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: (a) \(\mathrm{CH}_{3} \mathrm{CONH}_{2}\) (b) \(\mathrm{CH}_{2} \mathrm{NCO}\) (c) \(\left(\mathrm{NH}_{2}\right)_{2} \mathrm{CO}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CONH}_{2}\)

Short answer

The compound is (c) \((\mathrm{NH}_{2})_{2} \mathrm{CO}\).

Step-by-step solution

Determine the Oxygen Percentage

To find the percentage of oxygen, we subtract the sum of the percentages of carbon, hydrogen, and nitrogen from 100%. Thus, Oxygen percentage = 100% - (20% + 6.67% + 46.67%) = 26.66%.

Calculate Relative Moles of Each Element

Next, we calculate the moles of each element based on their percentages and molar masses: - Carbon: \(\frac{20}{12}\) - Hydrogen: \(\frac{6.67}{1}\) - Nitrogen: \(\frac{46.67}{14}\) - Oxygen: \(\frac{26.66}{16}\)

Simplify Molar Ratios to Find Empirical Formula

We divide the moles of each element by the smallest number of moles calculated to find the simplest ratio. After calculation: - C: approximately 1.67 - H: approximately 6.67 - N: approximately 3.33 - O: approximately 1.67 Simplifying gives us a ratio close to 1:4:2:1.

Determine the Empirical Formula

After evaluating the simplified ratios, we find the empirical formula as \(\mathrm{CH}_{4}\mathrm{N}_{2}\mathrm{O}\).

Match with Given Molecular Mass

Calculate the empirical formula mass: \(12+4+28+16 = 60\). The empirical formula mass matches the molecular mass given, confirming the compound formula is \(\mathrm{CH}_{4}\mathrm{N}_{2}\mathrm{O}\).

Identify the Compound from Options

The calculated formula \(\mathrm{CH}_{4}\mathrm{N}_{2}\mathrm{O}\) matches urea, \((\mathrm{NH}_{2})_{2} \mathrm{CO}\). From the given options, this corresponds to option (c). Additionally, the ammonia release and reaction with alkaline copper sulphate confirm it.

Key concepts

Empirical Formula Calculation

An empirical formula represents the simplest ratio of the elements in a compound. When given the percentage compositions of different elements, the first step is often to convert those percentages into a formula that shows the relative number of atoms of each element. This involves converting the percentage by mass into moles by dividing by the atomic mass of each respective element. For example, let's break down the process for the following elements:

• For Carbon (C), with a percentage of 20%, divide by its atomic mass, 12, to get \(rac{20}{12}\).
• Hydrogen (H) at 6.67% uses its atomic mass of 1: \(rac{6.67}{1} \).
• Nitrogen (N), 46.67%: divide by 14, \(rac{46.67}{14}\).
• Oxygen (O), calculated as 26.66% from the rest of the mass, use its atomic mass of 16: \(rac{26.66}{16}\).

Next, simplify the mole ratios by dividing each by the smallest number in your set, aiming for whole numbers that reflect the simplest ratio. This will give the empirical formula, as seen from calculating to achieve \( ext{CH}_{4} ext{N}_{2} ext{O}\). Understanding empirical formulas helps chemists predict chemical behaviors and interactions.

Molecular Mass Determination

The molecular mass or molecular weight of a compound is the sum of the atomic masses of all the atoms in its molecular formula. This is crucial in aligning the empirical formula to a particular compound. Once you have the empirical formula, calculate its mass to see if it matches a given molecular mass. This determines if your empirical formula is indeed the molecular formula. For the compound in our exercise, \( ext{CH}_{4} ext{N}_{2} ext{O}\), the steps to calculate are straightforward:

• Carbon (C): 12 units
• Hydrogen (H): 1 unit per atom times 4, equals 4 units
• Nitrogen (N): 14 units per atom times 2, equals 28 units
• Oxygen (O): 16 units

All summed up, the formula mass is 60 units. When the empirical formula mass matches the molecular mass provided, it confirms that the empirical formula is also the compound’s molecular formula. Calculating molecular mass helps in identifying specific substances based on weight and formula alignment.

Elements Percentage Composition

Percentage composition of a compound shows the percentage by mass of each element within. By knowing this, we can infer the amount of each atom present relative to the entire compound, a crucial step in empirical formula calculations. The sum of the percentage compositions of all elements in a compound should equal 100%. Let’s apply this understanding: - For Carbon: 20% - For Hydrogen: 6.67% - For Nitrogen: 46.67% Calculate what percentage of the compound is made up of oxygen by subtracting the sum of these from 100%, arriving at 26.66%. Let's discuss why understanding percentage composition is vital:

• Helps in determining the empirical formula as shown.
• Essential for analyzing and confirming chemical reactions, processes, and compound purities.

Recognizing and calculating percentage composition provides deep insights into chemical identities and molecular structures, forming a foundational tool in organic chemistry.