‘Calculate the mass percent of acetic in the vinegar sample. You may want t0 break acid the calculation down t0 the following steps: Using the volume and molarity of NaOH; find the number of moles of NaOHithat were used in this titration. Use this tO get the number of moles of acetic acid in the vinegar sample that was titrated, Convert moles of acetic acid to grams of acetic acid for the vinegar sample thatoas (Note: HOAc is NOT titrated. To do this, you will need acetic acid” $ molar mass: the formula_ the actual formula is HCzH;Oz) After converting grams to moles; you now have the mass of acetic acid in the vinegar sample. Now find the mass of the entire vinegar sample that was titrated You may do this by assuming that the density ofthe vinegar solution is 1.02 glmL. Finally, find the percent acetic acid in the vinegar sample.’
The Correct Answer and Explanation is:
Correct Answer:
To calculate the mass percent of acetic acid in a vinegar sample, the following step-by-step process applies:
Given Data (assumed typical values for a vinegar titration):
- Volume of NaOH used: 25.00 mL = 0.02500 L
- Molarity of NaOH: 0.5000 mol/L
- Volume of vinegar titrated: 10.00 mL
- Molar mass of acetic acid (HC₂H₃O₂): 60.05 g/mol
- Density of vinegar: 1.02 g/mL
Step 1: Calculate moles of NaOH used
Moles of NaOH=Molarity×Volume (in L)=0.5000 mol/L×0.02500 L=0.01250 mol\text{Moles of NaOH} = \text{Molarity} \times \text{Volume (in L)} = 0.5000 \, \text{mol/L} \times 0.02500 \, \text{L} = 0.01250 \, \text{mol}Moles of NaOH=Molarity×Volume (in L)=0.5000mol/L×0.02500L=0.01250mol
Step 2: Use mole ratio to find moles of acetic acid
From the balanced chemical equation:NaOH+HC₂H₃O₂→NaC₂H₃O₂+H₂O\text{NaOH} + \text{HC₂H₃O₂} \rightarrow \text{NaC₂H₃O₂} + \text{H₂O}NaOH+HC₂H₃O₂→NaC₂H₃O₂+H₂O
This shows a 1:1 molar ratio. Therefore:Moles of acetic acid=Moles of NaOH=0.01250 mol\text{Moles of acetic acid} = \text{Moles of NaOH} = 0.01250 \, \text{mol}Moles of acetic acid=Moles of NaOH=0.01250mol
Step 3: Convert moles of acetic acid to grams
Mass=Moles×Molar Mass=0.01250 mol×60.05 g/mol=0.7506 g\text{Mass} = \text{Moles} \times \text{Molar Mass} = 0.01250 \, \text{mol} \times 60.05 \, \text{g/mol} = 0.7506 \, \text{g}Mass=Moles×Molar Mass=0.01250mol×60.05g/mol=0.7506g
Step 4: Find mass of vinegar sample titrated
Mass of vinegar=Volume×Density=10.00 mL×1.02 g/mL=10.20 g\text{Mass of vinegar} = \text{Volume} \times \text{Density} = 10.00 \, \text{mL} \times 1.02 \, \text{g/mL} = 10.20 \, \text{g}Mass of vinegar=Volume×Density=10.00mL×1.02g/mL=10.20g
Step 5: Calculate mass percent of acetic acid
Mass percent=(0.7506 g10.20 g)×100%≈7.36%\text{Mass percent} = \left( \frac{0.7506 \, \text{g}}{10.20 \, \text{g}} \right) \times 100\% \approx 7.36\%Mass percent=(10.20g0.7506g)×100%≈7.36%
Conclusion:
The mass percent of acetic acid in the vinegar sample is approximately 7.36%.
Explanation:
In this titration-based analysis, the mass percent of acetic acid in vinegar is determined using stoichiometric relationships and conversion factors. The titration involves a reaction between sodium hydroxide (NaOH), a strong base, and acetic acid (HC₂H₃O₂), a weak monoprotic acid. The chemical equation demonstrates a 1:1 molar ratio between these two substances. Thus, the number of moles of NaOH used directly equals the number of moles of acetic acid present in the sample.
Using the volume and molarity of the NaOH solution, the number of moles is calculated by multiplying the molarity (mol/L) by the volume in liters. This value represents the quantity of acetic acid in moles due to the 1:1 reaction ratio. To obtain the mass of acetic acid, this mole value is multiplied by its molar mass, which is known to be 60.05 grams per mole.
Next, the total mass of the vinegar sample used in the titration is determined by multiplying its volume by its density. The density is assumed to be 1.02 grams per milliliter, a typical value for household vinegar. With the mass of acetic acid and the mass of the entire sample available, the mass percent is found by dividing the former by the latter and multiplying the result by 100.
This process yields the mass percentage concentration of acetic acid in the vinegar sample.
