Victoria Puckett, Anthony Mba, Walker Lierman
UA Summary
05/24/18
Summary Questions
1. List the vial number of your unknown acid
2. What is your calculated molar mass? Show this calculation for one of your trials
1 molUA
0.055L×0.1M =0.0055molNaOH × × 0.0055molUA
1 molNaOH
0.308gUA
=56g/mol(molar massof unknownac
0.0055moUA
56 g x 2= 112 mol UA
3. Did you choose to omit any titration trials in the calculation of your average molar mass?
If so, why?
a. No, we did not omit any trials because all trials were needed in calculating the
molar mass
4. Based on your titration results, what is the identity of your group’s unknown acid?
a. Malonic Acid
5. What is the percent error in your molar mass?
a. 0.2%
6. What is the melting point range you observed?
a. 132-136 degrees fahrenheit
7. Based on your melting point results, what is the identity of your group’s unknown acid?
Does this agree with your conclusion from your titration results. If not, explain.
a. Malonic Acid, yes it agrees with our conclusion from our titration results because
when doing the titration we ended up getting the same unknown acid. Our melting
point started at 132 degrees fahrenheit, which matched up perfectly with malonic
acid.
Procedure
The first thing that needed to be done was gathering all materials. Once all materials were
gathered we measured out .3 grams of our unknown acid. After measuring out the unknown acid
place it into a beaker. We then measured out 25 mL of DI water and placed that water into the
same beaker. Mix those together well until all powder is dissolved and then place the drops into
the beaker. Mix well. Fill a burette with NaOH and slowly start adding the NaOH into the beaker.
Keep adding in NaOH until the solution changes to a light light pink. Once pink stop adding in
NaOH and record how much was used. Record all data in a chart and repeat this experiment 2
more times but with .15 grams, and .2 grams of the unknown acid.
Specific Results
Our unknown acid ended up being Malonic acid. Our calculations were as followed:
1 molUA
0.055L×0.1M =0.0055molNaOH × × 0.0055molUA
1 molNaOH
UA Summary
05/24/18
Summary Questions
1. List the vial number of your unknown acid
2. What is your calculated molar mass? Show this calculation for one of your trials
1 molUA
0.055L×0.1M =0.0055molNaOH × × 0.0055molUA
1 molNaOH
0.308gUA
=56g/mol(molar massof unknownac
0.0055moUA
56 g x 2= 112 mol UA
3. Did you choose to omit any titration trials in the calculation of your average molar mass?
If so, why?
a. No, we did not omit any trials because all trials were needed in calculating the
molar mass
4. Based on your titration results, what is the identity of your group’s unknown acid?
a. Malonic Acid
5. What is the percent error in your molar mass?
a. 0.2%
6. What is the melting point range you observed?
a. 132-136 degrees fahrenheit
7. Based on your melting point results, what is the identity of your group’s unknown acid?
Does this agree with your conclusion from your titration results. If not, explain.
a. Malonic Acid, yes it agrees with our conclusion from our titration results because
when doing the titration we ended up getting the same unknown acid. Our melting
point started at 132 degrees fahrenheit, which matched up perfectly with malonic
acid.
Procedure
The first thing that needed to be done was gathering all materials. Once all materials were
gathered we measured out .3 grams of our unknown acid. After measuring out the unknown acid
place it into a beaker. We then measured out 25 mL of DI water and placed that water into the
same beaker. Mix those together well until all powder is dissolved and then place the drops into
the beaker. Mix well. Fill a burette with NaOH and slowly start adding the NaOH into the beaker.
Keep adding in NaOH until the solution changes to a light light pink. Once pink stop adding in
NaOH and record how much was used. Record all data in a chart and repeat this experiment 2
more times but with .15 grams, and .2 grams of the unknown acid.
Specific Results
Our unknown acid ended up being Malonic acid. Our calculations were as followed:
1 molUA
0.055L×0.1M =0.0055molNaOH × × 0.0055molUA
1 molNaOH
