Question

Arsenic(III) sulfide sublimes readily, even below its melting point of \(320^{\circ} \mathrm{C}\) . The molecules of the vapor phase are found to effuse through a tiny hole at 0.28 times the rate of effusion of Ar atoms under the same conditions of temperature and pressure. What is the molecular formula of arsenic (III) sulfide in the gas phase?

Short answer

The molecular formula of arsenic (III) sulfide in the gas phase is As2S3.

Step-by-step solution

Write Graham's law of effusion formula

Graham's law of effusion states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Mathematically, it can be expressed as: \( \frac{Rate_1}{Rate_2} = \sqrt{\frac{M_2}{M_1}} \) Here, Rate1 and Rate2 are the rates of effusion for gas1 and gas2, respectively. M1 and M2 are their respective molar masses.

Substitute the given values into the formula

We are given that the rate of effusion of arsenic (III) sulfide (AS) to that of the argon (Ar) is 0.28 (or 28%). The molar mass of argon gas is 39.95 g/mol. We can rearrange the formula to find the molar mass of arsenic (III) sulfide. \( \frac{Rate_{AS}}{Rate_{Ar}} = 0.28 = \sqrt{\frac{M_{Ar}}{M_{AS}}} \)

Calculate the molar mass of arsenic (III) sulfide

Now, we can calculate the molar mass of arsenic (III) sulfide by squaring both sides of the equation and rearranging it: \( M_{AS} = \frac{M_{Ar}}{0.28^2} \) \( M_{AS} = \frac{39.95\,g/mol}{0.0784} \) \( M_{AS} ≈ 509.49\,g/mol \)

Determine the molecular formula of arsenic (III) sulfide

Arsenic has a molar mass of approximately 74.92 g/mol and sulfur has a molar mass of approximately 32.07 g/mol. Let's assume the molecular formula to be AsxSy, where x and y are the numbers of arsenic and sulfur atoms, respectively. We can express the molar mass of arsenic (III) sulfide as: \( 509.49 \approx 74.92x + 32.07y \) Since this is arsenic (III) sulfide, the oxidation state of arsenic is +3 and each sulfur atom has an oxidation state of -2. This means that for each arsenic atom, there will be 3/2 sulfur atoms in the formula. Therefore, we get: \( y = \frac{3}{2} x \) Now we can substitute this expression into the molar mass equation: \( 509.49 \approx 74.92x + 32.07\left(\frac{3}{2}x\right) \) Solving for x, we find: \( x \approx 2 \) Since y = 3/2x: \( y \approx 3 \)

Write the molecular formula

Based on our calculations, the molecular formula of arsenic (III) sulfide in the gas phase is: As2S3

Key concepts

Graham's Law of Effusion

Graham's Law of Effusion is a crucial concept in chemistry that helps us understand how gas molecules move through small openings. This law states that the rate of effusion of gas molecules is inversely proportional to the square root of their molar masses. In simpler terms, lighter gases effuse faster than heavier ones. For example, if you have a balloon filled with hydrogen ( ext{H}_2) and another with carbon dioxide ( ext{CO}_2), the hydrogen will effuse faster through a hole due to its lower molecular mass. The formula for Graham's Law of Effusion is as follows:\[ \frac{Rate_1}{Rate_2} = \sqrt{\frac{M_2}{M_1}} \]Here, \(Rate_1\) and \(Rate_2\) are the rates of effusion for two different gases, and \(M_1\) and \(M_2\) are their respective molar masses. To practically apply this law, simply substitute the known values and manipulate the equation to solve for the unknown. This principle is valuable in identifying the molecular formulas of unknown gases by comparing their effusion rates to known gases.

Molar Mass Calculation

Calculating the molar mass of a compound is fundamental in chemistry, as it enables us to determine the formula mass of the compound's molecular units. The molar mass is essentially the mass of one mole of a substance, measured in grams per mole (g/mol). For example, the molar mass of carbon ( ext{C}) is \(12.01 \text{ g/mol}\), while that of oxygen ( ext{O}) is \(16.00 \text{ g/mol}\). To determine the molar mass of a compound, you add up the molar masses of all the atoms present in a single molecule of the compound.The calculation follows this structure:- Identify all the elements in the compound.- Determine the number of each type of atom in the formula.- Use the molar masses of these elements to compute the total molar mass.For example, in the step-by-step solution, the exercise used the known effusion rates to deduce the molar mass of arsenic (III) sulfide in gas form. Using the given rate comparisons and the molar mass of argon, the calculation provided a clear path to estimate the molar mass of unknown substances.

Chemical Oxidation States

Oxidation states are a key concept for determining how atoms interact in a molecule. They offer insights into the electron distribution among the atoms in a chemical compound. The oxidation state, often referred to as oxidation number, describes the total number of electrons that an atom either gains or loses to form a chemical bond.For example, in arsenic (III) sulfide, arsenic is shown to have an oxidation state of +3, while sulfur has an oxidation state of -2. This guides us in balancing the compound's stoichiometry and aids in formulating the correct chemical formula. The rules for determining oxidation states are straightforward:- The oxidation state of an element in its natural state is 0.- For ions, the oxidation state is equal to the charge of the ion.- Oxygen usually has an oxidation state of -2, and hydrogen typically is +1, except when bonded differently (e.g., in peroxides or hydrides).These rules allow us to set up equations to find unknown quantities in chemical formulas. Applying these principles to the arsenic (III) sulfide problem helped identify the correct ratios for As and S atoms, resulting in the derivation of the molecule's formula as As\(_2\)S\(_3\). Understanding oxidation states is integral to mastering the creation and interpretation of chemical formulas.