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?

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.

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.

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 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.