Question

A 3.592 (g) sample of hydrated magnesium bromide, MgBr2 {xH2O}, is dried in an oven. when the anhydrous salt is removed from the oven, its mass is 2.263 g. What is the value of x

Answer 1

Task:

A 3.592 (g) sample of hydrated magnesium bromide, MgBr2 (xH2O), is dried in an oven. when the anhydrous salt is removed from the oven, its mass is 2.263 g. What is the value of x

Solution:

The mass has changed because the water has vaporized.

$$\mathrm{MgBr_2} \cdot \mathrm{xH_2O} = \mathrm{MgBr_2} + \mathrm{xH_2O}$$

The mass of water is

$$\mathrm{m(H_2O)} = \mathrm{m_1} - \mathrm{m_2}$$

$\mathrm{m_1}$ – mass before drying (g)

$\mathrm{m_2}$ – mass after drying (g)

$$\mathrm{m(H_2O)} = 3.592 - 2.263 = 1.329 \mathrm{~g}$$

The mass of $\mathrm{MgBr_2}$ is 2.263 g

The molar weight of $\mathrm{MgBr_2}$ is

$$\mathrm{MW(MgBr_2)} = \mathrm{MW(Mg)} + 2 \cdot \mathrm{MW(Br)} = 24 + 2 \cdot 80 = 184 \mathrm{~g/mol}$$

The number of moles of $\mathrm{MgBr_2}$ is

$$\mathrm{n(mol)} = \mathrm{m(g)} / \mathrm{MW(g/mol)}$$

$\mathrm{n(MgBr_2)} = 2.263 / 184 = 0.0123 \mathrm{~mol}$

The number of moles of $\mathrm{H_2O}$ is

$$\mathrm{n(H_2O)} = 1.329 / 18 = 0.0738$$

The ratio of indexes is equal to the ratio of moles

$$1 : x = \mathrm{n(MgBr_2)} : \mathrm{n(H_2O)} = 0.0123 : 0.0738 = 1 : 6$$ $$x = 6$$

The formula of hydrated magnesium bromide is

$$\mathrm{MgBr_2} \cdot 6\mathrm{H_2O}$$

Answer: $x = 6$