CHEM 312 Lecture Notes - Lecture 28: Nanometre, Nanomedicine, Graphene

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bluecheetah583

5 May 2019

School

University of Illinois

Department

Chemistry

Course

CHEM 312

Professor

C Murphy

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Document Summary

Chem 312 lecture 28 proteins and nucleic acids - inorganic perspectives. When you make the semiconductors really small, the bands of electron energies become less of a band and more of discrete energy levels. The quantum dots have more distinct energy levels that can create different colors that are observed. This is why different quantum dots that are different colors in this diagram, it shows the absorbance of the quantum dots and how that absorbance directly corresponds to the bandgap energy of the dot. Unusual properties of metals on the nm scale. When gold is spread out, that means that the colors can change. This is how the stained glass can occur. A plasmon is the collective oscillation of charged particles. There are free electrons in metals that allow the transport of the electric current and this is where the plasmons exist.

Related Questions

Synthesis of Copper(I) Oxide Particles with Variable Color:Demonstrating Size-Dependent Optical Properties

Introduction
Cluster based materials exhibit considerable advantages over bulk materials including surface features, electrostatic fields and reactivity that can differ and/or be varied as a function of both composition and cluster size. Nanocrystals formed from semiconductor materials with diameters of 1 â 30 nm are often referred to as quantum dots. Quantum dots are already in commercial use in sensors, LEDS and lasers including those used as readers for HD-DVD and Blu-ray high-definition DVDs. The spectral properties of quantum dots arise from a quantum confinement effect. When an electron in these semiconducting materials is excited into the conduction band a hole is created in the valence band. The physical distance between the location of the excited electron and the hole it came from is called the Bohr exciton radius (rB). In a bulk semiconductor crystal rB is small compared to the overall size of the crystal and the electron can migrate freely through the lattice. However, in the nanoscale quantum dot, rB is close to the diameter of particle and the electron is trapped in molecular orbitals rather than bands of orbitals. The energy spacing of these molecular orbitals depends on the number of atoms and hence the size of the crystal. Consequently, the excitation or absorption energy changes with the size of the quantum dot. As the dot gets bigger the absorption energy moves to lower frequency and the colour of light reflected changes from blue to red.

In this experiment you will synthesise micro- and nanosized copper(I) oxide particles with size-dependent optical properties by reducing alkaline copper(II)âcitrate complex (Benedictâs reagent) with glucose. The Cu2O particles varies from hundreds of nanometers to micrometers depending on concentrations of initial reagents and their colour varies from yellow to dark red, accordingly. Shown below are the electron-half-equations (a, b) and summary redox reaction (c) between copper(II) 25 ions and Î±-hydroxy-ketone group of reducing sugars (e.g., fructose) in open-chain form. Redox reaction is activated by transformation of Î±-hydroxy-ketone group to enediol via keto-enol tautomerization in alkali solutions.
a) Oxidation: R1CH(OH)COR2 + 2 OH â R1COCOR2 + 2 H2O + 2 eâ
b) Reduction: 2Cu(in complex)2+ + 2 OHâ + 2 eâ â Cu2O + H2O
c) Summary: R1CH(OH)COR2 + 2Cu2+ + 4 OHâ â R1COCOR2 + Cu2O + 3H2O

Question1:

Provide a brief summary of three research articles covering Cu₂O semiconductors (cite refs)?

Question2:

In this experiment you use Benedictâs reagent to synthesis copper (I) oxide nanoparticles. Research the scientific literature and provide a brief description of two alternative applications of Benedictâs reagent. (cite refs)?

The conduction of electricity inmaterials is explained in terms of band theory. The band that isformed from the filled bonding molecular orbitals is called thevalence band, and the band that is formed from the emptyantibonding molecular orbitals is called the conductionband. The energy difference between these bands is known as theband gap energy, or the band gap.

Click on the image below to explore the activity, which will help you to identify the insulator and thetypes of semiconductors based on band gap energy. Blue-coloredbands indicate conduction bands, and red-colored bands indicatevalence bands.

When you click on each box, you will observe the changes in bandgap energy for that material.

Part A

In the interactive activity, the bandshown at the top is the conduction band, and the band shown at thebottom is the valence band. The valence band consists of bondingmolecular orbitals that are filled with electrons, and theconduction band consists of vacant antibonding molecular orbitals.The red color indicates that the orbitals in that band are filledwith electrons, and the blue color indicates that the orbitals inthat band are empty. The space between the two bands shows theenergy difference between them and is known as the band gapenergy, or the band gap.

Which of the following statements are correct with respect tothe band structure and the band gap of different materials?

