Question

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

Short answer

The value of \( n \) is 10 (option d).

Step-by-step solution

Understand the Problem

We need to determine the value of \( n \) in the hydrated salt \( \mathrm{Na}_{2} \mathrm{CO}_{3} \cdot \mathrm{n}\, \mathrm{H}_{2} \mathrm{O} \) given that the salt loses 63% of its mass on heating to become anhydrous \( \mathrm{Na}_{2} \mathrm{CO}_{3} \). This means we need to find the number of water molecules \( n \) attached to this salt molecule.

Calculate Initial and Remaining Masses

Let the molar mass of \( \mathrm{Na}_{2} \mathrm{CO}_{3} \) be \( M_1 = 106\, \mathrm{g/mol} \), and the molar mass of \( \mathrm{H}_{2} \mathrm{O} \) be \( M_2 = 18\, \mathrm{g/mol} \). The initial mass of the hydrated salt is \( M_i = 106 + n \times 18 \). After heating, the remaining anhydrous mass is \( M_r = 106 \).

Use the Percentage Loss in Mass

The hydrated mass loss turns out to be 63%: \( M_i - M_r = 0.63 \times M_i \). Rearrange to find: \[ M_r = M_i - 0.63 \times M_i = 0.37 \times M_i \].

Set Up the Equation

From \( M_r = 0.37 \times M_i \), we substitute \( M_r = 106 \) and \( M_i = 106 + 18n \) to get: \[ 106 = 0.37 \times (106 + 18n) \].

Solve the Equation for \( n \)

Substitute and solve: \[ 106 = 0.37 \times (106 + 18n) \] \[ 106 = 39.22 + 6.66n \] Move 39.22 to the other side to isolate terms with \( n \): \[ 106 - 39.22 = 6.66n \] \[ 66.78 = 6.66n \] Divide both sides by 6.66 to solve for \( n \): \[ n = \frac{66.78}{6.66} \approx 10 \].

Interpret the Result

The calculated value of \( n \) is approximately 10. Thus, the number of water molecules in the hydrated salt is 10.

Key concepts

Mass Loss Calculation

When a hydrated salt is heated, it loses water molecules and ultimately transforms into an anhydrous salt. To understand how to calculate the mass loss upon heating, consider the hydrated salt \( \mathrm{Na}_{2} \mathrm{CO}_{3} \cdot \mathrm{n} \mathrm{H}_{2} \mathrm{O} \). The problem states that this salt loses 63% of its mass, indicating a substantial reduction due to loss of water (i.e., water of crystallization).
The initial mass of the hydrated salt includes both the mass of \( \mathrm{Na}_{2} \mathrm{CO}_{3} \) (sodium carbonate) and the mass of water molecules that are part of this compound. When the salt is heated, the water evaporates, and the remaining mass consists only of the anhydrous \( \mathrm{Na}_{2} \mathrm{CO}_{3} \).
The calculation involves determining the percentage loss: the initial mass minus the remaining mass is equal to 63% of the initial mass, represented as \( M_i - M_r = 0.63 \times M_i \). Rearranging gives \( M_r = 0.37 \times M_i \), which is crucial to find the initial mass when the mass of the anhydrous salt \( M_r \) is known.

Molar Mass Determination

Determining the molar mass is essential in chemistry, especially when dealing with compounds like hydrated salts. In this exercise, the molar masses provided are for \( \mathrm{Na}_{2} \mathrm{CO}_{3} \) and \( \mathrm{H}_{2} \mathrm{O} \), which are 106 g/mol and 18 g/mol, respectively.
To find the molar mass of the hydrated salt \( \mathrm{Na}_{2} \mathrm{CO}_{3} \cdot \mathrm{n} \mathrm{H}_{2} \mathrm{O} \), you combine the molar mass of anhydrous \( \mathrm{Na}_{2} \mathrm{CO}_{3} \) with the total mass of the water molecules: \( M_i = 106 + n \times 18 \).
Precision in calculating molar masses is vital as small errors can lead to significant differences in understanding and interpreting chemical reactions or compositions.
Keep in mind, molar mass is all about weighing one mole of the molecules, and by knowing the individual masses of elemental components, the total mass for complex compounds can be calculated.

Water of Crystallization

Water of crystallization refers to the water molecules that are an integral part of the crystalline structure of a compound, like our hydrated sodium carbonate (\( \mathrm{Na}_{2} \mathrm{CO}_{3} \cdot \mathrm{n} \mathrm{H}_{2} \mathrm{O} \)). These molecules are crucial for maintaining the structural integrity and properties of the hydrated form.
When the compound is heated, these water molecules are released as vapor, leading to a decrease in mass. In our example, each molecule of \( \mathrm{Na}_{2} \mathrm{CO}_{3} \) is associated with \( n \) water molecules, and these are termed as physico-chemical bonded water molecules.
Understanding the concept of water of crystallization is key in analyzing reactions involving hydrated salts. Removing these water molecules changes the physical appearance and sometimes the chemical properties of the compound, turning it from hydrated to anhydrous.
Hydrated salts are not only fascinating in academia but also incredibly useful in industries, such as in detergents and construction, where control over drying and hardening rates is required.

Anhydrous Salt Formation

Anhydrous salts form when hydrated salts, such as \( \mathrm{Na}_{2} \mathrm{CO}_{3} \cdot \mathrm{n} \mathrm{H}_{2} \mathrm{O} \), lose their water of crystallization due to heating. The term 'anhydrous' simply means 'without water'.
The transition from hydrated to anhydrous is a physical change, not involving the breaking of ionic or covalent bonds beyond the release of water. This process is usually reversible, where the anhydrous salt can absorb moisture again from the environment under suitable conditions.
For our specific example, heating leads to a mass change where the final anhydrous form is 37% of the original mass. As a result, only the "dry" form of \( \mathrm{Na}_{2} \mathrm{CO}_{3} \) remains. This change in form is often accompanied by visible changes, like alterations in color or texture.
Anhydrous salts have practical applications, particularly as drying agents or desiccants, because they readily absorb water. This ability makes them useful in laboratory experiments and in preserving materials sensitive to moisture.

• It’s crucial to understand that the formation of anhydrous salts helps determine formulas of unknown hydrates.
• This knowledge is especially useful in chemical manufacturing and in understanding natural cycles where moisture plays a pivotal role.