A student has a sample of CaSO4 hydrate and it weighs 0.4813 grams. He heats it strongly to drive off the water of hydration, and after subsequent heatings, the student finds the anhydrous compound has a constant mass of 0.3750 grams. Find the formula of the hydrate.(2 points)

CaSO4•3H2O.

Explanation:

Let the compound be CaSO4.xH2O.

The following data were obtained from the question:

Mass of hydrated compound (CaSO4.xH2O) = 0.4813g

Mass of anhydrous compound (CaSO4) = 0.3750g

Next we shall determine the mass of the water is the hydrated compound.

This is illustrated below:

Mass of water = mass of hydrated – mass of anhydrous.

Mass of water = 0.4813 – 0.3750

Mass of water = 0.1063g

Next, we shall determine the number of mole of the anhydrous compound and the number of mole of the water present in the compound. This is illustrated below:

Molar mass of anhydrous CaSO4 = 63.5 + 32 + (16x4) = 159.5g/mol

Mass of anhydrous CaSO4 = 0.3750g

Mole of anhydrous CaSO4 =...?

Mole = mass /Molar mass

Mole of anhydrous CaSO4 = 0.3750/159.5 = 2.35×10¯³ mole

Molar mass of H2O = (2x1) + 16 = 18g/mol

Mass of H2O = 0.1063g

Number of mole of H2O =.?

Mole = mass /Molar mass

Mole of H2O = 0.1063/18 = 5.91×10¯³ mole

Next we shall determine the ratio of number of mole of anhydrous CaSO4 to that of H2O. This is illustrated below:

Mole of anhydrous CaSO4 = 2.35×10¯³ mole

Mole of H2O = 5.91×10¯³ mole

Ratio of anhydrous CaSO4 to H2O =>

CaSO4 : H2O => 2.35×10¯³ /5.91×10¯³

CaSO4 : H2O => 1 : 3

Therefore, for 1 mole of the anhydrous CaSO4, there are 3 moles of H2O.

Therefore, the formula for the hydrate compound CaSO4.xH2O => CaSO4•3H2O.

a pan containing water is placed in a positio n where it is exposed to a 10 mile per hour wind. the pan contains a uniform depth of 1.27 cm of water, is quite wide, and has a length of 2.44 m in the wind direction. the water in the pan is at a constant temperature of 15.5 °c.