If a 3.300 g sample of an unknown monoprotic acid is dissolved in 2.0 x 10 2 mL
ID: 1039584 • Letter: I
Question
If a 3.300 g sample of an unknown monoprotic acid is dissolved in 2.0 x 102 mL of water and titrated with a 0.500 M NaOH solution, 25.00 mL of the NaOH is required to reach the equivalence point.
If you were to calculate moles of OH- used to reach the equivalence point, how many significant figures would your answer have?
How many moles of the compound were in the analyte sample? Show all work. Give your answer with the correct number of significant figures.
What is the molar mass of the unknown monoprotic acid? Show all work. Give your answer with the correct number of significant figures.
The unknown compound has the empirical formula C3H4O3.What is the molecular formula of the compound? Show all work.
Explanation / Answer
Number of moles of NaOH , n= Molarity x volume in L
= 0.500 M x 25.00 mL x10-3 L/mL
= 0.0125 moles
So number of moles of OH- are 0.0125 moles
Since the acid is monoprotic so
1 mole of NaOH reacts with 1 mole of acid
0.0125 moles of NaOH reacts with 0.0125 moles of acid ----> 3 significant figures
Number of moles , n' = mass/molar mass
0.0125 mol = 3.300 g / M
M = 264 g/mol
The emperical mass of C3H4O3 = ( 3xAt.mass of C) + ( 4xAt.mass of H)+(3 x At.mass of O)
= ( 3x12)+ ( 4x1)+(3x16)
= 88
So n = Molar mass / emperical mass = 264 / 88 = 3
So Molecular formula = n(Emperical formula)
= 3( C3H4O3)
= C9H12O9