Problem 35 - Chemistry and Chemical Reactivity

Question

An unknown compound has the formula \(\mathrm{C}_{x} \mathrm{H}_{y} \mathrm{O}_{z}\) You burn 0.0956 g of the compound and isolate \(0.1356 \mathrm{g}\) of \(\mathrm{CO}_{2}\) and \(0.0833 \mathrm{g}\) of \(\mathrm{H}_{2} \mathrm{O} .\) What is the empirical formula of the compound? If the molar mass is \(62.1 \mathrm{g} / \mathrm{mol},\) what is the molecular formula?

Step-by-Step Solution

Answer

Empirical formula: CH₃O. Molecular formula: C₂H₆O₂.

Step 1: Determine Moles of Carbon

First, calculate the moles of carbon in the compound. Carbon is found in CO₂. The molar mass of CO₂ is 44.01 g/mol, so use this to find moles of carbon:\[ ext{Moles of } ext{CO}_2 = \frac{0.1356 ext{ g CO}_2}{44.01 ext{ g/mol}} = 0.00308 ext{ mol CO}_2 \]Since each molecule of CO₂ contains one atom of carbon, moles of carbon is equal to moles of CO₂:\[ ext{Moles of Carbon} = 0.00308 ext{ mol}\]

Step 2: Determine Moles of Hydrogen

Next, determine the moles of hydrogen. Hydrogen is found in H₂O. The molar mass of H₂O is 18.02 g/mol:\[ ext{Moles of } ext{H}_2 ext{O} = \frac{0.0833 ext{ g H}_2 ext{O}}{18.02 ext{ g/mol}} = 0.00462 ext{ mol H}_2 ext{O}\]Each molecule of water contains two atoms of hydrogen, so:\[ ext{Moles of Hydrogen} = 2 imes 0.00462 ext{ mol} = 0.00924 ext{ mol} \]

Step 3: Determine Mass and Moles of Oxygen

The total mass of the compound is 0.0956 g, made up of carbon, hydrogen, and oxygen. Subtract the mass due to carbon and hydrogen to find the mass of oxygen:\[ ext{Mass of Carbon} = 0.00308 ext{ mol} imes 12.01 ext{ g/mol} = 0.0370 ext{ g}\]\[ ext{Mass of Hydrogen} = 0.00924 ext{ mol} imes 1.01 ext{ g/mol} = 0.00933 ext{ g}\]\[ ext{Mass of Oxygen} = 0.0956 ext{ g} - 0.0370 ext{ g} - 0.00933 ext{ g} = 0.0493 ext{ g}\]Convert mass to moles of oxygen:\[ ext{Moles of Oxygen} = \frac{0.0493 ext{ g}}{16.00 ext{ g/mol}} = 0.00308 ext{ mol}\]

Step 4: Determine Empirical Formula

Determine the mole ratio of carbon, hydrogen, and oxygen by dividing each by the smallest number of moles (0.00308):\[ ext{Ratio for Carbon} = \frac{0.00308}{0.00308} = 1\]\[ ext{Ratio for Hydrogen} = \frac{0.00924}{0.00308} = 3\]\[ ext{Ratio for Oxygen} = \frac{0.00308}{0.00308} = 1\]Thus, the empirical formula is \( ext{CH}_3 ext{O}\).

Step 5: Determine Molecular Formula

The empirical formula mass of CH₃O is:\[ ext{Molar mass of CH}_3 ext{O} = 12.01+ (3 \times 1.01) + 16.00 = 31.04 ext{ g/mol}\]Divide the molar mass of the compound (62.1 g/mol) by the empirical formula mass to find the ratio:\[\frac{62.1 ext{ g/mol}}{31.04 ext{ g/mol}} = 2\]Therefore, multiply the subscripts in CH₃O by 2 to get the molecular formula:\( ext{C}_2 ext{H}_6 ext{O}_2\).

Key Concepts

Molecular Formula

A molecular formula reveals the actual number of each type of atom present in a molecule of a compound. This formula is derived from the empirical formula by using the molar mass of the compound.
When given the molar mass, you compare it to the empirical formula mass to find a ratio. To calculate the empirical formula mass, sum the atomic masses of the atoms in the empirical formula.
For example, an empirical formula may be \( \text{CH}_3\text{O} \), and its molar mass is calculated as \( 31.04 \text{ g/mol} \). If the actual molar mass of the compound is \( 62.1 \text{ g/mol} \), divide this by the empirical mass to find:

• \( \frac{62.1}{31.04} = 2 \)

Multiply each atom in the empirical formula by this number to find the molecular formula:

• In this case, \( \text{C}_2\text{H}_6\text{O}_2 \).

Combustion Analysis

Combustion analysis is a technique used to determine the elemental composition of a compound. In this method, a known mass of a compound is burned in an excess of oxygen. The carbon and hydrogen in the compound are converted into \( \text{CO}_2 \) and \( \text{H}_2\text{O} \), respectively.
By measuring the amounts of \( \text{CO}_2 \) and \( \text{H}_2\text{O} \) produced, the initial amounts of carbon and hydrogen in the compound can be determined.
This process provides a way to trace the elements in the compound accurately, which is essential for determining empirical and molecular formulas.

• The carbon in the compound corresponds to the carbon in the \( \text{CO}_2 \), while the hydrogen in the water corresponds to the hydrogen in the compound.

Moles Calculation

Calculating moles is vital in determining empirical formulas. Moles relate the mass of a sample to the number of atoms or molecules it contains.
To calculate moles, divide the mass of a substance by its molar mass (molecular weight from the periodic table).
For example:

• For \( \text{CO}_2 \): \( \frac{0.1356 \text{ g}}{44.01 \text{ g/mol}} = 0.00308 \text{ mol} \)
• Every molecule of \( \text{CO}_2 \) contains one carbon atom; thus, moles of carbon is \( 0.00308 \) mol.

This process is repeated for all components (carbon, hydrogen, and oxygen) in the compound to find the moles required for further calculations.

Elemental Composition

The elemental composition of a compound refers to the percentage or amount of each element within that compound.
When conducting an analysis, you first determine the moles of each element present. Using the mass from measurements and the moles calculated, you can determine the composition of the compound.
For instance, in the provided analysis, the masses of carbon, hydrogen, and oxygen are calculated and converted into moles. Then, the proportions of these elements are used to derive the empirical formula.

• Calculate each element's mass and transform it into moles.
• Determine the total mass of all elements.
• Compare each element's mole value to this total to find their relative composition.

These detailed calculations allow chemists to understand the makeup of compounds, crucial for applications in new compound synthesis and analysis.