Problem 113 - Chemistry The Central Science In Si Units

Question

A compound composed of \(6.7 \% \mathrm{H}, 40.0 \% \mathrm{C},\) and \(53.3 \% \mathrm{O}\) has a molar mass of approximately \(60 \mathrm{~g} / \mathrm{mol}\). (a) What is the molecular formula of the compound? (b) What is its Lewis structure if the two \(\mathrm{O}\) are bonded to \(\mathrm{C} ?(\mathbf{c})\) What is the geometry and hybridization of the \(\mathrm{C}\) atom that is bonded to \(2 \mathrm{O}\) atoms? (d) How many \(\sigma\) and how many \(\pi\) bonds are there in the molecule?

Step-by-Step Solution

Answer

(a) The molecular formula is \(\text{C}_2\text{H}_4\text{O}_2\). (b) Lewis structure has C bonded to two O atoms, one being an OH and the other as C=O. (c) Geometry around C is trigonal planar, with \(sp^2\) hybridization. (d) There are 6 \(\sigma\) bonds and 1 \(\pi\) bond.

Step 1: Convert Percentages to Grams

Assume we have 100 g of the compound, so we have 6.7 g of H, 40.0 g of C, and 53.3 g of O. This simplifies further calculations using moles.

Step 2: Calculate Moles of Each Element

Use the molar mass of each element to find the number of moles: \(\text{moles of H} = \frac{6.7}{1.01}, \text{moles of C} = \frac{40.0}{12.01}, \text{moles of O} = \frac{53.3}{16.00}\).

Step 3: Determine the Empirical Formula

Divide each of the mole values calculated in Step 2 by the smallest number of moles among the three elements and round to the nearest whole number if necessary to find empirical formula subscripts.

Step 4: Find the Empirical Formula Molar Mass

Calculate the empirical formula molar mass using the atomic weights and determine how many times the empirical formula's mass fits into the given molar mass of the compound, 60 g/mol.

Step 5: Derive the Molecular Formula

Multiply the subscripts in the empirical formula by the factor found in Step 4 to determine the molecular formula of the compound.

Step 6: Draw Lewis Structure

Using the molecular formula, draw the Lewis structure ensuring each atom follows the octet rule. Given two O atoms are bonded to C and H is likely to bond to O forming alcohol or ether-like structures.

Step 7: Identify Geometry and Hybridization of C

For part (c), analyze C bonded to two O atoms to determine the geometry, which is linear or bent, and the sp/\(sp^2\)/\(sp^3\) hybridization based on the number of bonds.

Step 8: Count \(\sigma\) and \(\pi\) Bonds

Identify all the \(\sigma\) bonds (single bonds) and \(\pi\) bonds (double bonds) in the molecule based on its Lewis structure.

Key Concepts

Empirical Formula

The empirical formula of a compound is the simplest whole-number ratio of atoms of each element in the compound. To determine it, you begin by converting the mass percentages into grams. It's simplest to assume you have 100 grams of the compound, turning percentages directly into grams. For instance,

• 6.7 g of hydrogen (H)
• 40.0 g of carbon (C)
• 53.3 g of oxygen (O)

The next step is to convert these masses into moles using the atomic masses of the elements:

• Moles of H = \( \frac{6.7}{1.01} \)
• Moles of C = \( \frac{40.0}{12.01} \)
• Moles of O = \( \frac{53.3}{16.00} \)

Next, you identify the smallest number of moles from these calculations. You divide the number of moles of each element by this smallest value to find the simplest ratio. This ratio gives you the empirical formula. Ensure you round to the nearest whole number if necessary, since an empirical formula consists of whole numbers only.

Lewis Structure

Lewis structures are graphical representations that show how atoms in a molecule are bonded. They use dots to represent electrons and lines to represent bonds. Let's use the compound whose molecular formula we derived earlier as an example. Given the condition that the two oxygen atoms are bonded to carbon, we draw carbon in the center; it typically has four bonding sites. Hydrogen atoms will link to oxygens or directly to carbon since they can form only single bonds. Each line represents a pair of shared electrons in a covalent bond. The primary goal when drawing a Lewis structure is to ensure all atoms (except hydrogen) meet the octet rule, which is having 8 electrons around them in total. For this compound, the oxygen atoms will likely form double bonds with carbon, resulting in a structure similar to formaldehyde derivatives. Always ensure to allocate remaining electrons to satisfy the octet rule for each atom, starting with the most electronegative elements, like oxygen.

Hybridization

Hybridization is a concept used to explain the geometry and bonding properties of an atom in a molecule. It involves mixing atomic orbitals to create new hybrid orbitals. For a carbon atom bonded to two oxygen atoms, we evaluate its environment to determine the correct hybridization. If carbon has two groups attached (in this case, oxygen atoms), it will exhibit sp hybridization. This is because it requires two orbitals for bonding, formed by mixing one s and one p orbital. An sp hybridized carbon is typically part of a linear arrangement, providing 180-degree bond angles. However, in different scenarios where the carbon is part of a bent or trigonally planar arrangement, such as in carboxylic groups, it might employ sp2 hybridization. This involves using one s and two p orbitals, resulting in a trigonal planar geometry with 120-degree bond angles.

Sigma and Pi Bonds

In any molecule, bonds can be classified as sigma (\(\sigma\)) or pi (\(\pi\)) bonds. Sigma bonds are the first bond formed between two atoms. They are single covalent bonds created by the head-on overlap of atomic orbitals and are stronger due to direct overlap.In contrast, pi bonds are formed by the side-to-side overlap of two orbitals and usually exist in conjunction with sigma bonds in multiple bonds (double or triple bonds). In a double bond, there is one sigma bond and one pi bond.Using our current example of a molecule with carbon and two oxygens, each double bond between carbon and oxygen can be broken down into:

• One \(\sigma\) bond
• One \(\pi\) bond

This means if the molecular structure features two double bonds with oxygen, then within such a compound, there would be two \(\sigma\) bonds and two \(\pi\) bonds in total. Understanding how to differentiate and count these bonds helps decipher the molecule's stability and reactivity.