Problem 48 - 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 Mass

First, calculate the empirical formula mass of \(\mathrm{C}_2 \mathrm{H}_4 \mathrm{NO}\) by adding the atomic masses of all atoms in one empirical formula unit: \( (2 \times 12.01) + (4 \times 1.01) + (14.01) + (16.00) = 58.07 \ \mathrm{g/mol} \).

Step 2: Determine the Ratio of Molar Mass to Empirical Formula Mass

To find the ratio, divide the compound's given molar mass by the empirical formula mass: \( \frac{116.1}{58.07} \approx 2 \). This gives us the factor by which the empirical formula must be multiplied to obtain the molecular formula.

Step 3: Calculate the Molecular Formula

Multiply each subscript in the empirical formula \(\mathrm{C}_2 \mathrm{H}_4 \mathrm{NO}\) by the factor determined in the previous step (2). Therefore, the molecular formula is \((\mathrm{C}_2 \mathrm{H}_4 \mathrm{NO})_2 = \mathrm{C}_4 \mathrm{H}_8 \mathrm{N}_2 \mathrm{O}_2\).

Key Concepts

Empirical Formula

The empirical formula is a simplified way of representing the smallest whole-number ratio of elements in a compound. It is often the first step in finding a compound’s molecular formula. For example, the empirical formula for a substance might be

• \(\text{C}_2\text{H}_4\text{NO}\)

This formula tells us that in the compound, the ratio of carbon (C) atoms to hydrogen (H), nitrogen (N), and oxygen (O) atoms is 2:4:1:1.
To reach the molecular formula, we need more information about the compound’s molar mass. The empirical formula gives us only a ratio, not the actual number of atoms.

Molar Mass

Molar mass refers to the mass of one mole of a substance, typically in grams per mole (\(\text{g/mol}\)). Knowing the molar mass is crucial for determining the molecular formula from the empirical formula.
The molar mass of an organic compound might be given as 116.1 \(\text{g/mol}\).

• This value is used to calculate how many times the empirical formula unit is repeated in the molecular formula.

To find this relationship, we divide the given molar mass by the empirical formula's molar mass to get a multiplier, giving us a clearer picture of the entire molecule.

Atomic Mass

Atomic mass is the mass of a single atom of a chemical element, expressed in atomic mass units (\(\text{u}\)). Each element has a specific atomic mass, for instance:

• Carbon (C): 12.01 u
• Hydrogen (H): 1.01 u
• Nitrogen (N): 14.01 u
• Oxygen (O): 16.00 u

These values are used to calculate the empirical formula mass by multiplying the atomic mass of each element by the number of times the element appears in the formula (subscript).
Summing these gives the total empirical mass, essential for the molar mass ratio calculation.

Ratio Calculation

Ratio calculation is key to transitioning from an empirical to a molecular formula. It involves using the molar mass of the compound and the empirical formula mass to find a multiplier.

• First, divide the compound's molar mass by the empirical formula mass.
• The result is a whole number or a close approximation, like 2, which tells us how many empirical units fit into the actual molecular formula.

This ratio helps in multiplying the subscripts of the empirical formula to find the molecular formula. If the ratio is approximately 2, as seen in our example, each subscript in the empirical formula \(\text{C}_2\text{H}_4\text{NO}\) is doubled to become \(\text{C}_4\text{H}_8\text{N}_2\text{O}_2\).
This step is crucial for understanding the composition of the entire molecule.