Question

A compound containing \(\mathrm{C}, \mathrm{H}, \mathrm{N},\) and \(\mathrm{O}\) is burned in excess oxygen. The gases produced by burning \(0.1152 \mathrm{g}\) are first treated to convert the nitrogen-containing product gases into \(\mathrm{N}_{2},\) and then the resulting mixture of \(\mathrm{CO}_{2}, \mathrm{H}_{2} \mathrm{O}, \mathrm{N}_{2},\) and excess \(\mathrm{O}_{2}\) is passed through a bed of \(\mathrm{CaCl}_{2}\) to absorb the water. The \(\mathrm{CaCl}_{2}\) increases in mass by \(0.09912 \mathrm{g} .\) The remaining gases are bubbled into water to form \(\mathrm{H}_{2} \mathrm{CO}_{3},\) and this solution is titrated with \(0.3283 \mathrm{M} \mathrm{NaOH} ; 28.81 \mathrm{mL}\) is required to achieve the second equivalence point. The excess \(\mathbf{O}_{2}\) gas is removed by reaction with copper metal (to give CuO). Finally, the \(\mathrm{N}_{2}\) gas is collected in a 225.0 -mL. flask, where it has a pressure of \(65.12 \mathrm{mm} \mathrm{Hg}\) at \(25^{\circ} \mathrm{C} .\) In a separate experiment, the unknown compound is found to have a molar mass of \(150 \mathrm{g} / \mathrm{mol}\). What are the empirical and molecular formulas of the unknown compound?

Short answer

The empirical formula is \( \text{C}_3\text{H}_7\text{NO} \) and the molecular formula is \( \text{C}_6\text{H}_{14}\text{N}_2\text{O}_2 \).

Step-by-step solution

Determine moles of water formed

The change in mass of the \( \text{CaCl}_2 \) is attributed to the absorption of water. Given that \( \text{CaCl}_2 \) increased by 0.09912 g, this represents the mass of \( \text{H}_2\text{O} \) absorbed. The molar mass of water (\( \text{H}_2\text{O} \)) is around 18.02 g/mol. Thus, the moles of water are calculated as follows: \( \frac{0.09912 \, \text{g}}{18.02 \, \text{g/mol}} = 0.0055 \, \text{mol} \). This tells us there were 0.0055 moles of \( \text{H}_2\text{O} \) formed, meaning there were 0.011 moles of hydrogen atoms in the compound since each \( \text{H}_2\text{O} \) molecule contains 2 hydrogen atoms.

Calculate moles of carbon from titration

When \( \text{H}_2\text{CO}_3 \) is formed by bubbling it through water, it requires two moles of \( \text{NaOH} \) per mole of \( \text{CO}_2 \) to reach the second equivalence point because \( \text{H}_2\text{CO}_3 \) is a diprotic acid. With \( 28.81 \text{ mL} \) of \( 0.3283 \text{ M} \) \( \text{NaOH} \), the moles of \( \text{NaOH} \) are \( 0.3283 \times \frac{28.81}{1000} \), which equals \( 0.00945 \text{ mol} \). Since two moles of \( \text{NaOH} \) react with one mole of \( \text{CO}_2 \), the moles of \( \text{CO}_2 \) are \( \frac{0.00945}{2} = 0.004725 \text{ mol} \). Thus, there are 0.004725 moles of carbon atoms.

Calculate moles of nitrogen from gas properties

Using the ideal gas law \( PV = nRT \), we can find the moles of \( \text{N}_2 \). The pressure \( P \) is \( 65.12 \text{ mm Hg} \) or 0.0857 atm \( \left(\frac{65.12}{760}\right) \), the volume \( V \) is 225.0 mL or 0.225 L, \( R \) is the ideal gas constant \( 0.0821 \text{ L·atm/(mol·K)} \), and the temperature \( T \) is 298 K \( (25+273) \). Solving for moles \( n \): \( n = \frac{PV}{RT} = \frac{0.0857 \times 0.225}{0.0821 \times 298} = 0.000785 \text{ mol} \) of \( \text{N}_2 \), equivalent to 0.00157 mol of nitrogen atoms since each molecule is \( \text{N}_2 \).

