Introduction to the Mole Concept

To work with hydrates and other chemical formulas, chemists use the mole concept. A mole is a counting unit, much like a dozen, but instead of 12, it represents 6.022 × 10²³ particles (Avogadro’s number). These “particles” can be atoms, molecules, or formula units depending on the substance. The mass of one mole of a substance, in grams, is equal to its molar mass, which is obtained by adding the atomic masses of all the atoms in the formula. The mole concept allows you to relate a measured mass in the lab to the actual number of particles present and to the ratios in a chemical formula. In this lab, you will use the mole concept to compare the number of moles of water lost to the number of moles of anhydrous salt remaining after heating, which will let you determine x in CuSO₄·xH₂O.

When a hydrate is heated, the water of hydration is released as vapor, leaving behind the anhydrous compound, which means “without water.” This process can be represented as:

$$\text{hydrate}\stackrel{\text{heat}}{\to }\text{anhydrous salt}+\text{water vapor}$$

For example, barium chloride dihydrate (BaCl₂·2H₂O) decomposes when heated to produce barium chloride (BaCl₂) and water vapor:

$$\text{BaCl}_2\cdot 2\text{H}_2\text{O (s)} \rightarrow \text{BaCl}_2\text{ (s)} + 2\text{H}_2\text{O (g)}$$

In this case, one mole of the hydrate produces one mole of anhydrous salt and two moles of water vapor.

The percent by mass of water in a hydrate can be calculated in two ways:

Method 1 uses the known chemical formula and molar masses. First, calculate the molar mass of the entire hydrate by adding together the molar mass of the salt and the molar mass of the water molecules. Then, divide the molar mass of just the water portion by the total molar mass of the hydrate, and multiply by 100 to get the percent water. For example, BaCl₂·2H₂O has a total molar mass of 244.26 g/mol, and the water portion has a molar mass of 36.03 g/mol. The percent water is:

$$\frac{36.03}{244.26}\times 100=14.75\mathrm{\%}$$

Method 2 uses experimental data. First, weigh the hydrate, heat it to remove the water, and then weigh the anhydrous salt. The difference in mass is the water lost. Divide the mass of water by the original mass of the hydrate and multiply by 100 to find the percent water. For example, if 1.000 g of BaCl₂·2H₂O is heated to yield 0.853 g of BaCl₂, then the water mass is:

$$1.000\text{g}-0.853\text{g}=0.147\text{g}$$

The percent water is:

$$\frac{0.147}{1.000}\times 100=14.7\mathrm{\%}$$

Once you know the masses of water and anhydrous salt, you can use the mole concept to determine the formula of the hydrate. Convert each mass to moles using their respective molar masses, then divide the moles of water by the moles of anhydrous salt to find x, the number of water molecules per formula unit. Using the above data, 0.853 g of BaCl₂ is:

$$0.853\text{g}\times \frac{1\text{mol}}{208.23\text{g}}=0.004096\text{mol salt}$$

The water mass of 0.147 g is:

$$0.147\text{g}\times \frac{1\text{mol}}{18.00\text{g}}=0.00817\text{mol water}$$

The ratio is:

$$\frac{0.00817}{0.004096}\approx 2.00$$

Thus, the formula is BaCl₂·2H₂O.

In this lab, you will apply the same process to a hydrated version of copper sulfate. Upon heating, it will release its water of hydration and turn into a white anhydrous compound. By weighing the sample before and after heating, using the mole concept, and applying percent composition calculations, your goal will be to determine the value of x in CuSO₄·xH₂O and write the correct chemical formula for the hydrate.