Problem 1 - Chemical Principles in the Laboratory

Question

A student is given a sample of a blue nickel sulfate hydrate. She weighs the sample in a dry covered crucible and obtains a mass of 23.711 g for the crucible, cover, and sample. Earlier she had found that the crucible and cover weighed \(21.594 \mathrm{g}\). She then heats the crucible to drive off the water of hydration, keeping the crucible at red heat for about 10 minutes with the cover slightly ajar. She then lets the crucible cool, and finds it has a lower mass; the crucible, cover and contents then weigh \(22.840 \mathrm{g}\). In the process the sample was converted to greenish-yellow anhydrous Niso \(_{4}\) a. What was the mass of the hydrate sample? _____ g hydrate b. What is the mass of the anhydrous \(\mathrm{NiSO}_{4} ?\) ______ \(\mathrm{g} \mathrm{NiSO}_{4}\) c. How much water was driven off? _____ \(\mathrm{g} \mathrm{H}_{2} \mathrm{O}\) d. What is the percent by mass of water in the hydrate? \(\%\) water \(=\frac{\text { mass of water in sample }}{\text { mass of hydrate sample }} \times 100 \%\) ______ \(\%_{\text {mass }} \mathrm{H}_{2} \mathrm{O}\) e. How many grams of water would there be in 100.0 g hydrate? How many moles? ______ \(g H_{2} O\) ______ moles \(\mathrm{H}_{2} \mathrm{O}\) f. How many grams of NiSO, are there in \(100.0 \mathrm{g}\) hydrate? How many moles? (What percentage of the hydrate is \(\mathrm{NiSO}_{4} ?\) Convert the mass of \(\mathrm{NiSO}_{4}\) to moles. The molar mass of \(\mathrm{NiSO}_{4}\) is 154.8 g. _____ \(\mathrm{g} \mathrm{NiSO}_{4}\) _____ moles \(\mathrm{NiSO}_{4}\) g. How many moles of water are present per mole \(\mathrm{NiSO}_{4} ?\) ______ h. What is the formula of the hydrate? _____

Step-by-Step Solution

Answer

a. 2.117 g hydrate b. 1.246 g NiSO4 c. 0.871 g H2O d. 41.15% H2O e. 41.15 g H2O, 2.284 mol H2O f. 58.85 g NiSO4, 0.38 mol NiSO4 g. 6 moles H2O per mole NiSO4 h. NiSO4·6H2O

Step 1: Find Mass of Hydrate

To find the mass of the hydrate sample, subtract the mass of the crucible and cover from the mass of the crucible, cover, and sample: \[ 23.711 \text{ g} - 21.594 \text{ g} = 2.117 \text{ g} \].

Step 2: Find Mass of Anhydrous NiSO4

After heating, the mass of the crucible, cover, and anhydrous NiSO4 is \(22.840 \text{ g} \). Subtract the mass of the crucible and cover to find the mass of NiSO4: \[ 22.840 \text{ g} - 21.594 \text{ g} = 1.246 \text{ g} \].

Step 3: Calculate Mass of Water Lost

The mass of water driven off is the difference between the mass of the hydrate and the mass of the anhydrous compound: \[ 2.117 \text{ g} - 1.246 \text{ g} = 0.871 \text{ g} \].

Step 4: Calculate Percent by Mass of Water in Hydrate

Use the formula \( \frac{\text{mass of water}}{\text{mass of hydrate sample}} \times 100\% \) to find the percentage of water: \[ \frac{0.871 \text{ g}}{2.117 \text{ g}} \times 100\% \approx 41.15\% \].

Step 5: Calculate Grams and Moles of Water in 100 g Hydrate

If 41.15% of the hydrate is water, then in 100 g of hydrate, \(41.15 \text{ g} \) is water. Convert this to moles: \( \frac{41.15 \text{ g}}{18.015 \text{ g/mol}} \approx 2.284 \text{ mol}\).

