Question

a student places 2.500 g of their hydrated alum crystals in a crucible. after heating to a constant mass, the mass of the remaining anhydrous solid is 1.360 g. what mass of water was in the original sample? select one: 1.140 g 3.860 g 2.500 g 1.838 g

Explanation:

Step1: Recall the mass relationship

The mass of water in the original sample is the difference between the mass of the hydrated alum crystals and the mass of the anhydrous solid.
Let \(m_{hydrated}\) be the mass of hydrated alum (\(m_{hydrated}=2.500\ g\)) and \(m_{anhydrous}\) be the mass of anhydrous solid (\(m_{anhydrous} = 1.360\ g\)). The formula for the mass of water \(m_{water}\) is \(m_{water}=m_{hydrated}-m_{anhydrous}\).

Step2: Calculate the mass of water

Substitute the values into the formula: \(m_{water}=2.500 - 1.360\).

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Answer:

1.140 g