Check all that apply.

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Part B

When you add a controlled amount ofsubstance known as an impurity to a pure semiconductor, itsconductivity increases. This process is known as doping. Theaddition of an impurity can increase or decrease the number ofavailable electrons. If the number of electrons is increased, theextra electrons conduct electricity. If the number of electrons isdecreased, a vacant hole is created. The adjacent electron entersthe vacant hole and conducts electricity. Thus, it seems like theholes act as charge carriers. They are considered to be positivecharge carriers. Depending upon the charge carriers that conductelectricity, there are two types of semiconductors: p-typesemiconductors and n-type semiconductors.

Which of the following statements are correct with respect tosemiconductors?

Check all that apply.

	The undoped semiconductor has a partiallyfilled conduction band.
	The n-type semiconductor has apartially filled valence band.
	The p-type semiconductor has apartially empty conduction band.
	The p-type semiconductor has apartially empty valence band.
	The band gap of a semiconductor is lessthan that of an insulator.
	The conduction band for insulators isalways vacant.

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Part C

Germanium is a group 4A semiconductor. Theaddition of a dopant atom (group 3A element) that has fewer valenceelectrons than the host atom results in a p-typesemiconductor. The addition of a dopant atom (group 5A elements)that has more valence electrons than the host atom results in ann-type semiconductor.

Which of the following elements when used for doping germaniumwill yield p-type semiconductors, and which elements whenused for doping germanium will yield n-typesemiconductors?

Drag the appropriate elements totheir respective bins.

	The conductivity of semiconductors can beincreased by the doping process.
	The addition of group 3A elements to group4A semiconductors decreases the conductivity of the group 4Asemiconductors.
	As the temperature increases, theconductivity of semiconductors decreases.
	Holes are charge carriers in p-typesemiconductors.
	The addition of group 5A elements to group4A semiconductors increases the conductivity of the group 4Asemiconductors.
	Electrons are charge carriers inn-type semiconductors.

SubmitHintsMy AnswersGive UpReviewPart

Part D

Observe the band structure for fourdifferent substances in the following figure. The band shown at thetop is the conduction band, and the band shown at the bottom is thevalence band. The red color indicates that the orbitals in thatband are filled with electrons, and the blue color indicates thatthe orbitals in that band are empty. Identify which band structurecorresponds to which substance.

Drag the appropriate labels totheir respective targets.

Decomposition of Malachite

Here is what Malachite is Made Out Of

Malachite is a famous and very popular semi-precious stone. Its banded light and dark green bands are one-of-a-kind, and give it a unique ornamental quality unlike that of any other stone. The light and dark green bands are so distinctive that malachite may be one of the most easily recognized minerals by the general public. Malachite is popular in jewelry, especially Native American southwestern jewelry. The stones inlayed in silver make a nice alternative to traditional turquoise jewelry.

Malachite is a form of copper carbonate with the molecular formula Cu₂CO₃(OH)₂. When heated to around 200 ï¿½C, malachite decomposes into copper (II) oxide (CuO) and other products.

Reaction Stoichiometry

The Law of Conservation of Mass states that matter can neither be created nor destroyed. Therefore, the number of atoms of each element on the reactants side must be same as the number of atoms on the products side. This quantitative study of chemical reactions is called stoichiometry. The substances in a balanced reaction are related to each other by stoichiometric molar ratios. For example, magnesium reacts with hydrochloric acid in a 1:2 stoichiometric ratio in the balanced equation below.

Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g)

The molar ratios can be used as conversion factors to find the amount (in moles) of one substance given the amount of another. For example, given the number of moles of HCl, the amount of MgCl₂ can be determined as shown below.

nMgCl2=nHCl x (1mol MgCl2) / (2mol HCL)

The molar ratios are used to determine the amount in moles. However, the amount of reactant or product is often measured in grams. Mass can be converted to moles and vice versa as shown below.

n = m / (MM)

m = n x MM

where n is the number of moles, m is the mass of the product in g, and MM is its molar mass in g/mol.

To observe the Law of Conservation of Mass, you can perform different types of chemical reactions. For example, you can use a decomposition reaction to observe one compound being broken down into two or more elements or new compounds. For this reaction to take place, you must break existing bonds within the starting compound. Different sources of energy such as heat, light, or electricity may be used. Through a decomposition reaction, you can observe that the atoms in the original compound are not destroyed but are rearranged to form simpler substances, such as elements or compounds.

