I also need help in this please:
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? Explain.
is it low? if so, can you explain in some detail because I'm not understanding.
thank you!
It isn't clear as to when the sample is contaminated; i.e., before the original sample is weighed or after it is weighed.
a. If the sample is weighed, then contaminated, the %H2O in the hydrated salt will be unaffected. For example, say we take a sample, weigh it at 15 g, then contaminate with an unknown anhydrous salt (any amount). Upon heating it loses 10 g H2O so percent H2O is (10 g H2O/15 g sample)*100 = about 67%. No weights have changed and the anhydrous salt gives off no water so no harm done.
b. If the sample is contaminated BEFORE weighing the 15 g sample, then that 15 g is partly hydrated salt and partly anhydrous salt. Then %H2O = (loss in mass/15)*100 = ? But that loss in weight will be less (since there is less of the hydrated salt there) and the result will be low.
If the original sample is unknowingly contaminated with a second anhydrous salt (a salt without any water molecules), the reported percent water in the hydrated salt will be too high. Let's break down why this happens:
1. Hydrated salt sample: When we originally analyze a sample that contains a hydrated salt, we determine the percent water in the salt by measuring the mass loss when it is heated. The loss in mass corresponds to the water molecules being driven off.
2. Contamination with anhydrous salt: If the sample is contaminated with an anhydrous salt, it means that there are additional salts present in the sample, which do not contain any water molecules. These salts do not contribute to the mass loss when heated.
3. Resulting impact: When we heat the contaminated sample to determine the percent water, the mass loss will be the sum of the water molecules from the hydrated salt and the anhydrous salt. Since the anhydrous salt does not contain any water, its mass will not contribute to the mass loss measurement. As a result, the measured mass loss will be lower than expected, leading to a higher reported percent water content in the hydrated salt.
To summarize, if the original sample is unknowingly contaminated with an anhydrous salt, the reported percent water in the hydrated salt will be too high because the additional anhydrous salt does not contribute to the mass loss when heated.
If the original sample is unknowingly contaminated with a second anhydrous salt, the reported percent water in the hydrated salt will be too low. This is because the presence of the second anhydrous salt will decrease the overall mass of water molecules present in the sample, leading to a lower reported percent water.
To explain this further, let's consider an example. Let's say the original sample is a hydrated salt, known as salt A, which contains water molecules within its crystal structure. The formula of salt A could be, for example, A·xH2O, where A represents the cation and x represents the number of water molecules.
Now, if this original sample is contaminated with a second anhydrous salt, known as salt B, it means that the sample now contains a mixture of salt A and salt B together. The anhydrous salt B does not contain any water molecules within its crystal structure.
When the sample is heated to remove the water content and determine the percent water, both salt A and salt B will be heated. However, upon heating, only the water molecules in salt A will be released. The anhydrous salt B will not contribute any water to the measured mass.
Therefore, the mass of water obtained in the experiment will be lower than the actual mass of water present in the original hydrated sample. As a result, the reported percent water in the hydrated salt will be too low due to the presence of the second anhydrous salt.
In summary, contamination with a second anhydrous salt will decrease the actual mass of water obtained in the experiment, resulting in a lower reported percent water.