A sample of vinegar was titrated with 1.054M NaOH. 21.6mL of NaOH were required to reach the endpoint. Find the mass of...
A sample of vinegar was titrated with 1.054M NaOH. 21.6mL of NaOH were required to reach the endpoint. Find the mass of HC2H3O2 in grams that were in the sample of vinegar.
Hint: answer included 3 sig figs
Homework Answers
Balanced chemical equation is:
NaOH + C2H4O2 ---> NaC2H3O2 + H2O
lets calculate the mol of NaOH
volume , V = 21.6 mL
= 2.16*10^-2 L
use:
number of mol,
n = Molarity * Volume
= 1.054*2.16*10^-2
= 2.277*10^-2 mol
According to balanced equation
mol of C2H4O2 reacted = (1/1)* moles of NaOH
= (1/1)*2.277*10^-2
= 2.277*10^-2 mol
This is number of moles of C2H4O2
Molar mass of C2H4O2,
MM = 2*MM(C) + 4*MM(H) + 2*MM(O)
= 2*12.01 + 4*1.008 + 2*16.0
= 60.052 g/mol
use:
mass of C2H4O2,
m = number of mol * molar mass
= 2.277*10^-2 mol * 60.05 g/mol
= 1.367 g
Answer: 1.37 g
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