Problem 96 - Chemistry The Central Science In Si Units

Question

An organic compound was found to contain only \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{Cl}\). When a \(1.50-g\) sample of the compound was completely combusted in air, \(3.52 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) was formed. In a separate experiment, the chlorine in a \(1.00-\mathrm{g}\) sample of the compound was converted to \(1.27 \mathrm{~g}\) of AgCl. Determine the empirical formula of the compound.

Step-by-Step Solution

Answer

The empirical formula of the compound is \(\text{C}_9\text{H}_{25}\text{Cl}\).

Step 1: Determine Moles of Carbon

First, find the moles of carbon in the given amount of \(\text{CO}_2\). Since each molecule of \(\text{CO}_2\) contains one atom of carbon, use the molar mass of \(\text{CO}_2\) (44.01 g/mol) to convert grams to moles:\[\text{Moles of } \text{CO}_2 = \frac{3.52 \text{ g}}{44.01 \text{ g/mol}} = 0.0800 \text{ mol}\]Since 1 mole of \(\text{CO}_2\) gives 1 mole of \(\text{C}\), the moles of \(\text{C}\) is 0.0800 mol.

Step 2: Calculate Mass of Carbon

Now calculate the mass of carbon, knowing that the atomic mass of carbon is approximately 12.01 g/mol:\[\text{Mass of C} = 0.0800 \text{ mol} \times 12.01 \text{ g/mol} = 0.961 \text{ g}\]

Step 3: Analyze Chlorine Content

In a separate 1.00-g sample, chlorine is precipitated as \(\text{AgCl}\). Calculate the moles of \(\text{AgCl}\) using its molar mass (143.32 g/mol):\[\text{Moles of } \text{AgCl} = \frac{1.27 \text{ g}}{143.32 \text{ g/mol}} = 0.00886 \text{ mol}\]Since \(\text{AgCl}\) contains one mole of \(\text{Cl}\) per mole of \(\text{AgCl}\), the moles of \(\text{Cl}\) are also 0.00886 mol.

Step 4: Calculate Mass of Chlorine

Determine the mass of chlorine by multiplying the moles of chlorine by its atomic mass (35.45 g/mol):\[\text{Mass of Cl} = 0.00886 \text{ mol} \times 35.45 \text{ g/mol} = 0.314 \text{ g}\]

Step 5: Determine Mass of Hydrogen

Calculate the mass of hydrogen in the initial 1.50-g sample from the total mass minus the masses of carbon and chlorine:\[\text{Mass of H} = 1.50 \text{ g} - 0.961 \text{ g (C)} - 0.314 \text{ g (Cl)} = 0.225 \text{ g}\]

Step 6: Calculate Moles of Hydrogen

Find moles of hydrogen using its atomic mass (1.008 g/mol):\[\text{Moles of H} = \frac{0.225 \text{ g}}{1.008 \text{ g/mol}} = 0.223 \text{ mol}\]

Step 7: Find Empirical Formula Ratios

Now, calculate the ratio of moles of each element. The number of moles we've calculated are:- C: 0.0800 mol- H: 0.223 mol- Cl: 0.00886 molDivide each by the smallest number of moles (Cl = 0.00886):\[\text{Ratio of C} = \frac{0.0800}{0.00886} \approx 9.03 \approx 9\]\[\text{Ratio of H} = \frac{0.223}{0.00886} \approx 25.17 \approx 25\]\[\text{Ratio of Cl} = \frac{0.00886}{0.00886} = 1\]

Step 8: Write Empirical Formula

Convert the ratios to the nearest whole numbers to get the empirical formula: \(\text{C}_9\text{H}_{25}\text{Cl}\).

Key Concepts

Combustion Analysis

Combustion analysis is a critical technique used in chemistry to determine the elemental composition of a compound. In the exercise provided, we see its application in determining the amount of carbon present in an organic compound. When a compound is completely combusted, it reacts with oxygen to produce carbon dioxide (\( \text{CO}_2 \)) and often water (\( \text{H}_2\text{O} \)) if hydrogen is present as well. This technique leverages the fact that each molecule of \( \text{CO}_2 \) produced contains one atom of carbon from the original organic compound.
By measuring the mass of \( \text{CO}_2 \) formed, we can calculate the number of moles of carbon in the sample. Use the equation:
\[\text{Moles of } \text{CO}_2 = \frac{\text{mass of } \text{CO}_2}{\text{molar mass of } \text{CO}_2}\]
In this exercise, you measured a yield of \( 3.52 \text{ g} \) of \( \text{CO}_2 \), and after calculating, you found there were \( 0.0800 \text{ mol} \) of carbon.
This forms the basis for further calculations, which can then be utilized to understand the complete elemental composition of the compound.

Elemental Composition

Understanding the elemental composition is essential for determining the empirical formula of a compound. It involves calculating the amount of each element present in a substance. In the exercise, after calculating the moles of carbon from the combustion analysis, we turned our attention to finding the number of moles of hydrogen and chlorine.
To find chlorine, a portion of the sample was reacted to form \( \text{AgCl} \), a compound where each molecule corresponds to one atom of chlorine. By measuring \( 1.27 \text{ g} \) of \( \text{AgCl} \), you calculated \( 0.00886 \text{ mol} \) of chlorine, showing how elemental composition extends beyond just carbon and hydrogen analysis.
Calculating hydrogen involved subtracting the mass of the other elements from the total mass and then converting to moles using hydrogen's atomic mass. This approach is a cornerstone in empirical formula calculations, allowing for deduction of molecular structures based on mass percentages or weights observed after reactions.

Organic Chemistry

Organic chemistry studies compounds predominantly made of carbon and hydrogen, possibly with other elements, such as chlorine, sulfur, or nitrogen. In organic chemistry, the empirical formula is a fundamental concept representing the simplest ratio of the elements in a molecule.
The exercise shows you determining the empirical formula from the elemental composition derived through combustion and other quantitative analyses. For a compound containing carbon, hydrogen, and chlorine, understanding and interpreting chemical data through stoichiometry is key.
Calculating the empirical formula involves dividing the number of moles of each element by the smallest number of moles present (in this case, chlorine), then adjusting these ratios to the smallest whole numbers. This gives \( \text{C}_9\text{H}_{25}\text{Cl} \), representing the fundamental nature of the compound's structure within the framework of organic chemistry.
Moreover, understanding the principles behind elemental determination and translating that into molecular insight further encapsulates the essence of organic chemistry.