Question

The hydrated salt \(\mathrm{Na}_{2} \mathrm{CO}_{3} \mathrm{n} \mathrm{H}_{2} \mathrm{O}\) undergoes \(63 \%\) loss in mass on heating and becomes anhydrous. The value of \(n\) is: (a) 4 (b) 6 (c) 8 (d) 10

Short answer

The value of \(n\) is 10.

Step-by-step solution

Understand the Problem

We have a hydrated salt, which loses water upon heating. The mass loss is 63%, which is due to the water content evaporating, resulting in an anhydrous salt. The goal is to determine the number of water molecules (n) in the hydrated salt.

Define the Ratio and Formula

The sample starts as \(\mathrm{Na}_{2} \mathrm{CO}_{3} \cdot n \mathrm{H}_{2} \mathrm{O}\) and becomes \(\mathrm{Na}_{2} \mathrm{CO}_{3}\). Let the molecular weight of anhydrous \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) be \(M_0 = 106\) g/mol, and each water molecule has a molar mass of 18 g/mol. The mass of the hydrated form is \(106 + 18n\).

Calculate the Anhydrous Mass

The anhydrous portion consists of only \(\mathrm{Na}_{2} \mathrm{CO}_{3}\), and after 63% weight loss, it constitutes 37% of the original mass. Therefore, we equate 37% of the initial mass to the mass of the anhydrous salt: \(0.37 \times (106 + 18n) = 106\).

Solve for n

Solve the equation \(0.37 \times (106 + 18n) = 106\). First, divide both sides by 0.37 to get: \(106 + 18n = \frac{106}{0.37}\). Calculate \(\frac{106}{0.37} \approx 286.49\), and rearrange to solve for \(n\): \(18n = 286.49 - 106\), yielding \(18n = 180.49\). Dividing by 18 gives \(n \approx 10\).

Conclusion

Since \(n\) must be a whole number, and our calculation yields approximately 10, it indicates that there are 10 water molecules in the hydrated salt.

Key concepts

Water of Crystallization

Water of crystallization is the water molecules that are part of the crystal structure of a hydrated salt. These water molecules are essential for maintaining the shape and stability of the crystal.
In the chemical formula of a hydrated salt, such as \(\text{Na}_2\text{CO}_3 \cdot n\text{H}_2\text{O}\), \(n\) represents the number of water molecules associated with each formula unit of the salt. This is crucial in influencing several properties of the salt.

• Water of crystallization is weakly bonded to the salt's crystalline structure, making it easy to lose through heating.
• Upon heating, the hydrated salt can transition to an anhydrous form, as the water evaporates.
• In the given problem, the salt initially contains water of crystallization, which constitutes a significant part of its overall mass – resulting in a 63% loss upon heating.

Chemical Formulas

Chemical formulas help in identifying the composition of substances. They illustrate the relative number of each type of atom in a compound.
In hydrated salts, the formula is presented in a way that showcases both the main ionic component and the number of water molecules.

• The chemical formula \(\text{Na}_2\text{CO}_3 \cdot n\text{H}_2\text{O}\) tells us it contains sodium, carbon, oxygen, and water.
• This formula also allows us to calculate the mass of the compound by adding the atomic masses of each constituent.
• In chemical reactions, these formulas depict how the substances interact, such as when heated and water is lost.

This knowledge is key in understanding the transformation of compounds in various physical or chemical processes.

Molar Mass

Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It is an essential concept in stoichiometry and chemical calculations.

• The molar mass of a compound is found by adding up the atomic masses of all the atoms in the compound's chemical formula. For example, \(\text{Na}_2\text{CO}_3\) has a molar mass of 106 g/mol, derived from its components.
• For hydrated salts like \(\text{Na}_2\text{CO}_3 \cdot n\text{H}_2\text{O}\), the molar mass includes both the anhydrous salt and the water molecules, calculated as \(106 + 18n\).
• Molar mass plays a pivotal role when calculating the amount of substance transformed during reactions, like losing crystallization water upon heating.

This concept helps in quantifying how much of each substance is involved in a chemical process.

Anhydrous Compounds

Anhydrous compounds are those that do not contain water molecules; they are essentially the dry forms of hydrates. When the water of crystallization is removed, be it through heat or another process, the remaining substance is anhydrous.
In the context of our exercise, heating the hydrated salt \(\text{Na}_2\text{CO}_3 \cdot n\text{H}_2\text{O}\) leads to its conversion into an anhydrous state: \(\text{Na}_2\text{CO}_3\).

• Anhydrous compounds usually weigh less than their hydrated forms since the mass of the water is no longer part of the structure.
• This loss of water—63% in our problem—is accounted for when the salt's water content evaporates through heating, leading to a decrease in mass.
• Knowing how to form anhydrous compounds is important in various industrial and laboratory processes, where the removal of water is necessary.

This understanding helps to analyze the original compound and determine the number of water molecules it contained.