Chem 01LA Lab Final

Lab 1 - Density of a Liquid

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How to properly read the initial and final volumes on a buret

Initial and final volumes on a buret should be recorded to the nearest 0.01mL from the bottom of the meniscus at eye level.

Determine the density of a liquid from volume and mass; or if given the density, be able to calculate the volumes or mass.

D=M/V

density of liquid = mass of liquid/ volume of liquid.

mass of liquid = mass of container and liquid - mass of container

density of water at given temperature

rules and calculations involved with significant figures, mean, and standard deviation

Sig Figs

When subtracting/adding, the answer follows the number of decimal places of the value with the lowest amount of decimal places.

When multiplying/dividing, the answer follows the value with the least amount of significant figures.

When rounding the result of a calculation, drop digit if lower than 5, round up if greater than 5, and round to the nearest even number if value is 5. (132.75 -> 132.8 and 132.65 -> 132.6)

When taking the logarithm of a number, the result should have the same number of digits to the right of the decimal point as there are significant digits in the original number (log 2.998 x 10^8 is 8.4768)

(0.100 has three sigfigs; 100 has 1 sigfigs; 100. has three sigfigs; 100.0 has four sigfigs; 0.01 has 1 sigfig)

Mean

mean = sum of values / number of values

calculate sigfigs of sum of values using subtracting/adding rules then calculate sigfigs of mean using multiplying/dividing sigfig rules.

Standard Deviation

s = sq root ( (value-mean)^2 + (value-mean)^2 / number of values - 1)

the mean and standard deviation should both end at the same decimal place.

Lab 2 - Emission of Light from Hydrogen and Metal Atoms

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Relationship between energy, frequency, and wavelength of electromagnetic radiation

Short Wavelength = High Energy and Frequency

Long Wavelength= Low Energy and Frequency

High Energy = High Frequency

High Energy to Low Energy

Ultraviolet > Visible > Infrared

High Frequency to Low Frequency

Ultraviolet > Visible > Infrared

Long Wavelength to Short Wavelength

Infrared > Visible > Ultraviolet

Lyman = Ultraviolet and nl=1

Balmer = Visible and nl=2

Paschen = Infrared and nl=3

Use calibration curve to find the true wavelength on a spectroscope scale

The calibration curve is constructed by graphing the scale readings versus the known wavelength values. The X-axis is scale readings; Y-axis is known wavelength values.

Make calibration curve from scale readings and known wavelength from known element (Helium). Plug the scale readings of Hydrogen into the x-value of the equation of the line to solve for the y-value, which is the true wavelength of Hydrogen.

Assign electron transitions in the hydrogen atom based on wavelength of observed lines in the visible spectrum.

Visible spectrum has to be a transition to a lower principal quantum number of 2. The lowest wavelength 410 nm corresponds with the highest-energy transition. The 3->2 transition emits the least energy.

Given wavelengths 708.924, 659.7498, 487.158, and 410, assign transitions 3->2, 4->2, 5->2, and 6->2 to hydrogen atom.

708.924 has the longest wavelength, so the energy/frequency will be the lowest. The 3->2 transition has the least energy. Therefore the 708.924 wavelength has an electron transition of 3->2.

The wavelength 410 has the shortest wavelength, so the energy/frequency will be the highest. The 6->2 transition emits the most energy. Therefore, the 410 wavelength has an electron transition of 6->2.

Be able to calculate the change in energy for each transition

Change in energy = hc/wavelength

1/wavelength = RH(1/nl^2 - 1/nh^2)

Change in energy= hc RH(1/nl^2 - 1/nh^2)

Ex: Calculate the change in energy for 3->2 transition

*h, c, and RH values are given*

Change in energy = 6.62610^-34 Js (3.0010^8m/s)(1.1010^7m^-1)(1/2^2 - 1/3^2) = 3.0310^-19 J

Be able to calculate ionization energy for the hydrogen atom

The ionization energy is the energy necessary to take an electron from its most stable state (nl=1 for hydrogen), to a state infinitely far from the nucleus.

nl = 1

nh= infinity

Use change in energy = hc RH(1/nl^2 - 1/nh^2) equation

Change in energy = (6.626x10^-34 Js)(3.0010^8m/s)(1.1010^7m^-1)(1/1^2 - 1/infinity^2)

= 2.18*10^-18J(1/1 - 0)

=2.18*10^-18 J energy necessary to take an electron from nl=1 to nh=infinity.

Lab 3 - Density of a Solid

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Be able to calculate the volume of a container from the density of water used to fill up the container.

Volume=mass/density

To calculate the volume of a container, you first need to calculate the mass of water in the flask by subtracting mass of empty flask from mass of filled flask.

