Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 9: Problem 140

Determine the mass percent of each element in aluminum sulfate.

Short Answer

Expert verified
The mass percent of each element in aluminum sulfate (Al₂(SO₄)₃) is: Aluminum (Al): 15.77%, Sulfur (S): 28.10%, and Oxygen (O): 56.13%.

Step by step solution

01

Find the chemical formula of aluminum sulfate

The chemical formula of aluminum sulfate is Al₂(SO₄)₃.
02

Calculate the molar mass of aluminum sulfate

To do this, first find the atomic masses of each element present in the formula, and then multiply their atomic masses by the number of atoms. Add these values together for the total molar mass. - Aluminum (Al): 26.98 g/mol - Sulfur (S): 32.06 g/mol - Oxygen (O): 16.00 g/mol Molar mass of aluminum sulfate: = 2 * 26.98 (Al) + 3 * (4 * 16.00 (O) + 32.06 (S)) = 2 * 26.98 + 3 * (64 + 32.06) = 53.96 + 3 * 96.06 = 53.96 + 288.18 = 342.14 g/mol
03

Calculate the mass percent of each element

Now, divide the mass of each element in one mole of aluminum sulfate by the molar mass of aluminum sulfate and multiply by 100%. 1. Mass percent of aluminum (Al): = (53.96 g/mol (Al) / 342.14 g/mol (Al₂(SO₄)₃)) * 100% = (0.1577) * 100% = 15.77% 2. Mass percent of sulfur (S): = (3 * 32.06 g/mol (S) / 342.14 g/mol (Al₂(SO₄)₃)) * 100% = (96.18 g/mol (S) / 342.14 g/mol) * 100% = (0.2810) * 100% = 28.10% 3. Mass percent of oxygen (O): = (12 * 16.00 g/mol (O) / 342.14 g/mol (Al₂(SO₄)₃)) * 100% = (192.00 g/mol (O) / 342.14 g/mol) * 100% = (0.5613) * 100% = 56.13%
04

Final Answer

The mass percent of each element in aluminum sulfate is: - Aluminum (Al): 15.77% - Sulfur (S): 28.10% - Oxygen (O): 56.13%.

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.

Start your free trial

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

Consider the following unbalanced chemical equation: \(\mathrm{H}_{2}+\mathrm{N}_{2} \rightarrow \mathrm{NH}_{3}\) (a) To run this reaction in a balanced fashion, how much nitrogen is required if you start with \(10.0 \mathrm{~g}\) of \(\mathrm{H}_{2} ?\) (b) How many grams of ammonia \(\left(\mathrm{NH}_{3}\right)\) will you produce if you run the reaction with the masses calculated in part (a)? (c) How many molecules of ammonia will you produce?

A compound is \(91.77 \%\) by mass \(\mathrm{Si}\) and \(8.23 \%\) by mass \(\mathrm{H}\) and has a molar mass of approximately \(122 \mathrm{~g} / \mathrm{mol}\). What is its molecular formula?

The compound ibuprofen, used in some pain relievers, has the molecular formula \(\mathrm{C}_{13} \mathrm{H}_{18} \mathrm{O}_{2}\). If \(0.250 \mathrm{~g}\) of ibuprofen is burned in oxygen, how many grams of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) will be produced?

Consider the unbalanced chemical equation \(\mathrm{As}_{4} \mathrm{~S}_{6}+\mathrm{O}_{2} \rightarrow \mathrm{As}_{4} \mathrm{O}_{6}+\mathrm{SO}_{2}\) (a) How many grams of \(\mathrm{O}_{2}\) are needed to react completely with \(58.9 \mathrm{~g}\) of \(\mathrm{As}_{4} \mathrm{~S}_{6}\) ? (b) If \(41.2 \mathrm{~g}\) of \(\mathrm{SO}_{2}\) is produced, what is the percent yield for the reaction?

The amino acid arginine is made up of \(41.37 \%\) by mass \(\mathrm{C}, 8.10 \%\) by mass \(\mathrm{H}, 32.16 \%\) by mass \(\mathrm{N}\), and \(18.37 \%\) by mass \(\mathrm{O}\). If the molar mass of arginine is approximately \(174 \mathrm{~g} / \mathrm{mol}\), what is its molecular formula?
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Kinetics

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Reactions

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.

Sign-up for free