Question

A 7.028 gram sample of a sodium sulfate hydrate is heated. After driving off all the water, 3.100 grams of the anhydrous salt is left. What is the name of the hydrate?

Answer 1

Answer:

$NaSO_4.10H_2O$

Explanation:

Given that:-

Mass of the hydrated salt = 7.028 g

Mass of the anhydrous salt = 3.100 g

Mass of water eliminated = Mass of the hydrated salt - Mass of the anhydrous salt = 7.028 - 3.100 g = 3.928 g

Moles of water:

Mass of water = 3.928 g

Molar mass of $H_2O$ = 18 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus, moles are:

$moles= \frac{3.928\ g}{18\ g/mol}$

$moles_{water}= 0.2212\ mol$

Moles of anhydrous salt:

Amount = 3.100 g

Molar mass of $NaSO_4$ = 142.04 g/mol

Thus, moles are:

$moles= \frac{3.100\ g}{142.04\ g/mol}$

$moles_{CaSO_4}= 0.02182\ mol$

The simplest ratio of the two are:

$NaSO_4:H_2O$ =0.02182 :  0.2212 = 1 : 10

Hence, the formula for hydrate is:- $NaSO_4.10H_2O$