Problem 63 - Chemistry Matter and Change

Question

A colorless liquid composed of \(46.68 \%\) nitrogen and \(53.32 \%\) oxygen has a molar mass of \(60.01 \mathrm{~g} / \mathrm{mol} .\) What is the molecular formula?

Step-by-Step Solution

Answer

The molecular formula is \( \text{N}_2\text{O}_2 \).

Step 1: Calculate Moles of Each Element

Suppose we have a 100 g sample of the liquid. Then, this sample contains 46.68 g nitrogen and 53.32 g oxygen. Calculate the moles of nitrogen: \( \frac{46.68 \text{ g}}{14.01 \text{ g/mol}} = 3.33 \text{ moles} \). Calculate the moles of oxygen: \( \frac{53.32 \text{ g}}{16.00 \text{ g/mol}} = 3.33 \text{ moles} \).

Step 2: Determine the Empirical Formula

Find the simplest whole number ratio of moles of nitrogen to moles of oxygen. Both are \(3.33\), giving a ratio of \(1:1\). Thus, the empirical formula is \(NO\).

Step 3: Calculate the Empirical Formula Mass

Calculate the molar mass of the empirical formula \(NO\). \(\text{Molar mass of } N = 14.01 \text{ g/mol} \) and \(\text{molar mass of } O = 16.00 \text{ g/mol} \), so \( \, 14.01 + 16.00 = 30.01 \text{ g/mol} \).

Step 4: Find the Molecular Formula

Divide the given molar mass by the empirical formula mass: \( \frac{60.01}{30.01} = 2\). Multiply the empirical formula by this factor to find the molecular formula: \( \text{N}_2\text{O}_2 \).

Key Concepts

Empirical Formula

The empirical formula of a compound is the simplest whole-number ratio of atoms of each element in the compound. It doesn't represent the actual number of atoms in a molecule, but rather the simplest reduced version of it.
For instance, in our exercise, the empirical formula is derived from the elemental composition of the substance: 46.68% nitrogen and 53.32% oxygen.

• We start by assuming a 100 g sample to work with percentages directly as grams.
• Given that we have 3.33 moles of nitrogen and oxygen each, the ratio is 1:1. This means that for every atom of nitrogen, there's an atom of oxygen.

This leads us to an empirical formula, which in this case is NO. It's crucial to understand that this formula indicates only the simplest ratio of the elements involved, not their actual counts in the molecule.

Molar Mass Calculation

Calculating the molar mass is an essential step in determining both empirical and molecular formulas.
To calculate the molar mass of an empirical formula, you add up the atomic masses of each element as per their ratio in the empirical formula.

• For the empirical formula NO, the molar mass is given by adding the atomic masses of nitrogen (14.01 g/mol) and oxygen (16.00 g/mol), summing up to 30.01 g/mol.

This calculated mass of the empirical formula helps us further compare it with the given molar mass to deduce the molecular formula. When dividing the actual molar mass by the empirical molar mass, the result guides us to how many empirical units comprise a molecule, leading to the complete molecular formula.

Elemental Composition

The elemental composition of a compound is what gives insight into how many atoms of each element are present and their respective weights in a specific quantity of the compound.
It is expressed in percentages and can often be used to derive both empirical and molecular formulas.

• In the given problem, the elemental composition is 46.68% nitrogen and 53.32% oxygen.
• This percentage directly translates into grams when assuming a 100 g sample, making calculations straightforward.
Understanding these percentages allows us to determine moles needed to find the empirical formula. Additionally, this composition tells us the dominant element or how they are bonded, reflecting the detailed nature of the compound's structure.