FOUND.OF COLLEGE CHEMISTRY
15th Edition
ISBN: 9781119234555
Author: Hein
Publisher: WILEY
expand_more
expand_more
format_list_bulleted
Videos
Question
Chapter 7.4, Problem 7.11P
Interpretation Introduction
Interpretation:
The empirical formula of a compound has to be given.
Concept Introduction:
Empirical Formula:
The empirical formula of a compound is the simplest whole number ratio of each type of atom in a compound. It can be the same as the compound’s molecular formula but not always. An empirical formula can be calculated from information about the mass of each element in a compound or from the percentage composition.
The steps for determining the empirical formula of a compound as follows:
- Obtain the mass of each element present in grams.
- Determine the number of moles of each atom present.
- Divide the number of moles of each element by the smallest number of moles.
- Convert the numbers to whole numbers. The set of whole numbers are the subscripts in the empirical formula.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Chapter 7 Solutions
FOUND.OF COLLEGE CHEMISTRY
Ch. 7.1 - Prob. 7.1PCh. 7.1 - Prob. 7.2PCh. 7.1 - Prob. 7.3PCh. 7.2 - Prob. 7.4PCh. 7.2 - Prob. 7.5PCh. 7.2 - Prob. 7.6PCh. 7.3 - Prob. 7.7PCh. 7.3 - Prob. 7.8PCh. 7.3 - Prob. 7.9PCh. 7.4 - Prob. 7.10P
Ch. 7.4 - Prob. 7.11PCh. 7.5 - Prob. 7.12PCh. 7 - Prob. 1RQCh. 7 - Prob. 2RQCh. 7 - Prob. 3RQCh. 7 - Prob. 4RQCh. 7 - Prob. 5RQCh. 7 - Prob. 6RQCh. 7 - Prob. 7RQCh. 7 - Prob. 8RQCh. 7 - Prob. 9RQCh. 7 - Prob. 10RQCh. 7 - Prob. 11RQCh. 7 - Prob. 12RQCh. 7 - Prob. 13RQCh. 7 - Prob. 14RQCh. 7 - Prob. 15RQCh. 7 - Prob. 17RQCh. 7 - Prob. 18RQCh. 7 - Prob. 19RQCh. 7 - Prob. 1PECh. 7 - Prob. 2PECh. 7 - Prob. 3PECh. 7 - Prob. 4PECh. 7 - Prob. 5PECh. 7 - Prob. 6PECh. 7 - Prob. 7PECh. 7 - Prob. 8PECh. 7 - Prob. 9PECh. 7 - Prob. 10PECh. 7 - Prob. 11PECh. 7 - Prob. 12PECh. 7 - Prob. 13PECh. 7 - Prob. 14PECh. 7 - Prob. 15PECh. 7 - Prob. 16PECh. 7 - Prob. 17PECh. 7 - Prob. 18PECh. 7 - Prob. 19PECh. 7 - Prob. 20PECh. 7 - Prob. 21PECh. 7 - Prob. 22PECh. 7 - Prob. 25PECh. 7 - Prob. 26PECh. 7 - Prob. 27PECh. 7 - Prob. 28PECh. 7 - Prob. 29PECh. 7 - Prob. 30PECh. 7 - Prob. 31PECh. 7 - Prob. 32PECh. 7 - Prob. 33PECh. 7 - Prob. 34PECh. 7 - Prob. 35PECh. 7 - Prob. 36PECh. 7 - Prob. 37PECh. 7 - Prob. 38PECh. 7 - Prob. 39PECh. 7 - Prob. 40PECh. 7 - Prob. 41PECh. 7 - Prob. 42PECh. 7 - Prob. 43PECh. 7 - Prob. 44PECh. 7 - Prob. 45PECh. 7 - Prob. 46PECh. 7 - Prob. 47PECh. 7 - Prob. 48PECh. 7 - Prob. 49PECh. 7 - Prob. 50PECh. 7 - Prob. 51PECh. 7 - Prob. 52PECh. 7 - Prob. 53AECh. 7 - Prob. 54AECh. 7 - Prob. 55AECh. 7 - Prob. 56AECh. 7 - Prob. 57AECh. 7 - Prob. 58AECh. 7 - Prob. 59AECh. 7 - Prob. 60AECh. 7 - Prob. 61AECh. 7 - Prob. 62AECh. 7 - Prob. 63AECh. 7 - Prob. 64AECh. 7 - Prob. 65AECh. 7 - Prob. 66AECh. 7 - Prob. 67AECh. 7 - Prob. 68AECh. 7 - Prob. 69AECh. 7 - Prob. 70AECh. 7 - Prob. 71AECh. 7 - Prob. 72AECh. 7 - Prob. 73AECh. 7 - Prob. 74AECh. 7 - Prob. 75AECh. 7 - Prob. 76AECh. 7 - Prob. 77AECh. 7 - Prob. 78AECh. 7 - Prob. 79AECh. 7 - Prob. 80AECh. 7 - Prob. 81AECh. 7 - Prob. 82AECh. 7 - Prob. 83AECh. 7 - Prob. 84AECh. 7 - Prob. 88AECh. 7 - Prob. 89CECh. 7 - Prob. 90CE
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.Similar questions
- Salvarsan was long thought to be a single substance. Recently, however, a mass spectrometry study of the compound shows it to be a mixture of two molecules with the same empirical formula. Each has the composition 39.37% C, 3.304% H, 8.741% O, 7.652% N, and 40.932% As. One has a molar mass of 549 g/mol and the other has a molar mass of 915 g/mol. What are the molecular formulas of the compounds?arrow_forwardCalculate the mass in grams of 3.839 moles of (a) hydrazine, a rocket propellant. (b) caffeine, C4H5N2O. (c) theobromine, C7H8N4O2, the alkaloid present in chocolate and cocoa.arrow_forwardThe active ingredient in Pepto-Bismo® (an over- the-counter remedy for an upset stomach) is bismuth sub-salicylate, C7H5BiO4. Analysis of a 1.7500-g sample of Pepto-Bismol yields 346 mg of bismuth. What percent by mass is bismuth subsalicylate in the sample? (Assume that there are no other bismuth-containing compounds in Pepto-Bismol.)arrow_forward
