Problem 53 - 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 \(2.558\) -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

Moles of anhydrous sodium carbonate = \(\frac{0.948 \, \text{g}}{105.99 \, \text{g/mol}} = 0.00894 \, \text{mol}\) Mass of water = 2.558 g - 0.948 g = 1.61 g Moles of water = \(\frac{1.61 \, \text{g}}{18.01 \, \text{g/mol}} = 0.0894 \, \text{mol}\) x = \(\frac{0.0894 \, \text{mol}}{0.00894 \, \text{mol}} = 10\) Therefore, the value of x is 10, which means there are 10 moles of water of hydration per mole of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) in the washing soda sample.

Step 1: Calculate the moles of anhydrous sodium carbonate

To find the moles of anhydrous sodium carbonate, we will use the mass given (0.948 g) and the molar mass of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\), which is 2(22.99 g/mol) + 12.01 g/mol + 3(16.00 g/mol) = 105.99 g/mol. Moles of anhydrous sodium carbonate = \(\frac{0.948 \, \text{g}}{105.99 \, \text{g/mol}}\)

Step 2: Calculate the mass of water in the sample

We know the initial mass of the washing soda sample (2.558 g) and the mass of anhydrous sodium carbonate left after heating (0.948 g). To find the mass of water, we subtract the mass of anhydrous sodium carbonate from the initial mass of the washing soda sample: Mass of water = 2.558 g - 0.948 g

Step 3: Calculate the moles of water in the sample

Now we will calculate the moles of water in the sample using the mass of water found in the previous step and the molar mass of water, which is 18.01 g/mol. Moles of water = \(\frac{\text{Mass of water}}{18.01 \, \text{g/mol}}\)

Step 4: Find the value of x - ratio of moles of water to moles of sodium carbonate

Finally, to find the value of x, we need to determine the ratio of moles of water to moles of sodium carbonate, which means dividing the moles of water by moles of anhydrous sodium carbonate. x = \(\frac{\text{Moles of water}}{\text{Moles of anhydrous sodium carbonate}}\) By substituting the values obtained from Step 1 and Step 3 in the formula above, we can find out the value of x. After calculating the value of x, you will have the number of moles of water of hydration per mole of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) in the washing soda sample.

Key Concepts

Molar Mass Calculation

Understanding the molar mass of a substance is essential in chemistry, especially when working with hydrate compounds. The molar mass is the weight of one mole of a substance, measured in grams per mole (g/mol). It is determined by summing the atomic masses of all the atoms in a molecule of the substance, which can be found on the periodic table. For instance, to find the molar mass of anhydrous sodium carbonate ewline (ummer{Na}_{2}ummer{CO}_{3})ewline, we add together the molar masses of sodium (Na, approximately 22.99 g/mol), carbon (C, 12.01 g/mol), and oxygen (O, 16.00 g/mol).

With this information, calculating moles of any substance becomes straightforward. For a given mass of a compound, you can calculate the number of moles by dividing the mass of the sample by its molar mass. This step is fundamental in stoichiometry, as it allows chemists to convert between mass and moles, facilitating the comparison of amounts of different substances involved in a reaction.

Stoichiometry

Stoichiometry is the segment of chemistry that deals with the relationships between reactants and products in a chemical reaction. It is like a recipe for a chemical reaction that tells you how much of each reactant you need to produce a desired amount of product. A key aspect of stoichiometry is balancing chemical equations, ensuring that the number of atoms of each element is the same on both sides of the equation. This balance reflects the conservation of mass.

In the context of hydrates, stoichiometry allows us to find out not just the molar ratio of the compound and the water it contains, but also the exact formula of the hydrate. By calculating the moles of the anhydrous compound and the moles of water separately, and then taking their ratio, stoichiometry provides the value of ewline x ewline in the chemical formula for a hydrate, which denotes the number of moles of water associated with each mole of the compound.

Chemical Formula

A chemical formula represents a substance using the symbols for its constituent elements and numerical subscripts to show the proportions of each element in the molecule. When dealing with hydrates, the chemical formula also indicates the number of water molecules associated with each unit of the compound. The formula for a hydrate includes the compound's formula followed by a dot and the number of water molecules, each represented as ewline (ummer{H}_{2}ummer{O})ewline. For example, for washing soda, the formula is ewline (ummer{Na}_{2}ummer{CO}_{3}ewline) · x ewline (ummer{H}_{2}ummer{O}) ewline, where ewline x ewline reveals how many water molecules are included per formula unit of the anhydrous salt.

Determining the value of ewline x ewline is vital since it affects the properties and uses of the hydrate. This process involves experimental procedures like heating to remove the water and analyzing the remaining anhydrous compound versus the original hydrated substance, illuminating the bridge between theoretical principles and practical laboratory techniques in the chemistry of hydrates.