Question

Experimental Data and Calculations Data Table 1 Mass (g) Pure Lauric Acid Mixture 1 _.12_g lauric acid _.11_g stearic acid Mixture 2 .20_g lauric acid _.10_g stearic acid Mixture 3 _.09_g lauric acid _.22_g stearic acid Pure Stearic Acid Melting point (°C) 50 degrees 54 degrees 52 degrees C C C Percent stearic acid by mass (%) 0 59 degrees 63 degrees C C 100 Calculations employed: Graph 1 Insert yo aph of melting temperature (°C) versus percent stearic acid by mass here.

          Experimental Data and Calculations
Data Table 1
Mass (g)
Pure Lauric
Acid
Mixture 1
_.12_g
lauric acid
_.11_g
stearic
acid
Mixture 2
.20_g
lauric acid
_.10_g
stearic
acid
Mixture 3
_.09_g
lauric acid
_.22_g
stearic
acid
Pure Stearic
Acid
Melting
point (°C)
50 degrees 54 degrees 52 degrees
C
C
C
Percent
stearic acid
by mass (%)
0
59 degrees 63 degrees
C
C
100
Calculations employed:
Graph 1
Insert yo aph of melting temperature (°C) versus percent stearic acid by
mass here.
        
Show more…
Experimental Data and Calculations
Data Table 1
Mass (g)
Pure Lauric
Acid
Mixture 1
.12g
lauric acid
.11g
stearic
acid
Mixture 2
.20g
lauric acid
.10g
stearic
acid
Mixture 3
.09g
lauric acid
.22g
stearic
acid
Pure Stearic
Acid
Melting
point (°C)
50 degrees 54 degrees 52 degrees
C
C
C
Percent
stearic acid
by mass (%)
0
59 degrees 63 degrees
C
C
100
Calculations employed:
Graph 1
Insert yo aph of melting temperature (°C) versus percent stearic acid by
mass here.

Added by Juan Luis R.

Close

Chemistry: Structure and Properties
Chemistry: Structure and Properties
Nivaldo Tro 2nd Edition
AceChat toggle button
Close icon
Ace pointing down

Please give Ace some feedback

Your feedback will help us improve your experience

Thumb up icon Thumb down icon
Thanks for your feedback!
Profile picture
WHAT IS THE PERCENT OF MASS FOR STEARIC ACID FOR ALL 3 MIXTURES Calculations employed: Graph 1 Insert yo 임 aph of melting temperature (deg C) versus percent stearic acid by mass here. Experimental Data and Calculations Data Table 1 Pure Lauric Acid Pure Stearic Acid Mixture 1 Mixture 2 Mixture 3 .12g lauric acid .20_g .09 g lauric acid lauric acid Mass (g) .11_-g -.10 g .22g stearic stearic stearic acid acid acid 50 degrees 54 degrees 52 degrees 59 degrees 63 degrees c c c c c Melting point(C) Percent stearic acid by mass (%) 100 Calculations employed: Graph 1 Insert yo aph of melting temperature (-c) versus percent stearic acid by mass here.
Close icon
Play audio
Feedback
Powered by NumerAI
David Collins Kathleen Carty
Danielle Fairburn verified

Ronald Prasad and 51 other subject Chemistry 101 educators are ready to help you.

Ask a new question

*

Labs

-

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

*

Key Concepts

-
Key Concept
Premium Feature
Explore the core concept behind this problem.
Play button
Key Concept
Premium Feature
Explore the core concept behind this problem.
Your browser does not support the video tag.

*

Recommended Videos

-
stearic-acid-leftmathrmc_18-mathrmh_36-mathrmo_2right-and-palmitic-acid-leftmathrmc_16-mathrmh_32-ma

Stearic acid $\left(\mathrm{C}_{18} \mathrm{H}_{36} \mathrm{O}_{2}\right)$ and palmitic acid $\left(\mathrm{C}_{16} \mathrm{H}_{32} \mathrm{O}_{2}\right)$ are common fatty acids. Commercial grades of stearic acid usually contain palmitic acid as well. A $1.115 \mathrm{g}$ sample of a commercial-grade stearic acid is dissolved in $50.00 \mathrm{mL}$ benzene $(d=0.879 \mathrm{g} / \mathrm{mL}) .$ The freezing point of the solution is found to be $5.072^{\circ} \mathrm{C}$ The freezing point of pure benzene is $5.533^{\circ} \mathrm{C},$ and $K_{\mathrm{f}}$ for benzene is $5.12^{\circ} \mathrm{C} \mathrm{kg} \mathrm{mol}^{-1} .$ What is the mass percent of palmitic acid in the stearic acid sample?