Mass of water = mass of filled flask- mass of empty flask

**Volume of a container = mass of water / density of water at its temperature (usually given)

Be able to calculate the density of a solid by displacement of water in a container of known volume

1. Find volume of container as shown previously.

2. Find mass of water displacement:

Mass of water displacement = mass of filled flask with metal - mass of empty flask with metal

3. Density of solid = mass of water displacement / volume of container

Rules and calculations involved with significant figures, mean, and standard deviation

same as with Lab 1

Lab 4 - Paper Chromatography

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Know the two phases involved in paper chromatography: mobile (liquid) and solid (paper)

Most chromatographic techniques achieve separation because the various components of the chemical mixtures differentially favor one of the two phases: mobile or stationary phase. The mobile (liquid) phase is a pre-selected solvent, while the stationary "solid" phase is the paper.

The chemical mixture is "spotted" onto the paper stationary phase and then placed into the solvent. As the solvent spreads through the paper, it creates a readily discernible solvent front followed by a trail of chemical components from the mixture. A finished "chromatogram" is obtained once the solvent flow is stopped and the paper is allowed to dry. The chromatogram is a final picture of what has happened to the individual solutes with time. The "spots" on a paper chromatogram may be directly visible or may require further treatment in order to achieve visualization. UV light and chemical sprays (developing agents) are commonly used to visualize paper chromatography spots.

Understand how compounds and ions are separated by paper chromatography (interaction between the chemical species and the mobile and stationary phases)

Separation of the components occurs because the components travel along the paper at different rates. Components that interact more strongly with the mobile (liquid) phase would travel up the paper strip at a faster rate than components that interact more strongly with the stationary (paper) phase. The paper is removed and allowed to dry when the solvent front reaches a few centimeters from the top of paper strip.

Most components are not colored and need to be detected by some other means (UV light, iodine vapor, or host of other chemical staining techniques).

Be able to calculate the Rf values for each spot on a chromatogram

Rf, or "ratio to front" value is the distance a component migrates

Rf = (distance component moved) / (distance solvent moved)

The distance the component moved is measured at the center of the spot.

The distance solvent moved is the solvent front.

Know the relationship between Rf value and the strength of the interactions with the mobile and solid phases (ex: if Rf is large, is the interaction of the chemical species greater toward the mobile or solid phase?)

If a component in a mixture interacts more strongly with the stationary (paper) phase, it will travel up the paper strip at a slower rate than a component that interacts more strongly with the liquid phase.

If a component in a mixture interacts more strongly with the mobile (liquid) phase, it will travel up the paper strip at a faster rate than a component that interacts more strongly with the stationary phase.

If Rf is large, the interaction of the chemical species will be greater toward the mobile phase.

If Rf is small, the interaction of chemical species will be greater toward the solid phase.

Be able to identify an unknown by comparison with Rf values for known species

Rf value will be the same if the the same solvent and stationary phase are used.

Lab 5 - Using Models to Predict Molecular Structure

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Octet rule and exceptions

The octet rule is the tendency of atoms to prefer to have eight electrons in the valance shell. When writing the lewis structures of molecules composed of C, N, O, or F atoms, do not ignore the octet rule.

Exceptions of the octet rule include hydrogen, beryllium and boron.

draw a valid Lewis Structure given the formula of a compound

Steps to Drawing a Lewis Structure:

1. Place atoms relative to each other, normally with the atom of lower group number in the center (usually least electronegative in the center). There can be more than one central atom.

2. Find the total number of valance electrons supplied by all the atoms in the molecule. For polyatomic cation, decrease the number of electrons by the charge of the ion. For an anion, increase the number of electrons by the charge of the ion.

3. Draw a single bond from the central atom to each of the surrounding atoms. Indicate electron-pair bonds by drawing dashes between the element symbols. Subtract two electrons for every single bond from the total number of valance electrons.

4. Arrange the remaining valance electrons in pairs to achieve octet (or two for hydrogen) around each atom in the structure. Lone pairs go on surrounding (more electronegative) atoms, then on the central atom(s).

5. If the central atom does not have an octet after doing step 4, make multiple bonds by changing a lone pair from a surrounding atom into a bonding pair.

6. molecules with beryllium and boron as the central atom are often electron-deficient (have fewer than eight electrons around the central atom). Some can accommodate more than 8 valance electrons. Central atom may use empty outer d orbitals. Elements in the first two rows do not have available d orbitals. Expanded valance shell only possible for periods 3 or higher.

Calculate formal charges and use them to determine best Lewis structure

Use formal charges of the atoms to decide between two or more structures that obey the octet. Structures with a lower magnitude of formal charge are preferred. In addition, the more negative formal charge should be on the more electronegative element. Like charges (both positive or both negative) on adjacent atoms lead to a less stable structure.

*Formal Charge (FC) = Valance Electrons - Number of nonbonding electrons - 1/2(total number of electrons shared in bonds)*

*Formal Charge (FC) = Valance Electrons - (Bonds + dots)*

Know the geometric families and possible molecular shapes

-Geometric families are linear, trigonal planar, tetrahedral, trigonal bipyramidal, and octahedral.

-Possible molecular shapes are linear, trigonal planar, bent, tetrahedral, trigonal pyramidal, trigonal bipyramidal, seesaw, T-shaped, octahedral, square pyramidal, and square planar.

