Question

Ethylene glycol, commonly used as automobile antifreeze, contains only carbon, hydrogen, and oxygen. Combustion analysis of a \(23.46 \mathrm{mg}\) sample yields \(20.42 \mathrm{mg}\) of \(\mathrm{H}_{2} \mathrm{O}\) and \(33.27 \mathrm{mg}\) of \(\mathrm{CO}_{2}\). What is the empirical formula of ethylene glycol? What is its molecular formula if it has a molecular weight of \(62.0 ?\)

Short answer

Empirical formula: CH₃O, Molecular formula: C₂H₆O₂.

Step-by-step solution

Convert Measurements to Moles

First, convert the masses of water ((H_2O)) and carbon dioxide ((CO_2)) generated into moles to determine the moles of hydrogen and carbon, respectively. The molar mass of (H_2O) is 18.02 g/mol and (CO_2) is 44.01 g/mol.Mass of water is 20.42 mg = 0.02042 g.\[\text{Moles of } \mathrm{H}_2\mathrm{O} = \frac{0.02042 \text{ g}}{18.02 \text{ g/mol}} \approx 0.00113 \text{ mol}\]Mass of carbon dioxide is 33.27 mg = 0.03327 g.\[\text{Moles of } \mathrm{CO}_2 = \frac{0.03327 \text{ g}}{44.01 \text{ g/mol}} \approx 0.000756 \text{ mol}\]

Determine Moles of Elements

Next, calculate the moles of carbon and hydrogen from the moles of carbon dioxide and water.Each mole of (CO_2) contains one mole of carbon, thus:\[\text{Moles of C} = 0.000756 \text{ mol}\]Each mole of (H_2O) contains two moles of hydrogen:\[\text{Moles of H} = 2 \times 0.00113 \text{ mol} = 0.00226 \text{ mol}\]

Calculate Mass of Oxygen

Calculate the total mass of carbon and hydrogen in the original sample, then find mass of oxygen by difference. Mass of carbon in \(23.46 \mathrm{mg}\) sample:\[\text{Mass of C} = 0.000756 \text{ mol} \times 12.01 \text{ g/mol} \approx 0.00908 \text{ g}\]Mass of hydrogen:\[\text{Mass of H} = 0.00226 \text{ mol} \times 1.008 \text{ g/mol} \approx 0.00228 \text{ g}\]Total mass of carbon and hydrogen:\[0.00908 \text{ g} + 0.00228 \text{ g} = 0.01136 \text{ g}\]Mass of oxygen in the sample (23.46 mg = 0.02346 g):\[\text{Mass of O} = 0.02346 \text{ g} - 0.01136 \text{ g} \approx 0.01210 \text{ g}\]

Convert Mass of Oxygen to Moles

Convert the mass of oxygen to moles to find its quantity in the sample using its molar mass, 16.00 g/mol.\[\text{Moles of O} = \frac{0.01210 \text{ g}}{16.00 \text{ g/mol}} \approx 0.000756 \text{ mol}\]

Determine Empirical Formula

Use the mole ratios to determine the empirical formula by dividing by the smallest number of moles.\[\text{C} = \frac{0.000756}{0.000756} = 1,\quad \text{H} = \frac{0.00226}{0.000756} \approx 3,\quad \text{O} = \frac{0.000756}{0.000756} = 1\]So, the empirical formula of ethylene glycol is (CH_3O).

Determine Molecular Formula

Now, to find the molecular formula, compare the empirical formula's molar mass with the given molecular weight of 62.0. The molar mass of (CH_3O) is 31.03 g/mol.\[\frac{62.0}{31.03} \approx 2\]Thus, multiply the entire empirical formula by 2: (C_2H_6O_2).So, the molecular formula is (C_2H_6O_2).

Key concepts

Empirical Formula

The empirical formula is the simplest way we can represent a compound's composition. It shows the ratio of each element in the compound, using the smallest whole numbers possible. To determine the empirical formula, you start with the combustion analysis data, often provided in terms of grams or milligrams of products like water and carbon dioxide.

In our example, the product water helps us find moles of hydrogen, and carbon dioxide helps in finding moles of carbon. After we calculate these using their respective molar masses, we aim to find the actual number of atoms by converting these moles into ratios. These mole ratios are then simplified by dividing by the smallest number of moles among the elements present. In our case, for ethylene glycol, this simplification leads to the empirical formula \( \text{CH}_3\text{O} \).

• The empirical formula gives no direct information about the number of atoms in a molecule.
• It represents underlying atomic relationships in the compound's composition.

Molecular Formula

Once an empirical formula is determined, the journey towards finding the molecular formula begins. The molecular formula reflects the actual number of atoms present in the molecule, which sometimes is the same as the empirical formula. To find the molecular formula, you need the compound's molar mass.

In the provided example, calculate the empirical formula's molar mass first. Then, see how this relates to the provided molecular weight. For ethylene glycol, the empirical molar mass (\( \text{CH}_3\text{O} \)) is\(31.03\, ext{g/mol}\). Given a molecular weight of\(62.0\, ext{g/mol}\), the factor we find is 2. This indicates the actual compound consists of twice the numbers represented by the empirical formula. So, multiplying all parts by this factor gives us the molecular formula, which is \( \text{C}_2\text{H}_6\text{O}_2 \).

• The molecular formula shows both the type and the exact number of atoms in a molecule.
• It's essential for understanding the structure and properties of the compound.

Ethylene Glycol

Ethylene glycol is a colorless, sweet-tasting liquid widely used as an antifreeze in automobile engines. Its chemical formula is \( \text{C}_2\text{H}_6\text{O}_2 \), reflecting its useful properties like being a good heat transfer agent.

This chemical compound also finds use in manufacturing polyester fibers and as a raw material in the production of various resins. It is essential to handle with caution due to its toxicity if ingested. Understanding its molecular structure helps people make informed decisions in its use and processing.

• Structure-wise, ethylene glycol consists of two hydroxyl groups attached to a two-carbon chain.
• This diol setup gives ethylene glycol its hygroscopic nature, meaning it can attract and retain moisture.

Chemical Stoichiometry

Chemical stoichiometry is the heart of quantitative chemistry, providing insight into the proportions of reactants and products involved in a chemical reaction. It is what allows us to precisely measure how much of each substance will be involved or produced in a reaction, based on balanced chemical equations.

Applying stoichiometry involves converting between masses, moles, and molecules. This is pivotal in analysis tasks like combustion analysis where we deduce compound formulas. With ethylene glycol, stoichiometry helped in converting mass obtained from combustion into moles of individual elements, eventually guiding us to empirical and molecular formulas. Understanding these ratios provides clarity for predictions about reaction yields and optimal reactant amounts.

• Stoichiometry is built around the conservation of mass; atoms are neither created nor destroyed.
• It is crucial in laboratory and industrial chemistry for making accurate chemical predictions and scaling reactions.