Question

1.520 g of hydroxide of a metal on ignition gave \(0.995\) \(\mathrm{g}\) of oxide. The equivalent weight of the metal is a. \(0.995\) b. 190 c. \(1.90\) d. 9

Short answer

The equivalent weight of the metal is 9.

Step-by-step solution

Understanding the Reaction

The reaction involves the conversion of metal hydroxide, with the formula \( M(OH)_2 \), to metal oxide, \( MO \). This reaction can be written as: \[ M(OH)_2 \rightarrow MO + H_2O \] The loss in mass (from 1.520 g to 0.995 g) represents the mass of water released.

Calculate Mass of Water

The mass of water produced is the difference between the initial mass of the hydroxide and the final mass of the oxide. \[ \text{Mass of } H_2O = 1.520 \, \text{g} - 0.995 \, \text{g} = 0.525 \, \text{g} \]

Molar Mass of Water

The molar mass of water \((H_2O)\) is approximately \(2 \times 1 + 16 = 18 \, \text{g/mol}\).

Moles of Water

Calculate the number of moles of water produced using its molar mass: \[ \text{Moles of } H_2O = \frac{0.525 \, \text{g}}{18 \, \text{g/mol}} = 0.0292 \, \text{mol} \]

Moles of Metal Hydroxide

In the reaction, 1 mole of \( M(OH)_2 \) produces 1 mole of \( H_2O \), so the moles of \( M(OH)_2 \) is also 0.0292 mol.

Molar Mass of Metal Hydroxide

The molar mass of \( M(OH)_2 \) is calculated by dividing the mass of metal hydroxide by the moles of \( M(OH)_2 \): \[ Molar\, mass\, of\, M(OH)_2 = \frac{1.520 \, \text{g}}{0.0292 \, \text{mol}} = 52.05 \, \text{g/mol} \]

Metal Equivalent Weight

The metal equivalent weight is calculated from the molar mass of the hydroxide and the known molar mass of \( OH \) (where \( \text{OH} \approx 17\)): \[ M + 2 \times 17 = 52.05 \ M = 52.05 - 34 = 18.05 \, \text{g/mol} \]

Conclusion

The equivalent weight of the metal is thus 18.05 g/mol, which corresponds to option (d), 9 g/mol, as the equivalent weight represents \( \frac{M}{2} \) because of the 2 hydroxide ions.

Key concepts

Metal hydroxide

Metal hydroxides are compounds that consist of a metal cation bonded to hydroxide ion groups often written as \( M(OH)_x \) where \( M \) represents a metal atom and \( x \) is the number of hydroxide ions. These compounds are generally crystalline, ionic solids. They are commonly found in nature and can be synthesized in laboratories through various chemical reactions.

• Metal hydroxides are typically alkaline and can neutralize acids to form corresponding metal salts and water.
• They are often involved in reactions where they decompose to form metal oxides and water upon heating, as shown in the problem.

Understanding metal hydroxides is crucial in chemistry, especially when studying equilibrium, reactions, and the periodic table's behavior.

Metal oxide

Metal oxides are compounds consisting of metal atoms bonded with oxygen. They are commonly formed through the oxidation of metals or by heating metal hydroxides. The general formula for a metal oxide produced from metal hydroxide is \( MO \), indicating a single metal atom bonded to one oxygen atom.

• Metal oxides can be basic, acidic, or amphoteric depending on the metal's position in the periodic table.
• Basic metal oxides like those from alkali metals and alkaline earth metals react with water to form metal hydroxides.

In the exercise, the metal hydroxide was heated and the water was released, resulting in the formation of metal oxide.

Moles calculation

Calculating moles is a fundamental skill in chemistry. A mole is a unit of measurement that represents a specific number of atoms, molecules, or particles, often denoted as \( 6.022 \times 10^{23} \) entities (Avogadro's number). The formula used is:\[\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} \]Steps to calculate moles typically involve:

• Identifying the mass of a substance in grams.
• Determining its molar mass in grams per mole, which can be found on the periodic table for each element.
• Using the formula to solve for moles.

In the exercise, we calculated the moles of water produced by understanding its mass and molar mass, enabling further calculations of the metal hydroxide's molar values.

Chemical reaction stoichiometry

Stoichiometry is the part of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It tells us how much of each substance is involved at the start of a reaction, how much is used, and how much is produced. In our example, stoichiometry was used to understand the conversion of metal hydroxide to metal oxide and water:

• The balanced chemical equation shows the relationship between the reactants and products.
• It also allowed us to deduce the moles of products formed per mole of reactants.

Using stoichiometry, we could determine that one mole of \( M(OH)_2 \) forms one mole of water and one mole of \( MO \), providing the foundation for calculating equivalent weight.