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: Determine the Empirical Formula Mass

Calculate the molar mass of the empirical formula \(\mathrm{C}_2\mathrm{H}_4\mathrm{NO}\). First, calculate the atomic masses of each element: \(\text{C} = 12.01\, \mathrm{g/mol}\), \(\text{H} = 1.008\, \mathrm{g/mol}\), \(\text{N} = 14.01\, \mathrm{g/mol}\), \(\text{O} = 16.00\, \mathrm{g/mol}\). Now, calculate the empirical formula mass: \[(2 \times 12.01) + (4 \times 1.008) + 14.01 + 16.00 = 58.06\, \mathrm{g/mol}.\]

Step 2: Calculate the Molecular Formula

Determine how many times the empirical formula mass fits into the molar mass. Divide the empirical formula mass into the given molar mass: \[\frac{116.1 \, \mathrm{g/mol}}{58.06 \, \mathrm{g/mol}} \approx 2.\] The empirical formula must be multiplied by 2 to get the molecular formula.

Step 3: Solve for the Molecular Formula

Double the subscripts in the empirical formula \(\mathrm{C}_2\mathrm{H}_4\mathrm{NO}\) because the factor is 2. Therefore, the molecular formula is \(\mathrm{C}_4\mathrm{H}_8\mathrm{N}_2\mathrm{O}_2\).

Key Concepts

Empirical Formula

In chemistry, an empirical formula represents the simplest whole-number ratio between the elements in a compound. It provides a basic snapshot of the composition but not the exact number of atoms found in a molecule of the compound. To determine the empirical formula of a compound, you'll often calculate the ratio based on the relative number of moles of each element present in a sample. Here’s how it works:

- **Step 1:** Calculate the number of moles of each element in the compound.
- **Step 2:** Divide the number of moles by the smallest number of moles calculated.
- **Step 3:** Round the ratios to the nearest whole number to find the subscripts of each element.

This method results in an empirical formula, which may differ from a molecular formula that provides the actual number of atoms of each element in a molecule. For instance, the empirical formula for glucose is CH₂O, even though its molecular formula is C₆H₁₂O₆.

Molar Mass Calculation

The molar mass of a compound is crucial for converting between moles and grams during calculations. It is the mass of all the atoms in a molecule of that substance, expressed in grams per mole (g/mol). Calculating molar mass involves:

- **Identifying Elemental Masses:** Use the periodic table to find the atomic mass of each element within the compound.
- **Summing Up Masses:** For each element, multiply the atomic mass by the number of atoms of that element in the molecule and sum up all these values.

As an example, the empirical formula of C₂H₄NO in our exercise results in an empirical formula mass of 58.06 g/mol. We calculated this by summing up the masses of two carbons, four hydrogens, one nitrogen, and one oxygen according to their atomic masses. Understanding molar mass is key for converting between different units and establishing the formula weight for more complex calculations.

Atomic Mass

The atomic mass of an element is the weighted average mass of the atoms in a naturally occurring sample. It's measured in atomic mass units (amu), which reflects isotopes and natural abundance. In our context, atomic masses are essential for calculating both empirical and molecular formulas.

The atomic mass of an element can be found using the periodic table and is crucial when performing calculations such as:

- **Empirical to Molecular Formula:** Knowing atomic masses helps when scaling up from the simplest ratio of elements in the empirical formula to the actual number of atoms present in the molecular formula.
- **Reaction Calculations:** You can also use atomic masses to balance equations and determine reactant and product masses.

For example, the atomic mass of carbon (C) is approximately 12.01 g/mol. In our exercise, we used this information to find the empirical formula mass of C₂H₄NO, helping to further derive the molecular formula from the given molar mass.