Para-cresol, a substance used as a disinfectant and in the manufacture of several herbicides, is a molecule that contains the elements carbon, hydrogen, and oxygen. Complete combustion of a \(0.345-\mathrm{g}\) sample of \(p\) -cresol produced 0.983 g carbon dioxide and 0.230 \(\mathrm{g}\) water. Determine the empirical formula for \(p\) -cresol.

Short Answer

Expert verified
The empirical formula of p-cresol is \(C_7H_8O\).

Step by step solution

01

Determine the moles of CO_2 and H_2O produced

Given: Mass of CO_2 produced = 0.983 g Mass of H_2O produced = 0.230 g Molar mass of CO_2 = 12.01 (Carbon) + 2 * 16.00 (Oxygen) = 44.01 g/mol Molar mass of H_2O = 2 * 1.01 (Hydrogen) + 16.00 (Oxygen) = 18.02 g/mol Moles of CO_2 produced = mass of CO_2 / molar mass of CO_2 = 0.983 g / 44.01 g/mol ≈ 0.0223 mol Moles of H_2O produced = mass of H_2O / molar mass of H_2O = 0.230 g / 18.02 g/mol ≈ 0.0128 mol
02

Determine the moles of C, H, and O in p-cresol

In one mole of CO_2, there is one mole of Carbon. Thus, moles of Carbon in p-cresol = moles of CO_2 produced = 0.0223 mol In one mole of H_2O, there are two moles of Hydrogen. Thus, moles of Hydrogen in p-cresol = 2 * moles of H_2O produced = 2 * 0.0128 mol = 0.0256 mol Now we need to find the moles of Oxygen in p-cresol. From the initial mass of p-cresol given (0.345 g), we can find the mass of Oxygen by subtracting the mass of Carbon and Hydrogen present: Mass of Carbon = moles of Carbon * molar mass of Carbon = 0.0223 mol * 12.01 g/mol ≈ 0.267 g Mass of Hydrogen = moles of Hydrogen * molar mass of Hydrogen = 0.0256 mol * 1.01 g/mol ≈ 0.0258 g Mass of Oxygen = mass of p-cresol - mass of Carbon - mass of Hydrogen = 0.345 g - 0.267 g - 0.0258 g ≈ 0.052 g Now we can find the moles of Oxygen: Moles of Oxygen = mass of Oxygen / molar mass of Oxygen = 0.052 g / 16.00 g/mol ≈ 0.00325 mol
03

Find the ratio of C: H: O moles

To find the empirical formula, we need to find the simplest whole number ratio of C: H: O atoms in the molecule. First, divide all the moles by the lowest number of moles: Carbon = 0.0223 / 0.00325 ≈ 6.86 Hydrogen = 0.0256 / 0.00325 ≈ 7.87 Oxygen = 0.00325 / 0.00325 ≈ 1 We observe that all the ratios are close to whole numbers. We can round them to the nearest whole number to obtain: C: H: O ≈ 7:8:1
04

Write the empirical formula for p-cresol

The empirical formula is written using the whole number ratio of atoms. Therefore, the empirical formula of p-cresol is: \(C_7H_8O\) And that is the final answer.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Zinc and magnesium metal each reacts with hydrochloric acid to make chloride salts of the respective metals, and hydrogen gas. A 10.00 -g mixture of zinc and magnesium produces 0.5171 g of hydrogen gas upon being mixed with an excess of hydrochloric acid. Determine the percent magnesium by mass in the original mixture.

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.

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.

The compound adrenaline contains \(56.79 \% \mathrm{C}, 6.56 \% \mathrm{H}\) , \(28.37 \% \mathrm{O},\) and 8.28\(\% \mathrm{N}\) by mass. What is the empirical formula for adrenaline?

Natural rubidium has the average mass of 85.4678 \(\mathrm{u}\) and is composed of isotopes \(^{85} \mathrm{Rb}(\mathrm{mass}=84.9117 \mathrm{u})\) and $^{87} \mathrm{Rb}\( . The ratio of atoms \)^{85} \mathrm{Rb} /^{87} \mathrm{Rb}$ in natural rubidium is \(2.591 .\) Calculate the mass of \(^{87} \mathrm{Rb}\) .

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free