Homework Help is Here – Start Your Trial Now!
Science
Chemistry
Empirical Formula Proof for C₂H302, a compound containing 57.838% carbon, 3.641% hydrogen, and 38.521% oxygen : CARBON 57.838% 79.887% 27.293% 40.002% ASSUME A 100 g SAMPLE 57.838% of 100: 0.57838 100 = 57.838 g C 3.641% of 100: 0.03641100 = 3.641 g H 38.521% of 100: 0.38521100= 38.521 g 0 Show the empirical formula proof for any other compound listed in the table provided: Percent Composition HYDROGEN 3.6405% 20.113% 6.7142% OXYGEN 38.521% 72.711% 53.284% Empirical Formula 57.838 g C x CONVERT EACH TO MOLES USING MOLAR MASS OF EACH ELEMENT 1 mole C= 4.8154 moles C 12.011 g C 3.641 g H x 1 mole H = 1.008 g H 38.521 g 0 x 1 mole 0 2.4077 moles O 15.999 g O C₂H₂O₂ CH₂ CO₂ CH₂O Molecular Formula C₂H₂O₁ C₂H₂ CO₂ C₂H₂O 3.6121 moles H DIVIDE ALL MOLES BY THE SMALLEST MOLE AMOUNT 4.8154 moles C/ 2.4077= 2 3.6121 moles H/2.4077 = 1.5 2.4077 moles O/2.4077= 1 C₂H₁.501 MULTIPLY & ROUND SUBSCRIPTS TO ENSURE WHOLE NUMBERS AS NECESSARY C₂H₁.50₁ I C₂H₂O₂ x 2

Empirical Formula Proof for C₂H302, a compound containing 57.838% carbon, 3.641% hydrogen, and 38.521% oxygen : CARBON 57.838% 79.887% 27.293% 40.002% ASSUME A 100 g SAMPLE 57.838% of 100: 0.57838 100 = 57.838 g C 3.641% of 100: 0.03641100 = 3.641 g H 38.521% of 100: 0.38521100= 38.521 g 0 Show the empirical formula proof for any other compound listed in the table provided: Percent Composition HYDROGEN 3.6405% 20.113% 6.7142% OXYGEN 38.521% 72.711% 53.284% Empirical Formula 57.838 g C x CONVERT EACH TO MOLES USING MOLAR MASS OF EACH ELEMENT 1 mole C= 4.8154 moles C 12.011 g C 3.641 g H x 1 mole H = 1.008 g H 38.521 g 0 x 1 mole 0 2.4077 moles O 15.999 g O C₂H₂O₂ CH₂ CO₂ CH₂O Molecular Formula C₂H₂O₁ C₂H₂ CO₂ C₂H₂O 3.6121 moles H DIVIDE ALL MOLES BY THE SMALLEST MOLE AMOUNT 4.8154 moles C/ 2.4077= 2 3.6121 moles H/2.4077 = 1.5 2.4077 moles O/2.4077= 1 C₂H₁.501 MULTIPLY & ROUND SUBSCRIPTS TO ENSURE WHOLE NUMBERS AS NECESSARY C₂H₁.50₁ I C₂H₂O₂ x 2

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
See similar textbooks
icon
Related questions
Question

