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-g sample of washing soda is heated at \(125^{\circ} \mathrm{C}\), all the water of hydration is lost, leaving \(0.948 \mathrm{~g}\) of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\). What is the value of \(x\) ?

Short answer

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

Step-by-step solution

Calculate mass of water removed

To find the mass of the water removed, we can subtract the remaining mass of the anhydrous sodium carbonate from the initial mass of the washing soda: Mass of water removed = Initial mass - Mass of anhydrous sodium carbonate Mass of water removed = 2.558 g - 0.948 g Mass of water removed = 1.61 g Now let's calculate the moles of anhydrous \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) and \(\mathrm{H}_{2} \mathrm{O}\).

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

To find the moles of anhydrous \(\mathrm{Na}_{2} \mathrm{CO}_{3}\), we will use the molar mass of sodium carbonate (105.99 g/mol): Moles of anhydrous \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) = Mass of anhydrous \(\mathrm{Na}_{2} \mathrm{CO}_{3}\)/Molar mass of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) Moles of anhydrous \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) = 0.948 g / 105.99 g/mol Moles of anhydrous \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) = 8.943 × 10^{-3} mol

Calculate moles of \(\mathrm{H}_{2}\mathrm{O}\)

To find the moles of \(\mathrm{H}_{2} \mathrm{O}\), we will use the molar mass of water (18.015 g/mol): Moles of \(\mathrm{H}_{2} \mathrm{O}\) = Mass of \(\mathrm{H}_{2} \mathrm{O}\)/Molar mass of \(\mathrm{H}_{2} \mathrm{O}\) Moles of \(\mathrm{H}_{2} \mathrm{O}\) = 1.61 g / 18.015 g/mol Moles of \(\mathrm{H}_{2} \mathrm{O}\) = 8.941 × 10^{-2} mol Finally, let's calculate the mole ratio between anhydrous \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) and \(\mathrm{H}_{2} \mathrm{O}\) to determine the value of \(x\).

Calculate mole ratio and determine the value of \(x\)

To find the value of \(x\), we need to divide the moles of \(\mathrm{H}_{2} \mathrm{O}\) by the moles of anhydrous \(\mathrm{Na}_{2} \mathrm{CO}_{3}\): \(x\) = Moles of \(\mathrm{H}_{2} \mathrm{O}\) / Moles of anhydrous \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) \(x\) = 8.941 × 10^{-2} mol / 8.943 × 10^{-3} mol \(x\) ≈ 10 The mole ratio of water in washing soda is x = 10, so the formula of washing soda is \(\mathrm{Na}_{2} \mathrm{CO}_{3} \cdot 10 \mathrm{H}_{2} \mathrm{O}\).

Key concepts

Stoichiometry

Stoichiometry is the section of chemistry that deals with the calculation of reactants and products in chemical reactions. In essence, it's the math behind chemistry. Fundamental to stoichiometry is the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. This means that atoms are rearranged in a reaction, but the total number of atoms of each element must remain the same before and after the reaction.

For stoichiometry to work, a balanced chemical equation is paramount as it provides the ratio in which the reactants combine and the products form. These proportional relationships are crucial when calculating how much of a reactant is required to produce a desired amount of product, or in our case, determining the ratio of water to sodium carbonate in a hydrate. Students may find it helpful to think of stoichiometry as a recipe, where one must mix ingredients in specific ratios to get the desired outcome, and understanding these ratios is key to solving problems in stoichiometry.

Molar Mass

Every element has a certain mass, known as the atomic mass, which is the mass of a single atom of that element. When we scale this up to a mole, which is a standard scientific unit for measuring large quantities of very small entities such as atoms, ions, or molecules, we get the molar mass of the element.

The molar mass is critical in chemistry because it bridges the gap between the atomic scale and the scale we can observe and measure in the lab. When solving problems involving chemical reactions, we often deal with moles rather than counting individual atoms or molecules because moles provide a manageable way to express quantities. Knowledge of the molar mass allows us to convert back and forth between the mass of a substance and the amount in moles, as demonstrated in the computation of the mole ratio of water to sodium carbonate in the given problem.

Hydrates

Hydrates are compounds that contain water molecules within their crystalline structure. These water molecules are not merely trapped mechanically; they are chemically bonded to the host molecule. However, they can often be removed through the application of heat, resulting in a process known as dehydration.

Understanding hydrates is essential in many areas of chemistry, including the problem we are discussing. The water is integral to the structure of the hydrate, and its presence can significantly alter the physical properties of the compound. The number of water molecules associated with each formula unit of the compound is indicated by the numerical prefix (e.g., 'monohydrate' for one water molecule, 'decahydrate' for ten). Learning about hydrates is not only fascinating but also has practical implications, such as determining the composition of compounds used in everyday products, like the washing soda in our exercise.

Chemical Formula

The chemical formula of a compound tells us the exact number of different atoms of each element that make up a single unit of that compound. It serves as a type of shorthand to convey significant amounts of information about the composition of the substance.

In both simple and complex compounds, the substances involved may combine in fixed ratios that can vary widely across different compounds. For hydrates, the formula includes not just the ionic or molecular components but also the number of water molecules associated with them, a crucial piece of information that needs to be considered in calculations. By correctly interpreting the chemical formula, one can deduce the ratio of elements and, as in the case of the given problem, the stoichiometry of a hydrated compound can be analyzed to find the value of an unknown variable, such as the number of moles of water per mole of the compound.