Question

The student is given a sample of a blue copper sulfate hydrate. She weighs the sample in a dry covered crucible and obtains a mass of \(21.587 \mathrm{g}\) for the crucible, cover, and sample. The mass of the emply crucible and cover had been found carlicr to be \(20.623 \mathrm{g}\). She then heats the crucible to drive off the water of hydration, kceping the crucible at red heat for 10 minutes with the cover slightly ajar. On cooling, she finds the mass of crucible, cover, and contents to be \(21.240 \mathrm{g}\). The sample was converted in the process to very light blue anhydrous \(\mathrm{CuSO}_{4}\) a. What is the mass of the hydrate sample? g hydrate b. What is the mass of the anhydrous \(\mathrm{CuSO}_{4} ?\) $$\mathrm{g} \mathrm{CuSO}_{4}$$ c. What is the mass of water driven off? $$\mathrm{g} \mathrm{H}_{2} \mathrm{O}$$ d. What is the per cent water in the hydrate? \% water \(=\frac{\text { mass of water in sample }}{\text { mass of hydrate sample }} \times 100 \%\) "water = mass of water in sample \(\times 100 \%\) $$\%$$ e. How many grams of water would there be in \(100.0 \mathrm{g}\) of hydrate? How many moles? $$\begin{aligned} &\text { grams } \mathrm{H}_{2} \mathrm{O}\\\ &\text { moles } \mathrm{H}_{2} \mathrm{O} \end{aligned}$$ f. How many grams of CuSO_are there in \(100.0 \mathrm{g}\) of hydrate? How many moles? (What per cent of thhydrate is \(\mathrm{CuSO}_{4}\) ? Convert the mass of \(\mathrm{CuSO}_{4}\) to moles. Molar mass of \(\mathrm{CuSO}_{4}=159.6 \mathrm{g} .\) ) $$\text { grams } \mathrm{CuSO}_{4}$$ g. How many moles of water are present per mole of \(\mathrm{CuSO}_{4} ?\) h. What is the formula of the hydrate?

Short answer

The formula of the hydrate is \(\mathrm{CuSO}_{4} \cdot 5\mathrm{H}_{2}\mathrm{O}\).

Step-by-step solution

Determine the mass of the hydrate sample

To find the initial mass of the hydrate, subtract the weight of the empty crucible and cover from the weight of the crucible, cover, and the sample: Mass of hydrate sample = (21.587 g) - (20.623 g) = 0.964 g

Determine the mass of anhydrous \(\mathrm{CuSO}_{4}\)

To find the mass of anhydrous \(\mathrm{CuSO}_{4}\), subtract the weight of the empty crucible and cover from the weight of the crucible, cover, and anhydrous \(\mathrm{CuSO}_{4}\) after being heated: Mass of anhydrous $\mathrm{CuSO}_{4} = (21.240 g) - (20.623 g) = 0.617 g

Determine the mass of water driven off

Subtract the mass of anhydrous \(\mathrm{CuSO}_{4}\) from the mass of the hydrate to find the mass of water driven off: Mass of $\mathrm{H}_{2}\mathrm{O} = (0.964 g) - (0.617 g) = 0.347 g

Calculate the percentage of water in the hydrate

Divide the mass of the water by the mass of the hydrate sample, and multiply the result by 100: Percentage of water = \(\frac{0.347 g}{0.964 g} \times 100 = 36.0 \%\)

Determine the mass and moles of water in 100 g of hydrate

Calculate the mass of water in 100 g of hydrate: Mass of water = 0.36 * 100 g = 36 g Calculate the moles of water: Moles of \(\mathrm{H}_{2}\mathrm{O} = \frac{36}{18.015} = 2 \text{ moles}\)

Determine the mass and moles of CuSO4 in 100 g of hydrate

Calculate the mass of CuSO4 in 100 g of hydrate: Mass of \(\mathrm{CuSO}_{4} = 100 - 36 = 64 \,g\) Calculate the moles of \(\mathrm{CuSO}_{4}\): Moles of \(\mathrm{CuSO}_{4} = \frac{64}{159.6} = 0.401 \, \text{ moles}\)

Determine the moles of water per mole of \(\mathrm{CuSO}_{4}\)

Divide moles of water by moles of CuSO4 to find the ratio between them: Moles of \(\mathrm{H}_{2}\mathrm{O} \text{ per mole of } \mathrm{CuSO}_{4} = \frac{2}{0.401} = 4.988 \approx 5\)

Determine the formula of the hydrate

Since we know that there are about 5 moles of water (\(\mathrm{H}_{2}\mathrm{O}\)) per mole of CuSO4, the formula of the hydrate is: \(\mathrm{CuSO}_{4} \cdot 5\mathrm{H}_{2}\mathrm{O}\)

Key concepts

Hydration Analysis

Hydration analysis is fundamental in understanding compounds that contain water molecules as an integral part of their structure. These compounds, known as hydrates, have a specific number of water molecules bound to them. In the context of a laboratory setting, determining the degree of hydration involves careful weighing and heating to remove the water of hydration, followed by further weighing.

For instance, in our exercise, to carry out a hydration analysis on a sample of copper sulfate hydrate, one would need to first find the mass of the hydrate as a whole and then heat it sufficiently to drive off the water. The remaining anhydrous copper sulfate is weighed, and the difference in mass indicates the amount of water that was present. It is a straightforward yet delicate process, as overheating can decompose the sample itself, leading to inaccurate results. Precise handling is essential to obtain a correct empirical formula for the hydrate.

Stoichiometry

Stoichiometry is a branch of chemistry that deals with the quantitative relationships of reactants and products in a chemical reaction. It provides the ratios by which compounds react with each other based on their molecular weights and the conservation of mass. In simple terms, it's the 'recipe' for a chemical reaction.

In our problem, stoichiometry comes into play when we interpret the molar ratio of water to copper sulfate in the hydrated compound. The conversion of grams to moles is an essential stoichiometric step that allows us to relate the mass of substances to the number of molecules present. This step is critical for determining the empirical formula, as stoichiometry helps us understand the proportion of components in a compound.

Molar Mass Calculations

Understanding molar mass is crucial when solving problems in chemistry. The molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol), and represents a bridge between the microscopic world of atoms and molecules and the macroscopic world we can measure.

In our laboratory scenario, calculating the molar mass allows us to convert between mass in grams and the number of moles. This is particularly useful when dealing with reactions that take place at the atomic or molecular level. For example, in this exercise, to find the number of moles of anhydrous copper sulfate and water, dividing the mass by their respective molar masses (159.6 g/mol for \(\mathrm{CuSO}_4\) and 18.015 g/mol for water) is necessary. Precision in molar mass calculation is vital for accurate stoichiometry and can impact the determination of an empirical formula.

Empirical Formula Determination

The empirical formula of a compound gives the simplest whole-number ratio of atoms of each element in the compound. This is often different from the compound's molecular formula, which shows the actual number of atoms of each element in a molecule. Determining an empirical formula requires careful mole ratio calculation, often through combustion analysis, titration, or in our case, dehydration analysis.

Using the data from the hydration analysis of the blue copper sulfate hydrate, we can deduce the empirical formula by finding the mole ratio between copper sulfate and water. For every mole of anhydrous \(\mathrm{CuSO}_4\), we found approximatively five moles of water. Therefore, the empirical formula of the hydrate is \(\mathrm{CuSO}_4 \cdot 5\mathrm{H}_2\mathrm{O}\). This result not only reflects the purity and composition of the sample but also is critical for understanding the substance's chemical properties and behavior.