General Chemistry: Principles and Modern Applications

how-does-the-melting-of-pure-stearic-acid-compare-to-that-of-the-mixture-of-myristic-acid-and-stearic-acid-how-does-the-melting-point-range-of-pure-stearic-acid-compare-to-that-of-the-mixtur-70478

Adi S.

stearic-acid-c18h36o2c18h36o2-and-palmitic-acid-c16h32o2c16h32o2-are-common-fatty-acids-commercial-grades-of-stearic-acid-usually-contain-palmitic-acid-as-well-a-1120-gg-sample-of-a-commerci-04713

Stearic acid (C18H36O2C18H36O2) and palmitic acid (C16H32O2C16H32O2) are common fatty acids. Commercial grades of stearic acid usually contain palmitic acid as well. A 1.120 −g−g sample of a commercial-grade stearic acid is dissolved in 50.00 mLmL benzene (dd = 0.879 g/mLg/mL). The freezing point of the solution is found to be 5.072 ∘C∘C. The freezing point of pure benzene is 5.533 ∘C∘C, and KfKf for benzene is 5.12 ∘Cm−1∘Cm−1 . What is the mass percent of palmitic acid in the stearic acid sample?

Nicole S.


*

Recommended Textbooks

-
Chemistry: Structure and Properties

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition
achievement 1,988 solutions
Chemistry The Central Science

Chemistry The Central Science

Theodore L. Brown 14th Edition
achievement 1,303 solutions
Chemistry

Chemistry

Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste 10th Edition
achievement 1,955 solutions

*

Transcript

-
00:03 Let's solve for the mass percent of palmetic acid and the stearic acid sample, given the information given.
00:10 So we can calculate from the information given, delta tf.
00:14 We're told that the freezing point of the solution, it is 5 .072 degrees celsius, and the freezing point of the solvent is 5 .533 degrees celsius.
00:28 So delta tf will work out to minus 0 .461 .1.
00:33 Degrees celsius.
00:37 Knowing delta tf, we can calculate the molality using this formula here, negative, minus .461 degrees celsius minus the constant here is going to be 5 .12 degrees celsius per moleal and we'll solve for the molality here.
00:58 So the molality will work out to point 0900.
01:05 So we're told that the solution contains 50 milliliters of benzene and we're given its density.
01:13 So let's solve for the mass of benzene.
01:23 So starting with 50 milliliters times the density.
01:28 0 .879 grams per milliliter.
01:31 Density given for benzene.
01:32 This will work out to 43 .9 grams or converted to kilogram.
01:42 0 .0439 kilograms.
01:44 We'll need this in kilograms as we bring in our molality.
01:50 So we can calculate the moles of the solute, which would be equal to the molality, 0 .0900 moles of the solute.
02:05 The solute would be the mixture of palmetic and stearic acids.
02:12 It's dissolved in benzene, so per kilogram benzene, times the mass of benzene.
02:19 0 .0439 kilograms of benzene will cancel, and this would leave us with 0 .00395 moles of the solute.
02:35 So the moles of solute is the mixture.
02:40 So the total mass is given as the total mass for the mixture is given as 1 .15 grams.
02:56 So let's let the mass of let mass of stearic acid equal to x and the mass of pulmonary.
03:20 Acid equal to y, therefore combining these together, x plus y would be equal to 1 .115 grams...
Need help? Use Ace
Ace is your personal tutor. It breaks down any question with clear steps so you can learn.
Start Using Ace
Ace is your personal tutor for learning
Step-by-step explanations
Instant summaries
Summarize YouTube videos
Understand textbook images or PDFs
Study tools like quizzes and flashcards
Listen to your notes as a podcast
Continue solving this problem
Create a free account to:
  • View full step-by-step solution
  • Ask follow-up questions with Ace AI
  • Save progress and study later
Continue Free
Join the community

18,000,000+

Students on Numerade


Trusted by students at 8,000+ universities

Numerade

Get step-by-step video solution
from top educators

Continue with Clever
or



By creating an account, you agree to the Terms of Service and Privacy Policy
Already have an account? Log In

A free answer
just for you

Watch the video solution with this free unlock.

Numerade

Log in to watch this video
...and 100,000,000 more!


EMAIL

PASSWORD

OR
Continue with Clever