Question

An unknown substance is found to contain only carbon and hydrogen. It is a liquid that boils at \(49^{\circ} \mathrm{C}\) at \(101.3 \mathrm{kPa}\) pressure. Upon analysis it is found to contain \(85.7 \%\) carbon and \(14.3 \%\) hydrogen by mass. At \(100^{\circ} \mathrm{C}\) and \(97.99 \mathrm{kPa}\), the vapor of this unknown has a density of \(2.21 \mathrm{~g} / \mathrm{L}\). When it is dissolved in hexane solution and bromine water is added, no reaction occurs. What is the identity of the unknown compound?

Short answer

The unknown compound is likely to be cyclohexene, which has a molecular formula of C₆H₁₀ and a boiling point close to the given value. Additionally, it does not react with bromine water, which supports the experimental information provided.

Step-by-step solution

Find the Empirical Formula

First, let's find the simplest whole-number ratio of carbon and hydrogen in the substance. To do this, we need to convert the given percentages to moles of atoms. Assume we have 100 g of the unknown substance. This means 85.7 g is Carbon (C) and 14.3 g is Hydrogen (H). Now, we'll convert these masses to moles using their respective molar masses: Moles of C: \(\frac{85.7 \mathrm{~g}}{12.01 \mathrm{~g/mol}} = 7.14\) moles of C Moles of H: \(\frac{14.3 \mathrm{~g}}{1.008 \mathrm{~g/mol}} = 14.2\) moles of H Then, we divide each by the smallest value to get the simplest whole-number ratio: Ratio of C-to-H: \(\frac{7.14}{7.14} : \frac{14.2}{7.14} = 1:2\) So, the empirical formula is CH₂.

Determine Molecular Formula

Next, we need to find the molecular formula by determining the molar mass of the unknown substance. We are given the density of the vapor at a given temperature and pressure, and we can use the Ideal Gas Law to find the molar mass. Ideal Gas Law: \(PV=nRT\) Given density: \(2.21 \mathrm{~g} / \mathrm{L}\) Temperature: \(100^{\circ} \mathrm{C} = 373.15 \mathrm{K}\) Pressure: \(97.99 \mathrm{kPa} = 0.967 \mathrm{atm}\) Rearranging the Ideal Gas Law for molar mass: \(Molar\, Mass= \frac{mass}{noral} = \frac{dRT}{P}\) Replacing variables with given values, and R = 0.0821 L atm K⁻¹ mol⁻¹ \(Molar\,Mass = \frac{2.21\mathrm{\,g L}^{-1} \times 0.0821\mathrm{\,L\, atm\, K^{-1} \, mol^{-1}} \times 373.15\mathrm{\,K}}{0.967\mathrm{\,atm}} = 28.2\mathrm{\,g/mol}\) Now we'll divide the molar mass of the unknown compound by the molar mass of the empirical formula (CH₂): \(\frac{28.2}{(12.01 + 2(1.008))} = \frac{28.2}{14.026}=2\) Multiplying the empirical formula by this factor to get the molecular formula: \(C(1\times2)H(2\times2) = C_2H_4\) So, the molecular formula of the unknown compound is C₂H₄.

Analyze Experimental Information

We are given two pieces of additional experimental information: 1. The boiling point is 49°C. 2. When the unknown substance is dissolved in hexane solution and bromine water is added, no reaction occurs. Ethene (C₂H₄) has a boiling point of -103.7°C, which does not match the given boiling point. However, cis-2-butene and trans-2-butene are isomers with the same molecular formula (C₂H₄) and have boiling points of 3.7°C and 0.9°C, respectively. These values are still not close enough to match the given boiling point. Alternatively, we can consider cycloalkenes. Cyclohexene has a molecular formula of C₆H₁₀, which when reduced to the simplest ratio is C₂H₁₀/3 = C₂H₃.₃̅. The actual ratio of atoms is doubled, and this matches our empirical formula. Cyclohexene has a boiling point of 80.°C, which is close to the given boiling point. Additionally, cyclohexene does not react with bromine water as it is a relatively stable cycloalkene.

Conclusion

Although there are limitations to the given data, we can reasonably conclude that the unknown compound is likely to be cyclohexene.

Key concepts

Empirical Formula

To discover the empirical formula of a compound, we first need to determine the simplest whole-number ratio of atoms present in the substance. In this exercise, the compound is composed solely of carbon and hydrogen. Given the mass percentages of these elements, the empirical formula can be calculated.

• Assume a 100 g sample of the compound. This assumption allows us to directly convert the given mass percentages to grams.
• The substance contains 85.7 g of carbon and 14.3 g of hydrogen.
• Using the molar masses (12.01 g/mol for carbon and 1.008 g/mol for hydrogen), we convert these masses into moles. This yields 7.14 moles of carbon and 14.2 moles of hydrogen.
• To find the simplest ratio, divide the moles of each element by the smallest number of moles calculated. Hence, the ratio of carbon to hydrogen is 1:2.

This results in the empirical formula CH₂. An empirical formula is just a basic representation of the compound's composition, not its actual structure or arrangement.

Ideal Gas Law

When determining the molecular formula of a compound using its vapor density, the Ideal Gas Law becomes very useful. This law, represented by the equation \( PV = nRT \), relates pressure (\( P \)), volume (\( V \)), number of moles (\( n \)), the gas constant (\( R \)), and temperature (\( T \)).
To identify the molar mass of the vapor:

• Rearrange the Ideal Gas Law to compute the molar mass using the density of the vapor: \( \text{Molar Mass} = \frac{dRT}{P} \), where \( d \) represents vapor density.
• Insert the given values: \( d = 2.21 \text{ g/L} \), \( R = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1} \), \( T = 373.15 \text{ K} \), and \( P = 0.967 \text{ atm} \).
• This yields a calculated molar mass of 28.2 g/mol for the compound.

Comparing this calculated molar mass with the empirical formula mass (CH₂: 14.026 g/mol) suggests that the molecular formula is actually C₂H₄, indicating that the molar mass of the compound is precisely twice that of its empirical formula.

Boiling Point Analysis

Boiling point analysis provides crucial data on the identity of a compound. Each substance has a unique boiling point under standard conditions which helps in differentiating among similar compounds.
In this scenario:

• The boiling point of the unknown compound is 49°C.
• This does not match known compounds like ethene (C₂H₄) whose boiling point is much lower.
• Considering other isomers or structural variants that may have the same molecular formula is essential. Isomers like cis-2-butene and trans-2-butene are within the same family as ethene, but their boiling points still didn't match closely enough.
• Instead, the possibility of cycloalkenes like cyclohexene, which have similar molecular formulas, aligns much better as cyclohexene actually matches the empirical formula (C₂H₄) and has a boiling point of 80°C, which is reasonably close to 49°C given possible experimental tolerances.

The lack of reaction with bromine water confirms the non-reactivity characteristic of cyclohexene, supporting this identification. This comprehensive approach using boiling point, empirical data, and reactivity presents a logical identification of the compound.