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Chapter 5: Problem 84

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

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

Step by step solution

01

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)
02

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
03

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.
04

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.

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Most popular questions from this chapter

A compound that contains only carbon, hydrogen, and oxygen is \(48.64 \%\) C and \(8.16 \%\) H by mass. What is the empirical formula of this substance?

A compound contains only carbon, hydrogen, nitrogen, and oxygen. Combustion of 0.157 g of the compound produced \(0.213 \mathrm{g} \mathrm{CO}_{2}\) and \(0.0310 \mathrm{g} \mathrm{H}_{2} \mathrm{O} .\) In another experiment, it is found that 0.103 g of the compound produces \(0.0230 \mathrm{g} \mathrm{NH}_{3}\) What is the empirical formula of the compound? Hint: Combustion involves reacting with excess \(\mathrm{O}_{2}\). Assume that all the carbon ends up in \(\mathrm{CO}_{2}\) and all the hydrogen ends up in \(\mathrm{H}_{2} \mathrm{O}\). Also assume that all the nitrogen ends up in the \(\mathrm{NH}_{3}\) in the second experiment.

A sample of urea contains \(1.121 \mathrm{g} \mathrm{N}, 0.161 \mathrm{g} \mathrm{H}, 0.480 \mathrm{g} \mathrm{C}\) and \(0.640 \mathrm{g}\) O. What is the empirical formula of urea?

Calculate the percent composition by mass of the following compounds that are important starting materials for synthetic polymers: a. \(\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{O}_{2}\) (acrylic acid, from which acrylic plastics are made) b. \(\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{2}\) (methyl acrylate, from which Plexiglas is made) c. \(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{N}\) (acrylonitrile, from which Orlon is made)

Tetrodotoxin is a toxic chemical found in fugu pufferfish, a popular but rare delicacy in Japan. This compound has an LD\(_{50}\) (the amount of substance that is lethal to \(50 . \%\) of a population sample) of \(10 . \mu g\) per \(\mathrm{kg}\) of body mass. Tetrodotoxin is \(41.38 \%\) carbon by mass, \(13.16\%\) nitrogen by mass, and \(5.37\%\) hydrogen by mass, with the remaining amount consisting of oxygen. What is the empirical formula of tetrodotoxin? If three molecules of tetrodotoxin have a mass of \(1.59 \times 10^{-21} \mathrm{g},\) what is the molecular formula of tetrodotoxin? What number of molecules of tetrodotoxin would be the \(\mathrm{LD}_{50}\) dosage for a person weighing 165 lb?
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