Question

Nickel forms a gaseous compound of the formula \(\mathrm{Ni}(\mathrm{CO})_{x} .\) What is the value of \(x\) given the fact that under the same conditions of temperature and pressure, methane \(\left(\mathrm{CH}_{4}\right)\) effuses 3.3 times faster than the compound?

Short answer

The value of \( x \) is 4.

Step-by-step solution

Analyze the Problem

We need to determine the value of \( x \) in the compound \( \mathrm{Ni}( ext{CO})_x \) using the information provided about effusion rates.

Understand Graham's Law of Effusion

Graham's Law of Effusion states that the rate of effusion is inversely proportional to the square root of the molar mass of the gas. Mathematically, this can be expressed as \( \frac{\text{rate}_1}{\text{rate}_2} = \sqrt{\frac{M_2}{M_1}} \), where \( M \) represents the molar mass.

Identify Variables and Known Values

From the problem, methane effuses 3.3 times faster than the nickel compound. Thus, \( \frac{\text{rate}_{\text{CH}_4}}{\text{rate}_{\text{Ni(CO)}_x}} = 3.3 \).

Molar Mass Calculation

First, determine the molar mass of methane, \( \mathrm{CH}_4 \), which is \( 12.01 + 4 \times 1.01 = 16.05 \) g/mol. Let the molar mass of \( \mathrm{Ni(CO)}_x \) be \( M_x \).

Setup the Equation Using Effusion Ratio

According to Graham's Law, \( 3.3 = \sqrt{\frac{M_x}{16.05}} \).

Solve for Molar Mass of Ni(CO)_x

Square both sides of the equation to remove the square root: \( 3.3^2 = \frac{M_x}{16.05} \). This simplifies to \( 10.89 = \frac{M_x}{16.05} \). Solve for \( M_x \) by multiplying both sides by 16.05: \( M_x = 10.89 \times 16.05 = 174.7045 \approx 174.7 \) g/mol.

Calculate the Value of x

The molar mass of \( \mathrm{Ni} \) is 58.69 g/mol, and \( \mathrm{CO} \) is 28.01 g/mol. Therefore, the molar mass of \( \mathrm{Ni}( ext{CO})_x \) is \( 58.69 + 28.01x = 174.7 \). Solve for \( x \): \( 28.01x = 174.7 - 58.69 \). This gives \( x = \frac{116.01}{28.01} \approx 4.14 \). Rounding to the nearest whole number, \( x = 4 \).

Key concepts

Molar Mass Calculation

Molar mass is a crucial concept in chemistry, as it helps us find out how much one mole of a specific element or compound weighs. Calculating it requires adding up the atomic masses of all the atoms in a molecule. Here's a simple way to think about it:
For a compound like methane (\[ \text{CH}_4 \]), the molar mass is determined by adding the mass of one carbon atom (approximately 12.01 g/mol) with four hydrogen atoms (each about 1.01 g/mol). Thus, the molar mass becomes:\[ 12.01 + 4 \times 1.01 = 16.05 \text{ g/mol} \]This helps in various calculations, such as determining how gases will behave under specific conditions. For example, if you know the molar mass, you can predict how fast a gas might move through a small opening, which leads us directly to our next concept.

Effusion Rate

Effusion is the process where gas particles pass through a tiny hole from a container. Graham's Law of Effusion allows us to compare the rates at which two gases effuse. It states that a gas with a lower molar mass will effuse faster than a gas with a higher molar mass. The relationship can be expressed by:\[ \frac{\text{rate}_1}{\text{rate}_2} = \sqrt{\frac{M_2}{M_1}} \]Here, \(\text{rate}_1\) and \(\text{rate}_2\) are the effusion rates of two gases, while \(M_1\) and \(M_2\) are their respective molar masses. From the exercise, methane effuses 3.3 times faster than the nickel compound. By knowing this and the molar mass of methane, we can solve for the molar mass of the nickel compound using the effusion formula.

• This allows us to compare and calculate effectively, providing insights into unknown compounds.

Chemical Compound Formula

Understanding a chemical compound's formula is fundamental to knowing its composition and properties. The formula tells us how many of each type of atom are present in a molecule. For instance, molecules like methane (\[ \text{CH}_4 \]) or nickel carbonyl (hypothetically represented as \[ \text{Ni}( ext{CO})_x \]) show us the number and type of atoms involved. In our exercise, we explored how to determine the value of \(x\) in \(\text{Ni}(\text{CO})_x\).

• We calculated the molar masses of \( \text{Ni} \) and \( \text{CO} \) to find the total molar mass of the compound.
• This involved utilizing the given formula, balancing known molar mass values, and simple arithmetic to solve for \(x\) (the number of \( \text{CO} \) groups).
• The final computation yielded \(x = 4\), showing how the compound's structure can be deciphered.

By interpreting the chemical formula correctly, we gain insights into how these elements can be combined and their potential applications.