Problem 58 - Introductory Chemistry Atoms First

Question

A 1.56-g sample of gas is contained in a \(250.0-\mathrm{mL}\) cylinder. Its pressure is \(1255.6 \mathrm{~mm} \mathrm{Hg}\), and its temperature is \(22.7{ }^{\circ} \mathrm{C}\). (a) What is the molar mass of the gas? (b) Combustion analysis reveals the empirical formula of this gas to be \(\mathrm{NO}_{2}\). What is the molecular formula?

Step-by-Step Solution

Answer

The molar mass of the gas is approximately \(44.44 \, \mathrm{\frac{g}{mol}}\), and the molecular formula of the gas is \(\mathrm{NO}_{2}\).

Step 1: Convert to SI units

First, convert the given pressure and volume to SI units: Pressure: 1 atm = 760 mm Hg, so \(1255.6 \, \mathrm{mm} \, \mathrm{Hg} \times \frac{1\, \mathrm{atm}}{760\, \mathrm{mm} \, \mathrm{Hg}} = 1.65 \, \mathrm{atm}\) Volume: \(250.0 \, \mathrm{mL} \times \frac{1\, \mathrm{L}}{1000\, \mathrm{mL}} = 0.250 \, \mathrm{L}\) Temperature: \(T_{K} = 22.7^\circ \mathrm{C} + 273.15 = 295.85 \, \mathrm{K}\)

Step 2: Calculate the number of moles

Use the ideal gas law equation to find the number of moles (n): \(PV = nRT \Rightarrow n = \frac{PV}{RT}\) Substituting the values into the equation: \(n = \frac{(1.65 \, \mathrm{atm})(0.250 \, \mathrm{L})}{(0.0821 \, \mathrm{\frac{L \cdot atm}{mol \cdot K}})(295.85 \, \mathrm{K})} \approx 0.0351 \, \mathrm{mol}\)

Step 3: Calculate molar mass

To get the molar mass, divide the given mass of the gas by the number of moles calculated above: Molar mass = \(\frac{1.56 \, \mathrm{g}}{0.0351 \, \mathrm{mol}} \approx 44.44 \, \mathrm{\frac{g}{mol}}\)

Step 4: Determine molecular formula

We are given that the empirical formula of the gas is \(\mathrm{NO}_{2}\), with a molar mass of 30.01 g/mol (14.01 g/mol for N and 16.00 g/mol for each O). Now divide the molar mass of the gas by the molar mass of its empirical formula, and round to the nearest whole number: \(\frac{44.44 \, \mathrm{\frac{g}{mol}}}{30.01 \, \mathrm{\frac{g}{mol}}} \approx 1.48 \approx 1\) Since the result is 1, the molecular formula of the gas is the same as its empirical formula: \(\mathrm{NO}_{2}\).

Key Concepts

The Ideal Gas Law and Molar Mass Calculation

The calculation of molar mass is a fundamental skill in chemistry, and the ideal gas law provides a reliable tool for this when we're dealing with gases. Let's recall the ideal gas law: it's given by the equation \(PV = nRT\), where \(P\) stands for pressure, \(V\) for volume, \(n\) for the number of moles, \(R\) for the universal gas constant, and \(T\) for temperature in Kelvin.

To determine the molar mass of a gas, we first need to convert given units to the International System of Units (SI). Pressure is converted to atmospheres (atm), volume to liters (L), and temperature to Kelvin (K). With these conversions, we then rearrange the ideal gas law to solve for the moles of the gas. Once we have the number of moles, the molar mass can be calculated by dividing the mass of the gas sample by the moles obtained from the equation.

Empirical vs. Molecular Formulas

Understanding the distinction between empirical and molecular formulas is vital for properly characterizing a compound. An empirical formula represents the simplest whole-number ratio of the elements within a compound, whereas a molecular formula shows the exact number of atoms of each element present in a molecule of the compound.

For example, if combustion analysis reveals the empirical formula of a gas to be \(NO_2\), this is the simplest ratio of nitrogen to oxygen in the compound. Determining the molecular formula involves comparing the molar mass of the compound to the molar mass of the empirical formula. \(NO_2\) has a molar mass of approximately 30.01 grams per mole. If the actual molar mass of the gas is 44.44 grams per mole, we'd expect a whole number ratio when we divide the molar mass of the gas by that of the empirical formula; this ratio reveals if the molecular formula contains multiple units of the empirical formula.

Gas Stoichiometry

Gas stoichiometry involves the calculation of the volume, pressure, mass, or moles of gases involved in a chemical reaction under conditions of a known temperature and pressure. It relies on the ideal gas law, and it's particularly useful when working with reactions that produce gases as products.

When solving gas stoichiometry problems, it's essential to follow a systematic approach: balance the chemical equation, convert known quantities into moles (using the molar mass for solid or liquid substances, or the ideal gas law for gases), apply the mole-to-mole conversion factors from the balanced equation, and finally use the ideal gas law again to convert moles back into the desired unit (e.g., volume or pressure of a gas). This process allows for the quantitative analysis of reactants and products in gaseous state.