Question

A hydrated nickel chloride compound was found to contain \(24.69 \% \mathrm{Ni}, 29.83 \% \mathrm{Cl}\), and \(45.48 \%\) water. Determine the empirical formula of this hydrated compound.

Short answer

NiCl₂·6H₂O

Step-by-step solution

- Calculate Moles of Nickel

Using the percentage composition, begin by assuming you have 100 grams of the hydrated nickel chloride compound. Therefore, the mass of nickel (Ni) is 24.69 grams. To find the moles, use the formula: moles of Ni = \( \frac{\text{mass of Ni}}{\text{molar mass of Ni}} \) The molar mass of Ni is 58.69 g/mol. moles of Ni = \( \frac{24.69}{58.69} \approx 0.421 \text{ moles} \)

- Calculate Moles of Chlorine

Next, find the moles of chlorine (Cl). The mass of chlorine is 29.83 grams. Using the molar mass of Cl, which is 35.45 g/mol: moles of Cl = \( \frac{29.83}{35.45} \approx 0.841 \text{ moles} \)

- Calculate Moles of Water

Now calculate the moles of water (H₂O). The mass of water is 45.48 grams. The molar mass of H₂O is 18.02 g/mol: moles of H₂O = \( \frac{45.48}{18.02} \approx 2.524 \text{ moles} \)

- Determine the Simplest Whole Number Ratio

Divide each of the moles by the smallest number of moles calculated. In this case, the smallest number of moles is 0.421 (moles of Ni): \[ \text{moles of Ni} = \frac{0.421}{0.421} = 1 \] \[ \text{moles of Cl} = \frac{0.841}{0.421} \approx 2 \] \[ \text{moles of H}_2\text{O} = \frac{2.524}{0.421} \approx 6 \]

- Write the Empirical Formula

Using the simplest whole number ratio of the elements, the empirical formula of the hydrated nickel chloride compound is: \[ \text{NiCl}_2 \cdot 6\text{H}_2\text{O} \]

Key concepts

mole concept

The mole concept is fundamental in chemistry. It relates amounts of chemical substances to the number of entities like atoms, ions, and molecules they contain.
To understand the mole concept, remember that one mole is exactly equivalent to Avogadro's number (\textapprox \(6.022 \times 10^{23}\)) of particles. This number helps in converting between atomic scale and real-world scale.
For example, in the original exercise, we calculated the moles of nickel (Ni) using its mass and molar mass. The molar mass, which is the mass of 1 mole of that element, was used in the formula:

• \text{moles of Ni} = \frac{\text{mass of Ni}}{\text{molar mass of Ni}}

This process connects macroscopic amounts (grams) with the microscopic world (moles). The same principle is applied for chlorine (Cl) and water (H\(_2\)O). The mole concept simplifies understanding chemical composition and reactions.

percent composition

Percent composition tells us the relative amounts of elements in a compound by percent.
It is crucial in empirical formula calculations. In our example, we used the percentages of nickel (Ni), chlorine (Cl), and water (H\(_2\)O). These percentages help us figure out how many grams of each element are present in 100 grams of the compound.
For instance:

• 24.69% of Ni in 100 grams means 24.69 grams of Ni.
  
• 29.83% of Cl means 29.83 grams of Cl.
  
• 45.48% of water means 45.48 grams of H\(_2\)O.

These masses were then converted to moles, allowing us to determine the simplest whole-number ratio, which represents the empirical formula. Understanding percent composition aids in analyzing the makeup of compounds and deriving formulas from experimental data.

hydrated compounds

Hydrated compounds, or hydrates, are compounds that include water molecules integrated into their structure.
These water molecules are crucial for the compound's properties, and they must be accounted for in calculations.
In our exercise, the hydrated nickel chloride compound was analyzed to find the amount of water present, which was 45.48% of the sample. Removing this water would yield the anhydrous (water-free) compound.
To capture the water content in the formula, we computed the moles of water and included it in the empirical formula:

• moles of H\(_2\)O = \(\frac{\text{mass of H}_2\text{O}}{\text{molar mass of H}_2\text{O}}\)

By determining the ratio of water molecules to the nickel chloride units, we arrived at the formula NiCl\(_2 \times 6\)H\(_2\)O. Hydrated compounds are common in chemistry and must be properly accounted for in empirical formulas.

chemistry problem solving

Effective chemistry problem solving requires a systematic approach.
Breaking down complex problems into simpler steps makes them manageable. Here is a structured way to solve empirical formula problems:

• Step 1: Assume a sample size (usually 100 grams for ease).
  
• Step 2: Convert percentage composition to mass of each element.
  
• Step 3: Use the molar mass to find the number of moles of each element.
  
• Step 4: Divide by the smallest number of moles to get the simplest whole-number ratio.
  
• Step 5: Write the empirical formula based on this ratio.

This method ensures consistency and accuracy in solving such chemistry problems. Practice and familiarity with these steps develop problem-solving efficiency, making chemistry exercises less daunting and more enjoyable.