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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: An Atoms First Approach
Chemistry: An Atoms First Approach
2nd Edition
ISBN:9781305079243
Author:Steven S. Zumdahl, Susan A. Zumdahl
Publisher:Steven S. Zumdahl, Susan A. Zumdahl
Chapter5: Stoichiometry
Section: Chapter Questions
Problem 138AE: Terephthalic acid is an important chemical used in the manufacture of polyesters and plasticizers....
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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
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