Question

A compound was analyzed and was found to contain the following percentages of the elements by mass: nitrogen, \(11.64 \% ;\) chlorine, \(88.36 \%\). Determine the empirical formula of the compound.

Short answer

The empirical formula of the compound is NCl3.

Step-by-step solution

Convert Percentages to Masses

First, assume that we have 100 grams of the compound. This will make it easier to work with the given percentages. Convert the percentages of nitrogen and chlorine to masses: - Nitrogen: \(11.64\%\) of 100 grams = 11.64 grams - Chlorine: \(88.36\%\) of 100 grams = 88.36 grams

Calculate the Moles of Nitrogen and Chlorine

Now, convert the masses of nitrogen and chlorine into moles by using their molar masses. - Molar mass of nitrogen (N) = 14.01 g/mol - Molar mass of chlorine (Cl) = 35.45 g/mol Moles of nitrogen = (mass of nitrogen) / (molar mass of nitrogen) = \(11.64 g / 14.01 g/mol = 0.831 mol\) Moles of chlorine = (mass of chlorine) / (molar mass of chlorine) = \(88.36 g / 35.45 g/mol = 2.492 mol\)

Find the Mole Ratio

To find the mole ratio, we'll divide the moles of each element by the smallest number of moles: Mole ratio of nitrogen (N) = \(0.831 mol / 0.831 mol = 1\) Mole ratio of chlorine (Cl) = \(2.492 mol / 0.831 mol = 2.998 \approx 3\)

Determine the Empirical Formula

The mole ratios found in step 3 give us the empirical formula of the compound: Empirical Formula = NCl3

Key concepts

Molar Mass

Molar mass is a fundamental concept in chemistry that helps us bridge the gap between the mass of a substance and the number of particles it contains. Simply put, molar mass is the mass of one mole of a given substance, usually measured in grams per mole (g/mol). This concept is key when we want to convert a given mass into moles — a basic calculation in stoichiometry and empirical formula determination.

To calculate the molar mass of an element, we look at the atomic mass given on the periodic table. For example:

• The molar mass of nitrogen (N) is 14.01 g/mol.
• The molar mass of chlorine (Cl) is 35.45 g/mol.

By using molar mass, we can easily convert the mass of each element in a compound to moles. In the original problem, we converted 11.64 grams of nitrogen to moles by using the formula:\[\text{Moles of nitrogen} = \frac{11.64 \, \text{g}}{14.01 \, \text{g/mol}} = 0.831 \, \text{mol}\]
Understanding molar mass helps us gain insights into the quantitative aspects of chemical reactions and compounds.

Mole Ratio

The mole ratio is a simple yet powerful tool that allows chemists to understand the proportions of elements within a compound. It is defined as the relative amount of moles of each element present in a compound. This concept is crucial for finding the empirical formula, which gives us the simplest whole-number ratio of elements in a compound.

To determine the mole ratio, one must first convert the mass of each element to moles using their respective molar masses. Then, as in the original exercise, divide the number of moles of each element by the smallest number of moles found among the elements:

• Moles of nitrogen (N) = 0.831 mol
• Moles of chlorine (Cl) = 2.492 mol

Next, divide each by the smallest mole number:\[\text{Mole ratio of nitrogen (N)} = \frac{0.831 \, \text{mol}}{0.831 \, \text{mol}} = 1\]\[\text{Mole ratio of chlorine (Cl)} = \frac{2.492 \, \text{mol}}{0.831 \, \text{mol}} \approx 3\]These mole ratios help create a clear picture of how elements combine at the atomic level to form compounds, leading to the empirical formula: NCl₃.

Percentage Composition

Percentage composition is a straightforward and practical concept in chemistry, which specifies the amount of each element in a compound by mass. It tells you what percentage of the entire compound's mass is contributed by each element.

For instance, in the problem, the compound has a percentage composition of:

• 11.64% nitrogen
• 88.36% chlorine

This percentage provides a quick overview of the proportion of each element. When calculating empirical formulas, these percentages are often converted to masses assuming a 100-gram sample, simplifying the math. For instance, 11.64% of nitrogen in 100 grams translates to 11.64 grams.

This simplified conversion step allows you to focus on finding how much of each element, in moles, is needed to express the compound in its simplest, whole-number ratio form, as demonstrated in empirical formula calculations. Understanding percentage composition offers valuable insights into the elemental makeup of a compound and is foundational for many chemical analyses.