Question

An herbicide is found to contain only \(\mathrm{C}, \mathrm{H}, \mathrm{N},\) and \(\mathrm{Cl} .\) The complete combustion of a 100.0 -mg sample of the herbicide in excess oxygen produces \(83.16 \mathrm{~mL}\) of \(\mathrm{CO}_{2}\) and \(73.30 \mathrm{~mL}\) of \(\mathrm{H}_{2} \mathrm{O}\) vapor expressed at STP. A separate analysis shows that the sample also contains \(16.44 \mathrm{mg}\) of \(\mathrm{Cl}\). (a) Determine the percentage of the composition of the substance. (b) Calculate its empirical formula. (c) What other information would you need to know about this compound to calculate its true molecular formula?

Short answer

(a) The percentage composition of the herbicide is calculated as follows: - C: Percentage = \(\frac{mass\;of\;C}{total\;mass\;of\;sample}\) * 100 - H: Percentage = \(\frac{mass\;of\;H}{total\;mass\;of\;sample}\) * 100 - N: Percentage = \(\frac{mass\;of\;N}{total\;mass\;of\;sample}\) * 100 - Cl: Percentage = \(\frac{mass\;of\;Cl}{total\;mass\;of\;sample}\) * 100 (b) The empirical formula is found by dividing the moles of each element by the smallest number of moles observed and rounding to the nearest whole number, resulting in CxHyNzClw. (c) To determine the true molecular formula, additional information is needed, such as the molar mass of the compound or structural information like spectroscopy data or a description of its structure. This would help us find the actual molecular formula by comparing it to the empirical formula.

Step-by-step solution

Calculate moles of C and H from combustion volumes.

Remember that STP conditions are 0°C (273.15 K) and 1 atm pressure. We can use the ideal gas law to calculate the moles of CO2 and H2O formed. The ideal gas law is given by: PV = nRT Where P is pressure, V is volume, n is the number of moles, R is the gas constant (0.0821 L*(atm)/(mol*K) for STP), and T is the temperature. Given the volumes of CO2 and H2O, we can calculate the moles of each compound and then determine the moles of carbon and hydrogen. For CO2: (1 atm) * \(83.16 \times 10^{-3} L = n \times 0.0821\frac{L.atm}{mol. K}\) * (273.15 K) For H2O: (1 atm) * \(73.30 \times 10^{-3} L = n \times 0.0821\frac{L.atm}{mol. K}\) * (273.15 K)

Find moles of Cl and calculate moles of N.

We are given the mass of Cl in the sample, i.e., 16.44 mg. To find the moles of Cl, we can use the molar mass of Cl (35.45 g/mol). Moles of Cl = \(\frac{16.44 \times 10^{-3} g }{35.45 g/mol}\). Next, since the sample contains only C, H, N and Cl, we can subtract the masses of C, H and Cl from the sample's total mass to find the mass of N. We can then use the molar mass of N (14.01 g/mol) to find the moles of N in the sample.

Determine the percentage composition.

We can now calculate the percentage of each element in the sample by dividing the mass of the element by the total mass of the sample, and then multiplying it by 100. Percentage composition = \(\frac{mass\;of\;element}{total\;mass\;of\;sample}\) * 100 Calculate the percentage for C, H, N, and Cl.

Calculate the empirical formula.

To find the empirical formula, divide the moles of each element by the smallest number of moles observed. C: moles of C / smallest number of moles H: moles of H / smallest number of moles N: moles of N / smallest number of moles Cl: moles of Cl / smallest number of moles Round each calculated value to the nearest whole number to find the empirical formula (CxHyNzClw).

Discuss the information needed for the true molecular formula.

To find the true molecular formula of the compound, we need additional information, such as the molar mass of the compound or some other structural information, like spectroscopy data or a description of its structure. This information would help us to determine the actual molecular formula by comparing it to the empirical formula obtained in Step 4. The true molecular formula is a multiple of the empirical formula.

Key concepts

Percentage Composition

Percentage composition is a way of expressing the composition of a compound by stating the percentage of each element present by mass. It's an important concept in chemistry as it gives insight into the proportions of elements in any given substance. For the herbicide in the problem, we're asked to determine the percentage of elements like carbon, hydrogen, nitrogen, and chlorine.

• To find the percentage composition, you first calculate the mass of each element.
• Then you divide the mass of each particular element by the total mass of the compound, in this case, the herbicide sample.
• Lastly, multiply the result by 100 to express it as a percentage.

Percentage composition is crucial as it helps us understand the make-up of a compound, which can, in turn, relate to its properties and reactivity.

Ideal Gas Law

The ideal gas law is a crucial equation that relates the variables of a gas (pressure, volume, temperature, and moles). It is expressed as \(PV = nRT\). Understanding each of these variables is essential:

• \(P\): Pressure of the gas
• \(V\): Volume of the gas
• \(n\): Number of moles of the gas
• \(R\): Ideal gas constant, typically \(0.0821\, \text{L} \cdot \text{atm} / \text{mol} \cdot \text{K}\)
• \(T\): Temperature in Kelvin

In the exercise, the ideal gas law is utilized to convert the volume measurements of gases produced during combustion, CO₂ and H₂O, into moles. This transformation is crucial, as it links gas volumes to amounts of substance, offering a pathway to uncover the composition of the herbicide.

Molar Mass

Molar mass is the mass of one mole of a substance (element or compound) and is expressed in grams per mole (g/mol). It plays a pivotal role in helping us connect the microscopic world of atoms and molecules to the macroscopic world of grams and liters.

• To find the moles of an element in a compound, you divide the element’s mass by its molar mass.
• For example, chlorine’s molar mass is 35.45 g/mol, a value used in calculating the moles of chlorine in the herbicide.

In essence, knowing the molar mass allows us to calculate how much of a given element is present in a sample by translating tiny atomic mass units into tangible amounts manageable in the lab.

Molecular Formula

The molecular formula of a compound provides the actual number of atoms of each element in a molecule. It's often a multiple of the empirical formula, which represents the smallest whole-number ratio of the elements in the compound.
To ascertain the true molecular formula, additional information is crucial:

• Molar Mass: By comparing the molar mass to the empirical unit mass, we can find the multiplication factor needed to derive the molecular formula from the empirical one.
• Structural Information: Techniques like spectroscopy can offer insights into the actual arrangement and number of atoms in a molecule.

Overall, while the empirical formula gives a glimpse of the elemental ratio, the molecular formula paints a complete picture of the compound's structure.