Question

A substance is analyzed and found to contain \(85.7 \%\) carbon and \(14.3 \%\) hydrogen by mass. A gaseous sample of the substance is found to have a density of \(1.87 \mathrm{~g} / \mathrm{L}\) at STP. (a) Calculate the molar mass of the compound. (b) Determine the empirical and molecular formulas of the compound. (c) Draw two possible Lewis structures for molecules of the compound.

Short answer

Empirical formula: CH2, Molecular formula: C3H6, Molar mass: 41.89 g/mol.

Step-by-step solution

Determine Empirical Formula

Firstly, we convert the given percentages of elements to grams based on an assumption of a 100 g sample. Therefore, the sample contains 85.7 g of carbon and 14.3 g of hydrogen. Next, convert these masses to moles:- Moles of carbon: \( \frac{85.7}{12.01} = 7.14 \text{ moles} \)- Moles of hydrogen: \( \frac{14.3}{1.008} = 14.19 \text{ moles} \)Determine the simplest whole number ratio of moles by dividing by the smallest number of moles:- Carbon: \( \frac{7.14}{7.14} = 1 \)- Hydrogen: \( \frac{14.19}{7.14} = 1.99 \approx 2 \)The empirical formula is thus \( \text{CH}_2 \).

Calculate Molar Mass from Density

Given the density of the gas at STP is 1.87 g/L, use the molar volume of a gas at STP, which is 22.4 L/mol to find the molar mass.Multiply the molar volume by the density:\[ \text{Molar Mass} = 1.87 \frac{\text{g}}{\text{L}} \times 22.4 \frac{\text{L}}{\text{mol}} = 41.89 \text{ g/mol} \]

Determine Molecular Formula

Compare the empirical formula mass to the molar mass. The empirical formula mass of \( \text{CH}_2 \) is \( 12.01 + 2 \times 1.008 = 14.03 \text{ g/mol} \).Calculate how many empirical units fit into the molar mass:\[ n = \frac{41.89}{14.03} \approx 3 \] Thus, the molecular formula is \( \text{C}_3\text{H}_6 \).

Draw Lewis Structures

The compound \( \text{C}_3\text{H}_6 \) could be a few isomers. Two possible Lewis structures are:1. **Propene (an Alkene):** A carbon double bonded to another carbon: CH\text{2} = CH - CH\text{3}.2. **Cyclopropane (a Cycloalkane):** A triangle of carbon atoms, each end bonded to two hydrogens.

Key concepts

Molecular Formula

The molecular formula of a compound tells us the exact number of each type of atom present in one molecule of the compound. To determine it, we start by finding the **empirical formula**, which is the simplest positive integer ratio of atoms in the compound. In this case, the empirical formula was found to be \( \text{CH}_2 \). To find the molecular formula, we need to know the compound's **molar mass**. By comparing the empirical formula mass to the molar mass, we can determine how many empirical units are present in a molecule of the compound. In the example provided, the empirical formula mass is \( 14.03 \ \text{g/mol} \). The calculated molar mass was \( 41.89 \ \text{g/mol} \). Dividing the molar mass by the empirical formula mass yields approximately 3, indicating that the molecular formula consists of three empirical units, thus \( \text{C}_3\text{H}_6 \). This step is crucial because understanding the exact composition helps in predicting physical and chemical properties.

Lewis Structures

Lewis structures are diagrams that showcase the bonds between atoms in a molecule and any lone pairs of electrons present. They are essential for visualizing the molecule's structure, understanding potential bonding patterns, and predicting molecular shape. For \( \text{C}_3\text{H}_6 \), there are multiple possible structures. Two common ones include propene and cyclopropane. - **Propene**: This structure features a double bond between two carbon atoms, denoted as \( \text{CH}_2 = CH-CH_3 \). Double bonds are represented by double lines, showing shared electrons. This configuration is typical for alkenes, which include at least one carbon-carbon double bond.- **Cyclopropane**: Cyclopropane is a ring structure where the carbon atoms form a triangle, each bonding to two hydrogen atoms. This configuration creates a ring strain due to the 60-degree angles between the bonds, making it distinct within the alkane family, where typically single bonds form straight or branched chains.Understanding different possible structures enables predicting reaction outcomes and assessing the molecule’s stability, as well as its possible isomers.

Molar Mass

Molar mass is the mass of one mole of a substance and is typically expressed in grams per mole (g/mol). This concept is central to converting between mass and moles of a substance, which is vital in chemical calculations and experimentation. To compute the molar mass of a gaseous substance, you can use its density at standard temperature and pressure (STP). STP is defined as a temperature of 0 °C (273.15 K) and a pressure of 1 atm, where one mole of any ideal gas occupies 22.4 liters.For the given compound, the density was provided as \( 1.87 \, \text{g/L} \). Using the molar volume of \( 22.4 \, \text{L/mol} \) at STP, the molar mass is calculated by multiplying the density by the molar volume: \[\text{Molar Mass} = 1.87 \, \text{g/L} \times 22.4 \, \text{L/mol} = 41.89 \, \text{g/mol}. \]Calculating the molar mass is essential because it helps identify the molecular formula, as shown through the ratio between the actual molar mass and the empirical formula mass. Such conversions are especially useful in quantifying reactants and products in chemical reactions.