Predict the shape of a molecule from VSEPR considerations

VSEPR theory states that electron pairs, whether bonding or nonbonding, attempt to move as far apart as possible. Geometric identity is based solely on the number of electron groups about the central atom. Bonding and nonbonding pairs of electrons are counted as individual electron groups. Double and triple bonds only count as one electron group.

To determine molecular geometry, count the number of nonbonding pairs of electrons.

Geometric Family:

A molecule with 2 electron groups is linear, 3 electron groups is trigonal planar, 4 electron groups is tetrahedral, 5 electron groups is trigonal bipyramidal, and 6 electron groups is octahedral.

Molecular structure:

The number of nonbonding pairs of electrons affects the molecular structure. If there are no nonbonding pairs, then the structure is the same as the geometric family.

Write the lewis structure of isomers if molecule is capable of forming them (trans and cis isomers for molecules with double bonds)

Isomers are compounds that contain exactly the same number of atoms (they have exactly the same empirical formula) but differ from each other by the way in which the atoms are arranged. It is where the octet rule can be satisfied in several different atomic arrangements.

Cis isomers are where the surrounding atoms of the same kind are on the same side of the double bond. Trans bond is where the surrounding atoms of the same kind are on alternating sides of the double bond.

Identify polar bonds; determine whether a molecule is polar or nonpolar

Covalent bonds between atoms having significantly different electronegativities are polar because the bonding pair is not shared equally. If the electronegativity difference is small or zero, the bond is nonpolar.

Electronegativity differences of less than or equal to 0.4 are nonpolar.

Electronegativity differences of greater than 0.4 are polar.

A molecule is polar if it possesses a net (overall) dipole moment. Molecules with polar bonds can cancel out if equal in magnitude but in opposite directions.

assign hybridization in a molecule based on the number of electrons groups

-2 electron groups are sp

-3 electron groups are sp^2

-4 electron groups are sp^3

-5 electron groups are sp^3d

-6 electron groups are sp^3d^2

Lab 6 - Water of Hydration

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Write the formula of a hydrate when given the name and vice versa.

Ex:

CuCl2 * nH2O = Copper (II) Chloride hydrate

Calculate the moles of anhydrous salt and moles of water in a hydrate

Moles of anhydrous salt = mass of anhydrous salt / molar mass of compound

Moles of water = water loss( or water driven off)/molar mass of water (18.02g/mol)

Calculate the percentage of water in a hydrate

% of water in hydrate = mass of water in sample(water driven off) / mass of hydrate sample x 100

Determine the formula for a hydrate from mass data

1. Calculate the mass of hydrate, anhydrate, and water loss.

2. Calculate the moles of water (moles of water = mass of water loss / moles of water)

3. Calculate the moles of anhydrous (moles of anhydrous = mass of anhydrous sample / molar mass of compound sample)

4. Divide moles of anhydrous from the moles of water

Moles of water present per mole of anhydrous = moles of water/moles of anhydrous

5. This number will be the value n in the hydrate formula of X * nH2O

Lab 7 - Spectrophotometry and Beer's Law

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Know the mathematical expression for the Beer-Lambert (Beer's) Law, the relationship among the variables A, b, C, and ϵ, and the units for each.

A = bcϵ

A= absorbance of the substance at a wavelength, A=log(Io/I)

b= path line of solution (cm)

c= concentration of solution (mole/liter)

ϵ= epsilon, molar absorptivity of the substance at the wavelength the measurement is made. (liter/mole-cm)

Know the relationship between absorbance and transmittance, and how to convert between the two.

A= -log(%T/100)

A+log(%T/100)=0

log(%T/100)=-A

%T/100 = 10^-A

%T= 10^(-A)*100

Use a color wheel to predict the observed color of a solution from the wavelength of absorbed light (or vice versa).

Observed color is opposite to the wavelength of absorbed light

Understand the difference between an absorbance spectrum (absorbance vs wavelength) and calibration curve (absorbance vs concentration)

Absorbance spectrum is the graph of the amount of light absorbed by a solution. This tells the maximum wavelength of substance. This is a curve

Calibration curve is the absorption spectrum of the compound when the compound, concentration, and the path length are fixed. Use equation to find concentration when given absorbance or vice versa. This is a line.

Calculate the concentration of a solution by using calibration curve data

Solve for x in the linear equation and plug in absorbance for substance for y.

Calculate the percent by mass of an absorbing species (such as Cu) in a sample using calibration curve data

Mass of Cu in sample calculated by multiplying concentration of Cu (moles/L) by volume by molar mass of Cu (g/mole)

Percent Cu in penny calculated by dividing mass of Cu in sample by total mass of the sample x 100

Calculations involving dilution of solutions of known molarity (such as preparation of calibration curve solutions from a stock solution)

M1V1=M2V2

M1 is concentration in molarity of the concentrated solution

V1 is volume of concentrated solution

M2 is concetration in molarity of dilute solution

V2 is the volume of dilute solution

Volumetric measurements and dilutions

Appendix D