/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 131 Anabolic steroids are performanc... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 131

Anabolic steroids are performance enhancement drugs whose use has been banned from most major sporting activities. One anabolic steroid is fluoxymesterone \(\left(\mathrm{C}_{20} \mathrm{H}_{29} \mathrm{FO}_{3}\right) .\) Calculate the percent composition by mass of fluoxymesterone.

Short Answer

Expert verified
The percent composition by mass of fluoxymesterone is approximately 71.41% carbon, 8.70% hydrogen, 5.65% fluorine, and 14.27% oxygen.

Step by step solution

01

Determine the molar mass of fluoxymesterone

To find the molar mass of fluoxymesterone, we must first identify the molar mass of each individual element and then multiply each element's molar mass by its stoichiometric coefficient. The molar masses of the given elements are: - Carbon (C): \(12.01 \frac{g}{mol}\) - Hydrogen (H): \(1.01 \frac{g}{mol}\) - Fluorine (F): \(19.00 \frac{g}{mol}\) - Oxygen (O): \(16.00 \frac{g}{mol}\) Now, we can find the molar mass of fluoxymesterone: Molar mass = (Molar mass of C) 脳 (Number of C atoms) + (Molar mass of H) 脳 (Number of H atoms) + (Molar mass of F) 脳 (Number of F atoms) + (Molar mass of O) 脳 (Number of O atoms)
02

Calculate the molar mass of fluoxymesterone

Applying the molar masses and the stoichiometric coefficients to the formula: Molar mass = (20 脳 12.01) + (29 脳 1.01) + (19.00) + (3 脳 16.00) Molar mass = 240.20 + 29.29 + 19.00 + 48.00 Molar mass = 336.49 g/mol We now have the molar mass of fluoxymesterone.
03

Calculate the percentage composition by mass

Now, we need to find the fraction of each element's mass within the molar mass of fluoxymesterone, and then multiply that by 100 to get the percentage composition. Percentage of C = (\(\frac{Mass of C}{Molar mass of fluoxymesterone}\)) 脳 100 Percentage of H = (\(\frac{Mass of H}{Molar mass of fluoxymesterone}\)) 脳 100 Percentage of F = (\(\frac{Mass of F}{Molar mass of fluoxymesterone}\)) 脳 100 Percentage of O = (\(\frac{Mass of O}{Molar mass of fluoxymesterone}\)) 脳 100 Using these formulas: Percentage of C = (\(\frac{240.20}{336.49}\)) 脳 100 = 71.41% Percentage of H = (\(\frac{29.29}{336.49}\)) 脳 100 = 8.70% Percentage of F = (\(\frac{19.00}{336.49}\)) 脳 100 = 5.65% Percentage of O = (\(\frac{48.00}{336.49}\)) 脳 100 = 14.27% The percent composition by mass of fluoxymesterone is approximately 71.41% carbon, 8.70% hydrogen, 5.65% fluorine, and 14.27% oxygen.

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 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molar Mass Calculation
Understanding how to calculate molar mass is foundational to mastering stoichiometry and percent composition in chemistry. Molar mass is fundamentally the mass of one mole of a substance, measured in grams per mole (g/mol). To calculate the molar mass of a compound like fluoxymesterone (C20H29FO3), you would need to know the atomic masses of each element involved, typically found on the periodic table.

Here's a simple break down for fluoxymesterone:
  • Carbon (C) has an atomic mass of 12.01 g/mol.
  • Hydrogen (H) is 1.01 g/mol.
  • Fluorine (F) stands at 19.00 g/mol.
  • Oxygen (O) is 16.00 g/mol.
By multiplying each atomic mass by the number of atoms in a molecule of the compound and adding the products, we arrive at the compound's molar mass. This calculation method is a critical stepping stone for the subsequent analysis of a substance's composition and reactivity in chemical reactions.
Stoichiometry
Stoichiometry is the section of chemistry that provides tools for calculating relative quantities of reactants and products in chemical reactions. Essential for figuring out how much of each substance is involved in a reaction, stoichiometry is based on the law of conservation of mass, which states that matter cannot be created or destroyed.

In the context of the percent composition calculation, stoichiometry allows us to understand that each atom in the chemical formula contributes a specific part to the overall mass of the compound. Utilizing ratios derived from the balanced reaction or formula unit, chemists can calculate necessary amounts for desired chemical yields, which is handy not only in academic exercises but fundamental in industrial applications for creating everything from pharmaceuticals to plastics.
Chemical Formula
The chemical formula of a substance tells us the specific types and numbers of atoms present in a single unit of that compound. It serves as a type of shorthand that conveys a wealth of information to those with a knowledge of chemistry. For instance, fluoxymesterone's formula, C20H29FO3, indicates that each molecule consists of 20 carbon atoms, 29 hydrogen atoms, one fluorine atom, and three oxygen atoms. Understanding the chemical formula is key to not only figuring out molar mass but also to determining the percent composition, as shown in the workout problem. It's this formula that allows us to calculate the mass contribution of each element efficiently and can be used to infer properties about the compound, predict how it might react with other substances, and understand its role within a larger chemical process such as drug development.

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

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

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?

Boron consists of two isotopes, \(^{10} \mathbf{B}\) and \(^{11} \mathbf{B}\). Chlorine also has two isotopes, \(^{35} \mathrm{Cl}\) and \(^{37} \mathrm{Cl}\). Consider the mass spectrum of \(\mathrm{BCl}_{3}\) How many peaks would be present, and what approximate mass would each peak correspond to in the \(\mathrm{BCl}_{3}\) mass spectrum?

A compound with molar mass \(180.1 \mathrm{g} / \mathrm{mol}\) has the following composition by mass: $$\begin{array}{|ll|}\hline C & 40.0 \% \\\H & 6.70 \% \\\O & 53.3 \% \\\\\hline\end{array}$$ Determine the empirical and molecular formulas of the compound.

Some bismuth tablets, a medication used to treat upset stomachs, contain \(262 \mathrm{mg}\) of bismuth subsalicylate, \(\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{BiO}_{4},\) per tablet. Assuming two tablets are digested, calculate the mass of bismuth consumed.
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Ionic and Molecular Compounds

Read Explanation
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.