4. Part A.2. Suppose the original sample is unknowingly contaminated with a second anhydrous salt. Will the reported percent water in the hydrated salt be too high, too low, or unaffected by its presence?
The percent water in the hydrated salt would be too low because it would be from both the already anhydrous salt and the hydrous salt that has been heated. It would be considered too little water for such an amount of hydrous salt. The equation of part over whole would be used in this case. The part would still be the amount of water from both the hydrous salts but the weight which is the whole in the equation would have both the salts. This would affect it by being too low.
Part (water mass) / Whole (total mass of salt and water)
6. Part B.1. The hydrated salt is overheated and the anhydrous salt thermally decomposes, one product being a gas. Will the reported percent water in the hydrated salt be reported too high, too low, or be unaffected? Explain.
It will be reported too high because the gas released would be counted in with the evaporated water and it would be higher than supposed to. If the anhydrous salt is decomposed, then it will be released as water. Then the hydrated salt would decrease the weight of the anhydrous salt.
7. Part B. 1. Some of the hydrated salt splatters out of the crucible because of too rapid heating process. Will the reported percent water in the hydrated salt be reported too high, too low, or be unaffected? Explain.
The reported percent would be reported to be too low because the measurement of the crucible after the second heating would be lower than expected. This will increase the difference between the final mass of the fired crucible and hydrates salts subtracted from the mass of the fired crucible and anhydrous salt.