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 \(\mathrm{C} ?\) (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?

Short answer

The molecular formula of the compound is C2H4O2. Its Lewis structure has a central carbon (C2) double bonded to two oxygen atoms and connected to another carbon (C3) which is bonded to three hydrogen atoms. The geometry around C2 is linear with sp hybridization, and the geometry around C3 is tetrahedral with sp3 hybridization. The molecule has a total of 6 σ bonds and 2 π bonds.

Step-by-step solution

Calculate the Empirical Formula.

First, we'll use the given percentage composition to find the empirical formula. We will consider having 100 grams of the compound and calculate the moles of each element in it. For Hydrogen: \[6.7~g \left(\dfrac{1~mole}{1~g}\right) = 6.7~moles\] For Carbon: \[40.0~g \left(\dfrac{1~mole}{12.01~g}\right) \approx 3.3~moles\] For Oxygen: \[53.3~g \left(\dfrac{1~mole}{16.00~g}\right) \approx 3.3~moles\] Now, let's divide the moles of each element by their lowest value to find the ratio: \[C : H : O = 3.3 : 6.7 : 3.3\] Divide by 3.3: \[C : H : O = 1 : 2 : 1\] The empirical formula is CH2O.

Calculate the Molecular Formula.

Now, let's use the empirical formula and its molar mass to find the molecular formula. First, find the empirical formula molar mass. For CH2O: Molar Mass = 12.01 + 2 (1) + 16 = 30 g/mol Given that the molar mass of the compound is approximately 60 g/mol: Molecular Formula = Empirical Formula × n (where n is an integer) Let's find the value of n: \[n = \dfrac{60}{30} = 2\] Now, multiply the empirical formula by n to get the molecular formula: C(2)H(2×2)O(2) = C2H4O2

Draw the Lewis Structure.

Based on the given information, the two oxygen atoms are bonded to a central carbon atom. The Lewis structure for this molecule can be drawn as: O=C(2) || O=C ! H ! H /(3) C ! H ! H Here, the numbers inside the parentheses indicate the carbon atom number. Double bonds are shown with two lines.

Geometry and Hybridization of Carbon Atoms.

The central carbon atom (C2) is double bonded to two oxygen atoms. This arrangement suggests a linear geometry around this carbon. The hybridization for this carbon is sp, as it has two electron groups consisting of two double bonds. The geometry around carbon atom C3 is tetrahedral with a hybridization of sp3.

Determine the Number of σ and π Bonds.

To complete the exercise, we'll count the number of σ and π bonds in the molecule. - Carbon atom 2 (C2) has two σ bonds (one each to oxygen atoms) and two π bonds (one each in the double bonds with oxygen atoms). - Carbon atom 3 (C3) has four σ bonds (to carbon atom 2 and three hydrogen atoms). Total: 6 σ bonds and 2 π bonds.

Key concepts

Empirical Formula

The empirical formula is a simplified representation of a compound's elements, shown in the smallest whole number ratio. It's like the foundational blueprint in understanding molecular chemistry. To determine the empirical formula from given percentages, consider 100 grams of the compound. \(6.7\% \mathrm{H}\), \(40.0\% \mathrm{C}\), and \(53.3\% \mathrm{O}\) translate to this mass in grams when coming from a 100g sample. By dividing the mass of each element by its atomic mass, you determine the moles of each element. In this example:

• Hydrogen: \(6.7\, \text{g} = 6.7\, \text{moles}\)
• Carbon: \(40.0\, \text{g} \times \frac{1\, \text{mole}}{12.01\, \text{g}} \approx 3.3\, \text{moles}\)
• Oxygen: \(53.3\, \text{g} \times \frac{1\, \text{mole}}{16.00\, \text{g}} \approx 3.3\, \text{moles}\)

To find the simplest ratio of moles: divide each by the smallest value (\(3.3\, \text{moles}\)). Thus, you get \(C_1H_2O_1\). This is your empirical formula: CH₂O.

Molecular Formula

The molecular formula reveals the exact number of each type of atom in a molecule and is a multiple of the empirical formula. It's instrumental in understanding actual molecule size and mass in molecular chemistry. Given the empirical formula \(\text{CH}_2\text{O}\) with a molar mass of 30 g/mol, and knowing the compound’s molar mass is approximately 60 g/mol, you must calculate the molecular formula using this relation:\[ n = \frac{\text{Molar Mass of Compound}}{\text{Molar Mass of Empirical Formula}} \]In this case, \(n = \frac{60}{30} = 2\). Therefore, multiply the empirical formula by \(n = 2\). This results in the molecular formula: \(\text{C}_2\text{H}_4\text{O}_2\), meaning each mole of this compound contains two moles of carbon atoms, four moles of hydrogen atoms, and two moles of oxygen atoms.

Lewis Structure

The Lewis structure is a detailed diagram that shows the arrangement of atoms within a molecule, highlighting the bonding between atoms and lone pairs of electrons. It simplifies molecular geometry and electron sharing. For the compound \(\text{C}_2\text{H}_4\text{O}_2\), where two oxygen atoms bond with a single carbon atom, you represent it as:- Place carbon as the central atom with oxygens attached as \(\text{O}\, = \text{C} \cdots \text{O}\). - Add hydrogen atoms to complete the appropriate number of bonds per atom, taking into account that carbon typically forms four bonds and oxygen typically forms two.A double bond forms between carbon and oxygen due to the need for C to share additional electrons, thus meeting the octet rule for stability. The structure visually resembles a miniature model of the molecule's true electronic configuration.

Hybridization

Hybridization is the mixing of atomic orbitals to create new hybrid orbitals suitable for the pairing of electrons to form chemical bonds. Recognizing it helps understand molecular shape and chemical reactivity. For the central C atom in \(\text{C}_2\text{H}_4\text{O}_2\), bonded to two oxygens, the hybridization is \(\text{sp}\). This is because it forms two double bonds.- Each double bond signifies a \(\text{sp}\) hybridized orbital, associated with linear geometry surrounding the central C atom.- Another carbon in the molecule, bonded to three H atoms and one C atom, would typically exhibit \(\text{sp}^3\) hybridization due to four sigma bonds, indicating a tetrahedral geometry.This hybrid setup facilitates the appropriate angles between bonds and stability in molecular geometry.

Bonding

Understanding bonding within molecules involves identifying the types of bonds, such as sigma (σ) and pi (π) bonds, which influence molecular stability and interactions. In \(\text{C}_2\text{H}_4\text{O}_2\):- The carbon atom not double-bonded to oxygen forms \(\text{sp}^3\) hybridized \(\sigma\) bonds with three hydrogen atoms.- The central carbon (with \text{sp} hybridization) forms two \(\sigma\) bonds with oxygens and also involves \(\pi\) bonds from the double bonds.

• Total \(6 \sigma\) bonds: 2 for each carbon-oxygen single bond and the 4 remaining are from carbon-hydrogen bonds.
• Total \(2 \pi\) bonds: One for each carbon-oxygen double bond.

This distribution of bonds determines the extent of electron overlap, which is crucial for the compound’s structural and chemical properties.