Problem 94 - Chemistry and Chemical Reactivity

Question

Transition metals can combine with carbon monoxide (CO) to form compounds such as \(\mathrm{Fe}(\mathrm{CO})_{5}\) (Study Question 3.82 ). Assume that you combine 0.125 g of nickel with CO and isolate \(0.364 \mathrm{g}\) of \(\mathrm{Ni}(\mathrm{CO})_{x} .\) What is the value of \(x ?\)

Step-by-Step Solution

Answer

The value of \( x \) is 4.

Step 1: Determine the moles of nickel

First, find the moles of nickel by using its atomic mass. The atomic mass of nickel (Ni) is approximately 58.69 g/mol. Use the formula \( n = \frac{m}{M} \), where \( n \) is the number of moles, \( m \) is the mass of the sample, and \( M \) is the molar mass of nickel.\[ n = \frac{0.125 \text{ g}}{58.69 \text{ g/mol}} = 0.00213 \text{ mol}\]

Step 2: Determine the moles of Ni(CO)_x

Find the moles of the compound \( \mathrm{Ni}( ext{CO})_x \) using its mass and the formula \( n = \frac{m}{M} \), where \( M \) is the molar mass of the compound. Assume \( M = M_{\mathrm{Ni}} + x \times M_{\mathrm{CO}} \), where \( M_{\mathrm{CO}} \) is the molar mass of CO, approximately 28.01 g/mol.\[ n_{\mathrm{Ni}( ext{CO})_x} = \frac{0.364 \text{ g}}{58.69 \text{ g/mol} + x \times 28.01 \text{ g/mol}}\]

Step 3: Set up the equation

Assume all nickel atoms are converted into \( \mathrm{Ni}( ext{CO})_x \), so the moles of nickel equal the moles of \( \mathrm{Ni}( ext{CO})_x \):\[0.00213 = \frac{0.364}{58.69 + x \times 28.01}\]

Step 4: Solve for x

Multiply both sides to clear the fraction:\[0.00213 (58.69 + x \times 28.01) = 0.364\]Expand and solve for \( x \):\[0.12505 + 0.00213x \times 28.01 = 0.364\]\[0.00213x \times 28.01 = 0.364 - 0.12505\]\[0.00213x \times 28.01 = 0.23895\]\[x = \frac{0.23895}{0.00213 \times 28.01} \approx 4\]

Step 5: Verify the solution

To ensure the calculation is correct, substitute \( x = 4 \) back into the equation to check consistency:\[0.00213 = \frac{0.364}{58.69 + 4 \times 28.01} \approx 0.00213\]Since both sides match, the solution is verified.

Key Concepts

Mole Calculation

Understanding mole calculation is crucial in chemistry for converting mass into moles, which is a measure of quantity. Moles help connect the mass of a substance to the number of particles or atoms it contains. When calculating moles, you use the formula \( n = \frac{m}{M} \), where \( n \) is the number of moles, \( m \) is the mass of the substance in grams, and \( M \) is the molar mass of the substance in grams per mole.

For example, in the previous exercise, to find the moles of nickel, you take the mass of nickel, which is 0.125 grams, and divide it by the molar mass of nickel, which is 58.69 grams per mole. Calculations often require a calculator to ensure precision.

By converting mass to moles, we align measurements and connect them meaningfully to reactant and product amounts in chemical reactions. This consistency is vital for further computations such as chemical balance or compound formulation.

Nickel Carbonyl Compound

Nickel carbonyl is an interesting complex involving nickel and carbon monoxide. These types of compounds are formed when transition metals like nickel react with carbon monoxide. Nickel carbonyl, represented as \( \mathrm{Ni}(\mathrm{CO})_x \), showcases the unique ability of transition metals to bond with simple molecules like CO.

Such compounds are significant in chemistry due to their structural and bonding properties. They offer insights into how metals can form complex structures, either through direct bonding or coordinate covalent bonds. Risks such as toxicity make nickel carbonyl one of the more dangerous compounds, but it is also fascinating in industrial applications like purification of nickel through the Mond process.

Understanding the formation of nickel carbonyl requires comprehending how nickel atoms coordinate with carbon monoxide molecules. This drives the calculation of \( x \), showing how many CO molecules attach to a nickel atom to form the complete compound.

Chemical Equation Balancing

Balancing chemical equations is an essential skill for understanding reactions, ensuring the conservation of mass. When an equation is balanced, the amount of each element is the same on both sides. This process keeps the reaction's nature and the law of conservation of matter intact.

In the example problem, balancing involves ensuring that the moles of nickel correspond to the moles of the compound \( \mathrm{Ni}(\mathrm{CO})_x \). Here, the balanced equation represents how nickel atoms completely react with carbon monoxide to form the nickel carbonyl compound, thereby enclosing CO molecules.

This practice helps deepen understanding of both reactants and products in a reaction. It's crucial for predicting the outcome of reactions and for practical applications like synthesizing compounds in the lab. Perfecting this approach enables you to correctly derive coefficients of elements involved, aiding in accurate compound and reaction representation in chemistry.