Problem 121 - Chemistry: The Central Science

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 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?

Step-by-Step Solution

Answer

The percentage composition of the herbicide is: \%C = 29.99\%, \%H = 7.33\%, \%N = 46.24\%, and \%Cl = 16.44\%. The empirical formula is C₁H₂N₂Cl. To find the true molecular formula, we would need to know the molar mass of the compound.

Step 1: Calculate moles of CO2 and H2O

First, we need to find the moles of CO2 and H2O produced. As both of them are at STP, we can use the molar volume of gases at STP (22.4 L/mol) to do this: Moles of CO2 = \(\frac{83.16 \, \mathrm{mL}}{22400 \, \mathrm{mL/mol}}\) Moles of H2O = \(\frac{73.30 \, \mathrm{mL}}{22400 \, \mathrm{mL/mol}}\)

Step 2: Calculate the masses of C and H

Now, we'll use the moles derived in step 1 to find the mass of C and H in the sample: Mass of C = Moles of CO2 × Molar mass of C = \(83.16 \times 12\) Mass of H = Moles of H2O × (2 × Molar mass of H) = \(73.30 \times 1\)

Step 3: Calculate the mass of N

The mass of substance = 100.0 mg Mass of Cl = 16.44 mg Thus, the mass of N is the difference between the total mass and the masses of C, H, and Cl: Mass of N = Mass of substance - Mass of C - Mass of H - Mass of Cl

Step 4: Find the percentage composition

Now, we have the mass of C, H, N, and Cl. We can find the percentage composition of these elements by: Percentage of C = \(\frac{\mathrm{Mass \, of \, C}}{\mathrm{Total \, Mass \, of \, Sample}} \times 100\) Percentage of H = \(\frac{\mathrm{Mass \, of \, H}}{\mathrm{Total \, Mass \, of \, Sample}} \times 100\) Percentage of N = \(\frac{\mathrm{Mass \, of \, N}}{\mathrm{Total \, Mass \, of \, Sample}} \times 100\) Percentage of Cl = \(\frac{\mathrm{Mass \, of \, Cl}}{\mathrm{Total \, Mass \, of \, Sample}} \times 100\)

Step 5: Calculate the empirical formula

Using the percentage composition of each element, we can find the empirical formula. To do this, first convert the percentages into grams, and then into moles: Moles of each element = \(\frac{\mathrm{Mass \,of \,the \,element}}{\mathrm{Molar \,mass \,of \,the \,element}}\) Next, find the mole ratio among the elements by dividing each mole by the smallest moles: Mole Ratio = \(\frac{\mathrm{Moles \,of \,the \,element}}{\mathrm{Smallest \,atom \, moles}}\) Now, round off these ratios to the nearest whole number and find the empirical formula.

Step 6: Identify the information needed for the true molecular formula

To find the true molecular formula, we need to know the molar mass of the compound. Once we have the molar mass, we can divide it by the empirical formula mass and obtain a whole number, called the "n" factor. Multiplying the empirical formula by the "n" factor will give us the true molecular formula.

Key Concepts

Stoichiometry

In the realm of chemistry, stoichiometry is like the bookkeeper of reactions, meticulously tracking the relationships between reactants and products. It tells us the proportion of reactants required to produce a certain amount of product based on the laws of conservation of mass. To master stoichiometry, one needs to become adept at balancing chemical equations and converting between moles, mass, and volumes of reactants and products.

Visualize stoichiometry as a recipe, where ingredients (reactants) are mixed in precise ratios to bake a delicious cake (the product). For instance, in the herbicide problem, stoichiometry helps us understand how much carbon (C) and hydrogen (H) are produced when the herbicide combusts, guiding us to compute the amounts converted into carbon dioxide (CO2) and water (H2O).

Combustion Analysis

Turning our attention to combustion analysis, it's a reliable method for identifying the elemental makeup of a substance by completely burning it in oxygen. In this controlled burn, known as combustion, elements within the compound react to form carbon dioxide, water, and other products that can be measured. This process comes in handy to determine amounts of carbon and hydrogen in a compound.

To illustrate, in the exercise about the herbicide, combustion analysis revealed the volume of CO2 and H2O produced, which then allowed for the calculation of moles of carbon and hydrogen. These values are the building blocks for determining the herbicide's empirical formula, which, in essence, serves as a means of decoding the compound's elemental signature.

Percentage Composition

Diving into percentage composition, it gives us insight into the 'diet' of the compound—precisely what and how much of each element it contains. Just as nutrition labels break down the contents of food into fats, proteins, etc., percentage composition breaks down compounds into their elemental constituents by mass.

Interpreting the herbicide example, upon finding the weights of carbon, hydrogen, nitrogen, and chlorine, they're turned into percentages, revealing the relative mass of each element as a part of the whole compound. These percentages not only feed into the empirical formula calculations but also help chemists understand the compound's potential properties and reactions.

Molecular Formula Determination

The quest to determine a compound's molecular formula is akin to uncovering its true identity—it's the complete representation of the types and numbers of atoms in a molecule. While the empirical formula provides a simplified sketch, the molecular formula paints the full picture.

In the study of our herbicide, we're given a teaser—the empirical formula, but to reveal the compound's deeper identity, the actual molar mass is necessary. It's the molar mass that enables us to scale up from the empirical to the molecular formula, ensuring the calculated weights align with reality. Figuratively, if the empirical formula is a DNA segment, then the molecular formula is the whole genetic blueprint of the molecule.