Problem 110 - A Complete Resource Book in Chemistry for JEE Main

Question

An organic compound having molecular mass 60 is found to contain \(\mathrm{C}=20 \%, \mathrm{H}=6.67 \%\) and \(\mathrm{N}=\) \(46.67 \%\) while rest is oxygen. On heating, it gives \(\mathrm{NH}_{3}\) along with a solid residue. The solid residue give violet colour with alkaline copper sulphate solution. The compound is [2005] (a) \(\mathrm{CH}_{3} \mathrm{CONH}_{2}\) (b) \(\mathrm{CH}_{3} \mathrm{NCO}\) (c) \(\left(\mathrm{NH}_{2}\right)_{2} \mathrm{CO}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CONH}_{2}\)

Step-by-Step Solution

Answer

The compound is \((\mathrm{NH}_2)_2\mathrm{CO}\).

Step 1: Determine Percentage Composition of Elements

The problem provides percentage compositions for carbon (C), hydrogen (H), and nitrogen (N). We sum these percentages to find the remaining composition for oxygen (O). The sum of carbon, hydrogen, and nitrogen percentages is \(20 \% + 6.67\% + 46.67\% = 73.34\%\). Thus, the oxygen percentage is \(100\% - 73.34\% = 26.66\%\).

Step 2: Determine Empirical Formula by Moles

Find the number of moles of each element in terms of the lowest atomic mass (hydrogen). Using the percentages:- \( \text{C}: \frac{20}{12} \approx 1.67 \text{ mol} \)- \( \text{H}: \frac{6.67}{1} = 6.67 \text{ mol} \)- \( \text{N}: \frac{46.67}{14} \approx 3.33 \text{ mol} \)- \( \text{O}: \frac{26.66}{16} \approx 1.67 \text{ mol} \)Divide each element's moles by the smallest number of moles, which is found in carbon and oxygen, yielding their mole ratio.

Step 3: Identify the Empirical Formula

Divide all mole quantities by the smallest value found (1.67), resulting in an approximate ratio \( \text{C} : \text{H} : \text{N} : \text{O} = 1 : 4 : 2 : 1 \). This mole ratio suggests the empirical formula \(\mathrm{CH}_{4}\mathrm{N}_{2}\mathrm{O} \).

Step 4: Relate Empirical Formula to Molecular Formula

Calculate molecular weight of the empirical formula \[\mathrm{CH}_{4}\mathrm{N}_{2}\mathrm{O} \] \((12 + 1\times4 + 14\times2 + 16 = 60)\), which matches the given molecular mass in the problem, confirming the molecular formula is identical to the empirical formula.

Step 5: Interpretation of Heating Result and Identify the Compound

The problem states that upon heating, the compound gives off \(\mathrm{NH}_3\) and leaves a residue creating violet color with alkaline copper sulfate, which typically suggests biuret or urea presence. Given the empirical formula matches urea \((\mathrm{NH}_2)_2\mathrm{CO} \), this is the expected compound.

Key Concepts

Molecular Mass Determination

Understanding Molecular Mass Determination can be pivotal when it comes to analyzing chemical compounds and understanding their composition. Molecular mass is the sum of the atomic masses of all the atoms in a molecule. It provides critical insights when backed by empirical analysis, as it helps to identify unknown compounds when combined with the empirical formula.

For instance, when you have an empirical formula, you may determine the empirical formula weight by adding up the atomic weights of each constituent atom. In the solved exercise, the empirical formula of the compound was found to be \(\mathrm{CH}_{4}\mathrm{N}_{2}\mathrm{O}\). To ascertain if this empirical formula matches the molecular formula, you calculate its molecular mass: evaluating to \(60\).

Since it matches the given molecular mass exactly, you confirm the empirical formula is also the molecular formula. This reinforces how molecular mass determination helps verify the identity of unknown compounds, by matching detailed calculated values with experimental data.

Percentage Composition

Percentage Composition refers to the percentage by mass of each element in a compound, informing how much of each element is present. It plays a crucial role in determining empirical formulas and ultimately identifying compounds.

To calculate the percentage composition of a compound, you determine the mass each element contributes to the compound, divided by the total mass of the compound, times 100. In the given problem, the organic compound had given percentages for carbon, hydrogen, and nitrogen. Summing these percentages reveals the percentage for the remaining element, oxygen, ensuring the total reaches 100%.

• Carbon: \(20\%\)
• Hydrogen: \(6.67\%\)
• Nitrogen: \(46.67\%\)
• Oxygen: Calculated as \(26.66\%\)

This approach enables understanding of a compound's makeup before detailed stoichiometric calculations or other laboratory analysis are performed. This foundation leads into finding the mole ratios and further structural determination.

Mole Concept

The Mole Concept is a core chemical principle that quantifies the amount of substance. One mole corresponds to \(6.022 \times 10^{23}\) entities, such as atoms or molecules, allowing chemists to relate mass to moles, a fundamental step in chemical calculations.

In the solution provided, percentage compositions were converted to moles. By dividing the percentage composition of each element by its atomic mass, you obtain the moles of each element in the compound.

• Carbon: \(\frac{20}{12} \approx 1.67\) mol
• Hydrogen: \(\frac{6.67}{1} = 6.67\) mol
• Nitrogen: \(\frac{46.67}{14} \approx 3.33\) mol
• Oxygen: \(\frac{26.66}{16} \approx 1.67\) mol

These values are invaluable for deriving the empirical formula by taking mole ratios. Further dividing by the smallest mole value gives precise ratios correlating to the simple formula, \(\mathrm{CH}_4\mathrm{N}_2\mathrm{O}\). This step underlines how the mole concept transforms experimental data into meaningful chemical structures.