1C6H1206 + 602→6CO2 + 6H20 + 362.5 kJ If the theoretical yield of water is 354.3 grams and the actual yield is 320.3 grams, what is the percent error?
1C6H1206 + 602→6CO2 + 6H20 + 362.5 kJ If the theoretical yield of water is 354.3 grams and the actual yield is 320.3 grams, what is the percent error?
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...
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![**Chemical Reaction and Calculation of Percent Error**
**Chemical Reaction:**
The balanced chemical equation for the combustion of glucose is:
\[ \text{1C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2 \rightarrow 6\text{CO}_2 + 6\text{H}_2\text{O} + 362.5 \, \text{kJ} \]
**Problem Statement:**
If the theoretical yield of water is 354.3 grams and the actual yield is 320.3 grams, what is the percent error?
**Calculation of Percent Error:**
To find the percent error, use the formula:
\[ \text{Percent Error} = \left( \frac{\text{Theoretical Yield} - \text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\% \]
Plug in the values:
\[ \text{Percent Error} = \left( \frac{354.3 \, \text{g} - 320.3 \, \text{g}}{354.3 \, \text{g}} \right) \times 100\% \]
\[ \text{Percent Error} = \left( \frac{34.0 \, \text{g}}{354.3 \, \text{g}} \right) \times 100\% \]
\[ \text{Percent Error} \approx 9.6\% \]
This calculation shows that the percent error in the yield of water is approximately 9.6%.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F35844c14-8e0f-4ef2-9129-409849687b45%2Fb7f200e0-1f1d-4240-b256-f1f39a0d7cf3%2Fxuaneg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Chemical Reaction and Calculation of Percent Error**
**Chemical Reaction:**
The balanced chemical equation for the combustion of glucose is:
\[ \text{1C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2 \rightarrow 6\text{CO}_2 + 6\text{H}_2\text{O} + 362.5 \, \text{kJ} \]
**Problem Statement:**
If the theoretical yield of water is 354.3 grams and the actual yield is 320.3 grams, what is the percent error?
**Calculation of Percent Error:**
To find the percent error, use the formula:
\[ \text{Percent Error} = \left( \frac{\text{Theoretical Yield} - \text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\% \]
Plug in the values:
\[ \text{Percent Error} = \left( \frac{354.3 \, \text{g} - 320.3 \, \text{g}}{354.3 \, \text{g}} \right) \times 100\% \]
\[ \text{Percent Error} = \left( \frac{34.0 \, \text{g}}{354.3 \, \text{g}} \right) \times 100\% \]
\[ \text{Percent Error} \approx 9.6\% \]
This calculation shows that the percent error in the yield of water is approximately 9.6%.
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