Question

A colorless liquid composed of 46.68% nitrogen and 53.32% oxygen has a molar mass of 60.01 g/mol. What is the molecular formula?

Short answer

The molecular formula of the colorless liquid is N₂O₂.

Step-by-step solution

Convert percentage composition to grams

Since we know the percentage composition, we can assume we have 100 g of the compound. That means, we have 46.68 g of nitrogen and 53.32 g of oxygen.

Convert grams to moles

Next, we'll convert the grams of each element to moles using their atomic masses. The atomic mass of nitrogen (N) is 14.01 g/mol and that of oxygen (O) is 16.00 g/mol. Moles of nitrogen = \( \frac{46.68 \: g}{14.01 \: g/mol}=\) 3.33 moles Moles of oxygen = \( \frac{53.32 \: g}{16.00 \: g/mol}=\) 3.33 moles

Find the ratio of nitrogen atoms to oxygen atoms

Now that we have the moles of each element, we need to find the simplest whole number ratio of nitrogen atoms to oxygen atoms. Since both nitrogen and oxygen have equal moles (3.33), the simplest ratio N:O is 1:1.

Write the empirical formula

The empirical formula of this liquid is NO, which represents the simplest ratio of nitrogen to oxygen atoms in the compound.

Calculate the molecular formula

Next, we'll determine the molecular formula based on the empirical formula and the molar mass of the compound. First, find the empirical formula molar mass (EFM): Empirical formula molar mass (EFM) = 14.01 g/mol (N) + 16.00 g/mol (O) = 30.01 g/mol Now, divide the molar mass of the compound (60.01 g/mol) by the empirical formula molar mass (30.01 g/mol): \( \frac{60.01 \: g/mol}{30.01 \: g/mol} \approx 2 \) Multiply the empirical formula (NO) by this factor (2) to find the molecular formula: Molecular formula = N₂O₂ The molecular formula of this colorless liquid is N₂O₂.

Key concepts

Percentage Composition

Understanding the percentage composition of a compound is crucial when attempting to determine its molecular formula. Essentially, percentage composition refers to the percent by mass of each element present in the compound. For instance, in the exercise, a colorless liquid has 46.68% nitrogen (N) and 53.32% oxygen (O). This information tells us that for every 100 grams of this liquid, there are 46.68 grams of nitrogen and 53.32 grams of oxygen.

The concept might seem daunting, but imagine you have a 100-gram pie divided into two flavors, one taking up 46.68% and the other 53.32%. The pie chart analogy can help you visualize the percentage composition of a compound.

Molar Mass Calculation

Molar mass calculation is a fundamental concept in chemistry that involves determining the mass of one mole of a substance. The molar mass is typically expressed in grams per mole (g/mol) and can be found by adding up the atomic masses of all the atoms in the formula. In our exercise, the molar mass of nitrogen is 14.01 g/mol and oxygen is 16.00 g/mol.

It may help to think of molar mass as the 'weight' of one mole of marbles, where each marble represents an atom. When you sum up the individual 'weights,' you arrive at the total 'weight' of the mole of marbles. In practice, molar mass helps us convert between the mass of a substance and the number of moles, bridging the gap between the microscopic world of atoms and the macroscopic world where we can measure mass in grams.

Empirical Formula

The empirical formula represents the simplest whole number ratio of atoms of each element in a compound. It doesn't necessarily reflect the actual number of atoms in a molecule but gives us the proportional relationship. In our exercise, after converting the mass of each element to moles, we got the ratio of nitrogen to oxygen as 1:1, resulting in the empirical formula NO.

An analogy here can be made with a recipe; if a dish requires 1 cup of two ingredients, regardless of the amount you make, the basic recipe remains the same - one part of one ingredient to one part of another. Similarly, the empirical formula gives the basic 'recipe' of the compound’s composition.

Mole Conversion

Mole conversion involves switching between the number of moles and the mass of a substance using its molar mass as a conversion factor. In the given solution, we used the molar mass of nitrogen and oxygen to find the number of moles from the given mass (46.68 g of nitrogen and 53.32 g of oxygen).

To better understand mole conversion, imagine swapping out currency. If you know the exchange rate, you can convert dollars to euros and vice versa. Similarly, by knowing the molar mass (our 'exchange rate'), we can convert between grams (our 'currency') and moles. This step is critical because stoichiometry and chemical equations are mole-based, making mole conversions indispensable for quantitatively describing chemical reactions.