3. Explain how the following factors would affect the results of the lab. Give a specific direction of the effect on the formula you calculated and fully explain why this occurs.
a. You did not heat the sample to a constant mass.
If the sample was not heated to a constant mass, that means that there is still water trapped in the lattice structure of the hydrate. If the mass gradually increases after heating, this indicates water is being reabsorbed into the hydrate. Once the mass remained constant, it could be inferred all the water had evaporated out of the hydrate. If water was left in the hydrate, that means the final mass recorded is higher than it should be. This would skew the results of the %water calculations, since not all the water …show more content…
The mass lost from decomposition would be assumed to be mass of water that evaporated from the sample. This could potentially alter the proportions for the hydrate formula, as the aluminum sulfide to water molar ratio would be incorrect.
d. Some water reabsorbed into the salt after heating.
As stated in answer a, if water reabsorbs back into the salt after heating, the mass recorded would not be representative of the mass of just the salt. Also, the water that reabsorbed wouldn’t be accounted for when calculating the %water composition, causing the experimental yield to be less than the theoretical yield.
e. Some of the hydrate splatters out of the crucible while heating.
If some of the hydrate spatters out, then the final recorded mass of the hydrate will not coincide with the initial mass taken before the reaction took place. This will disrupt all of the calculations after as there will be a “missing” portion of the reaction’s mass. The %water composition calculation may or may not be affected, depending on whether or not all the water had been evaporated out of the hydrate prior to the spill. The water to Aluminum sulfide molar ratio would, in turn, show a higher water ratio as there would be led hydrate to compare
a. You did not heat the sample to a constant mass.
If the sample was not heated to a constant mass, that means that there is still water trapped in the lattice structure of the hydrate. If the mass gradually increases after heating, this indicates water is being reabsorbed into the hydrate. Once the mass remained constant, it could be inferred all the water had evaporated out of the hydrate. If water was left in the hydrate, that means the final mass recorded is higher than it should be. This would skew the results of the %water calculations, since not all the water …show more content…
The mass lost from decomposition would be assumed to be mass of water that evaporated from the sample. This could potentially alter the proportions for the hydrate formula, as the aluminum sulfide to water molar ratio would be incorrect.
d. Some water reabsorbed into the salt after heating.
As stated in answer a, if water reabsorbs back into the salt after heating, the mass recorded would not be representative of the mass of just the salt. Also, the water that reabsorbed wouldn’t be accounted for when calculating the %water composition, causing the experimental yield to be less than the theoretical yield.
e. Some of the hydrate splatters out of the crucible while heating.
If some of the hydrate spatters out, then the final recorded mass of the hydrate will not coincide with the initial mass taken before the reaction took place. This will disrupt all of the calculations after as there will be a “missing” portion of the reaction’s mass. The %water composition calculation may or may not be affected, depending on whether or not all the water had been evaporated out of the hydrate prior to the spill. The water to Aluminum sulfide molar ratio would, in turn, show a higher water ratio as there would be led hydrate to compare