- Phenol is a compound of carbon, hydrogen, and oxygen used commonly as a disinfectant. Combustion of a 175-mg sample of phenol yielded 491 mg CO2 and 100. mg H2O. Calculate the empirical formula of phenol. Identify what other information is needed to determine whether the empirical formula is the actual molecular formula.arrow_forwardCalculate the molar mass of the following substances. a. b. c. (NH4)2Cr2O7arrow_forwardA sample of a hydrocarbon (a compound consisting of only carbon and hydrogen) contains 2.59 1023 atoms of hydrogen and is 17.3% hydrogen by mass. If the molar mass of the hydrocarbon is between 55 and 65 g/mol, what amount (moles) of compound is present, and what is the mass of the sample?arrow_forward
- 1. Which of the following hydrocarbons has the highest mass percentage of carbon? methane, CH4 ethane, C2H6 propane, C3H8 butane, C4H10arrow_forwardA compound was analyzed and found to contain the following percentages of elements by mass: carbon, 79.89%; hydrogen, 20.11%. l type='a'> Determine the empirical formula of the compound. i>Which of the following could be a molar mass of the compound? 13.018 g/mol 16.042g/mol 28.052 g/mol 30.068 g/mol 104.23 g/rnolarrow_forwardNitrogen fixation in the root nodules of peas and other legumes occurs with a reaction involving a molybdenum-containing enzyme named nitrogenase. This enzyme contains two Mo atoms per molecule and is 0.0872% Mo by mass. Calculate the molar mass of the enzyme.arrow_forward
- A substance X2Z has the composition (by mass) of 40.0% X and 60.0% Z. What is the composition (by mass) of the compound XZ2?arrow_forward3.96 Methyl cyanoacrylate is the chemical name for the substance sold as Super Glue, and it has the chemical formula C5H5NO2. Calculate the number of molecules of this substance in a 1.0-ounce tube of Super Glue, assuming that the glue is 80% methyl cyanoacrylate by mass.arrow_forwardHydrogen cyanide is produced industrially from the reaction of gaseous ammonia, oxygen, and methane: 2NH3(g)+3O2(g)+2CH4(g)2HCN(g)+6H2O(g) If 5.00 103 kg each of NH3, O2, and CH4 are reacted, what mass of HCN and of H2O will be produced, assuming 100% yield?arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Chemistry: Matter and ChangeChemistryISBN:9780078746376Author:Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl WistromPublisher:Glencoe/McGraw-Hill School Pub Co
ChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage Learning
Chemistry: An Atoms First ApproachChemistryISBN:9781305079243Author:Steven S. Zumdahl, Susan A. ZumdahlPublisher:Cengage Learning
Chemistry & Chemical ReactivityChemistryISBN:9781133949640Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage Learning
Chemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage Learning
Chemistry: Matter and Change
Chemistry
ISBN:9780078746376
Author:Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl Wistrom
Publisher:Glencoe/McGraw-Hill School Pub Co
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Chemistry: An Atoms First Approach
Chemistry
ISBN:9781305079243
Author:Steven S. Zumdahl, Susan A. Zumdahl
Publisher:Cengage Learning
Chemistry & Chemical Reactivity
Chemistry
ISBN:9781133949640
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:Cengage Learning
Chemistry: The Molecular Science
Chemistry
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:Cengage Learning
GCSE Chemistry - Differences Between Compounds, Molecules & Mixtures #3; Author: Cognito;https://www.youtube.com/watch?v=jBDr0mHyc5M;License: Standard YouTube License, CC-BY