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Lab 22 – TLC of Universal Indicator

TLC of Universal Indicator
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In this lab we will perform a physical separation technique based on the solubility of the individual components in a mixture.

This is Lab will be written up on a google doc.  I would like for you to write a background on how TLC works and relative strength of the different IMF’s in the molecular compounds (chemical indicators) that make up a industry standard mixture called Universal Indicator.

 

Universal Indicator Acid/Base Indicator that has a unique color for each pH value:

 

 

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A lot of pool kits that measure pH use Universal Indicator!

Universal Indicator can accomplish this broad range of colors because it is a homogeneous solution of three Acid/base indicators (CHROMOPHORES!!!!!!!!!!!!) that change their amount of conjugation based on the concentration of available protons (H⁺) or hydroxides (OH^–).

 

Also do not forget the acid/base indicators ARE CONJUGATE ACID and CONJUGATE BASE solutions which are Buffer solutions.

 

                                              HIn                             —->                            H⁺           +         In^–

 

                             Conjugate Acid                                                                                      Conjugate Base

                                 one color                                                                                                 New color

 

  Phenolphthalein is an example:

There are slightly different types of Universal Indicators that use slightly different mixtures of indicators. We will using the BOGEN Universal Indicator mixture.

 

Bogen Universal Indicator Mixture:                      1.  Methyl Red                       pKa = 4.95

                                                                                         2.  Bromothymol Blue         pKa = 7.1

                                                                                         3.  Phenolphthalein             pKa = 9.4

 

So the colors of each pH from above are due to the mixture colors of each indicator at a certain pH levels:

                                                                                                 (negative theory of light)

 

Okay so the point of this lab it separate each of these 3 individual molecular substances by using TLC (Thin Layer Chromatography).  Thin Layer Chromatography is another physical way we can separate mixtures. 

 

– We have separated homogenous mixtures before using evaporation in the carbonate lab where we added HCl to an unknown white powder (that was either NaHCO₃ or Na2CO₃) and created NaCl (aq) and CO₂ that left as bubbles.  We boiled off the water (separated the H-Bonds of water) and obtained dry NaCl (s) which we used by stoichiometry to determine our unknown. Connections!!!!

 

– we also studied distillation (of Coca-Cola) as another physical separation technique of mixtures!

 

– we also used filtration of a precipitate to separate a mixture in our gravimetric titrations!

 

Remember that solutions are mixtures that are soluble that are held together by IMF’s!  When we separate a mixture it is a physical change because we are not breaking bonds we are ONLY BREAKING IMF’s between MOLECULES!!!

 

TLC just like all chromatography separates a mixtures based on the idea that in a mixture the individual molecules retain their individual properties since their chemical formulas are not changing in the process.  This means that each chemical indicator will retain their individual solubilities (their percentage of non-polarity due to the amount of predominant LDF forces or their percentage of polar nature due to the amount of dipolar and H-boding) in the mixture. 

 

Thus in TLC we will separate the 3 acid/base indicators by their solubility to either a mostly non- polar solvent (mobile phase – hexanes) or polar solvent (stationary phase – Silica Gel).  

 

We will spot a TLC plate with Universal Indicator in a starting position close to the bottom of a Silica TLC plate (stationary polar phase) and add a liquid solvent – hexanes (non-polar mobil phase) that move up the Silica TLC plate by capillary action.  When the solvent reaches the “spot” or mixture EACH component of the mixture, 

 

Methyl Red, Bromothymol Blue, and Phenolphthalein will have a choice to either stay with the polar Silica Gel on the TLC plate or move with the non-polar mobil hexanes.  BASED on their individual molecular structures (that define the IMF’s that they have), each chemical indicator will have a unique amount of polar or non-polar characteristics AND they will each move a unique distance up the TLC plate.

 

– If one of the chemicals is more polar (have stronger dipolar forces or have H-bonding) than the other three than they will prefer the Silica TLC plate and MOVE LESS than the other chemical indicators of the mixture that must have more non-polar characteristics.  

 

– The chemical indicator that is the most non-polar (have mostly LDF’s with less dipolar forces), would prefer to be attracted to the mobile non-polar hexane solvent and thus move up the plate more.

