Problem 25 - Chemistry and Chemical Reactivity

Question

A mixture of \(\mathrm{CuSO}_{4}\) and \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) has a mass of \(1.245 \mathrm{g}\). After heating to drive off all the water, the mass is only \(0.832 \mathrm{g}\). What is the mass percent of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) in the mixture? (See page \(98 .\) )

Step-by-Step Solution

Answer

The mass percent of \(\mathrm{CuSO}_{4} \cdot 5\mathrm{H}_{2} \mathrm{O}\) is approximately 91.79%.

Step 1: Determine the Mass of Water Lost

To find how much water was lost, we subtract the mass of the anhydrous mixture (0.832 g) from the total initial mass of the hydrated mixture (1.245 g). \[\text{Mass of water} = 1.245 \text{ g} - 0.832 \text{ g} = 0.413 \text{ g}\]

Step 2: Calculate the Molar Mass of Water and \(\mathrm{CuSO}_{4} \cdot 5\mathrm{H}_{2}\mathrm{O}\)

The molar mass of water (\(\mathrm{H}_{2}\mathrm{O}\)) is 18.015 g/mol. The molar mass of \(\mathrm{CuSO}_{4} \cdot 5\mathrm{H}_{2}\mathrm{O}\) is calculated as follows:\[96.06 + 64.55 + 4(15.999) + 5(18.015) = 249.685 \text{ g/mol}\]whereas the molar mass of \(\mathrm{CuSO}_{4}\) is:\[96.06 + 64.55 + 4(15.999) = 159.61 \text{ g/mol}\]

Step 3: Calculate Moles of Water Lost

Using the molar mass of water:\[\text{Moles of water} = \frac{0.413 \text{ g}}{18.015 \text{ g/mol}} \approx 0.0229 \text{ mol}\]

Step 4: Determine Moles of \(\mathrm{CuSO}_{4} \cdot 5\mathrm{H}_{2} \mathrm{O}\)

The moles of \(\mathrm{CuSO}_{4} \cdot 5\mathrm{H}_{2} \mathrm{O}\) is equal to moles of water divided by 5:\[\text{Moles of } \mathrm{CuSO}_{4} \cdot 5\mathrm{H}_{2} \mathrm{O} = \frac{0.0229}{5} = 0.00458 \text{ mol}\]

Step 5: Calculate Mass of \(\mathrm{CuSO}_{4} \cdot 5\mathrm{H}_{2} \mathrm{O}\)

Using the number of moles calculated, find the mass of \(\mathrm{CuSO}_{4} \cdot 5\mathrm{H}_{2} \mathrm{O}\):\[\text{Mass} = 0.00458 \text{ mol} \times 249.685 \text{ g/mol} \approx 1.143 \text{ g}\]

Step 6: Calculate Mass Percent of \(\mathrm{CuSO}_{4} \cdot 5\mathrm{H}_{2} \mathrm{O}\)

Finally, calculate the mass percent using the mass found and the initial total mass of the mixture:\[\text{Mass percent} = \left(\frac{1.143}{1.245}\right) \times 100 \approx 91.79 \%\]

Key Concepts

Mass Percent Calculation

Mass percent is a way of expressing a concentration in a mixture by comparing the mass of one component to the overall mass of the mixture. To calculate the mass percent of a specific component, follow these steps:

• Determine the mass of the component in question. In our example, this was found to be 1.143 g for the hydrated compound (\(\text{CuSO}_4 \cdot 5\text{H}_2\text{O}\)).
• Divide the mass of the component by the total mass of the mixture. Here, the mix has a mass of 1.245 g.
• Multiply the result by 100 to convert it to a percentage. Thus: \[\text{Mass percent} = \left(\frac{1.143}{1.245}\right) \times 100 = 91.79\%\]

This process allows scientists to communicate the composition of mixtures effectively, and is especially useful in chemistry when preparing solutions or analyzing unknown materials.

Molar Mass Determination

Molar mass is a calculation that gives the mass of one mole of a substance, making it a crucial concept in chemistry for converting between grams and moles. To determine molar mass:

• Add together the atomic masses of each element in the compound, using the periodic table as a reference.
• For (\(\text{CuSO}_4 \cdot 5\text{H}_2\text{O}\)), calculate as follows:
  • Copper (Cu) = 63.55 g/mol
  • Sulfur (S) = 32.07 g/mol
  • Oxygen (O), recalling it appears four times = 4 \(\times\) 15.999 g/mol
  • Hydration water added separately = 5 \(\times\) 18.015 g/mol
• Total it all together:\[\text{Molar mass of } \text{CuSO}_4 \cdot 5\text{H}_2\text{O} = 63.55 + 32.07 + 63.996 + 90.075 = 249.691\text{ g/mol}\]

Knowing the molar mass allows you to perform essential chemical calculations, such as finding the number of moles given a certain mass, which was used in our exercise when finding moles of water lost.

Chemical Reactions Involving Water Loss

During chemical reactions, compounds like hydrates can lose water through heating, causing shifts in mass we need to account for. This process completes multiple layers of learning about hydrates:

• Identify the initial mass of your hydrate sample. This was 1.245 g in the initial example.
• After heating, measure the remaining mass to find how much water was lost. This change happens due to the water being driven out of the crystalline structure during heating. In our example, 0.832 g is the mass after water evaporation.
• Estimate the mass of the water lost by subtracting the anhydrous mass from the initial mass. Thus, 0.413 g of water was lost.

Upon losing water, the physical properties of the compound might change, though the fundamental chemical structure of the anhydrous compound remains unaffected. This loss is crucial to calculating the extent of hydration in complex mixtures or identifying the quality and purity of a hydrated sample.