Question

Ethylene glycol is used as permanent antifreeze. If the molar mass is \(62 \mathrm{~g} / \mathrm{mol}\) and the percent composition is \(38.7 \% \mathrm{C}\), \(9.74 \% \mathrm{H},\) and \(51.6 \% \mathrm{O},\) what is the molecular formula of ethylene glycol?

Short answer

The molecular formula of ethylene glycol is C₂H₆O₂.

Step-by-step solution

Converting Percent Composition to Grams

Assume you have 100 grams of ethylene glycol. This assumption makes calculations easier, as the percent composition can be directly treated as grams. Therefore, you have: - 38.7 g of Carbon (C) - 9.74 g of Hydrogen (H) - 51.6 g of Oxygen (O).

Calculating Moles of Each Element

Calculate the moles of each element using their respective atomic masses:- Moles of Carbon: \(\frac{38.7 \text{ g}}{12.01 \text{ g/mol}} = 3.22 \text{ moles}\)- Moles of Hydrogen: \(\frac{9.74 \text{ g}}{1.008 \text{ g/mol}} = 9.67 \text{ moles}\)- Moles of Oxygen: \(\frac{51.6 \text{ g}}{16.00 \text{ g/mol}} = 3.23 \text{ moles}\)

Determining the Simplest Whole Number Ratio

Divide all mole quantities by the smallest number of moles calculated (3.22):- C: \(\frac{3.22}{3.22} = 1\)- H: \(\frac{9.67}{3.22} = 3\)- O: \(\frac{3.23}{3.22} = 1\)The empirical formula is thus CH₃O.

Determining the Empirical Formula Mass

Calculate the empirical formula mass of CH₃O: - C: 12.01 g/mol - H₃: 3 * 1.008 g/mol = 3.024 g/mol - O: 16.00 g/mol Adding these gives an empirical formula mass of 31.034 g/mol.

Comparing Molar Mass with Empirical Formula Mass

Given molar mass of ethylene glycol is 62 g/mol. Compare it with the empirical formula mass:\(\frac{62 \text{ g/mol}}{31.034 \text{ g/mol}} = 2\).This tells us that the molecular formula is 2 times the empirical formula.

Determining the Molecular Formula

Since the empirical formula is CH₃O and it needs to be doubled, the molecular formula of ethylene glycol is C₂H₆O₂.

Key concepts

Percent Composition

When we talk about percent composition, we're referring to the percentage, by mass, of each element in a compound. This tells us how much of each element is present in a compound compared to the total mass of the compound. For example, in the original problem, ethylene glycol is composed of 38.7% carbon, 9.74% hydrogen, and 51.6% oxygen. This means that in a 100g sample of ethylene glycol, you'll have:

• 38.7g of carbon
• 9.74g of hydrogen
• 51.6g of oxygen

This is a super handy way to break down compounds, especially when you want to convert percentage data to grams to make calculations easier.

Molar Mass

Molar mass is the mass of a given substance (chemical element or chemical compound) divided by the amount of substance (moles). It's represented as grams per mole (g/mol). For any element or compound, the molar mass allows chemists to convert between mass and moles—a valuable tool in chemical calculations.

In our problem, the molar mass of ethylene glycol is provided as 62 g/mol. This information is crucial when determining the molecular formula. By comparing this molar mass to the empirical formula mass, we can understand how many times the empirical unit fits into the actual compound, which ultimately helps us derive the molecular formula.

Empirical Formula

The empirical formula represents the simplest whole-number ratio of the atoms of each element in a compound. It gives us an idea about the relative number of atoms but does not necessarily indicate the actual number of atoms in a molecule, which is what the molecular formula provides.

To find the empirical formula, you begin with the percent composition of the compound. By translating these percentages into grams (assuming a 100g sample, where the percentages become equivalent to mass in grams), and then converting these masses into moles, you can determine the mole ratio of the elements.

The calculated mole values are then divided by the smallest mole value to get a whole-number ratio. For ethylene glycol, after these calculations, we determine the empirical formula to be CH extsubscript{3}O.

Chemical Calculations

Chemical calculations often involve using known quantities and relationships, such as percent composition, molar mass, and empirical formulas, to derive unknown quantities, like the molecular formula. These calculations are a critical part of chemistry because they allow chemists to predict the behavior and composition of chemical substances.

In our ethylene glycol example, once we have the empirical formula CH extsubscript{3}O, the next step is to calculate the empirical formula mass. With the empirical formula comprising carbon (12.01 g/mol), hydrogen (3.024 g/mol), and oxygen (16.00 g/mol), the empirical formula mass sums to 31.034 g/mol.

Given the molar mass of the compound is 62 g/mol, the empirical formula mass needs to be doubled to match it. Therefore, the molecular formula is determined to be C extsubscript{2}H extsubscript{6}O extsubscript{2}, signifying that the actual compound contains twice as many atoms as are represented in the empirical formula. Armed with the skills to perform these calculations, students can tackle a wide array of problems involving chemical compositions.