 

*Remember this a separation technique based on solubility. Solubility is based on how similar the IMF’s have with the mobile non-polar solvent (hexane) or the polar stationary solvent (TLC silica Plate). Look at the two solvent molecular structure above to identify what IMF are most similar with each solvent! 

 

The distance that the individual chemical indicators move compared to the original spot will be measured.  

 

 

Yellow layer – Carotenoids – 

 

 

Green layers – Chlorophyll A and Chlorophyll B – is one slightly more polar?

 

– Notice these pigments are chromophores!!!

 

– There were three layers separated in the Time Lapse Chromatography below.  Looking at the pigments separate in the video below, and the molecular structure above can you identify the chemicals in the chromatograph? Why was the yellow layer almost at the top while the 2 green layers barely moved at all?

 

– Also notice that we did not spot this TLC plate, we rolled a coin on a spinach leaf unto the TLC plate to make line. The green line was our starting point.               

 

So the assignment is to rank the polarity and non-polarity of the three chemicals in the mixture of Universal Indicator based on what you know of IMF’s AND Predict which chemicals will travel the greatest distance up the TLC plate and which chemicals with travel the shortest distance.  This discussion is the subject of your Background of the lab. Use these structural formulas in your Background.

THEN Write a hypothesis based on your background.

 

Hypothesis Example: Chemical A will have the largest Rate of Flow (move the farthest on the TLC plate) while the Chemical B will have the smallest Rate of flow.

 

Bogen Universal Indicator Mixture:                      1.   Methyl Red                      pKa = 4.95

                                                                                         2.  Bromothymol Blue         pKa = 7.1

                                                                                         3.  Phenolphthalein             pKa = 9.4

 

Methyl red:

Bromothymol Blue:

 

Phenolphthalein:

 

 

Thermodynamics: The end of it all!!!

Fire Syringe Demo – 

3: Hess Law Calculations – (slide 16 and beyond)

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HOMEWORK NOTES: 

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Lab 23 – Hess Law and Calorimetry 

Objectives:   

1.  To use constant pressure calorimetry to calculate the kJ of heat per mole reactant for three chemical reactions.

2. To calculate the ∆Hrxn for the third reaction by using Hess’s Law using the enthalpies determined through calorimetry for the first 2 reactions.

3. To dertermine the accuracy of your results by comparing your Hess law value of the the third reaction and the calorimetry value.

Time Lapse of the TLC of Universal Indicator – 

Thermodynamics: The end of it all!!!

2: Hess Law Introduction and Enthalpy derivation –I derive the ∆H  (ENTHALPY!) in the first 8 minutes.

Thermodynamics: The end of it all!!!

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Lab 23 – Hess Law and Calorimetry 

Objectives:   

1.  To use constant pressure calorimetry to calculate the kJ of heat per mole reactant for three chemical reactions.

2. To calculate the ∆Hrxn for the third reaction by using Hess’s Law using the enthalpies determined through calorimetry for the first 2 reactions.

3. To dertermine the accuracy of your results by comparing your Hess law value of the the third reaction and the calorimetry value.

Calculations in this Lab:

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TODAY’s NOTES: 

Remember that we only obtain qrxn values from calorimetry and it is equal and opposite in value from the qcal.                                                    qrxn = – qcal   Since most reactions occur in constant pressure conditions (NASA – Grade calorimeter) then qrxn values at constant pressure are equivalent to Enthalpy.                                                   qrxn = ∆H    ∆H is a state function that depends only on final conditions – initial conditions AND NOT THE pathway! Because ∆H is a state function than we can use its interconnectivity to solve for other ∆H values.   In the simple example to the left:  If we know the direct distance from North Riverhead to Westhampton and the direct distance from Westhampton to Northampton we can use both values to determine the Overall change of distance from North Riverhead to Northampton.   The interconnectivity of the state function ∆H is called Hess’s Law!   This law allows for chemists to calculate ∆H values without doing calorimetry that might be too impractical or expensive.

So ∆H of reactions that cancel out to give the overall reaction clearly support the interconnectivity of Hess’s Law but the question remains why?  

 

∆H is a state function yes but why does each reaction have a unique starting position  What explains its starting value?  We know the difference between the reactions creates an overall ∆H for the overall reaction as in the 3rd reaction above but it does not explain the positions in the energy diagram that creates the interconnectivity.  

