Problem 70 - Chemistry: An Atoms-Focused Approach

Question

The combustion of 40.5 mg of a compound extracted from the bark of the sassafras tree and known to contain \(\mathrm{C}, \mathrm{H}\) and O produces \(110.0 \mathrm{mg} \mathrm{CO}_{2}\) and \(22.5 \mathrm{mg} \mathrm{H}_{2} \mathrm{O} .\) The molar mass of the compound is 162 g/mol. What are its empirical and molecular formulas?

Step-by-Step Solution

Answer

2. What is the empirical formula of the compound? 3. What is the molecular formula of the compound?

Step 1: Convert the given masses of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2}\mathrm{O}\) produced into moles

Given that 110 mg of \(\mathrm{CO}_{2}\) is produced: Molar mass of \(\mathrm{CO}_{2}\) = 12.01 g/mol (C) + 2 × 16.00 g/mol (O) = 44.01 g/mol Moles of \(\mathrm{CO}_{2}\) = (110 mg × 1 g/1000 mg) / 44.01 g/mol = 0.00250 mol Given that 22.5 mg of \(\mathrm{H}_{2}\mathrm{O}\) is produced: Molar mass of \(\mathrm{H}_{2}\mathrm{O}\) = 2 × 1.01 g/mol (H) + 16.00 g/mol (O) = 18.02 g/mol Moles of \(\mathrm{H}_{2}\mathrm{O}\) = (22.5 mg × 1 g/1000 mg) / 18.02 g/mol = 0.00125 mol

Step 2: Determine the moles of Carbon and Hydrogen in the compound

Since each mole of \(\mathrm{CO}_{2}\) produces one mole of Carbon: Moles of Carbon = 0.00250 mol Since each mole of \(\mathrm{H}_{2}\mathrm{O}\) produces two moles of Hydrogen: Moles of Hydrogen = 0.00125 mol × 2 = 0.00250 mol

Step 3: Calculate the mass of Oxygen in the compound and convert it to moles

Initial mass of the compound = 40.5 mg Mass of Carbon = moles of C × molar mass of C = 0.00250 mol × 12.01 g/mol = 0.03003 g = 30.03 mg Mass of Hydrogen = moles of H × molar mass of H = 0.00250 mol × 1.01 g/mol = 0.002525 g = 2.525 mg Mass of Oxygen = initial mass - mass of Carbon - mass of Hydrogen = 40.5 mg - 30.03 mg - 2.525 mg = 7.945 mg Molar mass of Oxygen = 16.00 g/mol Moles of Oxygen = (7.945 mg × 1 g/1000 mg) / 16.00 g/mol = 0.000497 mol

Step 4: Determine the mole ratio between Carbon, Hydrogen, and Oxygen and find the empirical formula

Divide all mole values by the smallest value to get the mole ratio: Mole ratio of C:H:O = 0.00250 : 0.00250 : 0.000497 Mole ratio of C:H:O ≈ 5 : 5 : 1 (rounded to whole numbers) So, the empirical formula is \(\mathrm{C}_{5}\mathrm{H}_{5}\mathrm{O}\).

Step 5: Use the molar mass to find the molecular formula

To find the molecular formula, we need to determine the ratio between the molar mass of the compound (162 g/mol) and the molar mass of the empirical formula. Molar mass of empirical formula \(\mathrm{C}_{5}\mathrm{H}_{5}\mathrm{O}\) = 5 × 12.01 g/mol (C) + 5 × 1.01 g/mol (H) + 16.00 g/mol (O) = 81 g/mol Since the compound's molar mass is 162 g/mol, we can find the ratio as: Ratio = 162 g/mol / 81 g/mol = 2 Thus, the molecular formula is \(\mathrm{C}_{10}\mathrm{H}_{10}\mathrm{O}_{2}\) (each subscript in the empirical formula multiplied by the ratio).

Key Concepts

Combustion Analysis

Combustion analysis is a method used to determine the composition of a chemical compound by burning it and analyzing the products formed. Typically, when a compound containing carbon (C), hydrogen (H), and oxygen (O) is burned, it produces carbon dioxide (\( CO_2 \)) and water (\( H_2O \)). For instance, the combustion of a compound extracted from the sassafras tree results in measurable quantities of \( CO_2 \) and \( H_2O \) which help us determine the number of moles of carbon and hydrogen present in the original compound based on their distinct molar masses.

The process generally involves these steps:

• Burn the sample to fully convert C to \( CO_2 \) and H to \( H_2O \).
• Measure the masses of \( CO_2 \) and \( H_2O \) produced.
• Use the masses to calculate the moles of carbon and hydrogen originally present.

This method is particularly useful in organic chemistry for determining empirical formulas of organic substances.

Molar Mass Calculations

Molar mass is a critical concept in chemistry as it helps bridge between the mass of a substance and the amount of substance in moles. It represents the mass of one mole of a given substance, usually expressed in grams per mole (g/mol).

Here’s how to calculate the molar mass:

• Identify each type of atom in the molecule and determine the number of each atom present.
• Use the periodic table to find the atomic mass of each atom.
• Multiply the atomic mass by the number of atoms of that element.
• Add up all these values to obtain the total molar mass of the molecule.

For example, \( CO_2 \) has a molar mass of \( 44.01 \) g/mol because it comprises one carbon atom (\( 12.01 \) g/mol) and two oxygen atoms (each \( 16.00 \) g/mol). Knowing this helps in calculating how many moles correspond to a given mass, such as in combustion analysis where we start from actual masses of \( CO_2 \) and \( H_2O \).

Mole Ratios

Mole ratios are crucial for finding the simplest combination of atoms in a compound, often referred to as its empirical formula. With this approach, one seeks to compare the proportional amounts of elements present in a compound.

Here’s how you derive mole ratios:

• Begin with the calculated moles of each element.
• Identify the smallest number of moles among the elements.
• Divide the moles for each element by this smallest value to get a simple ratio.
• Round these ratios to the nearest whole numbers while considering practical chemistry constraints. These often serve as subscripts in the empirical formula.

For example, if the mole ratio of C: H: O initially comes out as roughly \( 5:5:1 \), then the empirical formula is \( C_5H_5O \). Calculating exact ratios is fundamental for understanding the chemical structure at a basic level, and for further steps like deriving the molecular formula.

Chemical Composition

Chemical composition refers to the types and amounts of atoms present in a chemical substance. It is essentially a breakdown of the components and their relative quantities.

For determining the chemical composition of an unknown compound:

• You need to calculate the mass percentages or mole fractions of each element.
• In combustion analysis, this is often handled by reacting the compound and using the resulting products to backtrack to the proportions of the original elements.
• An empirical formula provides the smallest whole-number ratio of the elements in a compound, while a molecular formula gives the actual number of each type of atom in a molecule.

Understanding the chemical composition helps chemists know precisely what a substance is made of, which is invaluable for predicting reactions, determining properties, and synthesizing new compounds. In our example, after calculating individual mole ratios and verifying with the given molar mass, the overall chemical composition identifies the compound found in the bark of the sassafras tree as \( C_{10}H_{10}O_2 \).