Hey everyone,
For Chem class we're finding the empiricle formula for Barium Chloride Hydrate.
The following information is what I got in the lab.
Mass of crucible Lid -- 29.90 Grams Mass Curcible lid and hydrate -- 32.01 grams. Mass / crucible lid anhydrate after our 1st burning -- 31.69. And the second burning came up to 31.68 grams.
Our Chem teacher told us that we had to get the Mass of Hydrate and Anhydrate, and then find the moles of the Barium Chloride and Anhydrate? ((This is where I get confused)) Then we have to find the mol ratio of the two Barium chloride and Anhydrate?
We also have the Error Anlysis, which I know equals out to 2:1, since there are 2H2O in Barium Chloride. TH Theoretical yiled and the calculated observed mass should equal out to 2:1
Anyhelp would be greatly appreciated!!!
One confused Chem student
2006-10-23 10:01:49 · 2 answers · asked by Jordan M 1 in Science & Mathematics ➔ Chemistry
I just ran a lab that was very similar to this. You weigh the crucible by itself and then weigh the crucible with the hydrated salt (Barium chloride hydrate). Subtract the two to get the mass of the barium chloride hydrate. THEN you burn off the water in the hydrate, since the formula will be BaCl2 .xH2O and by heating, you're taking the water off the molecule and leaving only the BaCl2 (the anhydride). You weigh the crucible after burning and then subtract empty crucible weight in order to get the resultant weight of the BaCl2 without its water (anhydride).
Since you're told that BaCl2 has 2 water molecules, you work out the molar weight % of water to BaCl2 by finding the formula weight of the hydrate and the dividing the % that is water. Since you have 2 H2O's per formula unit, you can get the portion of the formula weight that is water by multiplying 2 times the formula weight of water, which is 18.02 (O is 16 and H is 1.01). The total water per mole of compound would be 36.04 g/mole. The formula weight of the entire hydrated compound is the formula weight of barium chloride PLUS the water formula weight above. The accepted %H2O in the hydrate would be 36.04 divided by the hydrate's formula weight (multiply by 100 to get percent).
What you're doing experimentally is finding the weight lost when you heat the compound and taking that weight divided by the weight of the hydrated compound you started with to get the %H2O in the compound. To find the mole ratio of anhydride to water, you find the weight of water lost and determine how many moles of water that is. Then you get the weight of your dried salt (the anhydride) and find how many moles THAT is. (It'll be determined using the formula weight of BaCl2 alone, without the H2O molecules, since it is the anhydride). Divide the moles of water by the moles anhydride and you should come out with about 2:1 as you said.
In my lab, the error analysis was done with %H2O observed vs. %H2O accepted. My formula was (%observed-%accepted)/ %accepted (multiply by 100 to get percent error.)
I hope this helps!
2006-10-23 10:24:38 · answer #1 · answered by Black Dog 6 · 0⤊ 0⤋
what happens to the sample s reported% water if the salt decomposes yielding a volatile product?
2016-10-30 03:17:10 · answer #2 · answered by Saja 1 · 0⤊ 0⤋