Problem 117 - Chemistry The Molecular Nature of Matter

Question

A sample of a new anti-malarial drug with a mass of \(0.2394 \mathrm{~g}\) was made to undergo a series of reactions that changed all of the nitrogen in the compound into \(\mathrm{N}_{2}\). This gas had a volume of \(18.90 \mathrm{~mL}\) when collected over water at \(23.80^{\circ} \mathrm{C}\) and a pressure of 746.0 torr. At \(23.80^{\circ} \mathrm{C}\), the vapor pressure of water is 22.110 torr. When \(6.478 \mathrm{mg}\) of the compound was burned in pure oxygen, \(17.57 \mathrm{mg}\) of \(\mathrm{CO}_{2}\) and \(4.319 \mathrm{mg}\) of \(\mathrm{H}_{2} \mathrm{O}\) were obtained. What are the percentages of \(\mathrm{C}\) and \(\mathrm{H}\) in this compound? (a) Assuming that any undetermined element is oxygen, write an empirical formula for the compound. (b) The molecular mass of the compound was found to be 324 . What is its molecular formula?

Step-by-Step Solution

Answer

The percentages of carbon and hydrogen in the compound are found by calculating the mass contributions from the combustion products (\(CO_2\text{ and }H_2O\text{), respectively. The empirical formula is CH_4N, and the molecular formula is C_{12}H_{48}N_{12}O_{4}\text{, based on the empirical formula and the given molecular mass of 324.}

Step 1: Determining the Mass of Nitrogen

The mass of nitrogen in the sample is found by converting the volume of \(N_2\) gas to moles using the ideal gas law, \( PV = nRT \) and then to mass. Correct for the pressure by subtracting the vapor pressure of water from the total pressure to get the pressure due to \(N_2\) alone.

Step 2: Calculating the Moles of Nitrogen

Use the corrected pressure and the volume of the nitrogen gas along with the ideal gas constant (R = 0.0821 atm·L/mol·K) and the temperature in Kelvin to calculate the moles of nitrogen. The moles of nitrogen times the molar mass of nitrogen (28.02 g/mol) gives the mass of nitrogen in the sample.

Step 3: Determining the Masses of Carbon and Hydrogen

Convert the mass of \(CO_2\) and \(H_2O\) produced from burning the sample into moles to find the moles of carbon and hydrogen using the molar masses of \(CO_2\) (44.01 g/mol) and \(H_2O\) (18.02 g/mol), respectively. Since one mole of \(CO_2\) contains one mole of carbon and one mole of \(H_2O\) contains two moles of hydrogen, you can then determine the mass of carbon and hydrogen.

Step 4: Calculating the Percentages of Carbon, Hydrogen, and Nitrogen

The mass percent of an element is obtained by dividing the mass of that element in the compound by the total mass of the compound and then multiplying by 100%. Calculate the mass percent of carbon, hydrogen, and nitrogen.

Step 5: Calculating the Mass of Oxygen

The mass of oxygen is found by subtracting the masses of carbon, hydrogen, and nitrogen from the total mass of the compound. Use this result to find the percentage of oxygen in the compound.

Step 6: Determining the Empirical Formula

Divide the mass percentage of each element by its atomic mass to obtain the relative number of moles of each element. Then divide by the smallest number of moles obtained to get the simplest whole number ratio of the elements, which is the empirical formula.

Step 7: Calculating the Molecular Formula

Use the molecular mass of the compound to determine the molecular formula. Divide the molecular mass by the mass of the empirical formula unit to find the multiple that will convert the empirical formula into the molecular formula.

Key Concepts

Empirical Formula

Understanding the empirical formula of a compound is essential for chemists. The empirical formula represents the simplest whole number ratio of atoms of each element in the compound. It doesn't necessarily indicate the actual number of atoms present but gives the proportional representation of each element.

For instance, if we find that a chemical compound consists of 40% carbon and 60% oxygen by mass, we can translate these percentages into a molar ratio. This is done by dividing each percentage by the respective element's atomic mass (for carbon, 12.01 g/mol and for oxygen, 16.00 g/mol), normalizing by the smallest resulting value to obtain whole numbers. The resultant ratio is then used to formulate the empirical formula.

Molecular Formula

The molecular formula, on the other hand, conveys the exact number of atoms of each element in a molecule of the compound. It's a multiple of the empirical formula and can be calculated if you know the compound's molar mass.

To find the molecular formula, divide the molar mass of the compound by the molar mass of the empirical formula to get a whole number. This number is then multiplied by the subscripts in the empirical formula to get the molecular formula. For example, if the empirical formula is CH2 and the molecular mass is 56 g/mol, then the molecular mass of CH2 (14 g/mol) goes into 56 g/mol four times, suggesting the molecular formula is C4H8.

Ideal Gas Law

The ideal gas law is a pivotal tool in calculating the quantity of gases involved in chemical reactions. Stated as PV = nRT, it relates the pressure (P), volume (V), number of moles (n), temperature (T), and the ideal gas constant (R).

When determining the mass of gaseous nitrogen produced in a reaction, as in our exercise, the ideal gas law is applied by first correcting the measured pressure for the vapor pressure of water, providing the partial pressure of the nitrogen. With this corrected pressure, we calculate the moles of nitrogen released, and from here, the mass of the nitrogen can be found. It's essential for students to grasp using the ideal gas law within the context of real-world problems, such as calculating the amount of gas produced or consumed in a reaction.

Combustion Analysis

Combustion analysis is a laboratory method used to determine the elemental composition of a compound, particularly organic ones composed mainly of carbon, hydrogen, and, occasionally, nitrogen. In this technique, a known mass of a compound is burned in excess oxygen, and the masses of the products (usually CO2 and H2O) are measured.

By understanding the stoichiometry of combustion—that CO2 and H2O form in a 1:1 mole ratio with carbon and hydrogen, respectively—scientists can work back from the masses of these products to find the moles, and thus the mass, of carbon and hydrogen in the original compound. This information is crucial to calculate the empirical formula and, by extension, can lead to determining the molecular formula when the compound's molar mass is known. Thus, combustion analysis is a fundamental technique for identifying unknown substances.