Solve like the exmaple

Empirical Formula Proof for C4H30₂, a compound containing 57.838% carbon, 3.641% hydrogen, and 38.521% oxygen : CARBON 57.838% 79.887% 27.793% 40.002% ASSUME A 100 g SAMPLE 57.838% of 100: 0.57838*100 = 57.838 g C 3.641% of 100: 0.03641*100 = 3.641 g H 38.521% of 100: 0.38521*100 = 38.521 g O Show the empirical formula proof for any other compound listed in the table provided: Percent Composition HYDROGEN 3.6405% 20.113% 6.7142% OXYGEN 38.521% 72.7115 53.284% Empirical Formula CONVERT EACH TO MOLES USING MOLAR MASS OF EACH ELEMENT 57.838 g C x 1 mole C 12.011 g C 3.641 g H x 1 mole H 1.008 g H 38.521 g 0 x 1 mole 0 H 2.4077 moles 0 15.999 g 0 C₂H₂O₂ CH₂ CO₂ CH₂0 Molecular Formule C₂H₂0₁ C₂H₂ CO₂ C₂H₁ ₂00 = E 4.8154 moles C 3.6121 moles H DIVIDE ALL MOLES BY THE SMALLEST MOLE AMOUNT 4.8154 moles C/ 2.4077 = 2 3.6121 moles H/ 2.4077 = 1.5 2.4077 moles 0 / 2.4077 = 1 C₂H₁.501 MULTIPLY & ROUND SUBSCRIPTS TO ENSURE WHOLE NUMBERS AS NECESSARY C₂H₁.50₁ [x2 C4H₂O₂ cond proof
Transcribed Image Text:Empirical Formula Proof for C4H30₂, a compound containing 57.838% carbon, 3.641% hydrogen, and 38.521% oxygen : CARBON 57.838% 79.887% 27.793% 40.002% ASSUME A 100 g SAMPLE 57.838% of 100: 0.57838*100 = 57.838 g C 3.641% of 100: 0.03641*100 = 3.641 g H 38.521% of 100: 0.38521*100 = 38.521 g O Show the empirical formula proof for any other compound listed in the table provided: Percent Composition HYDROGEN 3.6405% 20.113% 6.7142% OXYGEN 38.521% 72.7115 53.284% Empirical Formula CONVERT EACH TO MOLES USING MOLAR MASS OF EACH ELEMENT 57.838 g C x 1 mole C 12.011 g C 3.641 g H x 1 mole H 1.008 g H 38.521 g 0 x 1 mole 0 H 2.4077 moles 0 15.999 g 0 C₂H₂O₂ CH₂ CO₂ CH₂0 Molecular Formule C₂H₂0₁ C₂H₂ CO₂ C₂H₁ ₂00 = E 4.8154 moles C 3.6121 moles H DIVIDE ALL MOLES BY THE SMALLEST MOLE AMOUNT 4.8154 moles C/ 2.4077 = 2 3.6121 moles H/ 2.4077 = 1.5 2.4077 moles 0 / 2.4077 = 1 C₂H₁.501 MULTIPLY & ROUND SUBSCRIPTS TO ENSURE WHOLE NUMBERS AS NECESSARY C₂H₁.50₁ [x2 C4H₂O₂ cond proof
Empirical Formula Proof for C4H30₂, a compound containing 57.838% carbon, 3.641% hydrogen, and 38.521% oxygen : 57.838% 79.887% 27.293% 40.002% Show the empirical formula proof for any other compound listed in the table provided: ASSUME A 100 g SAMPLE 57.838% of 100: 0.57838* 100 = 57.838 g C 3.641% of 100: 0.03641100 = 3.641 g H 38.521% of 100: 0.38521*100 = 38.521 g 0 Percent Composition HYDROGEN 3.6405% 20.113% 6.7142% 38.521% C 72.711% 53.284% Empirical Formula 27.293% of 100: 0.2793x100=27.293 g C₂H₂O₂ CH3 CO₂ CH₂O CONVERT EACH TO MOLES USING MOLAR MASS OF EACH ELEMENT 57.838 g C x 1 mole C 12.011 g C 3.641 g H x 1 mole H = 1.008 g H 38.521 g 0 x 1 mole 0 15.999 g O 72.711% of 100: 0.72711x100= 72.711 g 0 Molecular Formula C₂H₂O₁ C₂H₁ CO₂ C₂H₁₂0 = 4.8154 moles C 3.6121 moles H 2.4077 moles O DIVIDE ALL MOLES BY THE SMALLEST MOLE AMOUNT 4.8154 moles C/ 2.4077 = 2 3.6121 moles H/ 2.4077 = 1.5 2.4077 moles 0 / 2.4077 = 1 C₂H₁.501 27.293 g C x1 mole/ 12.011 g C= 2.272 moles C 72.711 g 0 x 1 mole O/15.999 g O=4.545 moles O MULTIPLY & ROUND SUBSCRIPTS TO ENSURE WHOLE NUMBERS AS NECESSARY C₂H₁.501 1x C4H₂O₂ x 2 4.545 moles O/ 2.272= 2 2.272 moles C/2.272=1 CO2 first proof
Transcribed Image Text:Empirical Formula Proof for C4H30₂, a compound containing 57.838% carbon, 3.641% hydrogen, and 38.521% oxygen : 57.838% 79.887% 27.293% 40.002% Show the empirical formula proof for any other compound listed in the table provided: ASSUME A 100 g SAMPLE 57.838% of 100: 0.57838* 100 = 57.838 g C 3.641% of 100: 0.03641100 = 3.641 g H 38.521% of 100: 0.38521*100 = 38.521 g 0 Percent Composition HYDROGEN 3.6405% 20.113% 6.7142% 38.521% C 72.711% 53.284% Empirical Formula 27.293% of 100: 0.2793x100=27.293 g C₂H₂O₂ CH3 CO₂ CH₂O CONVERT EACH TO MOLES USING MOLAR MASS OF EACH ELEMENT 57.838 g C x 1 mole C 12.011 g C 3.641 g H x 1 mole H = 1.008 g H 38.521 g 0 x 1 mole 0 15.999 g O 72.711% of 100: 0.72711x100= 72.711 g 0 Molecular Formula C₂H₂O₁ C₂H₁ CO₂ C₂H₁₂0 = 4.8154 moles C 3.6121 moles H 2.4077 moles O DIVIDE ALL MOLES BY THE SMALLEST MOLE AMOUNT 4.8154 moles C/ 2.4077 = 2 3.6121 moles H/ 2.4077 = 1.5 2.4077 moles 0 / 2.4077 = 1 C₂H₁.501 27.293 g C x1 mole/ 12.011 g C= 2.272 moles C 72.711 g 0 x 1 mole O/15.999 g O=4.545 moles O MULTIPLY & ROUND SUBSCRIPTS TO ENSURE WHOLE NUMBERS AS NECESSARY C₂H₁.501 1x C4H₂O₂ x 2 4.545 moles O/ 2.272= 2 2.272 moles C/2.272=1 CO2 first proof
Expert Solution
steps

Step by step

Solved in 4 steps with 1 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Knowledge Booster
Stoichiometry
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
  • SEE MORE QUESTIONS
Recommended textbooks for you
Chemistry
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education
Principles of Instrumental Analysis
Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning
Organic Chemistry
Organic Chemistry
Chemistry
ISBN:
9780078021558
Author:
Janice Gorzynski Smith Dr.
Publisher:
McGraw-Hill Education
Chemistry: Principles and Reactions
Chemistry: Principles and Reactions
Chemistry
ISBN:
9781305079373
Author:
William L. Masterton, Cecile N. Hurley
Publisher:
Cengage Learning
Elementary Principles of Chemical Processes, Bind…
Elementary Principles of Chemical Processes, Bind…
Chemistry
ISBN:
9781118431221
Author:
Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:
WILEY
SEE MORE TEXTBOOKS