About This Lab

In this lab, you will test if the relationship between malachite and copper (II) oxide is stoichiometric. To do this, you will heat a known mass of malachite to produce copper (II) oxide. You will then compare the amount of copper (II) oxide obtained experimentally with the amount of copper (II) oxide expected from the balanced equation for the decomposition reaction.

1.Take a crucible from the Containers shelf and place it on the workbench.

2.Take a balance from the Instruments shelf and place it on the workbench.

3.Move the crucible onto the balance. Record the mass of the empty crucible to the nearest 0.001g in your Lab Notes.

4.Move the crucible onto the workbench.

5.Take copper (II) carbonate hydroxide from the Materials shelf and add 10 g to the crucible.

6.Move the crucible onto the balance and record the mass of the crucible containing copper (II) carbonate hydroxide in your Lab Notes.

7.Take a Bunsen burner from the Instruments shelf and place it on the workbench.

8.Move the crucible onto the Bunsen burner.

9.Double-click on the crucible to open its properties. Select Show Contents and click OK. This gives you a cut-away view of the crucible's contents.

10.Turn the Bunsen burner on to low flame by clicking the knob at the bottom once. Clicking multiple times will increase the intensity of the flame until it is turned off.

11.Watch the crucible until the malachite fully decomposes. You will see a distinct color change.

12.Move the crucible onto the workbench to cool.

13.Turn off your Bunsen burner.

14.Weigh the crucible with its contents and record the mass to the nearest 0.001 g in your Lab Notes.

Clear your station.

Lab Notes:

Numbers below refer to numbers in the above âAbout this Labâ

3. Empty crucible = 88 g

5. Added 10 g of Copper (II) carbonate hydroxide to crucible

6. Crucible with Copper (II) carbonate hydroxide = 98 g

11. to Black

14. after heating crucible with its contents = 95.195

Questions I have answered:

How many grams of malachite did you add to the crucible? 10g

How many grams of product were in the crucible after heating? 7.195

By how much did the mass of the crucible's contents change after heating? -2.805

The molar mass of malachite is 221.1 g/mol. How many moles of malachite were present before the reaction? 0.04523 mol

The product remaining in the crucible after heating is copper (II) oxide. The molar mass of copper (II) oxide is 79.55 g/mol. How many moles of copper (II) oxide were produced during the reaction?

0.0945 mol

Please answer the following:

Q1 a. Which of the following represents the balanced chemical equation for this reaction?

Cu2CO3(OH)2=2CuO+H2O

Cu2CO3(OH)2=CuO+CO2

Cu2CO3(OH)2=2CuO+H2O+CO2

Cu2CO3(OH)2=CuO+H2O+CO2

b. Based on the number of moles of malachite that you started with, how many grams of water were produced? The molar mass of water is 18.0153 g/mol. Choose the closest answer.

1.815 g 0.8148 g 1.630 g 1.982 g

c. Based on the number of moles of malachite that you started with, how many moles of CO₂were produced? Choose the closest answer.

1.000 mol 0.04523 mol 0.9000 mol 0.09045 mol

d. Based on the number of moles of malachite that you started with, how many grams of CO₂were produced? The molar mass of CO₂ is 44.01 g/mol. Choose the closest answer.

1.991 g 0.4401 g 0.04523 g 3.981 g

Q2a. Based on the number of grams of CO₂ and H₂O produced during the reaction, does this compensate for the observed loss of mass?

A. No, because this violates the law of conservation of mass

B. yes, this compensates for the mass lost in the reaction

C. No, because more copper oxide was produced

D. No because mass is conserved

b. You observed that the products in the crucible weigh less than the reactants that you added. How is this possible?

A. Mass was lost due to experimental error

B. The Malachite absorbed CO2 from the air

C. CO2 and H2O were released as gases

D. Mass was converted to energy

c. What kind of a reaction took place upon heating?

A. decomposition

B. single displacement

C. synthesis

D. Double displacement

d. Baking soda is used as a leavening agent in baking. When heated, baking soda undergoes a decomposition reaction to form carbon dioxide. Given the data below, which of the following reactions occurs during baking? The molar mass of baking soda is 84.007 g/mol and the molar mass of carbon dioxide is 44.01 g/mol.

Mass of empty crucible (g) = 55.000

mass of crucible with baking soda (g) = 57.000

mass of crucible and contents after the reaction (g) =56.260

A. 2NaHCO3= 2Na+2CO2 +H2O

B. 2NaHCO3 = Na2O + 2CO2 +H2O

C. NaHCO3 = NaOH +CO2

D. 2NaHCO3 = Na2CO3 + CO2 +H2O

CHEM 312 Lecture Notes - Lecture 28: Nanometre, Nanomedicine, Graphene

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