Question

\(9.9 \mathrm{~g}\) of an amide with molecular formula \(\mathrm{C}_{4} \mathrm{H}_{5} \mathrm{~N}_{\mathrm{x}} \mathrm{O}_{y}\) on heating with alkali liberated \(1.7 \mathrm{~g}\) of ammonia. If the percentage of oxygen is \(32.33 \%\), then the ratio of \(\mathrm{N}\) and \(\mathrm{O}\) atoms in the compound is: (a) \(2: 1\) (b) \(1: 2\) (c) \(2: 5\) (d) \(2: 3\)

Short answer

The ratio of \( \text{N} \) to \( \text{O} \) atoms is 1:2, so answer (b).

Step-by-step solution

Determine the Mass of Oxygen in Compound

Calculate the mass of oxygen in the compound using its percentage by mass. Since the total mass of the compound is 9.9 g, the mass of oxygen is given by: \[\text{Mass of } O = \frac{32.33}{100} \times 9.9 = 3.19967 \text{ g}\]

Calculate Remaining Mass of Compound

Subtract the mass of oxygen from the total mass of the compound to find the mass of carbon, hydrogen, and nitrogen combined:\[\text{Remaining Mass} = 9.9 \text{ g} - 3.19967 \text{ g} = 6.70033 \text{ g}\]

Determine Mass of Nitrogen from Ammonia

Given that the mass of ammonia is 1.7 g and the molar mass of ammonia \((\text{NH}_3)\) is approximately 17 g/mol, calculate the mass of nitrogen in the ammonia:\[\text{Mass of } N = \frac{14}{17} \times 1.7 \text{ g} \approx 1.4 \text{ g}\]

Find Ratio of Nitrogen to Oxygen Atoms

Given the mass of nitrogen in the compound is approximately the same as the mass of nitrogen released, and recognizing that the mass of nitrogen in the amide also accounts for nitrogen atoms in the molecular formula, we calculate the moles of each:- Molar mass of \( \text{N} = 14g/mol \):\[\text{Moles of } N = \frac{1.4}{14} \approx 0.1 \text{ moles}\]- Molar mass of \( \text{O} = 16g/mol \):\[\text{Moles of } O = \frac{3.19967}{16} \approx 0.2 \text{ moles}\]The ratio of \( \text{N} \) to \( \text{O} \) atoms is therefore:\[\frac{0.1}{0.2} = \frac{1}{2}\]The ratio \( 1:2 \) corresponds to choice (b).

Key concepts

Molecular Formula Determination

Determining the molecular formula of a compound involves identifying the number and types of atoms present in it. In our exercise, the molecular formula is partially given as \(\mathrm{C}_{4}\mathrm{H}_{5}\mathrm{~N}_{x}\mathrm{O}_{y}\). To complete the formula, we need to find the exact number of nitrogen \( (N) \) and oxygen \( (O) \) atoms. This involves several steps such as analyzing the given data about the compound's mass, as well as using chemical principles like stoichiometry. We start by focusing on known quantities like the mass of ammonia released and the percentage of oxygen. By strategically calculating the mass of each element using the data, we eventually deduce the ratio of \(N\) to \(O\) atoms. Afterwards, we apply stoichiometric principles which allow us to translate these mass observations into molecular insights, leading us directly to the molecular formula.

Mass Percentage Calculation

Mass percentage is a way of expressing the concentration of an element in a compound. It is calculated using the formula: \( \text{Mass percentage of element} = \left(\frac{\text{Mass of element}}{\text{Total mass of compound}}\right) \times 100 \). In the given problem, the mass percentage of oxygen is provided as 32.33%. To find the mass of oxygen in 9.9 g of the compound, you multiply the total mass by the oxygen's mass percentage.

• Mass of oxygen = \( \frac{32.33}{100} \times 9.9 = 3.19967 \text{ g} \)
  

This simple calculation provides us a crucial piece of information to further analyze the compound's composition. It's notable that such calculations are foundational in determining how much of each element is present, laying the groundwork for further analysis or chemical reactions.

Stoichiometry

Stoichiometry is the discipline of chemistry that deals with the relative quantities of reactants and products in chemical reactions. In this exercise, it plays a central role in correlating the masses and ratios of the elements involved. For instance, we use stoichiometry to relate the information about released ammonia to determine the amount of nitrogen present in the compound.

• The molar mass of ammonia is approximately 17 g/mol, and nitrogen in ammonia is about 14 g/mol.
• From 1.7 g of ammonia, we calculate the nitrogen mass as approximately 1.4 g.
  

Next, using stoichiometry, we convert these masses into moles. Remember, moles allow chemists to equate the relative numbers of each type of atom. By knowing the moles of nitrogen (0.1 moles) and oxygen (0.2 moles), you can find the ratio of these elements in the compound. Understanding stoichiometry not only helps solve textbook problems but is also invaluable in industries that deal with batch processing, pharmaceuticals, and chemical manufacturing. It ensures accurate chemical formulations, leading to beneficial outcomes both scientifically and economically.