Quantitive Determination Of Emperical Formula
Submitted by cbelill02 on 06/30/2008 05:21 PM
- Category: Technology
- Words: 405
- Pages: 2
- Views: 7
- Popularity Rank: 28915
Quantitive Determination Of Emperical Formula
PROBLEM:
Find the mole ratio of tin and oxygen, and then predict the empirical formula for the product.
SAFETY:
Wear lab aprons and goggles at all times. Use forceps or tongs when using hot plates.
PROCEDURE:
Clean evaporating dish and watch cover, then heat for about two and a half minutes. After cooled, find the mass of dish and cover together and record. Measure to the nearest 0.01g, the mass of the two grams of tin placed in the watch glass-covered tin. Calculate the mass of tin used and record. Under the fume hood, place 5cm³ of nitric acid and replace the watch glass. Remove the dish and the watch glass from the heat source making sure there is no product loss. Break up the solid with a toothpick. Cover the mass of the dish, contents, and cover after the dish has been cooled. After heating the dish for a second time, again find the mass of the dish, contents, and cover. Repeat the previous steps until your last two masses are 0.01g from each other. Calculate the mass of the oxygen used up, and then calculate the moles of oxygen and tin reactants.
DATA and RESULTS:
1st Mass (g) 2nd Mass (g) 3rd Mass (g)
dish & cover together 84.46g ----- -----
dish and cover and tin 86.65g ----- -----
tin 2.19g ----- -----
dish, cover, and contents together 87.83g 87.70g 87.69g
oxygen 1.04g ----- -----
Moles of Oxygen: 0.0650mol O
Moles of Tin: 0.0185mol Sn
WORK:
Oxygen: 1.04gO · 1molO = 0.0650molO
16.00g
Tin: 2.19gSn · 1molSn = 0.0185molSn
118.69g
0.0650molO = (») 4O Ratio: 4O
0.0185molSn Sn Sn
Percent Composition:
SnO4 Sn: 118.71g
4O: + 64.00g (16.00·4)
182.71g
Percentage Sn: 118.71 g · 100 = 64.97%Sn...
You must Login to view the entire paper.
If you are not a member yet, Sign Up for free!
