A chemist has a sample of hydrated Li2SiF6 and it weighs 0.4813 grams. He heats it strongly to drive off the water of hydration, and after heating, he found that the anhydrous compound has a mass of 0.391 grams. What is the percent of water in the hydrate?

percentage of water = 18.76%

Explanation:

A chemist has a sample of hydrated Li2SiF6 and it weighs 0.4813 grams.

The overall weight of the compound is 0.4813 grams.

weight of hydrated sample = 0.4813 grams.

weight of anhydrous compound = 0.391 grams

percentage weight of water = mass of water/mass of the hydrated compound * 100

mass of water = mass of hydrated compound - mass of anhydrous compound

mass of water = 0.4813 - 0.391 = 0.0903 grams.

percentage of water = 0.0903/0.4813 * 100

percentage of water = 9.03/0.4813

percentage of water = 18.761687097

percentage of water = 18.76%

18.8%

Explanation:

Step 1:

The following data were obtained from the question.

Mass of hydrated Li2SiF6 = 0.4813 g

Mass of anhydrous Li2SiF6 = 0.391 g

Step 2:

Determination of the mass of the water in the hydrate.

Mass of water = (Mass of hydrated Li2SiF6) - (Mass of anhydrous Li2SiF6)

Mass of water = 0.4813 - 0.391

Mass of water = 0.0903 g

Step 3:

Determination of the percentage of water in the hydrate.

The percentage of water = Mass of water / mass of hydrate x 100

The percentage of water = 0.0903/0.4813 x 100

The percentage of water = 18.8%