A sample of a certain barium chloride hydrate, BaCl2.nH2O, has a mass of 7.62 g. When this sample is heated in a crucible, it is converted to anhydrous barium chloride and water, with the water escaping as steam during the process. The mass of the solid anhydrous residue that remains in the crucible is 6.48 g. Determine the formula of this hydrate

BaCl2.2H2O

Explanation:

Data obtained from the question include:

Mass of barium chloride hydrate (BaCl2.nH2O) = 7.62g

Mass of anhydrous barium chloride (BaCl2) = 6.48g

Next, we shall determine the mass of water in the barium chloride hydrate (BaCl2.nH2O). This can be obtained as follow:

Mass of water (H2O) = Mass of barium chloride hydrate (BaCl2.nH2O) – Mass of anhydrous barium chloride (BaCl2)

Mass of water (H2O) = 7.62 – 6.48

Mass of water (H2O) = 1.14g

Next, we shall write the balanced equation for the reaction. This is given below:

BaCl2.nH2O —> BaCl2 + nH2O

Next, we shall determine the masses of BaCl2.nH2O that decomposed and the mass H2O produced from the balanced equation.

Molar mass of BaCl2.nH2O = 137 + (35.5x2) + n[(2x1) + 16]

= 137 + 71 + n[2 + 16]

= (208 + 18n) g/mol

Mass of BaCl2.nH2O from the balanced equation = 1 x (208 + 18n) = (208 + 18n) g

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

Mass of H2O from the balanced equation = n x 18 = 18n g

Summary:

From the balanced equation above,

(208 + 18n) g of BaCl2.nH2O decomposed to produce 18n g of H2O.

Finally, we shall determine the formula for the hydrated compound as follow:

From the balanced equation above,

(208 + 18n) g of BaCl2.nH2O decomposed to produce 18n g of H2O.

But 7.62g of BaCl2.nH2O decomposed to produce 1.14g of H2O i.e

18n/(208 + 18n) = 1.14/7.62

Cross multiply

18n x 7.62 = 1.14(208 + 18n)

137.16n = 237.12 + 20.52n

Collect like terms

137.16n – 20.52n = 237.12

116.64n = 237.12

Divide both side by the coefficient of n i.e 116.64

n = 237.12/116.64

n = 2

Therefore the formula for the hydrate, BaCl2.nH2O is BaCl2.xH2O.

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.

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