Question

Anhydrous calcium chloride is a good drying agent because it will rapidly pick up water. Suppose you have stored some carefully dried \(\mathrm{CaCl}_{2}\) in a desiccator. Unfortunately, someone did not close the top of the desiccator tightly, and the \(\mathrm{CaCl}_{2}\) became partially hydrated. A \(150-\mathrm{g}\) sample of this partially hydrated material was dissolved in \(80 \mathrm{g}\) of hot water. When the solution was cooled to \(20^{\circ} \mathrm{C}, 74.9 \mathrm{g}\) of \(\mathrm{CaCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) precipitated. Knowing the solubility of calcium chloride in water at \(20^{\circ} \mathrm{C}\) is \(74.5 \mathrm{g} \mathrm{CaCl}_{2} / 100 \mathrm{g}\) water, determine the water content of the \(150-\mathrm{g}\) sample of partially hydrated calcium chloride (in moles of water per mole of \(\mathrm{CaCl}_{2}\) ).

Short answer

0.793 moles of water per mole of \(\mathrm{CaCl}_{2}\).

Step-by-step solution

Determine Mass of Soluble CaCl2

First, we need to find out how much \(\mathrm{CaCl}_{2}\) remains dissolved in the \(80 \, \mathrm{g}\) of water. Since we know the solubility at \(20^{\circ} \mathrm{C}\) is \(74.5 \, \mathrm{g} / 100 \, \mathrm{g}\) of water, we calculate the mass of \(\mathrm{CaCl}_{2}\) that remains in solution: \(74.5 \, \mathrm{g} \, \mathrm{CaCl}_{2} / 100 \, \mathrm{g} \, \mathrm{water} \times 80 \, \mathrm{g} \, \mathrm{water} = 59.6 \, \mathrm{g} \, \mathrm{CaCl}_{2}.\)

Calculate Total CaCl2 in Sample

Now, we find the total dry \(\mathrm{CaCl}_{2}\) initially present in the 150 g sample. The total \(\mathrm{CaCl}_{2}\) is the sum of the \(\mathrm{CaCl}_{2}\) that precipitated as the hexahydrate and the \(\mathrm{CaCl}_{2}\) that stayed in the solution:\(74.9 \, \mathrm{g} \, \mathrm{CaCl}_{2} \cdot 6 \mathrm{H}_{2}O \times \frac{1 \, \mathrm{CaCl}_{2}}{1 \, \mathrm{CaCl}_{2} \cdot 6 \mathrm{H}_{2}O} = 0.548 \, \mathrm{mol} \, \mathrm{CaCl}_{2}.\)0.548 mol \(\mathrm{CaCl}_{2}\) corresponds to the precipitated 74.9 g; remaining soluble \(\mathrm{CaCl}_{2}\) equals \(\frac{59.6 \, \mathrm{g} \, \mathrm{CaCl}_{2}}{111.0 \, \mathrm{g/mol}} = 0.537 \, \mathrm{mol} \, \mathrm{CaCl}_{2}.\)

Calculate Moisture Content

Since the mass of original \(\mathrm{CaCl}_{2}\) is the sum of the precipitated and remaining dissolved portion, we have 74.9 g (from precipitate) plus 59.6 g (soluble) = 134.5 g \(\mathrm{CaCl}_{2}\) in the original sample.This means that the moisture (water content): \(\text{water weight} = 150 \, \mathrm{g} - 134.5 \, \mathrm{g} = 15.5 \, \mathrm{g} \) is included in the originally hydrated material.

