Problem 120 - Chemistry The Central Science

Question

An herbicide is found to contain only \(\mathrm{C}, \mathrm{H}, \mathrm{N},\) and 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 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?

Step-by-Step Solution

Answer

The herbicide sample has a percentage composition of 44.5% C, 6.59% H, 32.47% N, and 16.44% Cl. The empirical formula is C8H14N5Cl. To find the true molecular formula, the molar mass of the actual compound is needed.

Step 1: Determine the moles of Carbon and Hydrogen

With the given volumes of CO2 and H2O, we first need to convert these volumes to moles. Recall that at STP, 1 mole of any gas occupies 22.4 L or 22,400 mL. Moles of CO2 = \(\frac{83.16}{22,400}\) moles = \(3.71\times10^{-3}\) moles As each mole of CO2 contains 1 mole of carbon, there are \(3.71\times10^{-3}\) moles of carbon. Moles of H2O = \(\frac{73.30}{22,400}\) moles = \(3.27\times10^{-3}\) moles As each mole of H2O contains 2 moles of hydrogen, there are \(2\times3.27\times10^{-3}=6.54\times10^{-3}\) moles of hydrogen.

Step 2: Calculate the masses of Carbon, Hydrogen, and Nitrogen

Using the molar masses of C (12.01 g/mol), H (1.008 g/mol), and N (14.01 g/mol), we calculate the masses of C, H, and N in the sample. Mass of C = \(3.71\times10^{-3}\) moles \(\times\) 12.01 g/mol = 0.0445 g Mass of H = \(6.54\times10^{-3}\) moles \(\times\) 1.008 g/mol = 0.00659 g We know that the sample is 100.0 mg, and we are given the mass of Cl (16.44 mg). Thus, we can calculate the mass of N. Mass of N = 100.0 mg - Mass of C - Mass of H - Mass of Cl Mass of N = 100.0 mg - 44.5 mg - 6.59 mg - 16.44 mg = 32.47 mg

Step 3: Determine the percentage composition of the substance

Now that we know the masses of C, H, N, and Cl, we can use their masses to determine the percentage of each element in the sample. % C = \(\frac{44.5}{100.0}\) × 100% = 44.5% % H = \(\frac{6.59}{100.0}\) × 100% = 6.59% % N = \(\frac{32.47}{100.0}\) × 100% = 32.47% % Cl = \(\frac{16.44}{100.0}\) × 100% = 16.44%

Step 4: Calculate the empirical formula

We first calculate the moles of each element, then divide each by the smallest value of moles to find the mole ratio. Moles of N = \(\frac{32.47\text{ mg}\times1\text{ g}}{14.01\text{ g/mol}\times1000\text{ mg/g}}\) = \(2.32\times10^{-3}\) moles As we already found the moles of C and H in Step 1, we now have: C: \(3.71\times10^{-3}\) moles H: \(6.54\times10^{-3}\) moles N: \(2.32\times10^{-3}\) moles Cl: We need to calculate the moles of Cl. Using the molar mass of Cl (35.45 g/mol), we find that there are \(4.64\times10^{-4}\) moles of Cl in the sample. Now we divide each molar amount by the smallest value (moles of Cl) to find the mole ratio: C: \(\frac{3.71\times10^{-3}\text{ moles}}{4.64\times10^{-4}\text{ moles}}\) ≈ 8 H: \(\frac{6.54\times10^{-3}\text{ moles}}{4.64\times10^{-4}\text{ moles}}\) ≈ 14 N: \(\frac{2.32\times10^{-3}\text{ moles}}{4.64\times10^{-4}\text{ moles}}\) ≈ 5 Cl: \(\frac{4.64\times10^{-4}\text{ moles}}{4.64\times10^{-4}\text{ moles}}\) ≈ 1 These values are very close to whole numbers, so the empirical formula is C8H14N5Cl.

Step 5: Information needed to find the true molecular formula

To find the true molecular formula, we would need information about the molar mass of the actual compound. This information could be obtained through methods such as mass spectrometry. Once the molar mass of the compound is found, divide the molar mass of the compound by the molar mass of the empirical formula to find the ratio between the two. If this ratio is a whole number, multiply each subscript in the empirical formula by that whole number to obtain the true molecular formula.

Key Concepts

Chemical Composition Analysis

Understanding the chemical composition of a substance is crucial for various applications in chemistry, including the formulation of chemical products like herbicides. The exercise involves a process called chemical composition analysis where a sample of herbicide was completely combusted to determine the amounts of carbon dioxide (CO2) and water (H2O) produced. These measurements allow us to backtrack and determine how much of each element—carbon (C), hydrogen (H), nitrogen (N), and chlorine (Cl)—was present in the original sample.

To achieve this, the volumes of the gases produced during combustion are measured and then converted into moles, because the amount of substance is more meaningful in the context of stoichiometry when expressed in this way. The analysis is based on principles such as the conservation of mass and the known stoichiometry of combustion reactions. For instance, knowing that CO2 is produced from C and O2 and that H2O is formed from H and O2, scientists can work out the amounts of C and H in the original compound. Additionally, because the conditions are standard temperature and pressure (STP), useful conversions like 1 mole of any gas occupies 22.4 liters can be applied. Later, calculating the percentage composition involves comparing the mass of each element to the total mass of the sample.

Stoichiometry

The core of solving this problem lies within the realm of stoichiometry, a section of chemistry that involves the quantitative relationships between the reactants and products in a chemical reaction. After combustion, stoichiometry helps us understand these relationships through the mole concept, demonstrating the bridge between the microscale world of atoms and molecules and the macroscale world we can measure.

Using the data from the herbicide's combustion, stoichiometry becomes the tool by which we interpret the volumes of gases released to determine the moles and subsequently the mass of each element in the sample. This is based on the stoichiometric coefficients found in balanced chemical equations, where, for example, one mole of CO2 always comes from one mole of carbon atoms. The ratio in which elements combine according to their moles—found by dividing the moles of each element by the smallest number of moles—is essential to determine the empirical formula. The empirical formula represents the simplest whole-number ratio between the elements in the substance, and breaking it down into this simplest form is an illuminating example of stoichiometry at work.

Molar Mass

Determining the molar mass of the individual elements and the compounds they form is a fundamental step in chemistry calculations. The molar mass is defined as the mass of one mole of a substance (in grams per mole), and it is a critical property used to convert between the mass of a substance and the amount in moles.

Knowing a substance's molar mass, as provided for C, H, N, and Cl, enables one to move effortlessly between the weight of a sample and the number of moles it contains. This conversion is pivotal in finding the empirical formula, as it allows for the derivation of mole ratios on a common basis. For instance, after calculating the moles of carbon from the volume of CO2 released, we used the molar mass of carbon to then find the mass of carbon in the original sample. In a similar way, a complete understanding of the substance, including its true molecular formula, requires knowing its total molar mass. To move from the empirical formula to the molecular formula, one must know the molar mass of the compound as a whole, which can be found using techniques such as mass spectrometry. The difference between the molar mass of the empirical formula unit and the actual molar mass of the compound can reveal how the empirical units stack up to form the true molecule.