Problem 114 - 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 percent composition of the substance. (b) Calculate its empirical formula.

Step-by-Step Solution

Answer

The percent composition of the herbicide is 44.56% C, 13.24% H, 25.76% N, and 16.44% Cl. The empirical formula is \(C_8H_{28}N_4Cl\).

Step 1: Calculate the mass of C and H from the combustion products

Using the given information, we can determine the masses of C and H from the CO₂ and H₂O production, respectively. By using the molar volume of gas at STP (22.4 L/mol), we can calculate the moles and then the mass of each element. From CO₂ production: \[\textrm{moles of }CO_{2}=\frac{83.16 \ \textrm{mL}}{1000 \ \textrm{mL/mol}\times22.4 \ \textrm{L/mol}}= 0.003712 \ \textrm{mol}\] Mass of C: \(0.003712 \ \textrm{mol}\times12.01 \ \textrm{g/mol}= 0.04456 \ \textrm{g} = 44.56 \ \textrm{mg}\) From H₂O production: \[\textrm{moles of }H_{2}O=\frac{73.30 \ \textrm{mL}}{1000 \ \textrm{mL/mol}\times22.4 \ \textrm{L/mol}}= 0.003275 \ \textrm{mol}\] Mass of H: \((0.003275 \ \textrm{mol}\times2.016 \ \textrm{g/mol}\times2)= 0.013238 \ \textrm{g} = 13.24 \ \textrm{mg}\)

Step 2: Calculate the mass of N

We have already found the masses of C, H, and Cl in the sample. We can now find the mass of N by subtracting the masses of these elements from the total mass. Mass of N: \(100.0 \textrm{mg} - 44.56 \textrm{mg} - 13.24 \textrm{mg} - 16.44 \textrm{mg} = 25.76 \textrm{mg}\)

Step 3: Determine the percent composition of the substance

Now we have the masses of all elements in the sample. We can divide each mass by the total mass and multiply by 100 to find the percent composition. Percent C: \(\frac{44.56 \ \textrm{mg}}{100.0 \ \textrm{mg}}\times100\%=44.56\%\) Percent H: \(\frac{13.24 \ \textrm{mg}}{100.0 \ \textrm{mg}}\times100\%=13.24\%\) Percent N: \(\frac{25.76 \ \textrm{mg}}{100.0 \ \textrm{mg}}\times100\%=25.76\%\) Percent Cl: \(\frac{16.44 \ \textrm{mg}}{100.0 \ \textrm{mg}}\times100\%=16.44\%\)

Step 4: Determine the moles of each element

To find the empirical formula, we must calculate the moles of each element in the sample. Moles of C: \(\frac{44.56 \ \textrm{mg}}{12.01 \ \textrm{mg/mol}}= 3.713 \ \textrm{mol}\) Moles of H: \(\frac{13.24 \ \textrm{mg}}{1.008 \ \textrm{mg/mol}}= 13.14 \ \textrm{mol}\) Moles of N: \(\frac{25.76 \ \textrm{mg}}{14.01 \ \textrm{mg/mol}}= 1.839 \ \textrm{mol}\) Moles of Cl: \(\frac{16.44 \ \textrm{mg}}{35.453 \ \textrm{mg/mol}}= 0.4636 \ \textrm{mol}\)

Step 5: Calculate mole ratios and empirical formula

To obtain the empirical formula, divide each number of moles by the smallest number of moles (0.4636 mol) to get the mole ratios. C: \(\frac{3.713 \ \textrm{mol}}{0.4636 \ \textrm{mol}}= 8.01\) H: \(\frac{13.14 \ \textrm{mol}}{0.4636 \ \textrm{mol}}= 28.35\) N: \(\frac{1.839 \ \textrm{mol}}{0.4636 \ \textrm{mol}}= 3.97\) Cl: \(\frac{0.4636 \ \textrm{mol}}{0.4636 \ \textrm{mol}}= 1\) Rounding each ratio to the nearest whole number, the empirical formula is \(C_8H_{28}N_4Cl\).

Key Concepts

Percent Composition

The percent composition of a compound represents the percentage by mass of each element within the compound. It's a crucial part of analyzing chemical substances since it allows chemists to determine the elemental makeup and purity of the sample.

To calculate the percent composition, you must first know the mass of each element in the compound. Once you have these masses, you use the formula:

• Percentage of Element = \(\frac{\text{Mass of Element}}{\text{Total Mass of Compound}} \times 100\%\)

In our exercise, we determined the mass of carbon, hydrogen, nitrogen, and chlorine in the herbicide. By dividing these individual masses by the total mass of the herbicide, and then multiplying by 100, we obtained the respective percentages: 44.56% for carbon, 13.24% for hydrogen, 25.76% for nitrogen, and 16.44% for chlorine. Understanding percent composition is an essential step in the larger process of identifying a substance’s empirical formula.

Combustion Analysis

Combustion analysis is a process used to determine the chemical composition of a compound containing carbon and hydrogen by burning it in excess oxygen, converting the elements into carbon dioxide (CO₂) and water (H₂O). This technique is particularly useful for organic compounds since it allows for precise measurement of the amounts of carbon and hydrogen in a sample.

During combustion, carbon is converted to CO₂ and hydrogen to H₂O. By measuring the amount of CO₂ produced, we can determine the mass of carbon present in the sample, and similarly, the amount of water vapor gives us the mass of hydrogen. For instance, in our exercise, using combustion analysis, the volumes of CO₂ and H₂O were used to calculate the moles, and further the mass of carbon and hydrogen.

• CO₂ Volume → Moles of CO₂ → Mass of Carbon
• H₂O Volume → Moles of H₂O → Mass of Hydrogen

One interesting aspect of combustion analysis is that it’s often combined with other techniques, such as separate analysis for elements not converted to gaseous products, like chlorine in our scenario.

Moles Calculation

Calculating the moles of substances is often the first step in chemical analysis, crucial for determining empirical formulas. A mole is a unit in chemistry used to express amounts of a chemical substance, defined as Avogadro's number of particles (atoms, molecules, etc.).

To find the number of moles of an element, you divide the mass of the element by its molar mass (the mass of one mole of atoms of an element). The equation is:

• \(\text{Moles} = \frac{\text{Mass of Element}}{\text{Molar Mass}} \)

In our exercise, we calculated the moles of carbon, hydrogen, nitrogen, and chlorine using their respective masses and molar masses. For example, the mass of carbon from the compound was divided by 12.01 g/mol (the molar mass of carbon) to get the moles of carbon. This systematic conversion allows us to compare chemical amounts in the simplistic form of moles, simplifying empirical formula calculations.

Molar Volume

Molar volume is a concept that comes into play when dealing with gases. It is defined as the volume occupied by one mole of a gas at standard temperature and pressure (STP) conditions, which are 0°C (273.15 K) and 1 atmosphere of pressure. At STP, one mole of any ideal gas occupies 22.4 liters.

Molar volume is leveraged in experiments like combustion analysis where gas volumes need to be converted into moles. In the earlier discussed exercise, we used the molar volume to calculate the number of moles of CO₂ and H₂O gases produced as a result of burning the sample. By dividing the given volume of gas by the molar volume (22.4 L/mol), we obtain the moles of the gas. Knowing this information is key to further calculations involving mass and mole ratios.

• Example: \(\text{Moles of gas} = \frac{\text{Volume of gas (at STP)}}{22.4 \text{ L/mol}} \)

This concept highlights how standard values in chemistry, such as molar volume, simplify the calculation processes across various scientific analyses.