Lab Questions Suppose water vapor condensed on the crucible after it was fired,

Question

Lab Questions Suppose water vapor condensed on the crucible after it was fired, but before it was initially weighed. The condensation caused the mass measurements of both the empty crucible, and the crucible with the hydrated salt, to be higher than actual. What happens when the hydrate mass is calculated by taking the difference between the measurements? 1. The condensed water (on the crucible in question 1) evaporates when the hydrate is heated in the crucible. Providing that the condensation does not recur on cooling, what now happens to the calculations for the lost water mass and the water percentage? 2. Suppose you touched the crucible with oily fingers after weighing it initially, but before the hydrate was weighed. What happens when the hydrate mass is calculated by taking the difference between the measurements? If the oil evaporates along with the water when the hydrate is heated, what now happens to the calculations for the lost water mass and the water percentage? 3. Suppose some of your hydrate sample is actually anhydrous. What does anhydrous mean? How will this affect the calculations for the lost water mass and the water 4. percentage? Suppose you do not heat your sample a second time to be sure that all the water is removed. How does this affect your final mass measurement for the anhyd How will this affect the calculations for the lost water mass and the water percentage? 5. rous salt?

Explanation / Answer

1) Hydrate mass=(mass of crucible +hydrated salt)-(mass of crucible)

So if error exists in both these masses then,

Hydrate mass=(mass of crucible +hydrated salt+error)-(mass of crucible +error)=(mass of crucible +hydrated salt)-(mass of crucible)

error does not add to the mass of the hydrated salt,it is accurate

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