Question

A compound containing only sulfur and nitrogen is \(69.6 \% \mathrm{S}\) by mass; the molar mass is \(184 \mathrm{g} / \mathrm{mol}\). What are the empirical and molecular formulas of the compound?

Short answer

The empirical formula of the compound is SN, and the molecular formula is S4N4.

Step-by-step solution

Determine the mass percentages of Sulfur and Nitrogen in the compound

Since the compound contains only sulfur and nitrogen, and we are given that it is 69.6% S by mass, we can easily determine the mass percentage of nitrogen as well. So, Mass percentage of Nitrogen (N) = 100% - Mass percentage of Sulfur (S) = 100% - 69.6% = 30.4%

Convert mass percentages to grams

Assume that we have 100 g of the compound. This will allow us to directly use the mass percentages as grams: Mass of Sulfur (S) = 69.6 g Mass of Nitrogen (N) = 30.4 g

Calculate the moles of Sulfur and Nitrogen in the compound

Next, we will find the moles of S and N using their respective atomic masses. Moles of S = \(\frac{\text{Mass of S}}{\text{Atomic Mass of S}} = \frac{69.6 g}{32.07 g/mol} = 2.17 mol\) (approximately) Moles of N = \(\frac{\text{Mass of N}}{\text{Atomic Mass of N}} = \frac{30.4 g}{14.01 g/mol} = 2.17 mol\) (approximately)

Find the empirical formula

Now, we have the moles of S and N in the compound, and we need to find their simplest whole number ratio to determine the empirical formula. Mole ratio of S:N = 2.17:2.17 = 1:1 (approximately) The empirical formula is therefore SN.

Determine the molar mass of the empirical formula

Next, we will find the molar mass of the empirical formula "SN" using the atomic masses of S and N. Molar mass of SN = Atomic mass of S + Atomic mass of N = 32.07 g/mol + 14.01 g/mol = 46.08 g/mol

Find the molecular formula

Finally, we will determine the molecular formula by dividing the given molar mass by the molar mass of the empirical formula and multiplying it by the empirical formula. \(\frac{\text{Molar Mass of Compound}}{\text{Molar Mass of Empirical Formula}} = \frac{184 g/mol}{46.08 g/mol} = 4\) Molecular Formula = Empirical Formula x 4 = (SN) * 4 = S4N4. The empirical formula of the compound is SN, and the molecular formula is S4N4.

Key concepts

Atomic Mass

Understanding atomic mass is fundamental when determining chemical formulas. It represents the mass of a single atom, usually expressed in atomic mass units (amu). For instance, sulfur (S) has an atomic mass of approximately 32.07 amu while nitrogen (N) has an atomic mass of 14.01 amu.

When calculating formulas, the atomic mass allows chemists to convert mass percentages into moles—an essential step for deriving the empirical and molecular formulas. This conversion is necessary because chemical formulas are based on the number of atoms, not their mass in grams. So, when we are given that a compound contains 69.6 g of sulfur, we divide this mass by the atomic mass of sulfur to find the number of moles of sulfur atoms present in the sample.

Mole Concept

The mole concept is a bridge that allows chemists to convert between the mass of a substance and the number of particles (atoms, molecules, ions, etc) it contains. One mole is equal to Avogadro's number, which is approximately 6.022 x 10^23 particles.

For example, to find the moles of sulfur and nitrogen given their masses, the formula is moles = mass (g) / atomic mass (g/mol). This step is crucial in our exercise since it normalizes the mass of the two elements, sulfur and nitrogen, allowing us to find the empirical formula through the stoichiometric ratio of moles. Here it simplifies to a 1:1 ratio, implying that for every one atom of sulfur, there is one atom of nitrogen, thus resulting in the empirical formula 'SN'.

Percent Composition

The percent composition of a compound reflects the proportion of each element's mass relative to the total mass of the compound. In our example, the compound was 69.6% sulfur and consequently, 30.4% nitrogen (since the compound only contains these two elements).

This percentage immediately tells us how much of each element would be present in any given sample mass of the compound. In the context of 100 g of the compound, it translates directly into 69.6 g of sulfur and 30.4 g of nitrogen. To understand a compound's formula, the percent composition is an indispensable piece of information, as it sets the stage for all subsequent calculations, leading us from mass to moles, and then from moles to the simplest ratio of atoms in the compound.