Question

Sam records the mass of his evaporating dish as 9.687 g. He records the mass of the evaporating dish and the sample of hydrate as 18.407 g. After heating the sample in the evaporating dish to constant weight, the mass of them combined is 14.007 g. How many moles of water were removed from the sample by the heating process?

Answer 1

Answer:

0.2544 moles of water were removed from the sample by the heating process.

Explanation:

Mass of empty evaporating dish = 9.687 g

Mass of evaporating dish and sample of hydrate = 18.407 g

Mass of hydrate  = 18.407 g - 9.687 g = 8.72 g

Mass of the evaporating dish and hydrate after heating = 14.007 g

14.007 g = mass of dish + mass of dehydrate

Mass of dehydrate = 14.007 g - 9.867 g = 4.14 g

Mass of water evaporated = hydrated sample - dehydrated sample

= 8.72 g - 4.14 g = 4.58 g

Moles of water evaporated :

$\frac{4.58 g}{18 g/mol}=0.2544 mol$

0.2544 moles of water were removed from the sample by the heating process.