Problem 111 - Introductory Chemistry Atoms First

Question

A \(1.25-g\) sample of a gas of unknown identity occupies \(2.50 \mathrm{~L}\). and exerts a pressure of \(0.400 \mathrm{~atm}\) at \(0^{\circ} \mathrm{C}\). (a) What is the molar mass of the gas? (b) Combustion analysis determines that the empirical formula of the gas is \(\mathrm{CH}_{2}\). What is its molecular formula?

Step-by-Step Solution

Answer

(a) The molar mass of the gas is 1.25 g/mol. (b) The molecular formula of the gas is CH2.

Step 1: Write down the Ideal Gas Law equation and all given information

Given: Mass (m) = 1.25 g Volume (V) = 2.50 L Pressure (P) = 0.400 atm Temperature (T) = 0°C = 273.15 K (converted to Kelvin) Ideal Gas Law equation: PV = nRT We will solve for the number of moles (n).

Step 2: Rearrange the Ideal Gas Law equation to find n

The Ideal Gas Law equation can be rearranged as follows: n = PV / RT Now let's plug in the given values: P = 0.400 atm V = 2.50 L R = 0.0821 L·atm/mol·K (Gas constant) T = 273.15 K

Step 3: Calculate the number of moles (n)

n = (0.400 atm)(2.50 L) / (0.0821 L·atm/mol·K)(273.15 K) n = 1.00 mol Now that we have the number of moles, we can calculate the molar mass.

Step 4: Calculate the molar mass (M)

Molar mass (M) is the mass of the sample divided by the number of moles: M = m / n M = 1.25 g / 1.00 mol M = 1.25 g/mol Answer (a): The molar mass of the gas is 1.25 g/mol.

Step 5: Calculate the molar mass of the empirical formula

Given empirical formula: CH2 Carbon: 1 atom x 12.01 g/mol = 12.01 g/mol Hydrogen: 2 atoms x 1.01 g/mol = 2.02 g/mol Total molar mass of the empirical formula = 12.01 g/mol + 2.02 g/mol = 14.03 g/mol

Step 6: Determine the molecular formula

Divide the molar mass of the unknown compound by the molar mass of the empirical formula: 1.25 g/mol / 14.03 g/mol ≈ 0.089 Since the molecular formula is a whole-number multiple of the empirical formula, we can assume this gas is just one unit of the empirical formula (CH2) as 0.089 is very close to 1. Answer (b): The molecular formula of the gas is CH2.

Key Concepts

Molar Mass Calculation

Understanding the molar mass of a substance is a fundamental concept in chemistry. It represents the mass of one mole of a substance and is expressed in grams per mole (g/mol). The molar mass can be calculated by summing the atomic masses of all atoms in a molecule. For example, the molar mass of water, H₂O, is calculated by adding the mass of two hydrogen atoms (1.01 g/mol each) to the mass of one oxygen atom (16.00 g/mol), giving a molar mass of approximately 18.02 g/mol.

In the exercise provided, the molar mass of a gas is determined by using the Ideal Gas Law and the mass of the gas sample. The mass is divided by the number of moles, which were found by reorganizing the Ideal Gas Law formula. This step is crucial because it lays the foundation for identifying the gas's identity through its molecular formula.

Empirical and Molecular Formulas

The empirical formula of a compound gives the simplest whole-number ratio of elements present, while the molecular formula indicates the actual number of atoms of each element in a molecule of the compound. For instance, the empirical formula for glucose is CH₂O, and its molecular formula is C₆H₁₂O₆. This tells us that glucose molecules contain six carbon atoms, twelve hydrogen atoms, and six oxygen atoms.

Within the provided exercise, the empirical formula is given as CH₂, suggesting a one-to-two ratio of carbon to hydrogen atoms. Determining the molecular formula involves comparing the molar mass of the empirical formula to the molar mass of the gas. In cases where the computed ratio approximates to one, as seen in the exercise, the empirical formula and the molecular formula are the same.

Gas Law Constants

When dealing with gas law calculations, understanding constants such as R (the ideal gas constant) is vital. The value of R depends on the units used for pressure, volume, and temperature, and it is imperative to use the correct value to ensure accurate calculations. In most cases, R is taken as 0.0821 L·atm/mol·K when pressure is in atmospheres and volume is in liters.

In the given exercise, the student must use the correct gas constant value to find the number of moles of gas. Using the appropriate constant is an essential practice in solving gas law problems. The consistency of these constants across different scenarios enables chemists to predict how gases will behave under varying conditions, emphasizing the importance of understanding and utilizing these gas law constants correctly.

Stoichiometry

Stoichiometry is the aspect of chemistry that involves calculating the amounts of reactants and products in chemical reactions. It is based on the law of conservation of mass and the concept of the mole. Stoichiometry allows chemists to make predictions about the outcomes of reactions, ensuring that materials are used efficiently without wastage.

In the textbook problem, stoichiometry is applied in reverse to deduce the formula of a gas from its combustion products. The analyst uses the mass and molar relationships to connect the empirical formula with the actual amount of gas present, demonstrating stoichiometry's practical application in real-world problem-solving. Being skilled in stoichiometry is fundamental for students as it is commonly used in areas like medicinal chemistry, environmental science, and materials engineering.