Question

A compound is \(85.6 \%\) carbon by mass. The rest is hydrogen. When \(10.0 \mathrm{g}\) of the compound is evaporated at \(50.0^{\circ} \mathrm{C},\) the vapor occupies \(6.30 \mathrm{L}\) at \(1.00 \mathrm{atm}\) pressure. What is the molecular formula of the compound?

Short answer

The molecular formula of the compound is \(C_{3}H_{6}\).

Step-by-step solution

Finding the ratio of Carbon to Hydrogen

Consider that the mass of the compound is 100g (to simplify the calculation). Given that it's 85.6% carbon, the mass of carbon in 100g of the compound is \( 85.6g \) . Since the rest of the compound is hydrogen, the mass of hydrogen is \( 100g - 85.6g = 14.4g \). Find the moles of carbon and hydrogen using their atomic masses, \( C = 12.01g/mol \), \( H = 1.008g/mol \). Moles of C = \( 85.6 ÷ 12.01 ≈ 7.12 \) and Moles of H = \( 14.4 ÷ 1.008 ≈ 14.29 \). The ratio of \( C : H \) is therefore approximately \( 7 : 14 \), or more simply, \( 1:2 \).

Determining Molar Mass using gas law

The ideal gas law is \( PV = nRT \) , where \( P \) is pressure in atm, \( V \) is volume in L, \( n \) is the amount of substance in moles, \( R \) is the gas constant 0.0821 \( atm \cdot L/mol \cdot K \) (at these units), and \( T \) is the absolute temperature in Kelvin. Convert the given Celsius temperature to Kelvin by adding 273.15: \( 50.0 + 273.15 = 323.15K \). Rearrange the gas law equation to solve for \( n = PV/RT \), then substitute given values: \( n = (1.00atm * 6.30L) / (0.0821 * 323.15K) ≈ 0.238 moles \). Molar mass of the compound is \(10.0g / 0.238 mol ≈ 42 g/mol.

Determining the Molecular Formula

The empirical formula found in step 1 was \(CH_{2}\) which has a molar mass of \(12.01 g/mol + 2 * 1.008 g/mol = 14.03 g/mol \). The ratio of the molecular mass to the empirical formula mass is \(42g/mol / 14.03g/mol ≈ 3 \). Therefore, the molecular formula is \(C_{3}H_{6}\).

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental principle in chemistry that connects the pressure, volume, temperature, and number of moles of a gas within a closed system. The equation is represented as \( PV = nRT \). Here:

• \( P \) is the pressure of the gas in atmospheres (atm).
• \( V \) is the volume the gas occupies, measured in liters (L).
• \( n \) is the number of moles of the gas.
• \( R \), the gas constant, is often \( 0.0821 \) \( atm \, L/mol \, K \) in ideal conditions.
• \( T \) is the temperature in Kelvin (K).

To apply the Ideal Gas Law, always ensure to convert temperatures from Celsius to Kelvin by adding 273.15. In our example, the compound's vapor occupies a volume at 50°C, which is 323.15 K. By using the rearranged form \( n = \frac{PV}{RT} \), you can find the moles of a gas, highlighting the versatility of this equation in calculating unknown variables in gas-related problems.

Empirical Formula

The empirical formula of a compound is the simplest integer ratio of atoms of each element in the compound. It may not represent the actual numbers of atoms in a molecule, but it does provide useful information about composition. For the compound in question, which contains carbon and hydrogen, we simplified the mass percentages into a ratio of moles:- Mass of carbon was calculated from 85.6%: 85.6 g.- Mass of hydrogen was derived from the rest of the mass: 14.4 g.To find the moles, we divide the mass by the atomic mass of each element:- Moles of Carbon (C): \( \frac{85.6}{12.01} \approx 7.12 \)- Moles of Hydrogen (H): \( \frac{14.4}{1.008} \approx 14.29 \)This gives us the mole ratio \( C:H = 1:2 \), implying the empirical formula is \( CH_2 \). This formula tells us that for every carbon atom, there are two hydrogen atoms, capturing the compound's simplest composition.

Molar Mass

Molar Mass is a critical property used to identify substances. It indicates the mass of one mole of a substance in grams. The process involves dividing the mass of the compound by the number of moles, as derived from the Ideal Gas Law.Using the example, the molar mass of the compound was found by calculating:- Mass of the compound = 10.0 g- Moles computed from \( PV = nRT = 0.238 \)The molar mass is then \( \frac{10.0 \text{g}}{0.238 \text{mol}} \approx 42 \text{g/mol} \). By understanding and using molar mass, one can easily identify or verify the molecular structure and consider the extensive applications in stoichiometric calculations.

Chemical Composition Analysis

Chemical composition analysis is the process of determining the elements that make up a substance and their proportions. In our exercise, it involved analyzing the given percentages of carbon and hydrogen and using this data to form empirical and molecular formulas.- The compound was reported with an 85.6% carbon composition.- The rest, 14.4%, is hydrogen, allowing us to calculate the ratio and moles for a 100 g sample for simplicity.These calculations help transition from an empirical formula \( CH_2 \) to understanding the real structure, the molecular formula \( C_3H_6 \). The molecular formula provides a true reflection of the numbers of atoms involved in the molecules of substances, which directly ties into the compound's molar mass, reaffirming the calculated values through the characterization and verification of its unique chemical identity.