Question

A compound contains only \(\mathrm{C}, \mathrm{H}\), and \(\mathrm{N}\). Combustion of \(35.0 \mathrm{mg}\) of the compound produces \(33.5 \mathrm{mg} \mathrm{CO}_{2}\) and \(41.1 \mathrm{mg} \mathrm{H}_{2} \mathrm{O}\). What is the empirical formula of the compound?

Short answer

The empirical formula of the compound is CH₆N₂.

Step-by-step solution

Find moles of Carbon in CO₂

Given that the mass of CO₂ produced is 33.5 mg, we can calculate the moles of carbon by using the molar mass of CO₂. Molar mass of CO₂ = \( 12.01 g\; C + 2 * 16.00 g\; O = 44.01\:g/mol\) Now, convert mg to g: \(33.5\:mg\:CO_{2} * \frac{1\:g}{1000\:mg} = 0.0335\:g\:CO_{2}\) To find the moles of Carbon in CO₂, divide the mass of CO₂ by its molar mass: \(\frac{0.0335\:g\:CO_{2}}{44.01\:g/mol} = 7.61\times10^{-4}\:mol\:CO_{2}\) Since there is one mole of Carbon in one mole of CO₂, the moles of carbon = moles of CO₂ Moles of Carbon = \(7.61\times10^{-4}\:mol\)

Find moles of Hydrogen in H₂O

Similarly, we can find the moles of hydrogen using the mass of H₂O produced (41.1 mg) and the molar mass of water. Molar mass of H₂O = \(2 * 1.01 g\; H + 16.00 g\; O = 18.02\:g/mol\) Convert mg to g: \(41.1\:mg\:H_{2}O * \frac{1\:g}{1000\:mg} = 0.0411\:g\:H_{2}O\) Now, find the moles of H₂O: \(\frac{0.0411\:g\:H_{2}O}{18.02\:g/mol} = 2.28\times10^{-3}\:mol\:H_{2}O\) Since there are 2 moles of Hydrogen in one mole of H₂O, the moles of hydrogen = 2 * moles of H₂O Moles of Hydrogen = \(2 * 2.28\times10^{-3}\:mol = 4.56\times10^{-3}\:mol\)

Calculate moles of Nitrogen

First, we need to find the mass of Carbon and Hydrogen in the compound using the moles calculated above and their molar masses. Mass of Carbon = Moles of Carbon * Molar mass of Carbon Mass of Carbon = \(7.61\times10^{-4}\:mol\; C * 12.01\:g/mol = 0.00913\:g\) Mass of Hydrogen = Moles of Hydrogen * Molar mass of Hydrogen Mass of Hydrogen = \(4.56\times10^{-3}\:mol\; H * 1.01\:g/mol = 0.00460\:g\) Now, subtract the mass of Carbon and Hydrogen from the total mass of the compound to get the mass of Nitrogen. Mass of Nitrogen = Total mass - (Mass of Carbon + Mass of Hydrogen) Mass of Nitrogen = \(0.035\:g - (0.00913\:g + 0.00460\:g) = 0.0213\:g\) Now, divide the mass of Nitrogen by its molar mass to find the moles of Nitrogen. Molar mass of Nitrogen = 14.01 g/mol Moles of Nitrogen = \( \frac{0.0213\:g\:N}{14.01\:g/mol} = 1.52\times10^{-3}\:mol\)

Find the simplest whole number ratio of elements

Now that we have the moles of Carbon, Hydrogen, and Nitrogen, we can find the empirical formula by calculating the simplest whole number ratio of elements. Divide the moles of each element by the smallest mole value to find the ratio: C: \(\frac{7.61\times10^{-4}\:mol}{7.61\times10^{-4}\:mol} = 1\) H: \(\frac{4.56\times10^{-3}\:mol}{7.61\times10^{-4}\:mol} \approx 6\) N: \(\frac{1.52\times10^{-3}\:mol}{7.61\times10^{-4}\:mol} \approx 2\) The empirical formula is hence CH₆N₂.

Key concepts

Combustion Analysis

Combustion analysis is a laboratory technique utilized to determine the elemental composition of a compound, particularly organic compounds containing carbon and hydrogen. During combustion, the compound reacts with oxygen to produce carbon dioxide (CO₂) and water (H₂O), with all carbon being converted to CO₂ and all hydrogen to H₂O. By measuring the masses of CO₂ and H₂O produced, we can calculate the amount of carbon and hydrogen in the original compound.

For example, when a sample of a compound is burned, and the resultant CO₂ and H₂O are collected, the data can be converted into moles using the molar masses of CO₂ and H₂O, providing insights into the mole ratio of carbon and hydrogen in the compound. From these moles, scientists can determine the part of the mass due to carbon and hydrogen, hence deducing the compound's empirical formula. This method is particularly valuable when unknown compounds need to be analyzed, offering a pathway to understanding their fundamental structures.

Mole Ratio

The concept of mole ratio is pivotal in deciphering the empirical formula of a compound from combustion analysis data. A mole ratio compares the number of moles of one substance to the number of moles of another substance in a chemical reaction. In determining the empirical formula, the mole ratio helps to find the simplest ratio of the elements in the compound.

This ratio is obtained by dividing the number of moles of each element found in the compound by the smallest number of moles of any of the elements present. If the ratios are not whole numbers, they may be multiplied by a common factor to yield the smallest whole numbers, which are then used to write the empirical formula of the compound. The mole ratio is instrumental as it reflects the stoichiometry of the elements in the compound—the underlying proportions that dictate its basic structure.

Molecular Mass Calculation

Molecular mass calculation is an essential step in quantitative chemical analysis, such as determining a compound's empirical formula from combustion analysis data. The molecular mass, or molecular weight, is the sum of the atomic masses of all atoms in a molecule and is traditionally expressed in atomic mass units (u) or grams per mole (g/mol).

In the context of finding the empirical formula, we calculate the mass of each element in the sample by multiplying the moles of each element by its atomic mass. This calculation tells us the fraction of the compound's total mass due to each component element. Understanding the molecular mass is not only critical for empirical formula determination but also has broader applications in calculating the quantities of reactants and products in chemical reactions, as well as in the preparation of solution concentrations in molarity terms. The ability to calculate molecular mass is thus a foundational tool across various fields of chemistry.