A certain headache relief tablet is composed of monoprotic acetylsalicylic acid(C9H8O4(s)) and an inert filler. A 4.00 g tablet was crushed and dissolved to make 40.0 mL of solution. The solution was then titrated with 0.900 mol/L NaOH(aq). The volume of NaOH(aq) needed to neutralize the dissolved tablet was 20.1 mL. The experimental value for the mass of C9H8O4(s) present in the tablet

was ___?

moles NaOH = M x L = ? moles NaOH

moles ASA = same since the reaction (you should write the equation) is 1 mole ASA to 1 mol NaOH.
moles ASA= grams ASA/molar mass ASA.
You have moles and molar mass, solve for grams ASA.
If you want percent, then
percent ASA = (g ASA/4.00)*100 = x

3.26

To find the experimental value for the mass of C9H8O4(s) present in the tablet, we need to use the information given in the problem.

First, let's calculate the number of moles of NaOH used in the titration:

Moles of NaOH = Molarity of NaOH × Volume of NaOH(aq) used
= 0.900 mol/L × 20.1 mL
= 0.01809 mol

Next, we'll use the balanced chemical equation to determine the stoichiometry between acetylsalicylic acid (C9H8O4) and NaOH:

C9H8O4(s) + 2NaOH(aq) -> NaC9H7O4(aq) + H2O(l)

From the equation, we can see that it takes 2 moles of NaOH to react with 1 mole of C9H8O4.

Therefore, the number of moles of C9H8O4 can be calculated as:

Moles of C9H8O4 = (Moles of NaOH) / 2
= 0.01809 mol / 2
= 0.009045 mol

Next, we find the molar mass of C9H8O4:

Molar mass of C9H8O4 = (9 × atomic mass of carbon) + (8 × atomic mass of hydrogen) + (4 × atomic mass of oxygen)
= (9 × 12.01 g/mol) + (8 × 1.01 g/mol) + (4 × 16.00 g/mol)
= 180.16 g/mol

Finally, we can calculate the experimental value for the mass of C9H8O4 present in the tablet:

Mass of C9H8O4 = Moles of C9H8O4 × Molar mass of C9H8O4
= 0.009045 mol × 180.16 g/mol
≈ 1.63 g

Therefore, the experimental value for the mass of C9H8O4 present in the tablet is approximately 1.63 grams.