Question

a student reported the percentage of water in a hydrate as 42%. the correct value for the percentage of water in the hydrate is 55%. which of the following is the most likely explanation for this difference? select an answer and submit. for keyboard navigation, use the up/down arrow keys to select an answer. a the dehydrated sample absorbed moisture after heating. your answer b strong initial heating caused some of the hydrate sample to spatter out of the crucible. c the amount of the hydrate sample used was too small. d excess heating caused the dehydrated sample to decompose. e the crucible was not heated to constant mass before use.

Explanation:

• Option a: If the dehydrated sample absorbed moisture after heating, the measured mass of the dehydrated sample would be higher than the actual mass. Using the formula for the percentage of water in a hydrate \( \text{Percentage of water}=\frac{\text{Mass of water}}{\text{Mass of hydrate}}\times100\%=\frac{\text{Mass of hydrate}-\text{Mass of dehydrated sample}}{\text{Mass of hydrate}}\times 100\% \), a higher mass of the dehydrated sample would lead to a lower calculated percentage of water.
• Option b: If some of the hydrate sample spattered out of the crucible during strong initial heating, the mass of the hydrate sample we think we are dehydrating is higher than the actual mass that was properly processed. Using the formula \( \text{Percentage of water}=\frac{\text{Mass of hydrate}-\text{Mass of dehydrated sample}}{\text{Mass of hydrate}}\times 100\% \), a lower actual mass of hydrate (due to spattering) with the same measured mass of dehydrated sample would lead to a higher calculated percentage of water.
• Option c: The amount of the hydrate sample used being too small does not systematically cause a lower percentage of water. As long as the procedure is followed correctly (heating to constant mass etc.), the percentage of water is a characteristic property of the hydrate and not dependent on the sample size (in a proper experiment).
• Option d: If the dehydrated sample decomposes due to excess heating, the mass of the final solid (after decomposition) is lower than the mass of the true dehydrated sample. Using the formula \( \text{Percentage of water}=\frac{\text{Mass of hydrate}-\text{Mass of dehydrated sample}}{\text{Mass of hydrate}}\times 100\% \), a lower mass of the “dehydrated” (actually decomposed) sample would lead to a higher calculated percentage of water.
• Option e: If the crucible was not heated to constant mass before use, it may have had some moisture. But when we heat the hydrate, we are measuring the change in mass. The initial moisture in the crucible would be driven off in the first heating (to constant mass for the crucible). So it should not affect the percentage of water in the hydrate calculation.

Answer:

A. The dehydrated sample absorbed moisture after heating.