An impure sample of benzoic acid (C 6H 5COOH, 122.12 g/mol) is titrated with 0.2099 M NaOH. A 1.021-g sample requires 35.73 mL of titrant to reach the endpoint. What is the percent by mass of benzoic acid in the sample? C6H5COOH(aq) + NaOH(aq) → NaC6H5COO(aq) + H2O(l)

The percent by mass of benzoic acid = 89.7 %

Explanation:

Step 1: Data given

Molarity of NaOH = 0.2099 M

Mass of the sample = 1.021 grams

Volume of NaOH = 35.73 mL = 0.03573 L

Step 2: The balanced equation

C6H5COOH(aq) + NaOH(aq) → NaC6H5COO(aq) + H2O(l)

Step 3: Calculate moles NaOH

Moles NaOH = molarity NaOH * volume NaOH

Moles NaOH = 0.2099 * 0.03573 L

Moles NaOH = 0.00750 moles

Step 3: Calculate moles benzoic acid

From your balanced equation - benzoic acid reacts in 1:1 molar ratio with NaOH  

The sample must have contained 0.00750 moles benzoic acid  

Mass benzoic acid in 0.00750 moles  = 0.00750 * 122.12 = 0.9159 g benzoic acid  

Mass % acid in sample =  0.9159/1.021 * 100 = 89.7% purity.

The percent by mass of benzoic acid = 89.7 %

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.