 = 87.09%
% 4.30/33.3 x 100 = 12.01%
2007-01-24 01:02:08 · answer #1 · answered by Tin 3 · 1⤊ 0⤋
1. Determine the formula of the resulting silver sulfide molecule and determine its molecular mass.
2. Divide the mass of S given by the molecular mass of your silver sulfide. Multiply this by 100 to get percent composition.
3. Repeat for Ag.
2007-01-24 08:43:32 · answer #2 · answered by ? 4 · 1⤊ 0⤋
I assume you mean molar composition. Find the number of moles that correspond to those weights:
Ag --> 29.0 g / 107.8682 g/mol = .2688 mol
S --> 4.30 g / 32.066 g/mol = .1341 mol
If you divide .2688 / .1341, you get 2. Therefore there are 2 moles of Ag to 1 mole of S
2007-01-24 08:41:54 · answer #3 · answered by . 4 · 1⤊ 0⤋
Sorry for my english but i'm italian.
We have to calculate the moles of Ag:
29.0/107.868 =0.268 moles
and now we have to calculate the moles of S
4.30/32.06=0.134 moles.
So we have Ag2S because the moles of Ag are twice the moles of S.
By,by from Italy
2007-01-24 08:41:21 · answer #4 · answered by Non più attiva su answers 7 · 1⤊ 0⤋