Question

A 1.58-g sample of \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{X}_{3}(g)\) has a volume of \(297 \mathrm{~mL}\) at \(769 \mathrm{~mm} \mathrm{Hg}\) and \(35^{\circ} \mathrm{C}\). Identify the element \(\mathrm{X}\).

Short answer

Answer: Element X is chlorine (Cl).

Step-by-step solution

Convert given units to SI units

First, we need to convert the given values to the SI units. Temperature should be in Kelvin (K), pressure should be in atmospheres (atm), and volume should be in liters (L). Temperature: \(35^{\circ}\mathrm{C} = 35 + 273.15 = 308.15\,\mathrm{K}\) Pressure: \(769\,\mathrm{mm}\,\mathrm{Hg} = \dfrac{769}{760} = 1.012\,\mathrm{atm}\) Volume: \(297\,\mathrm{mL} = \dfrac{297}{1000} = 0.297\,\mathrm{L}\)

Apply Ideal Gas Law

Now, using the ideal gas law, \(PV = nRT\), we can find the number of moles (n) of the substance. Given the values of pressure (P), temperature (T), and volume (V), we can calculate n. R: Gas constant = \(0.0821\,\mathrm{L}\,\mathrm{atm}\, (\mathrm{mol}\,\mathrm{K})^{-1} \) Solving for n: \(n = \dfrac{PV}{RT} = \dfrac{(1.012\,\mathrm{atm})(0.297\,\mathrm{L})}{(0.0821\,\mathrm{L}\,\mathrm{atm}\, (\mathrm{mol}\,\mathrm{K})^{-1})(308.15\,\mathrm{K})}\)

Calculate the number of moles

Calculate the value for n obtained in step 2: \(n = \dfrac{(1.012)(0.297)}{(0.0821)(308.15)} = 0.01245\,\mathrm{mol}\)

Calculate the molar mass

Using the mass of the sample and the number of moles calculated in step 3, determine the molar mass, M: M = \(\dfrac{\text{mass}}{\text{moles}} = \dfrac{1.58\,\mathrm{g}}{0.01245\,\mathrm{mol}} = 126.9\,\mathrm{g}\,\mathrm{mol}^{-1}\)

Determine the compound's empirical formula

Knowing that the empirical formula for the gas is \(\mathrm{C}_{2}\mathrm{H}_{3}\mathrm{X}_{3}\), and using the molar masses for carbon and hydrogen, calculate the molar mass for X. Molar mass of Carbon: \(12.01\,\mathrm{g}\,\mathrm{mol}^{-1}\) Molar mass of Hydrogen: \(1.01\,\mathrm{g}\,\mathrm{mol}^{-1}\) Sum of molar masses for \(\mathrm{C}_{2}\mathrm{H}_{3}\mathrm{X}_{3}\): \((2)(12.01) + (3)(1.01) + 3(\text{Molar mass of X}) = 126.9\)

Solve for the molar mass of X

Rearrange the formula in step 5 to solve for the molar mass of X: Molar mass of X \(= \dfrac{126.9 - (2)(12.01) - (3)(1.01)}{3} = 35\,\mathrm{g}\,\mathrm{mol}^{-1}\)

Identify element X

Now that we have the molar mass of element X, we can look for a corresponding element in the periodic table. The element with a molar mass close to 35 is chlorine (Cl) with a molar mass of \(35.45\,\mathrm{g}\,\mathrm{mol}^{-1}\). Therefore, element X is chlorine, and the gas compound is \(\mathrm{C}_{2}\mathrm{H}_{3}\mathrm{Cl}_{3}\) or trichloroethylene.

Key concepts

Stoichiometry in Chemistry

Understanding stoichiometry is fundamental for any chemistry student. It involves the calculation of reactants and products in chemical reactions. Stoichiometry is rooted in the conservation of mass where the number of atoms of each element must remain constant throughout a chemical reaction. This balance is achieved by adjusting the ratio of molecules (the coefficient in a chemical equation), to ensure that the same number of atoms is present on both sides.

When we work with stoichiometry, we often deal with the concept of 'mole'. One mole of any substance contains the same number of entities (atoms, molecules, ions, etc.) as there are atoms in exactly 12 grams of carbon-12, which is approximately \(6.022 \times 10^{23}\) entities (Avogadro's number). Thus, stoichiometry allows us to predict the outcomes of reactions, identify limiting reactants, calculate yields, and much more, by using this mole concept as the bridge between the macroscopic and microscopic worlds of chemistry.

Molar Mass Calculation

The molar mass is a crucial concept in stoichiometry as it connects the mass of a substance to the amount in moles. It is defined as the mass of one mole of a substance, typically expressed in grams per mole (g/mol). This value is particularly important for converting mass to moles and vice versa, which is a common practice in stoichiometric calculations.

To determine the molar mass of a compound, we sum up the molar masses of all the individual atoms in the formula. For example, in the case of water (\(H_2O\)), the molar mass would be calculated by adding the molar masses of hydrogen (\(1.01\) g/mol, and since there are two, it's \(2 \times 1.01\)) and oxygen (\(16.00\) g/mol), yielding a molar mass of about \(18.02\) g/mol for water.

Empirical Formula Identification

The empirical formula represents the simplest whole-number ratio of elements in a compound. It does not necessarily represent the actual numbers of atoms in a molecule, which is conveyed by the molecular formula, but it provides the proportional amounts of each type of atom.

Identifying an empirical formula often involves measuring the mass of each element in a sample, converting these masses to moles, and then finding the simplest ratio between these amounts. For instance, if we have a compound composed of carbon and oxygen and we measure 12 grams of carbon (equivalent to 1 mole of carbon) and 32 grams of oxygen (equivalent to 2 moles of oxygen), the empirical formula would be \(CO_2\).

In the study case provided, the process of determining the empirical formula also informs us of the identity of the unknown element by comparing the calculated molar mass with the standard values in a periodic table. This is an application of empirical formula identification that not only tells us the simplest ratio but also helps in solving a real-world chemistry puzzle.