Problem 57 - Chemistry The Central Science

Question

Washing soda, a compound used to prepare hard water for washing laundry, is a hydrate, which means that a certain number of water molecules are included in the solid structure. Its formula can be written as \(\mathrm{Na}_{2} \mathrm{CO}_{3} \cdot x \mathrm{H}_{2} \mathrm{O},\) where \(x\) is the number of moles of \(\mathrm{H}_{2} \mathrm{O}\) per mole of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\). When a \(2.558-\mathrm{g}\) sample of washing soda is heated at \(25^{\circ} \mathrm{C},\) all the water of hydration is lost, leaving \(0.948 \mathrm{~g}\) of \(\mathrm{Na}_{2} \mathrm{CO}_{3} .\) What is the value of \(x ?\)

Step-by-Step Solution

Answer

The value of \(x\) is 10, making the formula of washing soda hydrate \(\mathrm{Na}_{2} \mathrm{CO}_{3} \cdot 10\mathrm{H}_{2} \mathrm{O}\).

Step 1: Calculate moles of anhydrous \(\mathrm{Na}_{2} \mathrm{CO}_{3}\)

First, we need to calculate the moles of anhydrous \(\mathrm{Na}_{2} \mathrm{CO}_{3}\). The molar mass of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) is \(2(22.99 \ g/mol) + 12.01 \ g/mol + 3(16.00 \ g/mol)=105.99 \ g/mol\). The given mass of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) is \(0.948 \ g\). We can calculate the moles as follows: Moles of \(\mathrm{Na}_{2} \mathrm{CO}_{3}=\frac{0.948 \ g}{105.99 \ g/mol}=0.00894 \ mol\)

Step 2: Calculate the mass of water

Next, we need to calculate the mass of water that was lost during heating. We are given the mass of washing soda hydrate initially (2.558 g), and the mass of remaining \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) (0.948 g). Subtract the mass of the anhydrous \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) from the initial mass of washing soda hydrate to get the mass of water: Mass of water = Initial mass of washing soda hydrate - Mass of anhydrous \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) = \(2.558 \ g - 0.948 \ g = 1.610 \ g\)

Step 3: Calculate moles of water

Now, we need to calculate the moles of water. The molar mass of \(\mathrm{H}_{2} \mathrm{O}\) is \(2(1.01 \ g/mol) + 16.00 \ g/mol = 18.02 \ g/mol\). We can calculate the moles using the mass of water (1.610 g) as follows: Moles of \(\mathrm{H}_{2} \mathrm{O}=\frac{1.610 \ g}{18.02 \ g/mol} = 0.0894 \ mol\)

Step 4: Determine the value of \(x\)

Finally, we can determine the value of \(x\), which is the moles of water per mole of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\), by dividing the moles of water by the moles of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\): \(x = \frac{\text{moles of } \mathrm{H}_2 \mathrm{O}}{\text{moles of } \mathrm{Na}_2 \mathrm{CO}_3} = \frac{0.0894 \ mol}{0.00894 \ mol} = 10\) The value of \(x\) is 10. Therefore, the formula of washing soda hydrate is \(\mathrm{Na}_{2} \mathrm{CO}_{3} \cdot 10\mathrm{H}_{2} \mathrm{O}\).

Key Concepts

Chemical Formula of Hydrates

Hydrate compounds are a unique group of substances that incorporate water molecules into their molecular structure. The water is not merely attached to the compound; it is an integral part of its crystal lattice. The general formula used to represent hydrates is written as follows: \( \text{Compound} \cdot x\text{H}_2\text{O} \), where \( x \) signifies the number of water molecules (or moles of water) associated with each formula unit of the compound.

Hydrates can be found in various applications, from construction materials such as cement to household items like the washing soda mentioned in the exercise. Determining the correct value of \( x \), the number of hydrates, is crucial because it can significantly impact the substance's properties, such as its reactivity or stability.

During a dehydration process, hydrates lose water to become anhydrous, which is a critical step in calculating the water content. This process also played a central role in the given exercise, where heating washing soda led to the release of the water of hydration, allowing for the calculation of \( x \).

Molar Mass Calculation

The molar mass is a fundamental concept in chemistry, representing the mass of one mole of a substance. It's expressed in grams per mole (g/mol) and is equivalent to the molecular weight of a compound. For hydrates, the calculation of molar mass must include the mass of the bound water molecules as well as the anhydrous compound.

To calculate the molar mass, one would sum up the atomic masses of all atoms present in the formula unit, including the water molecules. For instance, in the exercise's case with washing soda \( \mathrm{Na}_2\mathrm{CO}_3 \cdot x\mathrm{H}_2\mathrm{O} \), the molar mass calculation first involved finding the mass of the anhydrous sodium carbonate and then accounting for the mass of water molecules corresponding to \( x \) hydrates.

Understanding molar mass is essential for converting between the mass of a substance and the number of moles, which is a fundamental step for stoichiometry and chemical calculations as shown in the provided step-by-step problem-solving method.

Stoichiometry

Stoichiometry is the section of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It relies heavily on the law of conservation of mass and the concept of moles. Stoichiometry comes into play when balancing chemical equations, calculating reacting masses, or in this exercise, finding the number of water molecules in a hydrate.

In the provided exercise, stoichiometry was used to solve for \( x \), the number of water molecules bound in the hydrate. This was done by employing the stoichiometric relationship between the mass of lost water upon heating and the mass of the remaining anhydrous compound. The ratio of the moles of water to the moles of anhydrous sodium carbonate directly gave the value of \( x \).

Through stoichiometry, one can understand how materials react and combine, how much product to expect from a reaction, and the composition of hydrates, which reinforces the principles of molar mass and hydrate formulas.