Problem 24 - Chemistry in 30 days

Question

A.gas is found to have the formula (CO) \(_{x}\). Its V.D. is 70 . The value of \(x\) must be (a) 7 (b) 4 (c) 5 (d) 6

Step-by-Step Solution

Answer

The value of \(x\) is 5, corresponding to option (c).

Step 1: Understanding the Problem

We know that the vapor density (V.D.) of a gas is half its molar mass. It is given that the formula of the gas is \((CO)_x\) and its vapor density is 70.

Step 2: Calculate Molar Mass using Vapor Density

Since the vapor density is 70, the molar mass \(M\) of the gas will be \(70 \times 2 = 140\). This molar mass represents the molecular weight of \((CO)_x\).

Step 3: Determine the Molar Mass of (CO)

To find how many units of CO are present in the compound, we first calculate the molar mass of a single CO unit. The molar mass of carbon (C) is 12 and of oxygen (O) is 16, so the molar mass of CO is \(12 + 16 = 28\).

Step 4: Find the Value of x

The molar mass of \((CO)_x\) is given by the equation \(28x = 140\). Solving for \(x\), we divide both sides by 28, resulting in \(x = \frac{140}{28} = 5\).

Step 5: Verification

To ensure our calculation is correct, verify that the molar mass of \((CO)_5\) is indeed 140: \(5 \times 28 = 140\). This confirms that our value of \(x\) is accurate.

Key Concepts

Vapor Density

Vapor density is an important concept in chemistry that helps us understand the properties of a gas. It is defined as half of the molar mass of a gas or vapor. This means that by knowing the vapor density, we can easily calculate the molar mass of a compound. Therefore, when the vapor density is provided, such as 70 in our exercise, it helps us deduce the molar mass by simply multiplying the vapor density by two, i.e., \(70 \times 2 = 140\).

Knowing the vapor density allows chemists to identify or confirm the molecular formula of a gaseous compound, which is crucial for understanding its chemical behavior. By this valuable property, we can quickly get insights about a gas without needing extensive equipment to measure molar mass directly.

Molecular Formula

The molecular formula of a compound represents the number and type of atoms present in a molecule. For instance, in our example, the molecular formula is expressed as \((CO)_x\). To find the complete molecular formula, we must determine the value of \(x\), which indicates the number of CO units in the compound.

The molecular formula is essential for distinguishing between compounds that might have the same empirical formula but different structures and properties. It's more specific than the empirical formula and provides details about each individual molecule.

• For the given vapor density of 70, the calculated molar mass was found to be 140.
• With the molar mass of a single CO unit being 28, the value of \(x\) can be calculated using the equation: \(28x = 140\).

This ultimately helps in deducing the molecular structure of the compound.

Step-By-Step Problem Solving

Problem-solving in chemistry often involves breaking down complex processes into a series of manageable steps. This approach is exhibitioned perfectly in solving the exercise.

Understanding the step-by-step methodology:

• Start by identifying what is known and what is needed. Vapor density was given as 70.
• The second step involved utilizing known relationships, such as the one that states vapor density is half of the molar mass, to find that \(M = 140\).
• Next, find auxiliary information like the molar mass of CO, which is 28, to resolve unknowns in the problem setup.
• Finally, execute simple arithmetic operations (e.g., division) to solve for \(x\), ensuring correctness through verification.

This systematic approach not only resolves the problem but also engenders confidence in solving similar types of problems.

Chemical Compound Calculations

Calculating properties of chemical compounds like their molar mass or molecular formula involves understanding fundamental concepts of chemistry and applying them accurately.

In the realm of chemical compound calculations, a few main steps are usually involved:

• First, identify the units involved and their molar masses, such as carbon (C) with a molar mass of 12 and oxygen (O) with a molar mass of 16.
• Sum these values to determine the molar mass of the compound's repeating unit, such as CO in our problem, resulting in 28.
• Use provided data (vapor density) to determine the molar mass of the entire compound, making it possible to quantify how many repeating units fit within the total molar mass.
• Finally, ensure accuracy by checking calculations against known chemical laws or previous steps.

By thoroughly grasping these components, students can confidently tackle a broad range of chemical calculation challenges.