Question

It was found from the chemical analysis of a gas that it has two hydrogen atoms for each carbon atom. At \(0^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\), its density is \(1.25 \mathrm{~g}\) per litre. The formula of the gas would be (a) \(\mathrm{CH}_{2}\) (b) \(\mathrm{C}_{2} \mathrm{H}_{4}\) (c) \(\mathrm{C}_{2} \mathrm{H}_{6}\) (d) \(\mathrm{C}_{4} \mathrm{H}_{8}\)

Short answer

Question: Determine the formula of a gas that contains 2 hydrogen atoms for each carbon atom in the compound. The density of the gas at \(0^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) is \(1.25 \mathrm{~g/L}\). Choose the correct answer from the following options: (a) \(\mathrm{CH}_{2}\) (b) \(\mathrm{C}_{2} \mathrm{H}_{4}\) (c) \(\mathrm{C}_{2} \mathrm{H}_{6}\) (d) \(\mathrm{C}_{4} \mathrm{H}_{8}\) Answer: (b) \(\mathrm{C}_{2} \mathrm{H}_{4}\)

Step-by-step solution

Convert density to molar mass

We are given the density of the gas at \(0^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\), which is \(1.25 \mathrm{~g}\) per litre. We can convert this density to a molar mass using the ideal gas equation, which in this case can be approximated as: \(PV=nRT\) Where: - \(P\) is the pressure (1 atm), - \(V\) is the volume (1 L), - \(n\) is the number of moles, - \(R\) is the ideal gas constant (0.08206 L atm K\(^{-1}\) mol\(^{-1}\)), and - \(T\) is the temperature (\(0^{\circ} \mathrm{C}\) or 273.15 K). Rearrange the equation to solve for \(n\): \(n=\frac{PV}{RT}\) And then substitute the given values: \(n=\frac{(1 \mathrm{~atm})(1 \mathrm{~L})}{(0.08206 \mathrm{ ~L\cdot atm\cdot K^{-1}\cdot mol^{-1}})(273.15 \mathrm{~K})} \approx 0.0446 \mathrm{~mol}\) Now we can calculate the molar mass by dividing the given mass (1.25 g) by the number of moles (0.0446 mol): Molar mass \(= \frac{1.25 \mathrm{~g}}{0.0446 \mathrm{~mol}} \approx 28.0 \mathrm{~g\cdot mol^{-1}}\) Since we know that the gas has 2 hydrogen atoms for each carbon atom, we will need to find a possible formula that fits the given molar mass and this specific ratio.

Determine the formula of the gas

Let's check the possible formulas from the multiple-choice options: (a) \(\mathrm{CH}_{2}\): In this case, the molar mass is equal to \(12.01 \mathrm{~g\cdot mol^{-1}}\) (C) \(+ 2 \times 1.00784 \mathrm{~g\cdot mol^{-1}}\) (H) \(= 14.03 \mathrm{~g\cdot mol^{-1}}\). This does not match the calculated molar mass of 28.0 g/mol. (b) \(\mathrm{C}_{2} \mathrm{H}_{4}\): In this case, the molar mass is equal to \(2 \times 12.01 \mathrm{~g\cdot mol^{-1}}\) (C) \(+ 4 \times 1.00784 \mathrm{~g\cdot mol^{-1}}\) (H) \(= 28.05 \mathrm{~g\cdot mol^{-1}}\). This is a very close match to the calculated molar mass. (c) \(\mathrm{C}_{2} \mathrm{H}_{6}\): In this case, the molar mass is equal to \(2 \times 12.01 \mathrm{~g\cdot mol^{-1}}\) (C) \(+ 6 \times 1.00784\mathrm{~g\cdot mol^{-1}}\) (H) \(= 30.07 \mathrm{~g\cdot mol^{-1}}\). This does not match the calculated molar mass. (d) \(\mathrm{C}_{4} \mathrm{H}_{8}\): In this case, the molar mass is equal to \(4 \times 12.01 \mathrm{~g\cdot mol^{-1}}\) (C) \(+ 8 \times 1.00784 \mathrm{~g\cdot mol^{-1}}\) (H) \(= 56.11 \mathrm{~g\cdot mol^{-1}}\). This does not match the calculated molar mass.

Select the correct formula

Based on the molar mass calculations, the correct formula for the gas is option (b) \(\mathrm{C}_{2} \mathrm{H}_{4}\), since it is the only one that matches the calculated molar mass of 28.0 g/mol and has the appropriate ratio of 2 hydrogen atoms per 1 carbon atom in the compound.

Key concepts

Chemical Analysis

Chemical analysis is a foundational concept in chemistry that involves determining the composition of substances. It's a process by which chemists characterize the elements or compounds within a sample. For our specific exercise, the chemical analysis identified that the gas has a 2:1 hydrogen to carbon atom ratio. This ratio is critical as it can help us narrow down the possible molecular formulas for the compound.

Understanding the empirical formula, which is the simplest whole number ratio of atoms in a compound, is a direct outcome of chemical analysis and is our first step towards identifying the compound in question. The empirical formula does not necessarily represent the exact number of atoms present but gives the proportional ratio, which for this gas was determined to be CH₂. This step is vital for deeper analysis and leads us to our next concept, the ideal gas equation.

Ideal Gas Equation

The ideal gas equation is a crucial tool in our toolkit for solving problems in gas chemistry. It's given by the formula:

\( PV = nRT \)
Where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature. Applying this equation allows for determining the molar mass when combined with the density of the gas.

In our exercise, the ideal gas equation was used to convert the gas's known density at standard temperature and pressure to find the number of moles. From there, the molar mass of the gas was determined. When you encounter problems involving gases, understanding how to manipulate and apply the ideal gas equation is essential to find the correct values needed for further calculations like molar mass, as we did here.

Molar Mass

Molar mass, measured in grams per mole (g/mol), is the mass of one mole of a substance. It is a physical property that is often used as a conversion factor in stoichiometry. In our problem, we used the information from the chemical analysis and the ideal gas equation to calculate the molar mass of the gas.

The molar mass is central to stoichiometry because it links the mass of a substance to the amount in moles, allowing for conversions between mass and mole amounts. In this gas compound's case, we calculated a molar mass of approximately 28.0 g/mol, which is essential for identifying the gas's empirical formula. Precision in calculating the molar mass is key to determining the correct molecular formula.

Stoichiometry

Stoichiometry is the study of quantitative relationships, or ratios, within chemical reactions. It involves calculations that use the principles of conservation of mass to relate quantities of reactants and products. In practical terms, stoichiometry comes down to the calculations concerning the masses of reactants and products in a chemical reaction.

In our exercise, we used stoichiometry to link the calculated molar mass of the gas to the possible molecular formulas provided. By comparing the molar mass of each molecule option with the given ratio of hydrogen to carbon atoms, we utilized stoichiometry to find the most probable molecular formula that corresponds to the calculated molar mass. This approach is part of a stoichiometric analysis, which paved the way to determine that the gas is indeed ethylene, \( \mathrm{C}_{2} \mathrm{H}_{4} \). It's an excellent example of how stoichiometry helps in deducing the correct chemical formula from empirical data.