Question

A compound contains only carbon, hydrogen, nitrogen, and oxygen. Combustion of 0.157 g of the compound produced \(0.213 \mathrm{g} \mathrm{CO}_{2}\) and \(0.0310 \mathrm{g} \mathrm{H}_{2} \mathrm{O} .\) In another experiment, it is found that 0.103 g of the compound produces \(0.0230 \mathrm{g} \mathrm{NH}_{3}\) What is the empirical formula of the compound? Hint: Combustion involves reacting with excess \(\mathrm{O}_{2}\). Assume that all the carbon ends up in \(\mathrm{CO}_{2}\) and all the hydrogen ends up in \(\mathrm{H}_{2} \mathrm{O}\). Also assume that all the nitrogen ends up in the \(\mathrm{NH}_{3}\) in the second experiment.

Short answer

The empirical formula of the compound is C4H3NO4.

Step-by-step solution

Convert the mass of CO2 and H2O to moles of carbon and hydrogen.

First, we need to figure out the moles of carbon and hydrogen in the compound. To do this, we will convert the mass of CO2 and H2O produced during combustion to moles of carbon and hydrogen, respectively. For CO2: Moles of CO2 = mass of CO2 produced / molar mass of CO2 Moles of CO2 = \(0.213 \textrm{g CO}_2 \) / \(44.01 \textrm{g/mol}\) = \(4.84 \times 10^{-3}\) moles of CO2 For H2O: Moles of H2O = mass of H2O produced / molar mass of H2O Moles of H2O = \(0.0310 \textrm{g H}_2\textrm{O}\) / \(18.015 \textrm{g/mol}\) = \(1.72 \times 10^{-3}\) moles of H2O Since each mole of CO2 contains 1 mole of carbon and each mole of H2O contains 2 moles of hydrogen, we can now find the moles of carbon and hydrogen in the compound as follows: Moles of carbon = moles of CO2 Moles of carbon = \(4.84 \times 10^{-3}\) moles Moles of hydrogen = 2 * moles of H2O Moles of hydrogen = 2 * \(1.72 \times 10^{-3}\) moles = \(3.44 \times 10^{-3}\) moles

Convert the mass of NH3 to moles of nitrogen.

Next, we will convert the mass of NH3 produced in the second experiment to moles of nitrogen, as all nitrogen ends up in NH3. Moles of NH3 = mass of NH3 produced / molar mass of NH3 Moles of NH3 = \(0.0230 \textrm{g NH}_3\) / \(17.031 \textrm{g/mol}\) = \(1.35 \times 10^{-3}\) moles of NH3 Since each mole of NH3 contains 1 mole of nitrogen, we can find the moles of nitrogen in the compound as follows: Moles of nitrogen = moles of NH3 Moles of nitrogen = \(1.35 \times 10^{-3}\) moles

Calculate the mass of oxygen in the compound.

Now that we have the moles of carbon, hydrogen, and nitrogen, we can find the mass of each element in the compound. Mass of carbon = moles of carbon * molar mass of carbon = \(4.84 \times 10^{-3}\textrm{ mol}\) * \(12.01 \textrm{g/mol}\) = \(0.0581 \textrm{g}\) Mass of hydrogen = moles of hydrogen * molar mass of hydrogen = \(3.44 \times 10^{-3}\textrm{ mol}\) * \(1.008 \textrm{g/mol}\) = \(0.00347 \textrm{g}\) Mass of nitrogen = moles of nitrogen * molar mass of nitrogen = \(1.35 \times 10^{-3}\textrm{ mol}\) * \(14.01 \textrm{g/mol}\) = \(0.0189 \textrm{g}\) Now, subtract the total mass of carbon, hydrogen, and nitrogen from the mass of the compound to find the mass of oxygen: Mass of oxygen = mass of the compound - (mass of carbon + mass of hydrogen + mass of nitrogen) Mass of oxygen =\(0.157\textrm{g}\) - (\(0.0581\textrm{g}\) + \(0.00347\textrm{g}\) + \(0.0189\textrm{g}\)) = \(0.0765\textrm{g}\)

Convert the mass of oxygen to moles.

Now, we will convert the mass of oxygen to moles: Moles of oxygen = mass of oxygen / molar mass of oxygen Moles of oxygen = \(0.0765 \textrm{g O}\) / \(16.00 \textrm{g/mol}\) = \(4.78 \times 10^{-3}\) moles of oxygen

Determine the empirical formula.

To determine the empirical formula of the compound, we will divide the moles of each element by the smallest number of moles among them and round the results to the nearest whole numbers. Divide by smallest moles (in this case, moles of nitrogen): Carbon: \(4.84 \times 10^{-3}\) / \(1.35 \times 10^{-3}\) = 3.6 ≈ 4 Hydrogen: \(3.44 \times 10^{-3}\) / \(1.35 \times 10^{-3}\) = 2.5 ≈ 3 Nitrogen: \(1.35 \times 10^{-3}\) / \(1.35 \times 10^{-3}\) = 1 Oxygen: \(4.78 \times 10^{-3}\) / \(1.35 \times 10^{-3}\) = 3.5 ≈ 4 So, the empirical formula of the compound is C4H3N1O4, which can be written as C4H3NO4.

Key concepts

Combustion Analysis

Combustion analysis is a laboratory procedure used to determine the elemental composition of a substance by completely burning the sample in excess oxygen. The products of such complete combustion are typically carbon dioxide (CO2) and water (H2O) for hydrocarbon samples. These products can then be measured to find out how much carbon and hydrogen were in the original compound.

For substances containing nitrogen, like in the example exercise, the nitrogen is usually converted into nitrogen-containing gases such as nitrogen oxides or ammonia (NH3), depending on the conditions of the analysis. By assuming all the carbon ends up in CO2, all the hydrogen in H2O, and all the nitrogen in NH3, we can convert the masses of these products into moles. From these calculations, we can infer the amount of each element present in the original sample.

Stoichiometry

Stoichiometry is a section of chemistry that involves using balanced chemical equations to calculate relative quantities of reactants and products in chemical reactions. It is founded on the law of conservation of mass where the total mass of the reactants equals the total mass of the products in a chemical reaction.

The step-by-step solution from the exercise displays stoichiometry by first converting the masses of CO2, H2O, and NH3 to their respective amounts in moles. Because the coefficients in a balanced chemical equation represent the mole ratios of the substances involved, stoichiometry allows for the direct comparison and ratio determination of the number of moles of each element in the initial compound. This ratio helps us to find the empirical formula of the compound by simplifying the mole ratios to the smallest whole numbers.

Molar Mass

The molar mass is a physical property defined as the mass of a given substance (chemical element or chemical compound) divided by the amount of substance. It is typically expressed in units of grams per mole (g/mol), and it corresponds to the mass of one mole of that substance. The molar mass of a compound can be determined by summing the molar masses of the individual elements found in the compound's formula.

In the given exercise, the molar mass is crucial for converting between the mass of a substance and the number of moles. The calculations in the step-by-step solution show how the mass of CO2, H2O, and NH3 are divided by their respective molar masses to find the number of moles of carbon, hydrogen, nitrogen, and oxygen, which is essential for determining the empirical formula.