Table of Contents:¶
| Experiment # | Topic |
|---|---|
| 1. | Measurement and Density |
| 2. | Determining Avogadro’s Number |
| 3. | The Empirical Formula of Selected Hydrates |
| 4. | Empirical Formula and Stoichiometry |
| 5. | Reactions and Solubility |
| 6. | Standardization of a base and Titration of a Vinegar Solution |
| 7. | Titration of an Unknown Acid – Determination of Molar Mass |
| 8. | Conductimetric Titrations |
| 9. | Using the Ideal Gas Law |
| 10. | Heats of Reaction and Solution |
| 11. | Solution Preparation and Beers Law |
1.¶
Experiment #1 Measurement and Density¶
This experiment presents us with the historical beauty of Archimedes principle or water displacement (250 BCE).
Vernier Callipers: Learn how to read a Vernier scale JavaScript Applet¶
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2.¶
Experiment #2 Determining Avogadro's Number¶
This lab is designed to help students build dimensional analysis skills as well as provide a brief introduction for the organization of molecules at the surface.
Calculations¶
Part A. Determine Avogadro's Number From the Density of Copper¶
Copper crystallizes into a face centered cubic (fcc) unit cell.
The atomic mass of Cu is 63.546 g/mol and the length of one edge in the unit cell is 361.2 pm.
There is 8 corners and 6 faces of the unit cell, where there exists a Cu atom (each of which are shared with other unit cells) at each of these positions.
I. Solve for b
$$ b^{2} = l^{2} + l^{2} = 2l^{2} $$
$$ b = \sqrt{2l^{2}} = 4r $$
II. Solve for l
$$ 2l^{2} = b^{2} $$
$$ l = \sqrt{\frac{b^{2}}{2}} = \frac{b}{\sqrt{2}} = \frac{4r}{\sqrt{2}} = 2\sqrt{2}r $$
1. Determine the volume, V of the unit cell.
$$ V_{cube} = l^{3}, \text{ where l is the edge length of the unit cell.} $$
2. Calculate the mass of the unit cell.
$$ Density = \rho = \frac{mass}{volume}$$
$$ Mass_{Cell} = \rho_{Calc}Volume_{Cell}$$
3. Determine the number of atoms that occupy the unit cell.
$$ \text{Atoms/Unit Cell } = (\frac{1}{8}\times \text{# of Corner Atoms}) + (\frac{1}{2} \times \text{# of Face Centered Atoms} )$$
$$ \text{Atoms/Unit Cell } = (\frac{1}{8}\times 8) + (\frac{1}{2} \times 6 ) = 4 $$
4. Use dimensional analysis to calculate Avogadro's number from the number of Cu atoms per unit cell, the mass per unit cell, and the molar mass of Cu.
$$ (\frac{\text{4 Cu Atoms}}{\text{1 Cu Unit Cell}})\times(\frac{\text{1 Cu Unit Cell}}{\text{Mass of Unit Cell}})\times(\frac{63.546 g}{\text{1 mol}}) = \text{# of Cu Atoms} $$
Part B. Stearic Acid Monolayer¶
Check out this video below to see what to look for during the experiment.¶
General idea of what it looks like at the atomic scale:
1. Determine the volume of stearic acid/hexane solution, $V_{SAH}$ using your calibration ratio (drops/mL), $\alpha$ and the average number of drops to form the monolayer.
$$ V_{SAH} = \alpha^{-1}\times(\text{Avg. Drops}) $$
2. Calculate the mass of stearic acid in the monolayer, $m_{stearic}$ using the concentration of stearic acid/hexane solution, $M_{SAH}$.
Note: Molar Mass of Stearic acid is 284.48 g/mol.
$$ m_{stearic} = (M_{SAH})\times(V_{SAH}) $$
3. Calculate the volume of stearic acid in the monolayer, $V_{Stearic}$
$$ V_{Stearic} = (\rho_{Steric Acid})^{-1}\times m_{stearic} $$
$$ $$
4. Calculate the thickness of the monolayer, h where the volume of stearic acid in the monolayer can be approximated by $V=\pi r^{2}h$.
$$ h = \frac{V}{\pi r^{2}} = \frac{V}{\pi(\frac{d}{2})^{2}} $$
5. Determine the volume of a carbon atom.
Note that h/18 can be used as a good approximation for the diameter of a carbon atom.
$$ V_{\text{Carbon atom}} = \frac{4}{3}\pi r^{3} = \frac{4}{3}\pi \frac{d}{2}^{3} = \frac{4}{3}\pi (\frac{h}{36})^{3} $$
6. Determine Avogadro's number.
$$ N_{A} = (\frac{\text{1 Carbon Atom}}{V_{\text{Carbon atom}}})\times(\rho_{\text{Carbon}})^{-1}\times(\frac{\text{12.0107 g Carbon}}{\text{1 mol Carbon}}) $$
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5.¶
Experiment #5 Reactions and Solubility¶
Gas forming reactions:
1.
$$2HCl_{(aq)} + K_{2}S \rightarrow H_{2}S (g) + 2KCl (aq)$$2.
$$2HCl_{(aq)} + K_{2}CO_{2} \rightarrow H_{2}O (l) + CO_{2}(g) + 2KCl (aq)$$3.
$$2HCl_{(aq)} + K_{2}SO_{2} \rightarrow H_{2}O (l) + SO_{2}(g) + 2KCl (aq)$$4.
$$NH_{4}Cl(aq) + KOH \rightarrow H_{2}O (l) + NH_{3}(g) + 2KCl (aq)$$5.
$$2HCl_{(aq}) + Zn(s) \rightarrow ZnCl_{2} (aq) + H_{2} (g)$$Weak electrolytes do not dissociate into ions, so another way gases can form in solution is through the decomposition of weak electrolytes. For example, $H_{2}CO_{3}$ readily decomposes into $H_{2}O$ and $CO_{2}$ gas.
