Question

\(0.21 \mathrm{~g}\) of an organic substance having \(\mathrm{C}, \mathrm{H} \mathrm{N}, \mathrm{O}\) on combustion gave \(0.462 \mathrm{~g} \mathrm{CO}_{2}, 0.1215 \mathrm{~g} \mathrm{H}_{2} \mathrm{O} .\) When \(0.104 \mathrm{~g}\) of it was distilled with \(\mathrm{NaOH}\) the evolved \(\mathrm{NH}_{3}\) was neutralised by \(15 \mathrm{ml}\) of \(0.05 \mathrm{~N} \mathrm{H}_{2} \mathrm{SO}_{4}\). The empirical formula is? (a) \(\mathrm{C}_{6} \mathrm{H}_{7} \mathrm{NO}_{2}\) (b) \(\mathrm{C}_{7} \mathrm{H}_{9} \mathrm{NO}_{2}\) (c) \(\mathrm{C}_{7} \mathrm{H}_{9} \mathrm{NO}\) (d) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NO}_{2}\)

Short answer

(b) \( \mathrm{C}_{7} \mathrm{H}_{9} \mathrm{NO}_{2} \) is the empirical formula.

Step-by-step solution

Calculate Moles of Carbon

Given 0.462 g of CO₂, calculate moles of Carbon. CO₂ contains one Carbon atom per molecule. Molar mass of CO₂ = 44.01 g/mol. Moles of CO₂ = \( \frac{0.462}{44.01} \approx 0.0105 \text{ mol} \). This is equal to moles of Carbon. Total mass of Carbon = \( 0.0105 \times 12.01 \approx 0.126 \text{ g} \).

Calculate Moles of Hydrogen

Given 0.1215 g of H₂O, calculate moles of Hydrogen. Each H₂O contains two Hydrogen atoms. Molar mass of H₂O = 18.02 g/mol. Moles of H₂O = \( \frac{0.1215}{18.02} \approx 0.00674 \text{ mol} \). Moles of Hydrogen = \( 2 \times 0.00674 \approx 0.0135 \text{ mol} \). Mass of Hydrogen = \( 0.0135 \times 1.008 \approx 0.0136 \text{ g} \).

Calculate Moles of Nitrogen

Given that 15 mL of 0.05 N H₂SO₄ neutralizes the NH₃. Equivalent weight of H₂SO₄ = 49 g/equiv, moles of H₂SO₄ (which equals moles of NH₃) = \( \frac{15}{1000} \times 0.05 \approx 0.00075 \text{ mol} \). Mass of Nitrogen = \( 0.00075 \times 14.01 \approx 0.0105 \text{ g} \).

Calculate Mass and Moles of Oxygen

Total mass of the substance = 0.21 g. Mass of other elements (C, H, N) = 0.126 g + 0.0136 g + 0.0105 g = 0.1501 g. Mass of Oxygen = 0.21 g - 0.1501 g = 0.0599 g. Molar mass of Oxygen = 16.00 g/mol. Moles of Oxygen = \( \frac{0.0599}{16.00} \approx 0.00374 \text{ mol} \).

Determine Empirical Formula

Find the simplest whole-number ratio of moles: C: 0.0105 mol / 0.00374 mol = 2.81 H: 0.0135 mol / 0.00374 mol = 3.61 N: 0.00075 mol / 0.00374 mol = 0.20 O: 0.00374 mol / 0.00374 mol = 1.00 Approximate the elements to closest whole numbers by proportionate multiplication: C : H : N : O = 7 : 9 : 1 : 2 Thus, the empirical formula is \( \text{C}_7\text{H}_9\text{NO}_2 \).

Key concepts

Combustion Analysis

Combustion analysis is an essential process used to determine the composition of organic compounds. By burning a substance and analyzing the resultant gases, chemists can deduce the amounts of different elements present in the initial compound. The main idea is that combustion transforms all the carbon to carbon dioxide and all the hydrogen to water. Then, by measuring these products, we can calculate the amounts of carbon and hydrogen in the original compound.

In a typical analysis like ours, the organic compound composed of carbon, hydrogen, nitrogen, and oxygen is combusted, resulting in the production of carbon dioxide and water. By using the molar masses of these byproducts—CO₂ and H₂O—we calculate the amount of carbon and hydrogen originally present. Following steps allow us to deduce the presence of other elements, like nitrogen and oxygen, by considering the remaining mass that is not accounted for by carbon and hydrogen.

Thus, combustion analysis, alongside stoichiometry and the knowledge of chemical reactions, helps in deriving empirical formulas for unknown organic compounds.

Mole Concept

The mole concept is a vital tool in chemistry for converting between atoms and grams. It signifies that the mole serves as a bridge, connecting microscopic particles to measurable quantities. In this scenario, for example, the mole concept is used to convert the grams of CO₂ and H₂O (the products of combustion) into moles to find how many moles of carbon and hydrogen those represent, respectively.

Each mole corresponds to Avogadro's number, approximately 6.022 x 10²³ entities, whether they are atoms, molecules, or ions. This allows chemists to handle practical quantities of substances. In calculating moles, the formula to use is:

• Number of moles = \( \frac{\text{mass of substance}}{\text{molar mass of substance}} \).

By consistently applying this formula, one can determine the number of moles present in substances involved in a chemical reaction. From here, these values can be used in stoichiometric calculations to determine empirical formulas or theoretical yields.

Organic Chemistry

Organic chemistry is the branch of chemistry that studies carbon compounds, primarily those containing carbons and hydrogens, but may also include heteroatoms like nitrogen and oxygen. Organic chemistry is foundational for understanding the composition and reactions of living organisms as well as synthetic compounds.

In the given problem, we dealt with an organic compound containing C, H, N, and O. These elements are common in organic molecules due to the unique bonding abilities of carbon, which can form stable covalent bonds with itself and other elements. This creates a versatile foundation for complex molecules like proteins, carbohydrates, lipids, and nucleic acids, essential for life.

Understanding organic chemistry involves grasping core concepts such as functional groups, molecular structure, and reactions like combustion. It allows for the prediction and explanation of molecular interactions, reactivity, and properties.

Empirical Formula

An empirical formula represents the simplest whole-number ratio of elements in a compound. To find an empirical formula, one must first determine the moles of each element present in the compound and then simplify the ratios between them.

• Calculation starts by finding the mass of each element in a sample, converting these masses into moles using their respective molar masses.
• The next step is finding the smallest number of moles in your data, which you then use to divide each element's mole amount, yielding the smallest ratio.

In our problem, after calculating moles for C, H, N, and O, we determined the ratios, leading us to the empirical formula \( \text{C}_7\text{H}_9\text{NO}_2 \).

The empirical formula is crucial because it gives insight into the compound's relative molecular composition, which is necessary for understanding the compound's identity and behavior in chemical reactions. Notably, while the empirical formula shows the relative proportions of elements, the molecular formula, which may require additional information, gives the actual number of each atom present in a molecule.