Problem 886 - Chemistry The Molecular Science

Question

A substance is analyzed and found to contain \(85.7 \%\) carbon and \(14.3 \%\) hydrogen by weight. A gaseous sample of the substance is found to have a density of \(1.87 \mathrm{~g} / \mathrm{L},\) and \(1 \mathrm{~mol}\) of it occupies a volume of \(22.4 \mathrm{~L}\). What are two possible Lewis structures for molecules of the compound? (Hint: First determine the empirical formula and molar mass of the substance.)

Step-by-Step Solution

Answer

The two possible Lewis structures are Propene (CH\(_2\)=CH-CH\(_3\)) and Cyclopropane (three-carbon ring).

Step 1: Convert Percentages to Grams

Assume we have 100 grams of the substance. This means that we have 85.7 grams of carbon and 14.3 grams of hydrogen.

Step 2: Calculate Moles of Each Element

Calculate the moles of each element using their atomic masses. For carbon (C): \[\text{Moles of C} = \frac{85.7 \text{ grams}}{12.01 \text{ g/mol}} \approx 7.14 \text{ mol}\]For hydrogen (H): \[\text{Moles of H} = \frac{14.3 \text{ grams}}{1.008 \text{ g/mol}} \approx 14.19 \text{ mol}\]

Step 3: Determine the Empirical Formula

Divide the moles of each element by the smallest number of moles to obtain the ratio of atoms. Here, the smallest is 7.14 mol:\[\text{C} = \frac{7.14}{7.14} = 1,\text{H} = \frac{14.19}{7.14} \approx 2\]Thus, the empirical formula is CH extsubscript{2}.

Step 4: Calculate the Empirical Formula Mass

Calculate the mass of the empirical formula unit, CH\(_2\):\[12.01 \text{ g/mol (C)} + 2 \times 1.008 \text{ g/mol (H)} = 14.026 \text{ g/mol}\]

Step 5: Determine the Molar Mass of the Substance

Use the density and volume information. Since 1 mole occupies 22.4 L:\[\text{Molar Mass} = 1.87 \text{ g/L} \times 22.4 \text{ L/mol} = 41.89 \text{ g/mol}\]

Step 6: Determine the Molecular Formula

Divide the molar mass by the empirical formula mass:\[\frac{41.89 \text{ g/mol}}{14.026 \text{ g/mol}} \approx 3\]This implies that the molecular formula is three times the empirical formula: C\(\text{H}_2\)\(_3\) or C\(_3\)H\(_6\).

Step 7: Draw Possible Lewis Structures

For C\(_3\)H\(_6\), the possible structures can be:1. Propene: CH\(_2\)=CH-CH\(_3\)2. Cyclopropane: A three-carbon ring structure with each carbon singly bonded to each other and two hydrogens per carbon.

Key Concepts

Lewis Structures

Lewis structures show how atoms in a molecule are bonded together and illustrate the arrangement of electrons. They are immensely useful in visualizing the structure of molecules and making predictions about the shapes and angles within them. When determining Lewis structures for a compound, like the molecules we have from C\(_3\)H\(_6\), you first need the molecular formula. This formula tells us the number and types of atoms in a molecule.

For C\(_3\)H\(_6\), a straightforward approach to draw possible structures is:

• Propene: In propene, the typical bonding involves a double bond between two carbon atoms and single bonds between the others. The structure is: CH\(_2\)=CH-CH\(_3\). Here, one carbon forms a double bond with another, highlighting a special type of bond called a "double bond." These bonds affect molecule shape and reactivity.
• Cyclopropane: Another possible structure is a cyclopropane ring. Cyclopropane is a cyclic structure with single bonds between each carbon atom forming a triangle. Each carbon in the ring holds two hydrogen atoms.

Understanding these structures exposes important aspects of the molecules such as possible reactions, polarity, and physical properties.

Molar Mass Calculation

Calculating the molar mass of a substance is key to converting between grams and moles, which is essential in chemistry for determining how much of a substance you have. Molar mass is basically the mass of one mole of a substance, expressed in grams per mole (g/mol).

To calculate the molar mass of our gaseous sample:

• Use the given density and volume occupied by one mole of the gas. We have that 1 mole occupies 22.4 L. Given the density is 1.87 g/L, the molar mass can be calculated as:

\[\text{Molar Mass} = 1.87 \, \text{g/L} \times 22.4 \, \text{L/mol} = 41.89 \, \text{g/mol}\]This molar mass matches the mass of the molecular formula C\(_3\)H\(_6\). Once you have this, it becomes easier to determine the molecular composition and analyze various physical properties and the reactivity of the substance.

Empirical Formula Calculation

The empirical formula represents the simplest whole-number ratio of elements in a compound. Obtaining it is one of the first steps in identifying a substance's composition. Here’s how you can calculate it:

1. Start with the percentage composition (like 85.7% carbon and 14.3% hydrogen).2. Assume you have 100 grams of the compound, making calculation easier by translating percentages directly to grams.3. Convert grams to moles by using atomic masses (12.01 g/mol for carbon and 1.008 g/mol for hydrogen).

• For carbon: \[\text{Moles of C} = \frac{85.7 \, \text{grams}}{12.01 \, \text{g/mol}} \approx 7.14 \, \text{mol}\]
• For hydrogen: \[\text{Moles of H} = \frac{14.3 \, \text{grams}}{1.008 \, \text{g/mol}} \approx 14.19 \, \text{mol}\]

Divide the moles of each element by the smallest number of moles to get the simplest ratio.

• \[\text{C} = \frac{7.14}{7.14} = 1,\, \text{H} = \frac{14.19}{7.14} \approx 2\]

This gives the empirical formula of CH\(_2\). From this simple ratio, you can build up to derive the molecular formula, especially when the compound's molar mass is known.