- Slides: 9
Cu. SO 4� 5 H 2 O Copper (II) Sulfate Pentahydrate Percent Composition by Mass Lab
The stuff we need… Scale Ring Stand Bunsen burner white EVAPORATING DISH Metal Scoop Blue Compound Cu. SO 4 � 5 H 2 O
Plan… This compound has a weird formula; it has five molecules of water attached to it. Cu. SO 4 � 5 H 2 O has one copper, one sulfur, four oxygens, and five water molecules. We will keep the water as water - not hydrogen and oxygen since it really is water here. 1. Heat up 3. 00 grams of this stuff in an evaporatin compound, evaporating it into the air (away). g dish, shaking the water out of the 2. While that is heating, calculate the molar mass of the compound, be sure to do “water” as water, not as hydrogen and oxygen separate. 3. Calculate the percent composition by mass of copper, sulfur, oxygen, and water too. 4. After heating the compound sufficiently, all water escapes, and a certain mass of water is no longer in the dish. We can measure this loss of water from the dish. 5. Compare the measured loss of water to the expected loss of water by comparing the percent comp by mass of water in the compound, to the percent lost in the lab. 6. Cool, right?
Evaporating dish mass was 30. 12 grams empty. With 3. 00 grams of the blue compound it is now 33. 12 grams. That means (duh!) that we are using exactly 3. 00 grams of the copper (II) sulfate pentahydrate.
The plan is to heat this up until ALL the water is shaken free of the compound. We want to do this: Cu. SO 4 � 5 H 2 O → Cu. SO hydrated compound 4 + 5 H 2 O dehydrated compound escaped water
Cu. SO 4 � 5 H 2 O → Cu. SO hydrated compound 3. 00 grams = 4 + 5 H 2 O dehydrated compound + escaped water (some grams of solid + mass of steam) Since the hydrated compound had mass of 3. 00 grams The dehydrated compound PLUS the mass of the water also equals 3. 00 grams (Matter cannot be created or destroyed in a chemical reaction or a physical change). We can get the molar mass of this funky compound, then calculate the percent composition by mass of water in it, then measure our results and compare what we know should happen to what did (by doing percent error).
Once heated long enough, the water all escapes, the new stuff in the dish is not copper (II) sulfate pentahydrate anymore. It’s been dehydrated into just copper (II) sulfate (no hydrate here anymore!) And it’s not blue anymore either.
After the first heating, the dish and the dehydrated compound mass was 32. 08 grams. After heating it again, the final mass was 32. 05 grams. You can’t really tell if all the water is out, unless you reheat the stuff, and see if the mass goes down. If all the water was out, the mass is constant. Here the mass dropped a bit, which means that after one heating we didn’t quite get all the water out all together. What does this mean? 32. 05 grams – 30. 12 grams (the dish) = 1. 93 grams of Cu. SO 4(S) So, 3. 00 g – 1. 93 g = 1. 07 grams “missing” water It’s not “gone” but it is in the air and not in the dish anymore. Our measured mass of dehydrated compound is 1. 93 grams Our measured mass of evaporated water is 1. 07 grams That totals the original 3. 00 grams we started with. (cool!)
To “end” the lab I put the evaporating dish on the black table, and sprinkled water from my fingers into the dish. Most of the Cu. SO 4 turned back into Cu. SO 4· 5 H 2 O. It started out as a hydrated ionic compound. We dehydrated it be heating it up with the Bunsen burner. Once done measuring, I rehydrated it by adding water back in. Blue to white, and back to blue again. + 5 H 2 O(G) invisible steam Cu. SO 4· 5 H 2 O starts Blue Cu. SO 4 dehydrates to white then Cu. SO 4 + 5 H 2 O → Cu. SO 4· 5 H 2 O white + water → blue again