Answer:
[tex]\boxed{x = 10}[/tex]
Explanation:
This is like an empirical formula question, except that you are finding the molar ratio of compounds instead of atoms.
Step 1. Gather the information in one place.
M_r: 142 18
Na₂SO₄·xH₂O(s) ⟶ Na₂SO₄(s) + xH₂O(g)
m/g: 3.22 1.42
Step 2. Calculate the mass of the water
Mass of H₂O = mass of Na₂SO₄·xH₂O – mass of Na₂SO₄
= 3.22 – 1.42 = 1.80 g
Step 3. Calculate the moles of each product
Na₂SO₄: [tex]n = \text{1.42 g} \times \dfrac{\text{1 mol}}{\text{142 g}} = \text{0.0100 mol}[/tex]
H₂O: [tex]n = \text{1.80 g} \times \dfrac{\text{1 mol}}{\text{18 g}} = \text{0.100 mol}[/tex]
Step 4. Calculate the molar ratios
[tex]\dfrac{\text{moles of Na$_{2}$SO$_{4}$}}{\text{moles of H$_{2}$O}} = \dfrac{0.0100}{0.100} = \dfrac{1}{10}[/tex]
[tex]\boxed{x = 10}[/tex] , so the formula of the compound is Na₂SO₄·10H₂O
