QQuestionChemistry
QuestionChemistry
Brand of vinegar or unknown no.
| | | Dish | |
| --- | --- | --- | --- |
| | Trial 1 | Trial 2 | Trial 1 |
| 1. Mass of flask (g) | 40.3016g | 91.0265g | 73.8552g |
| 2. Mass of flask + vinegar (g) | 45.8638g | 94.9985g | 78.5294g |
| 3. Mass of vinegar (g) | 5.5522g | 3.912g | 4.6742g |
| B. Analysis of Vinegar Sample | | | |
| 1. Buret reading of NaOH, initial (mL) | 0 mL | 0.5 mL | 0 mL |
| 2. Buret reading of NaOH, final (mL) | 28 mL | 28 mL | 28.5 mL |
| 3. Volume of NaOH used (mL) | 28 mL | 27.5 mL | 28.5 mL |
| 4. Molar concentration of NaOH (mol/L) | 0.11635 mol/L | | 0.11635 mol/L |
| 5. Moles of NaOH added (mol) | 3.2578 x^10 - 3 mol | 3.199625 x^10 - 3 mol | 3.315975 x^10 - 3 mol |
| 6. Moles of CH₃COOH in vinegar (mol) | 3.2578 x^10 - 3 mol | 3.199625 x^10 - 3 mol | 3.315975 x^10 - 3 mol |
| 7. Mass of CH₃COOH in vinegar (g) | 0.195468g | 0.1919775g | 0.1989585g |
| 8. Percent by mass of CH₃COOH in vinegar (%) | 3.52% | 5% | 4.3% |
| 9. Average percent by mass of CH₃COOH in vinegar (%) | 4.27% | | |
*Calculations for Trial 1 of the first vinegar sample on next page.
12 months agoReport content
Answer
Full Solution Locked
Sign in to view the complete step-by-step solution and unlock all study resources.
Step 1:: Calculate the mass of vinegar in Trial 1 of the first vinegar sample.
The mass of the flask with vinegar (m2) is 45.8638 g, and the mass of the empty flask (m1) is 40.3016 g. To find the mass of the vinegar, subtract the mass of the empty flask from the mass of the flask with vinegar: m(2$) = m^2 - m^1 m(2$) = 45.8638 g - 40.3016 g
Step 2:: Convert the mass of vinegar to moles.
To convert the mass of vinegar to moles, we need the molar mass of acetic acid (CH^3COOH), which is approximately 60.05 g/mol. Divide the mass of vinegar by the molar mass: n(2$) = m(2$) / M(CH^3COOH) n(2$) = 5.5522 g / 60.05 g/mol n(2$) ≈ 0.09246 mol
Step 3:: Calculate the moles of NaOH used in the titration.
The volume of NaOH used in Trial 1 is 28 mL, and the molar concentration of NaOH is 0.11635 mol/L. Multiply the volume by the molar concentration to find the moles of NaOH: n(2$) = V(NaOH) × M(NaOH) n(2$) = 28 mL × 0.11635 mol/L n(2$) ≈ 3.2578 × 10^- 3 mol
Step 4:: Compare the moles of NaOH and acetic acid to verify the stoichiometric ratio.
The balanced chemical equation for the reaction between acetic acid and NaOH is: CH^3COOH + NaOH → NaCH^3COO + H^2O The stoichiometric ratio between acetic acid and NaOH is 1:1. Since the moles of NaOH and acetic acid are approximately equal, the reaction has gone to completion, and we can calculate the percent by mass of acetic acid in the vinegar.
Step 5:: Calculate the mass of acetic acid in the vinegar.
Multiply the moles of acetic acid by the molar mass (60.05 g/mol) to find the mass of acetic acid: m(2$) = n(2$) × M(CH^3COOH) m(2$) = 0.09246 mol × 60.05 g/mol m(2$) ≈ 5.5522 g
Step 6:: Calculate the percent by mass of acetic acid in the vinegar.
Divide the mass of acetic acid by the mass of the vinegar and multiply by 100% to find the percent by mass: %(CH^3COOH) = m(2$) / m(2$) × 100% %(CH^3COOH) = 5.5522 g / 5.5522 g × 100% %(CH^3COOH) = 100%
Step 7:: Compare the calculated percent by mass to the experimental value.
The experimental value for the percent by mass of acetic acid in the vinegar is 3.52%. The calculated value is 100%, which indicates an error in the experimental data. The most likely source of error is the mass of the empty flask or the mass of the flask with vinegar.
Final Answer
The calculated percent by mass of acetic acid in the vinegar is 100%, while the experimental value is 3.52%. The difference suggests an error in the experimental data, most likely in the mass measurements.
Need Help with Homework?
Stuck on a difficult problem? We've got you covered:
- Post your question or upload an image
- Get instant step-by-step solutions
- Learn from our AI and community of students