Step 6: Calculate Grams and Moles of NiSO4 in 100 g Hydrate

If 100% - 41.15% = 58.85% of the hydrate is NiSO4, then in 100 g of hydrate, \(58.85 \text{ g} \) is NiSO4. Convert this to moles: \( \frac{58.85 \text{ g}}{154.8 \text{ g/mol}} \approx 0.38 \text{ mol}\).

Step 7: Determine Moles of Water per Mole NiSO4

The moles of water per mole of NiSO4 is found by \(\frac{2.284}{0.38} \approx 6.01\).

Step 8: Determine Formula of the Hydrate

The formula of the hydrate, based on the previous calculations, is \( \mathrm{NiSO}_{4} \cdot 6\mathrm{H}_{2} \mathrm{O} \).

Key Concepts

Mass Determination

Mass determination is the first step in analyzing hydrate samples. It involves calculating the mass of the sample and its components like the anhydrous salt and water.
First, you need to accurately measure the mass of the container (such as a crucible) with its contents before and after heating, to account for the sample and any changes it undergoes.

• Initial weight: Crucible + cover + hydrate.
• Weight post-heating: Crucible + cover + anhydrous salt.

By subtracting the mass of the crucible and cover from these weights, you derive the mass of the hydrate and the anhydrous compound separately.
For example, if the mass of the crucible, cover, and hydrate is 23.711 g and the mass of the empty crucible and cover is 21.594 g, the hydrate alone weighs 2.117 g:
\[ 23.711 \text{ g} - 21.594 \text{ g} = 2.117 \text{ g} \]
This straightforward subtraction is a crucial step in chemistry to ensure precise calculations further in the analysis.

Percent Composition

Percent composition is a useful calculation in chemistry to figure out the proportion of specific components within a compound. In a hydrate, you focus on the percentage of water present.
To find the percent composition of water, use the formula:
\[\frac{\text{mass of water}}{\text{mass of hydrate}} \times 100\%\]
Consider the mass of water from the hydrate sample. If a hydrate initially weighed 2.117 g before heating and had a water loss of 0.871 g during heating, the percentage of water in the hydrate would be:
\[\frac{0.871 \text{ g}}{2.117 \text{ g}} \times 100\% \approx 41.15\% \]
This method helps determine how much of the hydrate's mass was contributed by water, an essential factor when calculating formulas or reactions.

Molar Ratio

The molar ratio is a central concept in chemistry because it bridges the amount of each substance in moles. In this context, it helps to find the relationship between the anhydrous salt and water within a hydrate.
In calculating the molar ratio, you first convert the mass of each component into moles using their molar masses:

• Mole of anhydrous \( \text{NiSO}_4 \): \( \frac{58.85 \text{ g}}{154.8 \text{ g/mol}} \approx 0.38 \text{ mol} \)
• Mole of water: \( \frac{41.15 \text{ g}}{18.015 \text{ g/mol}} \approx 2.284 \text{ mol} \)

The molar ratio is then the ratio of moles of water to moles of anhydrous \( \text{NiSO}_4 \):
\[\frac{2.284}{0.38} \approx 6.01\]
This ratio suggests that for every mole of anhydrous nickel sulfate, there are about 6 moles of water, helping to determine the hydrate's chemical formula.

Chemical Formula

Determining the chemical formula of a hydrate involves combining concepts of mass determination, percent composition, and molar ratio. This analysis results in the empirical formula that reflects the actual compound structure.
For example, after establishing the molar ratio, where approximately 6 moles of \( \text{H}_2\text{O} \) are present for each mole of \( \text{NiSO}_4 \), the chemical formula can be written as:
\[ \text{NiSO}_4 \cdot 6\text{H}_2\text{O} \]
This formula indicates the hydrate contains six water molecules for every molecule of \( \text{NiSO}_4 \).
Such empirical formulas are critical for informing predictions about the behavior of hydrates in laboratory settings and real-world applications. Validating this formula through repeated experiments ensures accuracy in chemical characterization.