Problem 30 - Chemistry in 30 days

Question

In a compound \(\mathrm{C}, \mathrm{H}\) and \(\mathrm{N}\) atoms are present in \(9: 1: 3.5\) by weight. If molecular weight of the compound is 108 , then molecular formula of the compound is (a) \(\mathrm{C}_{2} \mathrm{H}_{0} \mathrm{~N}_{2}\) (b) \(\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{~N}\) (c) \(\mathrm{C}_{6} \mathrm{H}_{3} \mathrm{~N}_{2}\) (d) \(\mathrm{C}_{9} \mathrm{H}_{12} \mathrm{~N}_{3}\)

Step-by-Step Solution

Answer

The molecular formula is derived as \(\text{C}_6\text{H}_8\text{N}_2\), which matches best with the empirical formula \(\text{C}_3\text{H}_4\text{N}\) from option (b).

Step 1: Determine Moles Ratio

First, identify the given ratio of elements by weight: C : H : N = 9 : 1 : 3.5. Convert these weights into moles using the atomic weights (C = 12, H = 1, N = 14). This gives:- For C: \( \frac{9}{12} = 0.75 \)- For H: \( \frac{1}{1} = 1 \)- For N: \( \frac{3.5}{14} = 0.25 \)

Step 2: Calculate the Simplest Ratio

Divide all the values obtained in Step 1 by the smallest number of moles to find the simplest whole number ratio:- For C: \( \frac{0.75}{0.25} = 3 \)- For H: \( \frac{1}{0.25} = 4 \)- For N: \( \frac{0.25}{0.25} = 1 \)This means the simplest ratio of atoms is C : H : N = 3 : 4 : 1.

Step 3: Determine the Empirical Formula Mass

Calculate the empirical formula mass based on the simplest ratio. The empirical formula is \(\text{C}_3\text{H}_4\text{N}\):- Mass of C = 3 \times 12 = 36- Mass of H = 4 \times 1 = 4- Mass of N = 1 \times 14 = 14- Total empirical mass = 36 + 4 + 14 = 54.

Step 4: Find the Molecular Formula

Compare the molecular weight provided (108) with the empirical formula mass (54) by dividing the molecular weight by the empirical formula mass:\( \frac{108}{54} = 2 \).This indicates that the molecular formula is twice the empirical formula: \( (\text{C}_3\text{H}_4\text{N})_2 \), which results in \(\text{C}_6\text{H}_8\text{N}_2\).

Step 5: Choose the Correct Option

Compare the molecular formula \(\text{C}_6\text{H}_8\text{N}_2\) with the given options:- (a) \(\text{C}_2\text{H}_0\text{N}_2\)- (b) \(\text{C}_3\text{H}_4\text{N}\)- (c) \(\text{C}_6\text{H}_3\text{N}_2\)- (d) \(\text{C}_9\text{H}_12\text{N}_3\)None of the options correctly match \(\text{C}_6\text{H}_8\text{N}_2\).Assessing against the options, only (b) \(\text{C}_3\text{H}_4\text{N}\) corresponds to the correct empirical formula.

Key Concepts

Empirical Formula

The empirical formula is a way to represent the simplest ratio of elements in a compound. It gives you the base numbers of how many atoms of each type are present relative to one another. However, it doesn't necessarily tell you the actual number of atoms in a compound's molecule, just the simplest ratio.
For instance, given the weight ratios of carbon, hydrogen, and nitrogen as provided, you break it down to moles and discover the simplest integer ratio of atoms. This ratio forms the empirical formula. Therefore, for our specific problem, this is how we arrive at the empirical formula of \(\text{C}_3\text{H}_4\text{N}\).
This formula is particularly useful when you lack detailed information about the molecular structure but need a foundational idea of the compound's composition.

Atomic Weights

Atomic weights, sometimes referred to as atomic masses, are crucial for converting weights of elements to moles. Each element has its own atomic weight, representing the mass of one mole of atoms of that element (in grams).
For example, in our problem, carbon has an atomic weight of 12, hydrogen is 1, and nitrogen is 14. These values allow us to calculate how many moles of each element are present from the given weights.

• Carbon: \(\frac{9}{12} = 0.75\) moles
• Hydrogen: \(\frac{1}{1} = 1\) mole
• Nitrogen: \(\frac{3.5}{14} = 0.25\) moles

Understanding atomic weights simplifies the conversion from mass to the number of atoms, forming the bridge to finding a compound's empirical formula.

Mole Ratio

Mole ratio is an essential step for determining the empirical formula. It allows us to derive the proportion of elements in their simplest form. By using mole ratios, you can adjust the number of moles of each element to the smallest possible whole numbers.
This ratio is found by dividing all mole values by the smallest mole quantity present. In the provided example, nitrogen's smallest computed mole is 0.25.

• Carbon: \(\frac{0.75}{0.25} = 3\)
• Hydrogen: \(\frac{1}{0.25} = 4\)
• Nitrogen: \(\frac{0.25}{0.25} = 1\)

Thus, the simplest ratio of \(\text{C:H:N}\) is 3:4:1. Knowing how different elements relate via mole ratio is key to determining the empirical formula.

Molecular Weight

Molecular weight is the sum of the atomic weights of all atoms in a molecule's formula. It helps us relate the empirical formula to the actual measurable quantity of a substance.
To compare, we find the empirical formula mass by adding up the atomic masses of all the atoms in the empirical formula:

• Carbon: \(3 \times 12 = 36\)
• Hydrogen: \(4 \times 1 = 4\)
• Nitrogen: \(1 \times 14 = 14\)

The empirical formula mass is \(36 + 4 + 14 = 54\). The molecular weight is given as 108. By dividing the molecular weight by the empirical formula mass: \(\frac{108}{54} = 2\), we determine the molecular formula: \((\text{C}_3\text{H}_4\text{N})_2 = \text{C}_6\text{H}_8\text{N}_2\). It always pays off to check how empirical and molecular weights relate.

Chemical Composition

Chemical composition details the identity and proportion of elements present in a compound. It shapes the compound's characteristics and behavior. Knowing the chemical composition means understanding how many of each different atom are present.
In our example, the chemical composition is initially given by weight ratios: carbon, hydrogen, and nitrogen. By understanding weight to mole conversion and simplifying through empirical formulas, we can deeply understand what's in a compound. This is very informative when determining not just how a compound is written, but also how it might react in chemical processes.