Calcium Sulfate Hydrate: Determining Water Of Hydration

Hey guys! Ever wondered how much water is trapped inside a crystal? Let's dive into a fascinating chemistry problem involving calcium sulfate and figure out how many water molecules are hiding within its structure. We'll tackle this by looking at a scenario where calcium sulfate crystals lose weight upon heating, a classic example of how hydrates behave.

Understanding Hydrates: Water Trapped in Crystals

Before we jump into the math, let's quickly recap what hydrates are. Imagine a crystal structure as a tiny, organized cage. Sometimes, these cages can trap water molecules inside. These crystalline compounds that incorporate water molecules are called hydrates. The water molecules are chemically bonded within the crystal lattice, but they can be driven off as steam when the hydrate is heated. This process is known as dehydration.

The general formula for a hydrate is written as ${ \text{Salt} \cdot x\text{H}_2\text{O} }$, where "Salt" represents the ionic compound and x is the number of water molecules associated with each formula unit of the salt. This x is what we're trying to find in our calcium sulfate problem.

Calcium sulfate itself is a common ionic compound. In its anhydrous (water-free) form, it's used in various applications, including plaster and cement. However, calcium sulfate also readily forms hydrates, incorporating water molecules into its crystal structure. When we heat these calcium sulfate hydrates, the water molecules escape, leaving behind the anhydrous calcium sulfate. This weight loss is crucial for determining the value of x, the number of water molecules initially present.

The Problem: Heating Calcium Sulfate

Now, let's restate the problem we're trying to solve. We're given that when calcium sulfate crystals with the formula ( \text{CaSO}_4 cdot x\text{H}_2\text{O} ) are heated, they lose 21% of their weight. This weight loss corresponds to the water that is driven off during heating. Our mission is to figure out the value of x, which represents the number of water molecules associated with each formula unit of calcium sulfate in the hydrate.

To solve this, we'll need to use the molar masses of the compounds involved and set up a proportion based on the weight loss. We'll carefully analyze the chemical reaction that occurs during heating and use stoichiometry to connect the mass of water lost to the moles of water originally present in the hydrate. This problem beautifully illustrates the power of stoichiometry in understanding chemical reactions and the composition of compounds.

Step-by-Step Solution: Finding the Value of x

Okay, let's break down the solution step-by-step. This might seem tricky at first, but we'll make it super clear. We'll use a combination of stoichiometry (the math of chemical reactions) and some good ol' algebra to crack this problem.

1. The Chemical Reaction

First, let's write down the balanced chemical equation for the reaction that's happening when we heat the calcium sulfate hydrate:

${ \text{CaSO}_4 cdot x\text{H}_2\text{O} \text{(s)} \rightarrow \text{CaSO}_4 \text{(s)} + x\text{H}_2\text{O} \text{(g)} }$

This equation tells us that one mole of calcium sulfate hydrate (( \text{CaSO}_4 cdot x\text{H}_2\text{O} )) breaks down into one mole of anhydrous calcium sulfate (${ \text{CaSO}_4 }$) and x moles of water (${ \text{H}_2\text{O} }$). The key here is that x moles of water are released for every one mole of the hydrate that is heated.

2. Molar Masses: The Building Blocks

Next, we need the molar masses of the compounds involved. Remember, the molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). We'll need the molar masses of calcium sulfate (${ \text{CaSO}_4 }$) and water (${ \text{H}_2\text{O} }$).

• Molar mass of ${ \text{CaSO}_4 }$: Calcium (Ca) = 40.08 g/mol, Sulfur (S) = 32.07 g/mol, Oxygen (O) = 16.00 g/mol. So, ${40.08 + 32.07 + 4(16.00) = 136.15 \text{ g/mol}}$
• Molar mass of ${ \text{H}_2\text{O} }$: Hydrogen (H) = 1.01 g/mol, Oxygen (O) = 16.00 g/mol. So, ${2(1.01) + 16.00 = 18.02 \text{ g/mol}}$

3. The 21% Weight Loss: Setting Up the Proportion

The problem states that the hydrate loses 21% of its weight upon heating. This means that the mass of water lost is 21% of the mass of the original hydrate. Let's assume we start with 100 grams of the hydrate. This makes the math easier because 21% of 100 grams is simply 21 grams. So, 21 grams of water are lost, and the remaining 79 grams must be the anhydrous calcium sulfate.

Now, we can set up a proportion. The ratio of the mass of water lost to the mass of anhydrous calcium sulfate is equal to the ratio of their molar masses, adjusted by the x value:

${ \frac{\text{Mass of } x\text{H}_2\text{O}}{\text{Mass of } \text{CaSO}_4} = \frac{x imes \text{Molar mass of } \text{H}_2\text{O}}{\text{Molar mass of } \text{CaSO}_4} }$

Plugging in the values we have:

${ \frac{21 \text{ g}}{79 \text{ g}} = \frac{x imes 18.02 \text{ g/mol}}{136.15 \text{ g/mol}} }$

4. Solving for x: The Algebraic Finale

Now, it's just a matter of solving for x. Let's cross-multiply and simplify:

${ 21 imes 136.15 = 79 imes x imes 18.02 }$

${ 2859.15 = 1423.58x }$

Divide both sides by 1423.58:

${ x = \frac{2859.15}{1423.58} \approx 2.01 }$

Since x represents the number of water molecules, it should be a whole number. So, we round 2.01 to 2.

5. The Answer: x = 2

Therefore, the value of x is 2. This means the formula of the calcium sulfate hydrate is ( \text{CaSO}_4 cdot 2\text{H}_2\text{O} ), which is commonly known as calcium sulfate dihydrate or gypsum.

Conclusion: Hydrates Unveiled

So there you have it! By carefully analyzing the weight loss upon heating and using stoichiometry, we successfully determined the number of water molecules in calcium sulfate hydrate. This problem highlights the importance of understanding chemical formulas, molar masses, and the behavior of hydrates. It's a fantastic example of how chemistry helps us understand the composition of the world around us.

Remember, the key to solving these types of problems is to break them down into smaller, manageable steps. Identify the chemical reaction, determine the molar masses, set up the correct proportions, and then use your algebra skills to solve for the unknown. Keep practicing, and you'll become a pro at unraveling these chemical mysteries!

If you have any questions about hydrates or stoichiometry, don't hesitate to ask. Keep exploring the fascinating world of chemistry, guys! There's always something new to discover.