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Chapter 3: Problem 86

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 C6NH11O.

Step by step solution

01

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
02

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
03

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: \( \frac{0.884}{0.884} = 1 \)
04

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

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Most popular questions from this chapter

Which of the following statements about chemical equations is(are) true? a. When balancing a chemical equation, you can never change the coefficient in front of any chemical formula. b. The coefficients in a balanced chemical equation refer to the number of grams of reactants and products. c. In a chemical equation, the reactants are on the right and the products are on the left. d. When balancing a chemical equation, you can never change the subscripts of any chemical formula. e. In chemical reactions, matter is neither created nor destroyed so a chemical equation must have the same number of atoms on both sides of the equation.

Nitric acid is produced commercially by the Ostwald process, represented by the following equations: $$ \begin{aligned} 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}_{2}(g) \\ 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g) \end{aligned} $$ What mass of \(\mathrm{NH}_{3}\) must be used to produce $1.0 \times 10^{6} \mathrm{kg}\( \)\mathrm{HNO}_{3}\( by the Ostwald process? Assume 100\)\%$ yield in each reaction, and assume that the NO produced in the third step is not recycled.

A binary compound between an unknown element \(\mathrm{E}\) and hydrogen contains 91.27\(\% \mathrm{E}\) and 8.73\(\% \mathrm{H}\) by mass. If the formula of the compound is \(\mathrm{E}_{3} \mathrm{H}_{8},\) calculate the atomic mass of \(\mathrm{E}\)

a. Write the balanced equation for the combustion of isooctane \(\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)\) to produce water vapor and carbon dioxide gas. b. Assuming gasoline is \(100 . \%\) isooctane, with a density of 0.692 \(\mathrm{g} / \mathrm{mL}\) , what is the theoretical yield of carbon dioxide produced by the combustion of \(1.2 \times 10^{10}\) gal of gasoline (the approximate annual consumption of gasoline in the United States)?

Balance the following equations: a. $\operatorname{Cr}(s)+\mathrm{S}_{8}(s) \rightarrow \mathrm{Cr}_{2} \mathrm{S}_{3}(s)$ b. \(\operatorname{NaHCO}_{3}(s) \stackrel{\mathrm{Heat}}{\longrightarrow}\) $\mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)$ c. \(\quad \mathrm{KClO}_{3}(s) \stackrel{\mathrm{Heat}}{\longrightarrow}\) \(\mathrm{KCl}(s)+\mathrm{O}_{2}(g)\) d. $\operatorname{Eu}(s)+\mathrm{HF}(g) \rightarrow \operatorname{EuF}_{3}(s)+\mathrm{H}_{2}(g)$
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