Calculate Moles of Water

We can find the moles of water in the 15.5 g by using the molar mass of water (18.01528 \, \text{g/mol}): \(\frac{15.5 \, \mathrm{g}}{18.01528 \, \text{g/mol}} = 0.860 \, \text{moles of} \, \mathrm{H_{2}O}.\)

Calculate Molar Ratio

Knowing the moles of \(\text{water} = 0.860 \, \text{moles}\) and moles of original \(\mathrm{CaCl}_{2} = 0.548 \, \text{mol (precipitated)} + 0.537 \, \text{mol (soluble)} = 1.085 \, \text{mol}.\) Hence, the moles of water per mole of \(\mathrm{CaCl}_{2}\) can be calculated as: \\(\frac{0.860 \, \text{moles of} \, \mathrm{H_2O}}{1.085 \, \text{moles of} \, \mathrm{CaCl_{2}}} = 0.793.\)

Key concepts

Anhydrous Calcium Chloride

Anhydrous calcium chloride is a highly effective drying agent. It is referred to as 'anhydrous' because it means "without water".
This version of calcium chloride has no water molecules attached.
When exposed to air, it attracts and absorbs water quickly.
This property makes it extremely useful in keeping spaces dry and moisture-free.

• Used in desiccators to maintain an environment free of moisture
• Beneficial in laboratories and storage to prevent degradation of chemicals or materials
• Serves as a significant tool in preserving the dryness of substances

However, if exposed to any moisture unintentionally, as happens with loosely closed desiccators, calcium chloride can become partially hydrated, leading to changes in its expected performance.

Hydration of Salts

Hydration of salts occurs when water molecules integrate into the salt crystal structure.
For calcium chloride, it combines with water to form hydrates like \(\mathrm{CaCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\).
When water interacts with calcium chloride, it becomes either a dihydrate or hexahydrate, depending on the moisture availability.
These water molecules are not merely on the surface, but embedded in the lattice.
Here are a few points about hydration of salts:

• Hydration affects both the mass and physical appearance of a salt
• Changes in temperature can cause a hydrated salt to lose or gain more water
• Hydrogen bonds form between the water molecules and the calcium chloride molecules

Understanding the hydration process is crucial for predicting how much calcium chloride will need to dry substances or how much water it will absorb.

Solubility Calculations

Solubility calculations help determine the amount of a substance that can dissolve in a solvent at a specific temperature.
For calcium chloride, knowing the solubility in water at certain temperatures is vital for accurate predictions.
At 20°C, the solubility is given as 74.5 g per 100 g of water which helps in computing how much calcium chloride stays dissolved.
These calculations allow us to confidently find the mass of the \(\mathrm{CaCl}_{2}\) that could stay in solution after cooling tables.

• Identify temperature-specific solubility
• Calculate the correct mass that dissolves
• Predict how much substance remains in solution versus precipitating out

Accurate solubility calculations are essential for effective planning and utilization of calcium chloride in drying and precipitation tasks.

Molar Ratios

Molar ratios are an essential concept in chemistry.
They describe the relationship between the amounts (in moles) of reactants and products in a chemical reaction.
Essentially, it tells you how much of one substance relates to another on a molecular level.
To find the molar ratio, it's crucial to calculate the number of moles of calcium chloride and hydrated water.
Once we know the moles of calcium chloride in both precipitated and dissolved forms, the ratio can be established.
This involves:

• Finding the total amount of moles of each component
• Using the ratio to understand how much water each mole of \(\mathrm{CaCl}_{2}\) holds
• Relating these quantities to practical circumstances like drying or precipitating

This understanding ensures precise handling of chemicals and accurate predictions of chemical behavior based on the concentration of involved substances.

Precipitation Reactions

Precipitation reactions occur when two solutes in solution react to form an insoluble product, or precipitate.
In our context, after cooling, some of the dissolved calcium chloride \((\mathrm{CaCl}_{2})\) formed a solid precipitate in the form of its hydrated compound \((\mathrm{CaCl}_{2} \cdot 6\mathrm{H}_2\mathrm{O})\).
This reaction depends largely on the solubility limits of substances involved.
Understanding precipitation reactions is key in:

• Predicting and controlling the formation of solid products in a mixture
• Regulating conditions such as temperature and concentration to manage these reactions
• Applying in purification and separation techniques

Grasping this concept allows students and professionals alike to effectively precipitate desired materials and use the phenomenon in practical applications like salt extraction or waste treatment.