 

For instance from today’s Lab we get the following energy diagram for the 1st reaction using heats of formation values from the Thermo tables:

 

Heats of formation, ∆H_f ⁰ or the enthalpy change of heat that occurs when a compound is made from its elements.  Every compound is made from elements combining to form bonds between atoms thus every compound in EVERY chemical reaction is contains a certain potential energy associated with how much energy is released if forming that(those) bond(s). 

 

For example the ∆H_f ⁰  for HCl that is one of the reactants in reaction 1 from our lab is represented by the following synthesis (composition or formation) reaction:

 

                                                                                H₂   +      Cl₂   —>     2HCl

 

∆H_f ⁰   are always posted in tables in kJ/mol ( kilojoule per mol ).  Since the energy posted to form HCl from its elements is always PER 1 MOLE we reduce the above reaction to produce 1 mole of HCl.  Thus the reaction now becomes:

 

                                                                                 1/2 H₂   +     1/2 Cl₂   —>     HCl

 

When we look up the ∆H_f ⁰  for HCl in thermodynamic tables we obtain ∆H_f ⁰  = -167.2 kJ/mol.  The thermodynamic tables do not provide the reaction but only the product thus you need to be aware what the table is implying the energy change from the formation of the listed chemical from its elements. 

 

∆H_f ⁰  = -167.2 kJ/mol  for HCl means that -167.2 kJ/mol of energy is released when elements combine to form 1 mole of HCl.  Its negative because the energy is released as bonding of the elements creates increased stability. 

 

Notice the arrow moving down in the diagram below for Every chemical in the reaction for reaction 1. Some move farther down than others because some compounds release more energy when they are formed WHICH MEANS THEY ARE MORE STABLE and have LOWER potential energy than the chemicals that do not release as much energy when they are formed.

 

 

      So the starting point for the reactants in a chemical reaction is the SUM of the individual reactants heats of formations

 

     Notice the Sum of all of the ∆Hf of products – the Sum of all of the ∆Hf of products = ∆H _rxn

 

              That is the new skill tonight: determining the ∆H _rxn using ∆H f values.

Since ∆Hrxn is a state function we are not tied to just one way or pathway to calculates the change in Enthapy (∆Hrxn) of a chemical reaction.

                    END OF NOTES! 🙂 

 

Carbon Dioxide and Universal Indicator Demo –

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TODAY’s NOTES: 

 

I am focused on you being able to visualize the FAZ diagram that illustrates the ∆H_f ⁰  values that determine stability in the potential energy (stored energy of chemical bonds)  by how much energy is released to form individual compounds in a chemical reaction. 

 

Hess Law works because all compounds are tied to the same Baseline = F.A.Z.  (Free Atom Zone).  Every chemical in every chemical reaction has a unique potential energy because of how much energy is released in its formation from the same Baseline (FAZ)!   THAT is how ∆H’s are interconnected in reactions!!

 

To obtain a ∆Hrxn all you need to do sum all of the ∆Hf of the reactants you will obtain the total energy lost to form the reactants.  This will tell what potential Energy is REMAINING!!!!!!

 

Do the same for the products in the and compare remaining potential energy from the products and reactants AND YOU HAVE THE ∆H of the entire reaction.

 

THIS IS HESS LAW!  

 

So let’s revisit the reaction that we used to determine the enthalpy of the reactions below:      H_f⁰  is the GOAT !!!

 

If we consider that the energy lost to form each reaction from the FAZ (Free Atom Zone) we can determine the Potential energy of the reactants and products and develop the following potential energy diagram.  Remember that the level of potential energy is due to what is left after the energy is released to form the compound from its elements. 

 Now we can also use this potential energy diagram to identify a NEW way to obtain the ∆H of a chemical reaction which through average bond energies or average bond enthalpies.

Today I have introduced a new way to obtain ∆H of a chemical reaction using Average Bond Enthalpy values.  Through careful experimentation of many chemists we have a comprehensive list of the bond energy or bond enthalpy’s of all covalent (nonmetal to nonmetal bonds).   

