Question

An oxide of nitrogen contains 30.45 mass \(\%\) N. (a) What is the empirical formula of the oxide? (b) If the molar mass is \(90 \pm 5 \mathrm{~g} / \mathrm{mol},\) what is the molecular formula?

Short answer

The empirical formula is \(\text{NO}_2\). The molecular formula is \(\text{N}_2\text{O}_4\).

Step-by-step solution

Finding the Masses of Each Element

Assume 100 g of the compound. This means we have 30.45 g of nitrogen (N) and 69.55 g of oxygen (O), as the rest of the compound is oxygen.

Converting Mass to Moles

Convert the given masses to moles. Use the molar masses of nitrogen (14.01 g/mol) and oxygen (16.00 g/mol). \[ \text{Moles of } N = \frac{30.45 \text{ g}}{14.01 \text{ g/mol}} = 2.174 \text{ moles} \] \[ \text{Moles of } O = \frac{69.55 \text{ g}}{16.00 \text{ g/mol}} = 4.347 \text{ moles} \]

Finding the Simplest Ratio

Divide the moles by the smallest number of moles to find the simplest whole number ratio: \[ \text{Ratio of } N = \frac{2.174}{2.174} = 1 \] \[ \text{Ratio of } O = \frac{4.347}{2.174} = 2 \] So, the empirical formula is \(\text{NO}_2\).

Determining the Molecular Formula

First, find the molar mass of the empirical formula \( \text{NO}_2 \): \[ 14.01 \text{ g/mol (N)} + 2 \times 16.00 \text{ g/mol (O)} = 46.01 \text{ g/mol} \] Then, divide the given molar mass (90 g/mol) by the empirical formula mass (46.01 g/mol): \[ \frac{90 \text{ g/mol}}{46.01 \text{ g/mol}} \text{ approx } 2 \]

Finding the Final Molecular Formula

Multiply the empirical formula by this ratio: \[ \text{Molecular formula} = ( \text{NO}_2) \times 2 = \text{N}_2\text{O}_4 \]

Key concepts

Mass Percent

Mass percent is a way of expressing the concentration of an element in a compound. It tells us what percentage of the compound's total mass is made up of a particular element. For example, if we're given that an oxide of nitrogen contains 30.45 mass percent nitrogen, that means in 100 grams of this compound, there are 30.45 grams of nitrogen. The remaining mass (69.55 grams) is oxygen.

Calculating mass percent helps in determining the composition of compounds, which is crucial for figuring out their empirical formulas.

Molar Mass

The molar mass of an element or compound is the mass of one mole of that substance. For instance, the molar mass of nitrogen (N) is 14.01 grams per mole, and for oxygen (O), it is 16.00 grams per mole.

This value is essential in converting grams to moles, a fundamental step in deriving both empirical and molecular formulas. By knowing the molar masses, we can use the mass of each element given in a problem to find out how many moles of each we have.

Moles Conversion

To determine the empirical formula, we need to convert the mass of each element into moles. This involves dividing the mass of the element by its molar mass.

For example, in the given problem:

• Moles of N = \(\frac{30.45 \text{ g}}{14.01 \text{ g/mol}} = 2.174 \text{ moles}\)
• Moles of O = \(\frac{69.55 \text{ g}}{16.00 \text{ g/mol}} = 4.347 \text{ moles}\)

Converting mass to moles ensures that we are working with the number of atoms rather than their masses, which simplifies the comparison between different elements.

Simplest Ratio

Once we have the moles of each element, we can find the simplest ratio by dividing each mole value by the smallest number of moles from the values calculated.

In the exercise, the smallest mole value is 2.174. Thus:

• Ratio of N = \(\frac{2.174}{2.174} = 1\)
• Ratio of O = \(\frac{4.347}{2.174} = 2\)

This gives us the empirical formula \(\text{NO}_2\). The simplest ratio is vital because it shows the simplest whole-number ratio of elements in a compound.

Chemical Formula Calculation

To find the molecular formula, we first determine the molar mass of the empirical formula and then compare it to the given molar mass of the compound.

For \(\text{NO}_2\), the empirical formula mass is \(14.01 \text{ g/mol (N)} + 2 \times 16.00 \text{ g/mol (O)} = 46.01 \text{ g/mol}\). Given that the molar mass of the compound is about 90 g/mol, we can find the ratio by:

• \(\frac{90 \text{ g/mol}}{46.01 \text{ g/mol}} \approx 2\)

This ratio means the molecular formula is twice the empirical formula, leading to \(\text{N}_2\text{O}_4\). This process is critical for determining the exact formula of a compound, especially when the empirical and molecular formulas differ.