# Question 24b1e

Apr 7, 2017

The vinegar is 4.36 % acetic acid by mass.

#### Explanation:

Step 1. Write the balanced equation

$\text{CH"_3"COOH" + "NaOH" → "CH"_3"COONa" + "H"_2"O}$

Step 2. Calculate the moles of NaOH

$\text{Moles of NaOH" = "0.030 84" color(red)(cancel(color(black)("dm"^3color(white)(l) "NaOH"))) × "0.128 mol NaOH"/(1 color(red)(cancel(color(black)("dm"^3 color(white)(l)"NaOH")))) = "0.003 948 mol NaOH}$

Step 3. Calculate the moles of $\text{CH"_3"COOH}$

$\text{Moles of CH"_3"COOH" = "0.003 948" color(red)(cancel(color(black)("mol NaOH"))) × ("1 mol CH"_3"COOH")/(1 color(red)(cancel(color(black)("mol NaOH")))) = "0.003 948 mol CH"_3"COOH}$

Step 4. Calculate the mass of the $\text{CH"_3"COOH}$

$\text{Mass of CH"_3"COOH" = "0.003 948" color(red)(cancel(color(black)("mol CH"_3"COOH"))) × ("60.05 g CH"_3"COOH")/(1 color(red)(cancel(color(black)("mol CH"_3"COOH")))) = "0.2370 g CH"_3"COOH}$

Step 5. Calculate the mass percent of acetic acid.

"Mass percent" = "mass of acetic acid"/"mass of vinegar" × 100 % = (0.2370 color(red)(cancel(color(black)("g"))))/(5.441 color(red)(cancel(color(black)("g")))) × 100 %

= 4.36 %#