Determine mass and moles of oxygen

The total initial mass of the compound is 0.1152 g. The masses of elements are found using their moles and atomic weights from previous steps: carbon contributes \( 0.004725 \times 12.01 = 0.0567 \text{ g} \), hydrogen contributes \( 0.011 \times 1.008 = 0.0111 \text{ g} \), and nitrogen contributes \( 0.00157 \times 14.01 = 0.022 \text{ g} \). The mass of oxygen in the compound is then the remainder: \( 0.1152 - (0.0567 + 0.0111 + 0.022) = 0.0254 \text{ g} \). The moles of oxygen can be found using its molar mass (16.00 g/mol): \( \frac{0.0254}{16.00} = 0.0015875 \text{ mol} \).

Determine empirical formula

The empirical formula is determined by the moles of each element: C, H, N, and O. The mole ratio of C:H:N:O is 0.004725:0.011:0.00157:0.0015875. Dividing by the smallest ratio, the approximate whole number mole ratio is obtained: C=3, H=7, N=1, O=1. Hence, the empirical formula is \( \text{C}_3\text{H}_7\text{NO} \).

Determine molecular formula

The molar mass based on the empirical formula \( \text{C}_3\text{H}_7\text{NO} \) is \( 3(12.01) + 7(1.008) + 14.01 + 16.00 = 73.09 \text{ g/mol} \). Given the molecular weight is 150 g/mol, the ratio of molecular to empirical formula mass is \( \frac{150}{73.09} \approx 2.05 \). Rounding yields a factor of 2, indicating the molecular formula is \( \text{C}_6\text{H}_{14}\text{N}_2\text{O}_2 \).

Key concepts

Combustion Analysis

Combustion analysis is a fundamental technique used to determine the composition of a compound by burning it completely in excess oxygen. During the combustion process, elements in the compound such as carbon, hydrogen, and nitrogen are converted into simplified forms like carbon dioxide (CO₂), water (H₂O), and nitrogen gas (N₂). This helps in establishing the relative amounts of elements present in the compound.

If a compound contains oxygen, it must be calculated by difference since oxygen is both a part of the compound and contributes to the combustion process itself. By measuring amounts of CO₂ and H₂O produced, one can infer the amounts of carbon and hydrogen originally present. This data guides how we derive empirical formulas, which provide the simplest integer ratio of atoms present in the compound.

Molar Mass Calculation

To calculate the molar mass of a compound, you add up the atomic masses of all the atoms present in one mole of the compound's molecules. This is measured in grams per mole (g/mol) and is crucial in determining molecular formulas, especially when comparing with an empirical formula.

For example, if the empirical formula of a compound is found to be \(\text{C}_3\text{H}_7\text{NO}\), with a computed molar mass of 73.09 g/mol, and if we know the molar mass of the compound is 150 g/mol, we can determine that the actual number of each type of atom in the molecule is double that of the empirical formula. Thus, effectively illustrating how the empiri...

Ideal Gas Law

The ideal gas law provides a crucial relationship between the physical properties of gases. It is expressed as \( PV = nRT \), where \( P \) stands for pressure, \( V \) for volume, \( n \) for moles of gas, \( R \) is the universal gas constant, and \( T \) stands for temperature in Kelvin.

In the context of combustion analysis, the ideal gas law helps find quantities such as the moles of nitrogen gas produced during combustion. For instance, given the pressure, volume, and temperature of collected nitrogen gas from a combustion product, one can solve for the number of moles, providing insights into the nitrogen content of the original compound. Utilizing these values aids in constructing the empirical formula of the compound by defining precise atomic ratios.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions based on the balanced chemical equations. It is an essential aspect of chemistry that helps in understanding mass relationships within a reaction.

In a combustion analysis, stoichiometry enables the transition from the amount of a product, like CO₂ or H₂O, back to the amount of reactants, such as C, H, and O in the original compound. By knowing how these atoms combine in fixed ratios to produce other compounds in the reaction, we can back-calculate to determine how much of each element was present in the original sample.

• This involves using physical measurements from experiments, such as mass gain or volume changes, to infer chemical quantities.
• Such conversions and calculations help in formulating both empirical and molecular formulas, ultimately revealing the molecular structure and composition of the compound.