 

If we know the energy it takes to break a bond then we know the energy it takes to form a bond because:

                                                            q_sys    =     – q_surr

                                _{Energy Needed to break a bond     =       Energy needed to form a bond}

                                                                  _{Energy consumed  (+)     =        Energy released (-)}

 

 

In this reaction the ∆H_comb of propane (C₃H₈) the reaction is exothermic (-∆H ) because the energy needed to form the bonds of the STABLE products is greater than the energy needed to break the bonds of the reactants.

 

Remember that these values are Averages because bond lengths are constantly changing as they absorb and release infrared radiation.  Notice that all bond enthalpies (bond energy) are positive as it ALWAYS requires energy to break bonds which are the “covalent sharing electron conditions” that provide STABLE electron configurations for all atoms in the molecule! Notice the C=C bond energy is NOT double the bond energy of C-C because a double bond has sigma and a pi bond. If you remember a pi bond is weaker than the direct overlap of a sigma bond. Notice the *C-C bond energy is somewhere in between a single C-C bond and a C=C bond because of RESONANCE in Benzene (which is sp2 hybridized). Its Bond order is 1.5!

 Let’s not forget that the internuclear graph is plotted with potential energy! The potential energy drops as bonds are made because the attraction for the each other atoms valence electrons is provides stability as available orbitals are filled. In the case of Hydrogen below they will attain the stability of He when they bond diatonically.  Notice the potential energy drops because the ∆H _bond (bond enthalpy or bond Energy) is Negative as heat is released from the atoms when they bond.  Can you identify the Free atom zone?

 

What is the bond enthalpy of H2?  WHERE IS THE FAZ?

Bond Enthalpies will be given in small tables in AP questions when needed but remember that Bond Enthalpy is always positive to describe the Energy needed to be absorbed or consumed or BREAK an existing bond. Bonds being Broken always = +∆H Potential Energy increasing Bond being FORMED always = -∆H Potential Energy decreasing ∆H rxn = “bonds broken” – “bonds formed“.

SO if we have Bond Enthalpy:  

 

                                                     ∆H rxn =        ∑ bonds broken               –         ∑ bonds formed

                                                                                              [ ∑ bond enthalpy of reactants]  –                [ ∑ bond enthalpy of products ]  

 

This formula above for BOND ENTHAPy’s  is NOT provided in your reference tables.        

        

And if we have Heats of Formation (∆H_f ⁰ ):

 

                                                                     ∆H rxn =   [ ∑  ∆H_f ⁰ of the Products ]  –  [ ∑  ∆H_f ⁰ of the Products ]

 

This formula IS provided in your reference tables.        

Both of these equations ARE HESS LAW!!!

Since ∆Hrxn is a state function we are not tied to just one way or pathway to calculates the change in Enthapy (∆Hrxn) of a chemical reaction.

END OF NOTES! 🙂 

Thermite Micro Reaction –  Tremendous energy released due to the stability of aluminum oxide!

Thermite Reaction –  Tremendous energy released due to the stability of aluminum oxide!

Decomposition of Hydrogen Peroxide – 

 Heat of formation lecture- 

Bond Energy Lecture- 

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TODAY’s NOTES: 

Bond Enthalpy – revisted

 

                     ∆H rxn =        ∑ bonds broken                  –         ∑ bonds formed

                                              [ ∑ bond enthalpy of reactants]        –             [ ∑ bond enthalpy of products ]   

 

 

*So in every reaction there is absorption of energy to break bonds   and a release of energy to form new bonds.

 

In the example above we would write reactions as:

 

                                   C₃H₈   +   5O₂   →   3CO₂   +   4H₂O  +  2220 kJ      ∆H = -2220 kJ/mole

 

This represents an exothermic reaction Because there is a NET release of energy.

This means there was more energy released in forming new bonds than energy absorbed to break old bonds!

The heat is written on the product side to show the NET release of Heat.

The surroundings increase in energy and the Temperature increases.

 

 

                                     183 kJ   +   N₂    +    O₂    →   2NO        ∆H = +183 kJ/mole

 

This represents an endothermic reaction Because there is a NET absorption of energy.

This means there was more energy absorbed to break old bonds that energy released in forming new bonds!

The heat is written on the reactant side to show the NET absorption of Heat.

The surroundings decrease in energy and the Temperature decreases.

