Problem 153 - Chemistry and Chemical Reactivity

Question

If Epsom salt, \(\mathrm{MgSO}_{4} \cdot x \mathrm{H}_{2} \mathrm{O},\) is heated to \(250^{\circ} \mathrm{C},\) all the water of hydration is lost. On heating a \(1.687-\mathrm{g}\) sample of the hydrate, \(0.824 \mathrm{g}\) of \(\mathrm{MgSO}_{4}\) remains. How many molecules of water occur per formula unit of \(\mathrm{MgSO}_{4} ?\)

Step-by-Step Solution

Answer

There are 7 water molecules per formula unit of \(\mathrm{MgSO}_{4}\).

Step 1: Determine the mass of water lost

First, calculate the mass of water lost during heating. Subtract the mass of \( \mathrm{MgSO}_{4} \) left over after heating from the original mass of the hydrate: \[\text{Mass of water lost} = 1.687\,\mathrm{g} - 0.824\,\mathrm{g} = 0.863\,\mathrm{g}\]

Step 2: Calculate the moles of \(\mathrm{MgSO}_{4}\)

Next, find the number of moles of \( \mathrm{MgSO}_{4} \). Use the molar mass of \( \mathrm{MgSO}_{4} \) which is approximately \( 120.37\, \mathrm{g/mol} \):\[\text{Moles of } \mathrm{MgSO}_{4} = \frac{0.824\,\mathrm{g}}{120.37\,\mathrm{g/mol}} \approx 0.00684\,\mathrm{mol}\]

Step 3: Calculate the moles of water lost

Use the molar mass of water \( \mathrm{H}_{2}\mathrm{O} \), which is approximately \( 18.02\, \mathrm{g/mol} \), to find the moles of water that were lost:\[\text{Moles of } \mathrm{H}_{2} \mathrm{O} = \frac{0.863\, \mathrm{g}}{18.02\, \mathrm{g/mol}} \approx 0.0479\,\mathrm{mol}\]

Step 4: Calculate \(x\), the number of water molecules per formula unit

Divide the moles of water by the moles of \( \mathrm{MgSO}_{4} \) to find \( x \), the number of water molecules per formula unit of \( \mathrm{MgSO}_{4} \):\[x = \frac{0.0479\, \mathrm{mol}}{0.00684\, \mathrm{mol}} \approx 7\]Therefore, there are approximately 7 molecules of water per formula unit.

Key Concepts

Water of Hydration

When you hear about the water of hydration, it refers to the water molecules that are chemically bonded to a compound. In the case of Epsom salt, or magnesium sulfate hydrate \( \mathrm{MgSO}_{4} \cdot x \mathrm{H}_{2} \mathrm{O} \), these water molecules are an integral part of the solid's structure. The water of hydration is responsible for the crystalline state of the compound and its specific properties.
Heating such salts, like in our example, results in the loss of these water molecules. This process, known as dehydration, alters the structure and often the weight of the compound, leaving behind the anhydrous (water-free) form. Understanding how much water is lost helps determine the number of water molecules that were initially present.

Moles and Molar Mass

In chemistry, the mole is a unit of measurement used to express the amount of a chemical substance. It relates directly to the molecular scale, allowing us to count particles like atoms and molecules in practical amounts.
This measurement is closely tied to molar mass, which is the mass of one mole of a substance, expressed in grams per mole (g/mol). For our example, to find out how many molecules of water are present, it's crucial that we convert the masses given to moles using the molar masses of \( \mathrm{MgSO}_{4} \) and \( \mathrm{H}_2\mathrm{O} \).
Here's how it works in practice:

• First, we calculate moles of \( \mathrm{MgSO}_{4} \) using its molar mass.
• Then, we determine the moles of water from the mass lost during dehydration.

By accurately converting to moles, chemists can utilize a common language of measurement that effectively describes chemical reactions and relationships.

Chemical Formula Unit

The chemical formula unit represents the simplest number ratio of ions in the compound, giving the amount of each element present. In hydrated salt calculations, identifying this ratio is crucial, as it informs us about the composition of the compound.
For example:

• In \( \mathrm{MgSO}_{4} \cdot x \mathrm{H}_{2}\mathrm{O} \), the chemical formula unit indicates the ratio of magnesium sulfate to water molecules.
• The value of \( x \) reveals how many water molecules are bound in each unit of the compound.

This information helps in understanding the behavior and properties of the compound, like its reactivity and solubility, when subjected to various conditions such as heating or mixing with other substances.

Thermochemical Decomposition

Thermochemical decomposition involves breaking down a compound using heat. In the context of hydrates, like Epsom salt, this means removing the water of hydration, which leads to a change in the physical and chemical properties of the compound.
Heating such substances causes the water of hydration to evaporate, resulting in the formation of an anhydrous solid, in this case, dry \( \mathrm{MgSO}_{4} \). This process is crucial in many scientific applications:

• It helps determine the composition of the original hydrate.
• Utilized for purification and identification of compounds.

Understanding thermochemical decomposition gives insight into the thermal stability of materials and enables chemists to design processes to control or exploit these thermal effects.