The actual identity of the monoprotic acid is potassium hydrogen phthalate (KHCaH.O4), which has a molecular weight of 204 g/mol. Calculate the percent error using the average molecular weight from Calculation #2.
The actual identity of the monoprotic acid is potassium hydrogen phthalate (KHCaH.O4), which has a molecular weight of 204 g/mol. Calculate the percent error using the average molecular weight from Calculation #2.
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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![**Calculations:** *(Show all work and provide answers with correct significant figures.)*
1. **Determine the molecular weight of the unknown acid from:**
a. **Trial 1**
\[
0.12M \times 18.55mL = 2.226 \text{ mmol } \\
= 2.226 \times 10^{-3} \text{ mol}
\]
\[
0.47g / 2.226 \times 10^{-3} \text{ mol} = 211.1 \text{ g/mol}
\]
b. **Trial 2**
\[
0.12M \times 17.6mL = 2.112 \text{ mmol} \\
= 2.112 \times 10^{-3} \text{ mol}
\]
\[
0.46g / 2.112 \times 10^{-3} \text{ mol} = 213.8 \text{ g/mol}
\]
c. **Trial 3**
\[
0.12M \times 18.90mL = 2.268 \text{ mmol } \\
= 2.268 \times 10^{-3} \text{ mol}
\]
\[
0.48g / 2.268 \times 10^{-3} \text{ mol} = 211.6 \text{ g/mol}
\]
2. **Calculate the average molecular weight of the unknown monoprotic acid from Trials 1-3:**
\[
212.16 \text{ g/mol}
\]
3. **The actual identity of the monoprotic acid is potassium hydrogen phthalate (KHC8H4O4), which has a molecular weight of 204 g/mol. Calculate the percent error using the average molecular weight from Calculation #2.**](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5c5734d7-71be-4626-8cb5-227ca6c7fb69%2F7d250f64-de68-4bf9-b3df-acd30ac18cb0%2Fuzyr9o_processed.png&w=3840&q=75)
Transcribed Image Text:**Calculations:** *(Show all work and provide answers with correct significant figures.)*
1. **Determine the molecular weight of the unknown acid from:**
a. **Trial 1**
\[
0.12M \times 18.55mL = 2.226 \text{ mmol } \\
= 2.226 \times 10^{-3} \text{ mol}
\]
\[
0.47g / 2.226 \times 10^{-3} \text{ mol} = 211.1 \text{ g/mol}
\]
b. **Trial 2**
\[
0.12M \times 17.6mL = 2.112 \text{ mmol} \\
= 2.112 \times 10^{-3} \text{ mol}
\]
\[
0.46g / 2.112 \times 10^{-3} \text{ mol} = 213.8 \text{ g/mol}
\]
c. **Trial 3**
\[
0.12M \times 18.90mL = 2.268 \text{ mmol } \\
= 2.268 \times 10^{-3} \text{ mol}
\]
\[
0.48g / 2.268 \times 10^{-3} \text{ mol} = 211.6 \text{ g/mol}
\]
2. **Calculate the average molecular weight of the unknown monoprotic acid from Trials 1-3:**
\[
212.16 \text{ g/mol}
\]
3. **The actual identity of the monoprotic acid is potassium hydrogen phthalate (KHC8H4O4), which has a molecular weight of 204 g/mol. Calculate the percent error using the average molecular weight from Calculation #2.**
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