5: Composition of Molecules and Solutions
- Page ID
- 509304
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Lab 5: Percent Composition and Molecular Formula
This lab directly utilized information from:
Bonjour, J. L., Pitzer, J. M., & Frost, J. A. (2015). Introducing high school students to NMR spectroscopy through percent composition determination using low-field spectrometers. Journal of Chemical Education, 92(3), 529-533. https://doi.org/10.1021/ed500731y
- Students will...
- Students will...
Background Information
In class, we are beginning to look at the different ways atoms can bond together to form molecules and the relationships between formulas and covalent structures. One of the most significant ways covalent structures are determined for organic molecules is by the means of Nuclear Magnetic Resonance (NMR) Spectroscopy. 1H (Proton) NMR is often used to determine where hydrogen atoms are relative to each other and to other atoms. We will use Proton NMR in this lab to introduce you to the principles of the instrument and explore the relationships between structure, formula, and chemical signal.
Proton NMR gives information about each distinct proton (identical protons give identical signal) in three different ways
1. Shift (where the signal shows up based on what is near to the proton)
2. Integration (how many protons are giving the signal relative to other signals)
3. Splitting (how many different protons are nearby)
For the purposes of this lab, we only need to identify factors influencing shift and integration. We will use these two factors to separate signals we get from components in a solution. In this case we will be using rubbing alcohol, which is a mixture of isopropyl alcohol as the major component (solvent) and water as the minor component (solute). As these signals are based on the ratio of atoms, they can be used to give molar ratios of components. However, most common solutions are reported as percentages by mass or by volume (m/m% or v/v%), molarity (moles/liter), parts per million (ppm) or parts per billion (ppb).
Mouthwash, vinegar, and rubbing alcohol all report percentages by volume. We can use the technique in this lab to determine the percentage by volume for solutions that aren’t reported as well. For example, we have used this technique to verify ethanol and water ratios in imitation almond extract, and determine the relative concentrations of the benzaldehyde that makes it almond-y.
In this lab you will determine the concentration (as a percentage by mass and volume) of a sample of rubbing alcohol. You can find a variety of concentrations available in most pharmacies.
Terms to Remember
- Solution: A mixture in which a solid, liquid, or a gas dissolves in a liquid. Aqueous (aq) solutions are water-based solutions.
- Solvent: The majority component of a solution. In aq. solutions with concentrations < 50%, the solvent is water.
- Solute: The minority component of a solution. In aq. solutions with concentrations > 50%, the solute is water.
- Mass percent (m/m%) = mass solute / (mass solute + mass solvent) x 100
- Volume percent (v/v%) = volume solute / (volume solute + volume solvent) x 100
Procedure
1. Obtain an unknown sample from your instructor.
2. Follow the directions for acquiring a NMR spectrum of your sample.
3. In the processing of your spectrum, note the following:
a. Do all the peaks you predicted show up? How do you know which peak is which?
b. The split peaks will all correspond to the organic component of your solution. Are any of these peaks overlapping with other peaks? Which one is the most clearly separated? Integrate as your reference peak.
c. Integrate all remaining peaks as one region.
4. Take a picture of your spectrum to use for the analysis portion of the lab and insert it your spreadsheet.
Analysis – This lab workup is a continuation of the pre-lab questions and uses the same spreadsheet.
1. Indicate the location of the peak you chose as a reference peak in ppm units (B36).
2. Indicate the integration of the peak you chose as a reference peak (B37).
3. Indicate the # of hydrogens represented by this peak based on the structure (B38).
4. The integration of the reference peak divided by the number of hydrogen atoms it represents = the number of this type of molecule represented in this spectrum. Calculate the number of (organic) molecules represented by the spectrum using a formula in B39.
5. The number of organic hydrogens not represented is the total number of hydrogens in the formula for the organic molecule of interest minus those represented (B38). Enter this number into B40.
6. The organic hydrogens that aren’t in the reference peak show up as part of the other integrated area. This contribution is the number of organic hydrogens not represented (B40) times the number of organic molecules represented in the integration (B39). Calculate this using a formula in B41.
7. Enter the total integration of the other peaks in B43.
8. The contribution from water is the total integration of the other peaks minus the contribution from the main organic molecule. Calculate this in B44.
9. Enter the number of hydrogen atoms in water in B45.
10. The integration contribution from water divided by the number of hydrogens in water is the number of water molecules represented by this spectrum. Calculate this number using a formula in B46.
11. Calculate the ratio of organic molecules to water molecules using a formula in B47.
Work out the following conversions on a separate sheet of paper. You may do all the calculations on paper (show your work), or just work out the unit conversions on paper and do the math in Excel. Make sure to clearly label your work in either case.
12. The units for this ratio are \(\ce{\frac{molecules of organic component}{molecules of water}}\). Work out the unit conversions using dimensional analysis to \(\ce{\frac{moles of organic component}{moles of water}}\).
Hint: Avogadro’s number.
13. Next, work out the conversion from \(\ce{\frac{moles of organic component}{moles of water}}\) to \(\ce{\frac{grams of organic component}{moles of water}}\) and then to \(\ce{\frac{grams of organic component}{grams of water}}\).
Hint: What property from your pre-lab questions has units containing grams and moles?
14. You can envision this result as \(\ce{\frac{$x$ grams of organic component}{1 grams of water}}\), where the number you calculated in Question 13 is \(x\). Look at the definition of mass percent. How can you find the mass percent from using this number?
15. Next, work out the conversion from \(\ce{\frac{grams of organic component}{grams of water}}\) (from Question 13) to \(\ce{\frac{$mL$ of organic component}{$mL$ of water}}\).
16. You can envision this result as \(\ce{\frac{$x$ mL of organic component}{1 mL of water}}\), where the number you calculated in Question 15 is \(x\). Look at the definition of volume percent. How can you find the volume percent from using this number? This number will have some error because sometimes there is a change of volume upon mixing.
17. Compare your calculated value for volume percent to the reported value. What is the percent error?