Question

A compound made up of \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{Cl}\) contains 55.0 percent \(\mathrm{Cl}\) by mass. If \(9.00 \mathrm{~g}\) of the compound contain \(4.19 \times 10^{23} \mathrm{H}\) atoms, what is the empirical formula of the comnound?

Short answer

The empirical formula is \(\mathrm{C}_2\mathrm{H}_5\mathrm{Cl}\).

Step-by-step solution

Determine Mass of Chlorine

The compound contains 55.0% chlorine by mass. For a 9.00 g sample, the mass of chlorine is calculated as follows:\[ \text{Mass of Cl} = 55.0\% \times 9.00\,\text{g} = 4.95\,\text{g} \]

Determine Mass of Carbon and Hydrogen

Subtract the mass of chlorine from the total mass of the compound to find the combined mass of carbon and hydrogen:\[ \text{Mass of C + H} = 9.00\,\text{g} - 4.95\,\text{g} = 4.05\,\text{g} \]

Determine Moles of Chlorine

Calculate the number of moles of chlorine using its molar mass (approximately 35.45 g/mol):\[ \text{Moles of Cl} = \frac{4.95\,\text{g}}{35.45\,\text{g/mol}} \approx 0.1397\,\text{mol} \]

Determine Moles of Hydrogen

The number of hydrogen atoms is \(4.19 \times 10^{23}\). Since there are \(6.022 \times 10^{23}\) atoms in a mole (Avogadro's number):\[ \text{Moles of H} = \frac{4.19 \times 10^{23}}{6.022 \times 10^{23}} \approx 0.6959\,\text{mol} \]

Determine Mass and Moles of Carbon

Subtract the mass of hydrogen (using its moles) from the mass of C + H to calculate the mass of carbon. Assume the mass of hydrogen is:\[ \text{Mass of H} = 0.6959\,\text{mol} \times 1.008\,\text{g/mol} = 0.7018\,\text{g} \]\[ \text{Mass of C} = 4.05\,\text{g} - 0.7018\,\text{g} = 3.3482\,\text{g} \]Calculate the moles of carbon using its molar mass (approximately 12.01 g/mol):\[ \text{Moles of C} = \frac{3.3482\,\text{g}}{12.01\,\text{g/mol}} \approx 0.2789\,\text{mol} \]

Determine Empirical Formula

Calculate the simplest mole ratio by dividing each by the smallest value:- C: \( \frac{0.2789}{0.1397} \approx 2 \)- H: \( \frac{0.6959}{0.1397} \approx 5 \)- Cl: \( \frac{0.1397}{0.1397} = 1 \)The empirical formula is \( \mathrm{C}_2\mathrm{H}_5\mathrm{Cl} \).

Key concepts

Compound Composition

Understanding the composition of a compound is essential in chemistry. A compound is made up of two or more elements combined in definite proportions. For the given compound containing carbon (C), hydrogen (H), and chlorine (Cl), each element contributes to the overall mass of the compound based on its percentage by mass.
In the exercise, we know that chlorine (Cl) takes up 55% of the compound by mass. This means, in a sample weighing 9.00 grams, chlorine weighs 4.95 grams. The rest of the mass, comprising carbon and hydrogen, needs to be calculated by subtracting the chlorine mass from the total mass of the compound. They add up to 4.05 grams in this scenario.
By understanding each element's contribution, one can determine the proportional relationship among the elements, which is crucial for finding the empirical formula.

Moles Calculation

Moles are a fundamental concept in chemistry, representing a specific quantity of particles, such as atoms or molecules. It is calculated using the substance's molar mass, which links the mass to the number of moles.
In this problem, to find the moles of chlorine, we used its molar mass (approximately 35.45 g/mol). By dividing the mass of chlorine (4.95 g) by its molar mass, we find there are approximately 0.1397 moles of chlorine present.
This same technique is applied to hydrogen and carbon. By calculating how many moles of each element are in the compound, we gather valuable data needed to identify the empirical formula. This formula is a simplified representation showing the ratios of the constituent elements in the smallest whole number form.

Chlorine Mass Percent

Mass percent indicates the proportion of an element in a compound relative to its entire mass, expressed in percentage form. For chlorine, given as 55%, it plays a critical role in determining its amount in any given sample.
This percentage is useful when calculating the chlorine mass from a known compound mass. In the example, knowing 55% of the compound is chlorine, multiplying this percentage by the total mass gives us the actual mass amount of chlorine (4.95 g in this case).
The mass percent can be a key factor in diverse applications, including material formulation, health risk assessments, and quality control in manufacturing processes. By manipulating mass percentages, scientists can predict and control the properties and reactivity of compounds.

Avogadro's Number

Avogadro's Number is a cornerstone of chemistry, defining how many atoms or molecules are present in a mole of any substance. It is numerically equal to approximately \(6.022 \times 10^{23}\). This large constant allows chemists to work with manageable quantities at the macroscopic level when they are dealing with the microscopic scale of atoms or molecules.
In solving the problem, Avogadro's number was used to calculate moles of hydrogen from a given number of hydrogen atoms (\(4.19 \times 10^{23}\)). By dividing the number of atoms by Avogadro's Number, we approximated that there are 0.6959 moles of hydrogen in the compound.
This calculation is crucial in providing a bridge between the microscopic world of atoms and the macroscopic measurements we make in laboratories, enabling accurate reactions and formulations.