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 \(C_6N H_{11}O\).

Step-by-step solution

Convert Percentages to Grams

If we assume that we have 100g of nylon-6, then the percentage of each element can be directly converted to grams. - Carbon: \(63.68 \% \times 100g = 63.68g\) - Nitrogen: \(12.38 \% \times 100g = 12.38g\) - Hydrogen: \(9.80 \% \times 100g = 9.80g\) - Oxygen: \(14.14 \% \times 100g = 14.14g\)

Convert Grams to Moles

We can now convert the mass of each element to moles using their respective atomic weight. - Carbon: \( \frac{63.68g}{12.01g/mol} = 5.31 mol\) - Nitrogen: \( \frac{12.38g}{14.01g/mol} = 0.88 mol\) - Hydrogen: \( \frac{9.80g}{1.01g/mol} = 9.70 mol\) - Oxygen: \( \frac{14.14g}{16.00g/mol} = 0.88 mol\)

Divide Each Moles by the Lowest Moles

To find the simplest whole number ratio, we will divide the moles of each element by the lowest moles we obtained in the previous step, which is 0.88 mol. - Carbon: \( \frac{5.31 mol}{0.88 mol} = 6.03 \approx 6\) - Nitrogen: \( \frac{0.88 mol}{0.88 mol} = 1 \) - Hydrogen: \( \frac{9.70 mol}{0.88 mol} = 11.02 \approx 11\) - Oxygen: \( \frac{0.88 mol}{0.88 mol} = 1 \)

Determine the Empirical Formula

Now that we have the simplest whole number ratio of the elements in nylon-6, we can write the empirical formula. Empirical Formula: \(C_6N H_{11}O\) So, the empirical formula for nylon-6 is \(C_6N H_{11}O\).

Key concepts

Chemical Composition

Chemical composition refers to the types and quantities of atoms that make up a molecule or substance. For nylon-6, which is a common synthetic polymer, understanding chemical composition is crucial for determining its properties and uses.
In the given exercise, the chemical composition is expressed in terms of the percentage by mass of each element within nylon-6:

• Carbon: 63.68%
• Nitrogen: 12.38%
• Hydrogen: 9.80%
• Oxygen: 14.14%

These percentages help us understand the major contributing elements to the polymer's structure. By converting these percentages into grams, assuming a 100 g sample, we can then proceed to find the empirical formula. This empirical formula reveals the simplest ratio of the elements and is a fundamental step in understanding nylon-6's chemical makeup.

Molecular Structure

The molecular structure describes how atoms in a molecule are arranged and bonded together. This arrangement is critical in defining the characteristics of a substance.
The empirical formula of a molecule gives us insight into its molecular structure by showing the simplest whole number ratio of the constituent elements. In the case of nylon-6, the previously calculated empirical formula is \(C_6N H_{11}O\). This formula tells us:

• There are 6 carbon atoms.
• 1 nitrogen atom.
• 11 hydrogen atoms.
• 1 oxygen atom.

While the empirical formula gives a simplified view, the actual molecular structure is more complex with repeating units. The molecular structure, including the precise arrangement of atoms and bonds, significantly affects nylon-6's properties such as strength, flexibility, and chemical resistance.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions, based on their relationships derived from balanced chemical equations. It is fundamental in chemistry for scaling reactions and understanding formula compositions.
In calculating the empirical formula of nylon-6, stoichiometry is applied when converting the mass of each element into moles. By using the elements' atomic masses:

• Carbon: \( rac{63.68g}{12.01g/mol} = 5.31 \) moles.
• Nitrogen: \( rac{12.38g}{14.01g/mol} = 0.88 \) moles.
• Hydrogen: \( rac{9.80g}{1.01g/mol} = 9.70 \) moles.
• Oxygen: \( rac{14.14g}{16.00g/mol} = 0.88 \) moles.

These calculations are then simplified by dividing each element's mole quantity by the smallest value obtained, ensuring the ratio reflects the simplest form suitable for constructing the empirical formula.
Understanding stoichiometry allows us to accurately derive empirical and molecular formulas, key to comprehending chemical substances like nylon-6.