Question

An organic compound contains \(\mathrm{C}, \mathrm{H}, \mathrm{N},\) and \(\mathrm{O}\) . Combustion of 0.1023 \(\mathrm{g}\) of the compound in excess oxygen yielded 0.2766 \(\mathrm{g} \mathrm{CO}_{2}\) and 0.0991 \(\mathrm{g} \mathrm{H}_{2} \mathrm{O} .\) A sample of 0.4831 \(\mathrm{g}\) of the compound was analyzed for nitrogen by the Dumas method (see Exercise 137\() .\) At STP, 27.6 \(\mathrm{mL}\) of dry \(\mathrm{N}_{2}\) was obtained. In a third experiment, the density of the compound as a gas was found to be 4.02 \(\mathrm{g} / \mathrm{L}\) at \(127^{\circ} \mathrm{C}\) and 256 torr. What are the empirical and molecular formulas of the compound?

Short answer

The empirical formula of the compound is \(C_9H_{15}NO_3\) and the molecular formula is approximately \(C_{15}H_{25}N_{2}O_{5}\).

Step-by-step solution

Combustion and determination of CO2 and H2O

From the combustion of 0.1023 g of the compound, we get 0.2766 g of CO2 and 0.0991 g of H2O.

Find the mass of C and H

Using the molecular mass ratios in CO2 (1 mole of C produces 1 mole of CO2) and H2O (2 moles of H produce 1 mole of H2O). Mass of C in CO2 = 0.2766 g × \(\frac{12.01 g}{44.01 g}\) = 0.0756 g Mass of H in H2O = 0.0991 g × \(\frac{2.02 g}{18.02 g}\) = 0.0111 g

Find mass of N

Using the volume of N2 and the ideal gas law, we can find the mass of N in the sample by converting it to moles and then to grams. (0.0276 L × 1 mol/22.4 L) × (28.02 g/mol) = 0.0343 g of N

Mass of O and Empirical Formula

Find the mass of O in the compound by subtracting the masses of C, H, and N from the initial mass. Mass of O = 0.1023 g - (0.0756 g + 0.0111 g + 0.0343 g) = 0.0113 g. Convert these masses to moles: moles of C = \(\frac{0.0756}{12.01}\) = 0.00630 moles moles of H = \(\frac{0.0111}{1.01}\) = 0.0110 moles moles of N = \(\frac{0.0343}{14.01}\) = 0.00245 moles moles of O = \(\frac{0.0113}{16.00}\) = 0.0007063 moles _empirical formula_: \(CH_{\text{a}}N_{\text{b}}O_{\text{c}}\) Find the lowest whole number ratio of the constituent moles: Divide each mole by the smallest mole: a: 0.00630 moles / 0.0007063 moles = 8.92 b: 0.0110 moles / 0.0007063 moles = 15.6 c: 0.00245 moles / 0.0007063 moles = 3.47 Thus, the empirical formula is \(C_9H_{15}NO_3\).

Calculate Molecular Weight from Empirical Formula

Calculate the weight of the empirical formula: Empirical formula weight = \(9(12) + 15(1) + 1(14) + 3(16) = 236 \, \text{g/mol}\)

Molecular weight calculation

Use the density data to determine the molecular weight of the compound. Given density: 4.02 g/L Temperature: \(127^{\circ}C + 273.15\) K = 400.15 K Pressure: \(\frac{256 \, \text{torr}}{760 \, \text{torr/atm}}\) = 0.3368 atm Now we use the ideal gas law equation: \(n = \frac{PV}{RT}\) Where: n = number of moles P = pressure (0.3368 atm) V = volume (1 L) R = gas constant (0.08206 L atm/mol K) T = temperature (400.15 K) Solving for n, we get: n = \(\frac{0.3368 \times 1}{(0.08206)(400.15)}\) = 0.01024 moles Given that 4.02 g of the gas corresponds to 0.01024 moles, the molecular weight is: Molecular weight = \(\frac{4.02 \, \text{g}}{0.01024 \, \text{moles}}\) = 392.57 g/mol

Molecular Formula Calculation

Calculate the ratio between the molecular weight and the empirical formula weight to find the molecular formula. Ratio = \(\frac{392.57}{236}\) ≈ 1.66 Since the ratio is approximately equal to an integer, we multiply each subscript in the empirical formula by the ratio to find the molecular formula. Molecular formula: \(C_{9\cdot1.66}H_{15\cdot1.66}N_{1\cdot1.66}O_{3\cdot1.66}\) The molecular formula is approximately \(C_{15}H_{25}N_{2}O_{5}\).

