Problem 84 - Chemistry and Chemical Reactivity

Question

An organic compound has the empirical formula \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{NO} .\) If its molar mass is \(116.1 \mathrm{g} / \mathrm{mol},\) what is the molecular formula of the compound?

Step-by-Step Solution

Answer

The molecular formula is \( \mathrm{C}_{4} \mathrm{H}_{8} \mathrm{N}_{2} \mathrm{O}_{2} \).

Step 1: Calculate the Empirical Formula Molar Mass

The empirical formula given is \( \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{NO} \). Calculate the molar mass of this empirical formula by adding up the molar masses of each atom in the formula. Carbon (C) has a molar mass of \( 12.01 \ \mathrm{g/mol} \), hydrogen (H) has \( 1.01 \ \mathrm{g/mol} \), nitrogen (N) has \( 14.01 \ \mathrm{g/mol} \), and oxygen (O) has \( 16.00 \ \mathrm{g/mol} \). Thus, the empirical formula molar mass is:\[(2 \times 12.01) + (4 \times 1.01) + (14.01) + (16.00) = 58.06 \, \mathrm{g/mol}\]

Step 2: Determine the Ratio of Molar Masses

You are given the molecular molar mass of the compound as \( 116.1 \, \mathrm{g/mol} \). To find the ratio, divide the molecular molar mass by the empirical formula molar mass:\[\frac{116.1 \, \mathrm{g/mol}}{58.06 \, \mathrm{g/mol}} \approx 2\]This ratio indicates how many empirical formula units are present in the molecular formula.

Step 3: Calculate the Molecular Formula

Multiply all the subscripts in the empirical formula \( \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{NO} \) by the ratio calculated in Step 2. The ratio was 2, so multiply each subscript by 2:- For carbon \( C: 2 \times 2 = 4 \)- For hydrogen \( H: 4 \times 2 = 8 \)- For nitrogen \( N: 1 \times 2 = 2 \)- For oxygen \( O: 1 \times 2 = 2 \)Thus, the molecular formula is \( \mathrm{C}_{4} \mathrm{H}_{8} \mathrm{N}_{2} \mathrm{O}_{2} \).

Key Concepts

Empirical Formula

In chemistry, the empirical formula is the simplest way to express the ratio of different elements in a compound. It doesn't necessarily reflect the exact number of atoms in the molecule, but it shows the kind of atoms present and the simplest whole number ratio between them.
For example, if you have an empirical formula of \( \text{C}_{2}\text{H}_{4}\text{NO} \), this tells you that in the simplest ratio, there are two carbon (C) atoms, four hydrogen (H) atoms, one nitrogen (N) atom, and one oxygen (O) atom for every unit of the compound.
It's crucial to understand that the empirical formula represents the basic core of the composition. It is like the blueprint that guides you to determine how these elements are proportioned in the compound's simplest form.

• Simplest ratio of elements.
• Not necessarily the actual number of atoms in a single molecule.
• Derived from experimental data, usually from lab analyses.

Using this empirical formula, chemists can calculate the molecular formula, which provides more detailed information about the actual number of each type of atom in a molecule after considering its molar mass.

Molar Mass Calculation

Molar mass is the mass of one mole of a substance in grams. It is calculated by summing up the atomic masses of all the atoms in a formula. This value is significant because it helps us convert between the mass of a substance and the amount of substance in moles, which is a key concept in chemistry.
For any compound, getting the molar mass starts with its empirical or molecular formula. In our exercise, the empirical formula is \( \text{C}_{2}\text{H}_{4}\text{NO} \). The molar masses of individual atoms are as follows:

• Carbon (C): \( 12.01 \, \text{g/mol} \)
• Hydrogen (H): \( 1.01 \, \text{g/mol} \)
• Nitrogen (N): \( 14.01 \, \text{g/mol} \)
• Oxygen (O): \( 16.00 \, \text{g/mol} \)

The empirical formula molar mass is then calculated by:\[(2 \times 12.01) + (4 \times 1.01) + (14.01) + (16.00) = 58.06 \, \text{g/mol}\]This empirical value is used to find the ratio between the empirical molar mass and the known molecular molar mass to determine the molecular formula.

Organic Chemistry

Organic chemistry focuses on carbon-containing compounds and their reactions. Since carbon atoms can bond together in large chains and form rings, the discipline studies a variety of materials ranging from simple molecules to complex macrostructures.
The compound in our exercise is an organic compound because it contains carbon alongside hydrogen, nitrogen, and oxygen. Organic chemistry often involves studying functions and reactions inside biological systems where these organic compounds play crucial roles.

• Carbon's versatility: Can form multiple stable bonds, both single, double, and triple, with itself and other elements.
• Foundational in biochemistry: Organic compounds are crucial in life processes.
• Wide-ranging applications: Used in pharmaceuticals, plastics, fuels, and many more.

The molecular formula, like \( \text{C}_{4}\text{H}_{8}\text{N}_{2}\text{O}_{2} \) from the exercise, gives deeper insights into the structural configuration and potential reactivity of the compound in organic chemistry.

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships among substances in chemical reactions and compounds. It is essential for calculating the amounts of reactants needed or products formed in a chemical reaction.
Understanding stoichiometry allows chemists to work with chemical equations to determine the proportions of reactants and products. In our exercise of molecular formula determination, stoichiometry plays a role in ensuring that the calculated molecular formula aligns with the given molecular mass.

• Balancing chemical equations: Ensures that the mass and number of atoms are conserved during reactions.
• Calculating molar ratios: Derived from coefficients in balanced equations.
• Predicting product yields: From known amounts of reactants.

When you have the correct molecular formula, as determined in the exercise, you can accurately calculate the substance needed for any chemical reaction or experiment involving this compound.