Question

A compound contains only carbon, hydrogen, nitrogen, and oxygen. Combustion of \(0.157 \mathrm{~g}\) of the compound produced \(0.213 \mathrm{~g}\) \(\mathrm{CO}\), and \(0.0310 \mathrm{~g} \mathrm{H}_{2} \mathrm{O} .\) In another experiment, it is found that \(0.103 \mathrm{~g}\) 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 \(C_7H_5N_2O_7\).

Step-by-step solution

Calculate moles of Carbon based on CO2 produced

First, we need to find the moles of CO2 produced. Given 0.213 g of CO2 was produced, we can use the molar mass of CO2 (44.01 g/mol) to find the moles of CO2: moles of CO2 = 0.213 g / 44.01 g/mol = 0.00484 mol Next, knowing that one mole of CO2 contains one mole of Carbon, we conclude that the compound has 0.00484 mol of Carbon.

Calculate moles of Hydrogen based on H2O produced

Similar to the above step, we need to find the moles of H2O produced. Given 0.0310 g of H2O was produced, we can use the molar mass of H2O (18.02 g/mol) to find the moles of H2O: moles of H2O = 0.0310 g / 18.02 g/mol = 0.00172 mol Knowing that one mole of H2O contains two moles of Hydrogen, we conclude that the compound has 0.00172 * 2 = 0.00344 mol of Hydrogen.

Calculate moles of Nitrogen based on NH3 produced

Using the data from the second experiment, we will find the moles of NH3 produced. Given 0.0230 g of NH3 was produced, we can use the molar mass of NH3 (17.03 g/mol) to find the moles of NH3: moles of NH3 = 0.0230 g / 17.03 g/mol = 0.00135 mol Knowing that one mole of NH3 contains one mole of Nitrogen, we conclude that the compound has 0.00135 mol of Nitrogen.

Calculate moles of Oxygen

Since we know the mass of the compound (0.157 g) and the moles of Carbon, Hydrogen, and Nitrogen, we can find the moles of Oxygen by difference. First, we find the mass of Carbon, Hydrogen, and Nitrogen by multiplying their moles with their respective molar masses: mass of Carbon = 0.00484 mol * 12.01 g/mol = 0.0581 g mass of Hydrogen = 0.00344 mol * 1.008 g/mol = 0.00347 g mass of Nitrogen = 0.00135 mol * 14.01 g/mol = 0.0189 g Total mass of Carbon, Hydrogen, and Nitrogen = 0.0581 g + 0.00347 g + 0.0189 g = 0.0804 g Now we can find the mass of Oxygen: mass of Oxygen = mass of compound - total mass of Carbon, Hydrogen, and Nitrogen = 0.157 g - 0.0804 g = 0.0766 g Finally, moles of Oxygen can be calculated using the molar mass of Oxygen (16.00 g/mol): moles of Oxygen = 0.0766 g / 16.00 g/mol = 0.00479 mol

Find the simplest whole number ratio

We need to find the simplest whole number ratio for the number of each atom. First, we will find the least moles among the four elements in order to do so: Least moles = min(0.00484, 0.00344, 0.00135, 0.00479) = 0.00135 Now, we divide the moles of each element by least moles: C : 0.00484 / 0.00135 ≈ 3.6 H : 0.00344 / 0.00135 ≈ 2.5 N : 0.00135 / 0.00135 ≈ 1.0 O : 0.00479 / 0.00135 ≈ 3.6 To obtain whole number ratios, we will multiply these numbers by 2: C : 3.6 * 2 = 7.2 ≈ 7 H : 2.5 * 2 = 5.0 ≈ 5 N : 1.0 * 2 = 2.0 ≈ 2 O : 3.6 * 2 = 7.2 ≈ 7

Find the empirical formula

Now that we have the whole number ratios of each element, we can write the empirical formula of the compound: Empirical formula = C7H5N2O7

Key concepts

Stoichiometry

Stoichiometry is like the recipe for a chemical reaction. It tells us how to combine substances in the right amounts to get the desired products. In the context of empirical formula determination, stoichiometry helps us understand the quantitative relationship between the reactants used in a combustion analysis (like our compound containing C, H, N, and O) and the products formed (like CO₂ and H₂O).

Stoichiometry is grounded on the law of conservation of mass, which states that mass can neither be created nor destroyed in a chemical reaction. Therefore, the mass of the reactants must equal the mass of the products. Stoichiometric calculations often involve converting between mass, moles, and number of particles using the molar mass as a conversion factor.

To make stoichiometric calculations more intuitive, you can think of it like baking cookies: you need a specific number of cups of flour, sugar, and eggs. Similarly, a chemical reaction needs a specific number of moles of reactants to produce a certain number of moles of products.

Mole Concept

The mole concept is a bridge between the microscopic world of atoms and the macroscopic world we experience. One mole is 6.022 x 10²³ of something, whether it's atoms, molecules, or another particle. This number is also known as Avogadro's number.

Understanding the mole concept enables us to count particles by weighing them. For instance, in the given exercise, the number of moles of CO₂ and H₂O produced by combustion tells us how many moles of carbon and hydrogen were in the original compound. This is essential in determining the empirical formula because it provides a direct count of atoms based on their weights—something that's crucial when we're dealing with samples too small to see or count directly.

One effective study tip is to continuously practice converting between mass, moles, and number of particles, as this reinforces the mole concept and helps prevent mistakes in stoichiometry.

Combustion Analysis

Combustion analysis is a laboratory method used to determine the elemental composition of a compound. It involves burning a known mass of a substance and measuring the masses of products like carbon dioxide and water. From these measurements, we calculate the amount of each element in the original compound.

In this problem, combustion analysis is critical for finding the empirical formula. By assuming all carbon ends up in CO₂, all hydrogen ends up in H₂O, and all nitrogen ends up in NH₃, we're able to accurately determine the moles of each within the compound. Remember, these assumptions are standard in combustion analysis, as they simplify calculations and are based on how these elements typically react when burned in the presence of oxygen.

An additional tip here is to ensure complete combustion in a practical setting by providing excess oxygen. This ensures all combustible elements are fully converted to their respective oxides for accurate measurements.

Chemical Formulas

Chemical formulas are shorthand to express the composition of compounds. The empirical formula provides the simplest, whole-number ratio of atoms in a compound, while the molecular formula gives the actual number of atoms of each element in a molecule of the compound.

In the provided exercise, we determined the empirical formula C₇H₅N₂O₇ by deriving the simplest whole number ratio of atoms present. It's essential to realize that the empirical formula may not reflect the true number of atoms in the molecule but gives us valuable information on the ratio of the elements.

When working out these problems, it can be helpful to understand that the empirical formula is the reduced form of the molecular formula. To troubleshoot any issues while finding empirical formulas, ensure that you have correctly balanced your units of moles and double-checked your calculations for rounding errors. Moreover, confirming that the sum of the mass percentages equals approximately 100% can verify the accuracy of your determined formula.