Question

When a 3.22 g sample of an unknown hydrate of sodium sulfate, Na2SO4 . X H2O (s), is heated, H2O (molar mass 18 g) is driven off. The mass of the anhydrous Na2SO4 (s) (molar mass 142 g) that remains is 1.42g. the value of X in the hydrate is

Answer 1

`\boxed{x = 10}`

Step-by-step 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:
`n = \text{1.42 g} * \frac{\text{1 mol}}{\text{142 g}} = \text{0.0100 mol}`

H₂O:
`n = \text{1.80 g} * \frac{\text{1 mol}}{\text{18 g}} = \text{0.100 mol}`

Step 4. Calculate the molar ratios

`\frac{\text{moles of Na$_(2)$SO$_(4)$}}{\text{moles of H$_(2)$O}} = (0.0100)/(0.100) = (1)/(10)`

`\boxed{x = 10}` , so the formula of the compound is Na₂SO₄·10H₂O