Question

The most common form of nylon (nylon-6) is 63.68\(\%\) carbon, 12.38\(\%\) nitrogen, 9.80\(\%\) hydrogen, and 14.14\(\%\) oxygen. Calculate the empirical formula for nylon-6.

Short answer

The empirical formula for nylon-6 is C6NH11O.

Step-by-step solution

Convert percentages to grams

Imagine we have 100 grams of nylon-6, then the mass of each element will be equal to the percentage given: Carbon: 63.68 grams Nitrogen: 12.38 grams Hydrogen: 9.80 grams Oxygen: 14.14 grams

Convert grams to moles

Now, we will convert the mass of each element into moles by dividing by their respective atomic masses. Here are the atomic masses of the elements: Carbon (C): 12.01 g/mol Nitrogen (N): 14.01 g/mol Hydrogen (H): 1.01 g/mol Oxygen (O): 16.00 g/mol Moles of Carbon = \( \frac{63.68}{12.01} \) = 5.307 moles Moles of Nitrogen = \( \frac{12.38}{14.01} \) = 0.884 moles Moles of Hydrogen = \( \frac{9.80}{1.01} \) = 9.703 moles Moles of Oxygen = \( \frac{14.14}{16.00} \) = 0.884 moles

Find the simplest whole number ratio

In order to find the whole number ratio, we will divide all mole values by the smallest mole value. In this case, the smallest mole value is for Nitrogen and Oxygen, which are both 0.884 moles. Carbon: \( \frac{5.307}{0.884} \approx 6 \) Nitrogen: \( \frac{0.884}{0.884} = 1 \) Hydrogen: \( \frac{9.703}{0.884} \approx 11 \) Oxygen: \( \frac{0.884}{0.884} = 1 \)

Write the empirical formula

With the whole number ratios as 6, 1, 11, 1 for Carbon, Nitrogen, Hydrogen, and Oxygen, respectively, we can now write the empirical formula for nylon-6: Empirical Formula: C6N1H11O1 or simply C6NH11O

Key concepts

Nylon-6

Nylon-6, which is also called polycaprolactam, is a type of synthetic polymer. It is part of the larger nylon family but is unique because it is formed from a single type of monomer, caprolactam. This monomer undergoes a process called ring-opening polymerization to form the polymer chains of Nylon-6.

This material is quite popular due to its strong, elastic, and resilient nature. It's widely used in textiles, carpets, and even in automotive parts. Its ability to withstand wear and abuse while retaining its strengths makes it particularly useful in products that require durability.

For chemistry students, understanding the composition of Nylon-6 helps in visualizing how molecular combinations form practical materials. Analyzing its empirical formula reveals the makeup of the compound at its simplest whole number ratio, providing insight into its chemical makeup.

Chemistry Problem Solving

Solving chemistry problems often requires a methodical approach. With problems like finding an empirical formula, it is crucial to follow clear steps. Start by understanding what the question asks and gather all pertinent information.

In the case of the empirical formula calculation, we need to know the percentage composition of the compound. From this, we translate percentages into actual masses, often using the assumption of a 100-gram sample for simplicity.

• First, convert percentages to grams: Assume 100 g of the sample, making each percentage equal to the grams of each element.
• Second, transform grams to moles: Use the atomic masses of the respective elements for this conversion.

In the end, this organized methodology helps ensure each step leads smoothly to the next, ultimately resulting in the empirical formula.

Moles Calculation

Moles are a fundamental concept in chemistry, representing the amount of a substance. Calculating moles is a bridge between macroscopic amounts of substance and the microscopic particles they are made of—molecules or atoms.

To find the moles from a given mass, we need the atomic or molecular mass of the substance. This is done using the formula:
\[\text{Moles} = \frac{\text{Mass in grams}}{\text{Atomic mass in g/mol}}\]For example, if you have 63.68 g of carbon, you divide this value by the atomic mass of carbon, 12.01 g/mol, to find the moles: 5.307 moles as shown in the solution steps.

This calculation helps interpret amounts in chemistry precisely, allowing us to compare different elements and safely mix chemicals in a way that fosters chemical reactions.

Percentage Composition to Empirical Formula

Understanding the transition from percentage composition to the empirical formula is a key skill in chemistry. The empirical formula represents the simplest whole-number ratio of elements in a compound. Here's how to get there succinctly:

• Start by converting the percentage of each element to mass, assuming a 100 g sample size. This assumption simplifies calculations as it aligns percentages directly with grams.
• Then, convert these masses into moles using each element's atomic mass.
• Identify the smallest number of moles calculated and divide all mole values by this number to normalize the proportions.
• Once simplified, these ratios form the empirical formula, representing the basic structure of the compound.

For instance, in nylon-6, this process lets us understand its composition as C6NH11O, clearly breaking down how it is constructed at the molecular level.