Question

An organic compound was found to contain only C, \(\mathrm{H}\), and \(\mathrm{Cl}\). When a \(1.50-\mathrm{g}\) sample of the compound was completely combusted in air, \(3.52 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) was formed. In a separate experiment the chlorine in a 1.00-g sample of the compound was converted to \(1.27 \mathrm{~g}\) of \(\mathrm{AgCl}\). Determine the empirical formula of the compound.

Short answer

The empirical formula of the compound is \(C_9H_{25}Cl\).

Step-by-step solution

Determine the moles of C in the compound

We are given the mass of CO₂ formed when the compound is combusted in air as 3.52 g. We will first calculate the moles of C in the compound using the molar mass of CO₂. Moles of CO₂: \[\text{Moles of } CO_2 = \frac{3.52 \text{ g}}{44.01 \text{ g/mol}} = 0.0800 \text{ mol}\] Since there is one C atom in each CO₂ molecule, the moles of C in the compound are the same as the moles of CO₂. Moles of C: \(0.0800 \ \text{mol}\)

Determine the moles of Cl in the compound

We are given the mass of AgCl formed when the Chlorine in the compound is reacted as 1.27 g. We will calculate the moles of Cl in the compound using the molar mass of AgCl. Moles of AgCl: \[\text{Moles of } AgCl = \frac{1.27 \text{ g}}{143.32 \text{ g/mol}} = 0.00886 \text{ mol}\] Since there is one Cl atom in each AgCl molecule, the moles of Cl in the compound are the same as the moles of AgCl. Moles of Cl: \(0.00886 \ \text{mol}\)

Determine the mass of H in the compound

Since we know the total mass of the compound, the mass of H can be calculated by subtracting the masses of C and Cl from the total mass. Mass of C: \(0.0800 \ \text{mol} * 12.01 \ \text{g/mol} = 0.960 \ \text{g}\) Mass of Cl: \(0.00886 \ \text{mol} * 35.45 \ \text{g/mol} = 0.314 \ \text{g}\) Mass of H: \[1.50 \ \text{g} - 0.960 \ \text{g} - 0.314 \ \text{g} = 0.226 \ \text{g}\]

Determine the moles of H in the compound

Now we will calculate the moles of H in the compound using the mass of H and molar mass of H. Moles of H: \[\text{Moles of } H = \frac{0.226 \ \text{g}}{1.008 \ \text{g/mol}} = 0.224 \ \text{mol}\]

Find the empirical formula of the compound

To find the empirical formula, we'll first find the mole ratio of C, H, and Cl by dividing each of their moles by the smallest number of moles among them. Mole ratio of C: \(\frac{0.0800}{0.00886} \approx 9\) Mole ratio of H: \(\frac{0.224}{0.00886} \approx 25\) Mole ratio of Cl: \(\frac{0.00886}{0.00886} = 1\) Hence, the empirical formula of the compound is \(C_9H_{25}Cl\).

Key concepts

Combustion Analysis

Combustion analysis is a laboratory method commonly used to determine the elemental composition of a compound, particularly organic compounds containing carbon and hydrogen. During combustion, the substance is burned in excess oxygen to yield carbon dioxide and water as the primary products for carbon and hydrogen, respectively.

By measuring the mass of CO₂ and H₂O produced, the amounts of carbon and hydrogen in the original sample can be calculated. This method is especially useful for determining empirical formulas of unknown compounds. In the context of our exercise, combustion analysis enabled us to find the amount of carbon by converting it to CO₂ and measuring its mass.

Molar Mass Calculation

The molar mass of a substance is the mass of one mole of its particles, typically expressed in grams per mole (g/mol). It's a fundamental concept in chemistry, as it allows the conversion between the mass of a substance and the number of moles it contains.

Molar mass is especially critical when determining the empirical formula of a compound. By knowing the molar masses of CO₂ and AgCl, we were able to convert the measured masses of these compounds to moles, which then facilitated the calculation of the moles of C and Cl in the original compound.

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. In essence, it involves the use of balanced chemical equations to calculate the masses, moles, and percentages of compounds involved in chemical reactions.

In our exercise, stoichiometry is applied to find the ratio of each element within the original compound. By using the mole-to-mole conversion between the compound and the products of its combustion (CO₂ and AgCl), we determined the moles of carbon and chlorine. This step is crucial for deriving the mole ratio of elements and thereby determining the empirical formula of the compound.

Mole-to-Mass Conversion

Mole-to-mass conversion is a key skill in chemistry, involving the conversion of the number of moles of a substance to its mass, using the substance's molar mass. This process is pivotal when analyzing the composition of a compound, as it allows chemists to move from the abstract mole concept to concrete mass measurements that can be taken in a laboratory.

In our problem, we used mole-to-mass conversion to calculate the mass of carbon and chlorine within the sample by multiplying their mole quantities by their respective atomic masses. This is a vital step for determining how much hydrogen is present, which in turn is required for the final empirical formula calculation.