Question

An herbicide is found to contain only \(C, H, N\), and \(C 1\) The complete combustion of a \(100.0-\mathrm{mg}\) 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 at STP. A separate analysis shows that the sample also contains \(16.44 \mathrm{mg}\) of \(\mathrm{Cl}\). (a) Determine the percent composition of the substance. (b) Calculate its empirical formula.

Short answer

The percent composition of the herbicide is 44.68% C, 6.59% H, 32.29% N, and 16.44% Cl. The empirical formula of the herbicide is \(C_8H_{14}NCl\).

Step-by-step solution

Calculate moles of CO₂

Use the volume of CO₂ gas to calculate the number of moles, using the molar volume at STP (22.4 L/mol). Moles of CO₂ = \(\cfrac{83.16\,\text{mL}}{22.4\,\text{L/mol} * 1000\,\text{mL/L}}=3.72*10^{-3}\,\text{mol}\)

Calculate moles of H₂O

Similarly, use the volume of H₂O vapor to calculate the number of moles. Moles of H₂O = \(\cfrac{73.30\,\text{mL}}{22.4\,\text{L/mol} * 1000\,\text{mL/L}}=3.27*10^{-3}\,\text{mol}\) Step 2: Calculate the mass of C, H, and N

Calculate mass of C

Each mole of CO₂ contains 1 mole of C. Therefore, the mass of C in the sample is: Mass of C = 3.72*10^{-3}\,\text{mol} * 12.01\,\text{g/mol} = 44.68\,\text{mg}\)

Calculate mass of H

Each mole of H₂O contains 2 moles of H. Therefore, the mass of H in the sample is: Mass of H = 2 * 3.27*10^{-3}\,\text{mol} * 1.01\,\text{g/mol} = 6.59\,\text{mg}\)

Calculate mass of N

Given the mass of C, H, and Cl, we can calculate the mass of N as: Mass of N = 100.0\,\text{mg} - 44.68\,\text{mg} - 6.59\,\text{mg} - 16.44\,\text{mg} = 32.29\,\text{mg}\) Step 3: Find the percent composition of the substance

Calculate percent composition

Divide the mass of each element by the total mass of the sample and multiply by 100 to get the percent composition. Percent composition of C: \(\cfrac{44.68\,\text{mg}}{100.0\,\text{mg}}*100=44.68\,\%\) Percent composition of H: \(\cfrac{6.59\,\text{mg}}{100.0\,\text{mg}}*100=6.59\,\%\) Percent composition of N: \(\cfrac{32.29\,\text{mg}}{100.0\,\text{mg}}*100=32.29\,\%\) Percent composition of Cl: \(\cfrac{16.44\,\text{mg}}{100.0\,\text{mg}}*100=16.44\,\%\) (a) The percent composition of the herbicide is 44.68% C, 6.59% H, 32.29% N, and 16.44% Cl. Step 4: Calculate the empirical formula

Calculate moles of each element

Divide the mass of each element by the corresponding molar mass to get the moles. Moles of C = \(\cfrac{44.68\,\text{mg}}{12.01\,\text{g/mol} * 1000\,\text{mg/g}}=3.72*10^{-3}\,\text{mol}\) Moles of H = \(\cfrac{6.59\,\text{mg}}{1.01\,\text{g/mol} * 1000\,\text{mg/g}}=6.52*10^{-3}\,\text{mol}\) Moles of N = \(\cfrac{32.29\,\text{mg}}{14.01\,\text{g/mol} * 1000\,\text{mg/g}}=2.30*10^{-3}\,\text{mol}\) Moles of Cl = \(\cfrac{16.44\,\text{mg}}{35.45\,\text{g/mol} * 1000\,\text{mg/g}}=4.64*10^{-4}\,\text{mol}\)

Determine empirical formula

Divide all the moles by the smallest number of moles to find the ratio of atoms in the empirical formula. Ratio of C : H : N : Cl = \(\cfrac{3.72*10^{-3}}{4.64*10^{-4}}: \cfrac{6.52*10^{-3}}{4.64*10^{-4}} : \cfrac{2.30*10^{-3}}{4.64*10^{-4}} : \cfrac{4.64*10^{-4}}{4.64*10^{-4}}\) Ratio of C : H : N : Cl = 8 : N 14 : 1 Empirical formula: \(C_8H_{14}NCl\) (b) The empirical formula of the herbicide is \(C_8H_{14}NCl\).

Key concepts

Percent Composition

Understanding percent composition is fundamental when analyzing substances in chemistry. It represents the proportion of each element within a compound, relative to the total mass. To find the percent composition, you would divide the mass of each individual element by the total mass of the compound and then multiply by 100.

In the case of the herbicide described in the exercise, the percent composition was calculated by dividing the mass of each component (carbon, hydrogen, nitrogen, and chlorine) by the total mass of the sample. This calculation provides insight into the relative amounts of each element present and is essential for identifying the empirical formula of the compound.

Molar Volume at STP

The molar volume of a gas at Standard Temperature and Pressure (STP) is a constant value used in stoichiometry to convert between the volume of a gas and the amount of substance in moles. At STP (which is 0°C and 1 atm), one mole of any ideal gas occupies 22.4 liters. This is derived from the ideal gas law when applied to standard conditions.

By using the volume of CO₂ and H₂O vapor collected at STP from the combustion of the herbicide, the respective number of moles for each gas was calculated using the molar volume. This step is crucial for further stoichiometric calculations to determine the empirical formula of the compound.

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It is based on the conservation of mass and the concept of moles, allowing chemists to predict the amounts of substances consumed and produced in a reaction.

In the particular exercise, stoichiometry was used to convert the volume of gases produced during combustion into moles, and hence infer the moles of each element in the original sample. This was followed by a stoichiometric conversion to find the mass of each element, subtracting known masses to deduce the remaining components. Finally, the mole ratios were used to establish the simplest whole-number ratio of atoms in the compound, leading to the empirical formula.

Chemical Analysis

Chemical analysis is a broad field involving processes and techniques to identify and quantify the presence of different elements and compounds. In the case of the herbicide analysis, a combination of gravimetric and volumetric analysis was used to determine the percent composition and empirical formula.

Gravimetric analysis involves measuring the mass of an element or compound directly, which was the case with the chlorine present in the herbicide. Volumetric analysis, on the other hand, includes measurement of the volume to deduce the amount of substance, as seen with the carbon dioxide and water vapor measurements. These techniques, when used appropriately, provide a comprehensive understanding of a substance's chemical composition.