Question

Data Table 1
Mass of flask and oxalic acid (g)	117.43
Mass of empty flask (g)	116.93
Mass of oxalic acid (g)	0.5
Moles of oxalic acid (mol)
Final volume of NaOH (mL)	17
Initial volume of NaOH (mL)	5
Volume of NaOH used (mL)	12
Moles of NaOH (mol)
Molarity of NaOH (M)

Data Table 2
Mass of flask and vinegar (g)	126.61
Mass of empty flask (g)	121.63
Mass of vinegar (g)	4.98
Final volume of NaOH (mL)	27.5
Initial volume of NaOH (mL)	5
Volume of NaOH used (mL)	22.5
Mass of acetic acid (g)
Percentage of acetic acid in vinegar solution

.1. Measuring and Using Numbers. From the mass of oxalic acid used and the molar mass of oxalic acid, determine and record the number of moles of oxalic acid.

2. Applying Concepts. Write the equation for the reaction of oxalic acid (H2C2O4) with NaOH. What is the ratio of moles of NaOH to moles of H2C2O4?

3. Applying Concepts. Use the moles of oxalic acid calculated in question 1 and the mole ratio from question 2 to determine the moles of NaOH.

4. Measuring and Using Numbers. Convert the volume of NaOH used in mL to L of NaOH and determine the moles of NaOH per liter. Record your result in Data Table 1 as the molarity of NaOH (M).

5. Measuring and Using Numbers. Use the molarity of the NaOH solution and the volume of NaOH used in part B to determine the moles of NaOH used to titrate the acetic acid in the vinegar sample.

6. Applying Concepts. Write the equation for the neutralization of acetic acid (HC2H3O2). What is the ratio of moles of NaOH to moles of acetic acid? How many moles of acetic acid are in the vinegar sample?

7. Measuring and Using Numbers. Use the moles of acetic acid and the molar mass of acetic acid to calculate the mass of acetic acid in the vinegar sample.

8. Measuring and Using Numbers. Use the mass of acetic acid and the total mass of the vinegar sample to calculate the percent acetic acid in vinegar.

9. Error Analysis. Calculate the percent error of the experimental result using the actual value supplied by your teacher. Use the equation percent error = (deviation/correct answer) x 100. Explain what errors could have contributed to any deviation.

Answer 1

1. dns Foom the Data Talle t, man of 0xalie acud wod 0.1 xalie acid (4C204)-2x12.01 +4xis.99 + 2x1.008 g/m) mlar man = 90 g/m) mam mumlon of males of oale aud molan mam 0.59 0.0056 ms 2 bus equation fo the nealion of alio acud mth Nad4 + Nast (a) + Coalemus adion Jhe 0,(ag) + 2Nabt (a) Na3 209 (ag) a0 () Fuacts 2 Tmol of Na wth The atio ofmols Nab to mls 0g

3. Ausm of oxals acid 0.00 56 from the mle ralio Nadt and 04 Naslt 2 2 x 0.0056 mles NaoH 0.0112 mles mSles mdles of NaH =0.0112 4.Auy Vum NasH final-intial 12 mL IL=IMD ml 0.01a L numlaur of mdes of Nash MAlandy of Na me tho ston hto O.0112 0.012 0.93 M mAlaady of Nad -0.99m

5 An Mansty oNab 0.9%m a25 mL vaume of NabH was - 0.0225 L ne of mds Nab oP mdles of Nast (n) no 2) 0.92m= mdles of Na was 0.0 20 mes 6. Bus- eatim Po tho neutralienlon of acde aid (eč3 ,0,) + he lealanus reasuon n Nac, 4,p(a) + 502) + 1 mde of acti aud neacts wnth 1made of Nad Jhe ratio of mles ofNad to mols of ac aud 1:L umben ofmas fNa in eual to of males of acato acid

preant males of Nadt nu 0.020 s 0.020 mdlu So, numkw of les of acle aad .Aus maus of acelo acid, n 0.0a0 mSles 2x 12.01+ 4x1.008 +2 x1 999d ofHet,0a =60.03 g/m mam We lave, m molan mam nxmalan mam of aciti acud z) mam = 0.020 x 60.039 1-201 = mam of auth acud -1.20 nega4.98 mam of Pncunt acdee acid in 1.a0 * /0% 4.98 1 - 24.09%

9. AnsProm he loalanad eualtion acti acd contauns G0.05 $2.02 1 nnegan aeli aeid 60.03 4.98 7 4.987 ngan wll contain 32.02 auli aud 3. 649 empinmental nlue X1o Percent ero thwretial value 1.20 2.64 33.96 / duw to umfurutis. puont Cae trroi's alas lae due to vinegan miscaleulation Barinp w tho