Thanks for your help; I am still have a few problems, could you check my work:
Given: Volume of vinegar analyzed-5mL
Con'c of NaOH-0.09890M
Avg. Volume of NaOH from titration-43.75mL
Density CH3COOH: 1.049g/mL
Calculations:
1.Moles of NaOH required to reach equivalence point= M X L
(0.09890M)(0.04375L)= 0.004327 moles.
2.Moles of acetic acid 5.00mL of vinegar= M X L
(what do I do if the con'c of acetic isn't given)(0.0500L)=?
3. Mass of acetic acid in the 5.00mL sample of vinegar= Mol X MM (0.004327 mol)(60.01g/mol)=0.2529g
4. Volume of the acetic acid in the 5.00mL sample of vinegar
(1.049g/mL)(0.2597g)=0.2724mL
5. % by volume of acetic acid in the vinegar (0.2724mL)/(5.00mL)X 100%= 5.45%
Please let me know if this looks correct.
Let's go through your calculations step by step:
1. Moles of NaOH required to reach equivalence point: This calculation is correct. You multiplied the molarity of NaOH (0.09890 M) by the volume of NaOH used in the titration (43.75 mL converted to 0.04375 L) to get 0.004327 moles.
2. Moles of acetic acid in 5.00 mL of vinegar: If the concentration of acetic acid (CH3COOH) is not given, you won't be able to directly calculate the moles of acetic acid. The concentration is needed to determine the number of moles using the formula Molarity (M) = moles (mol) / volume (L). Without this information, you won't be able to proceed with the calculation.
3. Mass of acetic acid in the 5.00 mL sample of vinegar: Assuming you can find the moles from the previous step, you correctly multiplied the moles of acetic acid (0.004327 mol) by the molar mass of acetic acid (60.01 g/mol) to get 0.2597 g.
4. Volume of the acetic acid in the 5.00 mL sample of vinegar: You used the density of acetic acid (1.049 g/mL) and the mass of acetic acid (0.2597 g) to calculate the volume. However, it seems there was a typo in your calculations. It should be (0.2597 g)/(1.049 g/mL) = 0.2474 mL, not 0.2724 mL.
5. % by volume of acetic acid in the vinegar: You correctly divided the volume of the acetic acid (0.2474 mL) by the volume of the vinegar sample (5.00 mL) and multiplied it by 100% to calculate the percentage. However, using the corrected volume from step 4, the correct calculation would be (0.2474 mL)/(5.00 mL) × 100% = 4.948%.
Remember to double-check your calculation steps and ensure you have all the necessary information before proceeding with calculations.