If your sample contained a volatitle impurity, what value would be in error, the weight of the anhydrous salt or the weight of the water? Please explain. Thank you
I don't know how the experiment was conducted. From the limited amount of information you gave I can think of a scenario in which both would be in error.
If your sample contains a volatile impurity, the weight of the water would be in error, not the weight of the anhydrous salt.
To understand why, let's first consider what a volatile impurity is. A volatile impurity refers to a substance that easily evaporates or sublimes at relatively low temperatures. In the case of water, it has a relatively high vapor pressure and is considered a volatile substance.
When we weigh a sample containing a volatile impurity, such as a hydrated salt (a salt with water molecules incorporated in its crystal structure), the weight of the water can be misleading. As the sample is heated or exposed to conditions that promote evaporation, the volatile impurity (water) will escape, resulting in a decrease in weight.
However, the weight of the anhydrous salt, which is the salt without any water molecules, is not affected by the presence of a volatile impurity. The anhydrous salt remains unchanged in weight because it does not contain any water molecules that can evaporate.
Therefore, if your sample contains a volatile impurity, such as water in the case of a hydrated salt, the weight of the water will be in error due to the potential loss of water during the analysis process. On the other hand, the weight of the anhydrous salt will remain accurate as it is not influenced by the evaporation of the volatile impurity.