Problem 84 - Chemistry An Atoms First Approach

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.

Step-by-Step Solution

Answer

The empirical formula for nylon-6 is \(C_6NH_11O\).

Step 1: Convert percentages to grams

First, we can assume that we have 100 grams of the compound. This allows us to directly convert the percentages into grams. We have: 63.68% Carbon → 63.68 grams of Carbon (C) 12.38% Nitrogen → 12.38 grams of Nitrogen (N) 9.80% Hydrogen → 9.80 grams of Hydrogen (H) 14.14% Oxygen → 14.14 grams of Oxygen (O)

Step 2: Convert grams to moles

Using the molecular weight of each element, we can convert the grams to moles: - Carbon (C): 12.01 g/mol - Nitrogen (N): 14.01 g/mol - Hydrogen (H): 1.008 g/mol - Oxygen (O): 16.00 g/mol Moles of Carbon = 63.68 g / 12.01 g/mol = 5.306 moles Moles of Nitrogen = 12.38 g / 14.01 g/mol = 0.884 moles Moles of Hydrogen = 9.80 g / 1.008 g/mol = 9.722 moles Moles of Oxygen = 14.14 g / 16.00 g/mol = 0.884 moles

Step 3: Find the mole ratio

To find the empirical formula, we need to find the mole ratio between each element. Divide each element's mole value by the smallest mole value among the four elements: Mole value ratio of Carbon = 5.306 moles / 0.884 moles = 6.00 ≈ 6 Mole value ratio of Nitrogen = 0.884 moles / 0.884 moles = 1.00 ≈ 1 Mole value ratio of Hydrogen = 9.722 moles / 0.884 moles = 11.00 ≈ 11 Mole value ratio of Oxygen = 0.884 moles / 0.884 moles = 1.00 ≈ 1 Thus, the ratio is about 6:1:11:1 for Carbon, Nitrogen, Hydrogen, and Oxygen, respectively.

Step 4: Determine the empirical formula

The empirical formula is determined by using the mole value ratios as subscripts for each element: Empirical formula for nylon-6: C_6N_1H_11O_1, which can be simplified as C_6NH_11O. Therefore, the empirical formula for nylon-6 is C_6NH_11O.

Key Concepts

Chemical Composition

Understanding the chemical composition of a substance involves breaking it down into its constituent elements and determining the proportion or percentage of each element present. For nylon-6, the chemical composition is given as percentages: 63.68% carbon, 12.38% nitrogen, 9.80% hydrogen, and 14.14% oxygen. These values tell us the distribution of elements in the compound.

To interpret these percentages, we assume a sample size, commonly 100 grams, which simplifies percentages into direct grams. This makes it easy to visualize the exact amount of each element in a hypothetical sample. For instance, in a 100 gram sample of nylon-6:

• 63.68 grams would be carbon
• 12.38 grams nitrogen
• 9.80 grams hydrogen
• 14.14 grams oxygen

Breaking down compounds in this manner is a fundamental step in chemistry for understanding empirical formulas and examining molecular structures.

Mole Calculation

Mole calculation is an essential skill in chemistry used to convert masses to moles, which are a basic counting unit for atoms and molecules. The mole helps chemists create a bridge between the mass of a substance and the number of atoms or molecules it contains. To find the empirical formula of nylon-6, we use molar masses of each element to convert from grams to moles.

Molar masses used in our calculation include:

• Carbon (C): 12.01 g/mol
• Nitrogen (N): 14.01 g/mol
• Hydrogen (H): 1.008 g/mol
• Oxygen (O): 16.00 g/mol

This allows us to calculate:- Moles of Carbon: \( \frac{63.68}{12.01} = 5.306 \) moles- Moles of Nitrogen: \( \frac{12.38}{14.01} = 0.884 \) moles- Moles of Hydrogen: \( \frac{9.80}{1.008} = 9.722 \) moles- Moles of Oxygen: \( \frac{14.14}{16.00} = 0.884 \) moles

Converting grams to moles is a critical part of determining empirical formulas, enabling chemists to understand the simplest ratio of elements in a compound.

Nylon-6

Nylon-6 is a synthetic polymer widely used in fibers for textiles and for other industrial uses. Its name originates from the number of carbon atoms in the repeating unit of its polymer chain, denoted in its empirical formula. To arrive at this empirical formula, we focus on the idea of mole ratios, which helps in identifying the simplest whole-number ratio of the elements present.

For nylon-6, the mole ratios of each element are calculated as follows:

• Carbon: \( \frac{5.306}{0.884} \approx 6 \)
• Nitrogen: \( \frac{0.884}{0.884} \approx 1 \)
• Hydrogen: \( \frac{9.722}{0.884} \approx 11 \)
• Oxygen: \( \frac{0.884}{0.884} \approx 1 \)

This gives us the empirical formula \( C_6NH_{11}O \), showing the proportions of carbon, nitrogen, hydrogen, and oxygen in the nylon-6 compound. Understanding these proportions is vital for producing nylon materials with the desired properties and performance.