Answer to Question #184524 in General Chemistry for bubbles

Question

Glauber’s salt is a form of hydrated sodium sulfate, Na2SO4ᐧ xH2O(s), that is used medicinally as a mild laxative. Chemical analysis of this hydrate shows that it is 44.1% sodium sulfate. Determine the FORMULA and the NAME of this hydrate. 

Answer 1

Q184524

Glauber’s salt is a form of hydrated sodium sulfate, Na₂SO4ᐧ xH2O(s), that is used medicinally as a mild laxative. Chemical analysis of this hydrate shows that it is 44.1% sodium sulfate. Determine the FORMULA and the NAME of this hydrate. 

Solution :

If there is 44.1 % of sodium sulfate in the salt,

then the remaining (100 - 44.1 ) = 55.9 % will be H₂O.

Let us consider that there are 100 grams of hydrated sodium sulfate.

Since there is 44.1 % sodium sulfate, so 100 grams of hydrated sodium sulfate will contain

44.1 grams of sodium sulfate and 55.9 grams of H₂O.

Step 1: To find moles of Na₂SO₄ and H₂O present in 100 grams of hydrated salt.

Convert 44.1 grams of Na₂SO₄ and 55.9 grams of H₂O to moles by using their molar mass.

Molar mass of Na₂SO₄ = 142.041 g/mol

Molar mass of H2O = 18.0148 g/mol

$$moles \ of \ Na_2SO_4 = 44.1 \ \bcancel{g \ Na_2SO_4 } * \frac{1 \ mol \ Na_2SO_4 }{142.041 \bcancel{g \ Na_2SO_4 }}$$

$$= 0.3105 \ mol \ of \ Na_ 2SO_4$$

$$moles \ of \ H_2O= 55.9 \ \bcancel{g \ H_2O } * \frac{1 \ mol \ H_2O}{18.0148 \bcancel{g \ H_2O }}$$

$$= 3.1030 \ mol \ of \ H_2O$$

For finding the number 'x' . We will divide moles of both by the one with the least number of moles.

Here Na2SO4 has least number of moles.

$$For Na_2SO_4 , \ \ \ \ \ \frac{0.3105 \ mol }{0.3105 \ mol } = 1$$

$$For H_2O , \ \ \ \ \ \ \frac{3.1030 \ mol }{0.3105 \ mol } = 9.994 \ ≈ \ 10$$

So for each mole of Na₂SO₄, there is 10 mole of H₂O.

So x = 10.

The formula of the hydrated salt will be Na₂SO₄.10 H₂O.

There are 10 water molecules in this formula. For denoting the number of hydrated water molecules we will use the prefix ' deca' .

10 means deca.

So the name of the compound will be

Sodium sulfate decahydrate.