Key concepts

Combustion Analysis

Combustion analysis is a method used in chemistry to determine the elemental composition of hydrocarbons or organic compounds. By burning the substance in excess oxygen, the combustion products such as carbon dioxide ( \(\text{CO}_2\) ) and water ( \(\text{H}_2\text{O}\) ) are collected. From these products, the masses of carbon and hydrogen in the original compound can be calculated.

To analyze the combustion results, you start by weighing the amount of \(\text{CO}_2\) and \(\text{H}_2\text{O}\) produced:

• For \(\text{CO}_2\): The mass of carbon is determined by using the molar mass ratio of carbon to carbon dioxide (12.01 g/mol for C compared to 44.01 g/mol for COtwo).
• For \(\text{H}_2\text{O}\): The mass of hydrogen is derived using the molar mass ratio of hydrogen to water (2.02 g/mol for two hydrogen atoms in \(\text{H}_2\text{O}\) compared to 18.02 g/mol for water).

By calculating the moles of carbon and hydrogen, you can further understand the composition of the original compound. This technique is essential in deriving empirical formulas, which represent the simplest whole-number ratio of elements in a compound.

Dumas Method for Nitrogen Analysis

The Dumas method is a classic procedure used to measure the nitrogen content in organic compounds. In this method, the sample is combusted in the presence of oxygen, causing any nitrogen present in the compound to form nitrogen gas (\(\text{N}_2\)). The volume of this nitrogen gas can then be measured under specific conditions, often using the Ideal Gas Law to determine the mass of nitrogen.

At standard temperature and pressure (STP), one mole of nitrogen gas occupies 22.4 L. In our exercise, 27.6 mL of \(\text{N}_2\) was collected. By using this volume measurement, we can convert it to moles:

• First, convert the volume from mL to L (0.0276 L).
• Then, use the molar volume of a gas at STP to find moles: \(\frac{0.0276}{22.4}\) moles of \(\text{N}_2\)
• Finally, multiply the moles of gas by the molar mass of nitrogen (28.02 g/mol) to find the mass of nitrogen in the sample.

This method is not only used for determining nitrogen in unknown compounds, but also in food and agricultural analysis to measure protein content.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry that relates the pressure, volume, temperature, and moles of a gas. The formula is expressed as \(PV = nRT\), where:

• \(P\) represents the pressure of the gas.
• \(V\) is its volume.
• \(n\) denotes the amount of gas in moles.
• \(R\) is the universal gas constant (0.08206 L atm/mol K).
• \(T\) is the temperature in Kelvin.

In the context of the exercise, the Ideal Gas Law was applied to find the number of moles in the gaseous sample. Given certain conditions:- The density of the compound (4.02 g/L),
- Temperature (converted to Kelvin),
- Pressure (converted to atm),
You rearrange the formula to solve for \(n\). This value helps in determining the molecular weight by dividing the measured mass of one liter by the moles calculated.

Molar Mass Calculation

Molar mass is a key measurement in chemistry used to quantify the "weight" of a given substance's mole, expressed in grams per mole (g/mol). It is this value that connects the concept of moles to measurable mass in the laboratory.

In our exercise, the molar mass calculation is pivotal in determining both empirical and molecular formulas. Here's a simple breakdown:

• The empirical formula gives the simplest ratio of elements, whose combined atomic masses offer the empirical formula weight.
• To find the molecular formula, compare the compound's calculated molar mass (from the Ideal Gas Law and measured data) with that of the empirical formula.
• The molecular formula is derived by multiplying the subscripts in the empirical formula by this ratio (molar mass/empirical formula mass).

This step establishes the true representation of the compound, taking into account the entire account of atoms within the molecule.