 

3. Heat of dissolution/heat of formation review with CaCl₂ and NaCl

       _{Table I –}

 

 

First lets Review the concept of Lattice Energy WITH HESS LAW!

 

Lattice Energy – Born Haber Cycle:

 

Today was about linking the concepts of heat of formation (∆Hf)  with heat of dissolution (∆Hsol)

  Table I in the Regents Reference Tables lists ionization reactions and their ∆Hrxn.  Salts that dissolve water are endothermic (lower temperature of solution) or exothermic (increase temperature of solution). 

 

1:  Table I – Heat of dissolution of            NaCl demo – temperature fell (∆H dissolution = +4.2 KJ/mol)

                                                                   CaCl2 demo – temperature rose (∆H dissolution = -82.4 KJ/mol)

 

     Complete discussion of heat of formation into stability of bonds and ∆H dissolution.

*Connections – We can use heat of formations from our tables to calculate the ∆Hrxn . The ∆Hrxn has many names that describe the reaction.  In this case we are looking at a ionization reaction that occurs in water (aq) and thus we call this change in enthalpy (H) the ∆H_dissolution or ∆H_{sol (solution).}

 

The ∆H_dissolution for CaCl₂ (s) is exothermic (negative ∆H) and thus the solution temperature rises as heat flows from the chemical (CaCl₂ (s) ) to the water.  This means that the combined ions that result from the dissolving of CaCl₂ (s) ARE MORE STABLE in water than water H-bonding to itself.  This results in a release of energy!

 

Ion – dipole IMF’s

The IMF’s that soluble ions have with water (solvent) are called ion – dipole.  IF these ion- dipole IMF’s are more significant in strength than the combined                H – Bonding that water has for itself and the lattice energy of the salt then the ∆Hdissolution  = negative like the case of CaCl2 (aq). It has an exothermic heat of dissolution.    CaCl2 could be used in HOT packs. In the case of NaCl, the ion dipole IMF’s are less significant in strength than the combined H-Bonding that water has for itself and the lattice energy of the salt thus the ∆Hdissolution  = positive. NaCl(aq) has an endothermic heat of dissolution. NaCl could be used in Cold packs.

 We can describe whether a salt when dissolved is exothermic (raising the temperature of the solution) or endothermic (lowering the temperature of the solution )  by using our concepts of Bond Enthalpy!  

If it takes MORE energy to break the existing IMF’s that the water has for itself and the ionic bonds of the salt THAN the energy released to form new IMF’s (ion – dipole) between the water (solvent) and the ions (solute) then the process is ENDOTHERMIC.

If it takes LESS energy to break the existing IMF’s that the water has for itself and the ionic bonds of the salt THAN the energy released to form new IMF’s (ion – dipole) between the water (solvent) and the ions (solute) then the process is EXOTHERMIC.

 Also CaCl2 is called ice melt because of its exothermic Heat of Dissolution and its the preferred salt that is put on the roads because of its chemistry.

 

4. Started the Phase change enthalpy.pdf

     

* I really linked the idea of potential energy changes in the heating curve at the phase changes so that we better understand the Heat of Fusion and Heat of Vaporization that is used in Regents Chemistry.

 

This is an important concept as ∆H’s can also be used to determine the energy changes if you have a certain amount of the reactant and you know it’s ∆H . For instance if we calculate the ∆H fusion of water or Heat of Fusion (remember ∆H has a lot of names) from ∆H_f ⁰  values for the physical process of H₂O (s) —> H₂O (l) we get the amount of Energy needed to be consumed (or absorbed) per mole of H₂O (s).   We can use this value to determine calculate amount needed to melt (fusion) of a certain amount of ice.  We do this for chemical reactions as well.

 

We can determine the ∆H_vap (Heat of Vaporization) the same way with ∆H_f ⁰  values for the physical process

  H₂O (l) —> H₂O (g) and use it value to measure the energy it takes to vaporize the water with given amount of water in the liquid phase.

 

Phase change enthalpy.pdf
View Download

Heat of Dissolution Solution Classroom lecture- 

Ion Dipole or Molecule ion IMF of a Salt Solution- p can waters compete with the Ionic Bonds of the ions in the crystal lattice.  Is the Lattice energy too large?  

Heating Curve Demo- The potential Energy changes at the phase diagrams can be measured by enthalpy!