Aqueous Solutions; Precipitation Reactions; Acid-base Reactions¶
Use your knowledge of solubility rules: show the net ionic equation, provide the products and balance the chemical equation.¶
(1)
$${SrCl_{2} }_{(aq)} + {H_{2}SO_{4} }_{(aq)} \rightarrow {SrSO_{4}}_{(s)} + {2HCl }_{(aq)}$$$${Sr^{2+} }_{(aq)} + {SO_{4}^{2-}}_{(aq)} \rightarrow {SrSO_{4}}_{(s)}$$$$(H^{+} , Cl^{-} \text{ are spectator ions})$$(2)
$${2 NaOH }_{(aq)} + {H_{2}SO_{4} }_{(aq)} \rightarrow {Na_{2}SO_{4}}_{(aq)} + {2H_{2}O }_{(l)}$$$$2H^{+} + 2OH^{-} \rightarrow {2H_{2}O }_{(l)}$$$$(Na^{+} , SO_{4}^{2-} \text{ are spectator ions})$$(3)
$${(NH_{4})_{2}SO_{4} }_{(aq)} + {2NaOH}_{(aq)} \rightarrow {Na_{2}SO_{4} }_{(aq)} + {2NH_{3}}_{(g)} + {2H_{2}O }_{(l)}$$$${2NH_{4}^{+} }_{(aq)} + {2OH^{-} }_{(aq)} \rightarrow {2NH_{3}}_{(g)} + {2H_{2}O }_{(l)}$$$$(Na^{+} , SO_{4}^{2-} \text{ are spectator ions})$$(4) Balance the following equations and then write the net ionic equations.
(a)
$$(NH_{4})_{2}CO_{3}(aq) + Cu(NO_{3})_{2}(aq) \rightarrow CuCO_{3}(s) + NH_{4}NO_{3}(aq)$$Answer:
Balanced Eq:
$$(NH_{4})_{2}CO_{3}(aq)+ Cu(NO_{3})_{2}(aq) \rightarrow CuCO_{3}(s) + 2 NH_{4}NO_{3}(aq)$$Net:
$$Cu^{2+}(aq)+ CO_{3}^{2-}(aq) \rightarrow CuCO_{3} (s)$$(b)
$$Pb(OH)_{2}(s) + HCl(aq) \rightarrow PbCl_{2}(s) + H_{2}O(l)$$Answer:
Balanced Eq:
$$Pb(OH)_{2}(s) + 2 HCl(aq) \rightarrow PbCl_{2}(s) + 2 H_{2}O(l)$$Net:
$$Pb(OH)_{2}(s) + 2 H^{+}(aq)+ 2 Cl^{-}(aq) \rightarrow PbCl_{2}(s) + 2 H_{2}O(l)$$(c)
$$BaCO_{3}(s) + HCl(aq) \rightarrow BaCl_{2}(aq)+ H_{2}O(l) + CO_{2}(g)$$Answer: Balanced Eq:
$$BaCO_{3}(s) + 2 HCl(aq) \rightarrow BaCl_{2}(aq) + H_{2}O(l) + CO_{2}(g)$$Net:
$$BaCO_{3}(s) + 2H^{+}(aq) \rightarrow Ba^{2+}(aq) + H_{2}O(l) + CO_{2}(g)$$(d)
$$CH_{3}CO_{2}H(aq) + Ni(OH)_{2}(s) \rightarrow Ni(CH_{3}CO_{2})_{2}(aq) + H_{2}O(l)$$Answer:
Balanced Eq:
$$2 CH_{3}CO_{2}H(aq) + Ni(OH)_{2}(s) \rightarrow Ni(CH_{3}CO_{2})_{2}(aq) + 2 H_{2}O(l)$$Net:
$$2CH_{3}CO_{2}H(aq) + Ni(OH)_{2}(s) \rightarrow 2 CH_{3}CO_{2}^{-} (aq) + Ni^{2+}(aq) + 2H_{2}O(l)$$Jump to table of contents.¶
6.¶
Experiment #6 Standardization of a base and Titration of a Vinegar Solution¶
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7.¶
Experiment #7 Titration of an Unknown Acid – Determination of Molar Mass¶
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8.¶
Experiment #8 Conductimetric Titrations¶
This lab is continuation of the various application of titrations.
Where is conductivity found in environmental applications?
• Water treatment
• Aquacultures
• Irrigation/ Agriculture
Theory - a very brief Introduction to Electrochemistry¶
Ohm's Law: $V = IR$
V = Potential difference in voltz, V
I = Current in Amphere's, A
R = Resistance in ohms, $\Omega$
Conductance, G is the reciprocal of resistance and is measured in siemens(S). German inventor Werner Von Siemens (mid 1800's)
$$G \frac{1}{R} = \frac{\kappa A}{l} $$A = Area of electrode ($m^{2}$)
l = distance between electrodes (m)
Conductivity, $\kappa$ ($Sm^{-1}$) is then defined as $\kappa = \frac{l}{RA} $, where $\frac{l}{A}$ is typically called the cell constant, $c^{*}$.
Side Notes:
Although conductivity, $\kappa$ is not in terms of temperature in the equation above. Generally, ionic conductivity increases with increasing temperature. As the mobility of ions increase with temperature, so will conductivity.
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11.¶
Experiment #11 Solution Preparation and Beers Law¶
This lab was designed to help students develop the skills for determining the amount of a certain species inside some solution by linear regression, where the equation of a line is given by Beer's Law.
Theory:¶
Derivation of Beer's Law:¶
Here is an image of the situation we wish to model:
The density of particles, $\rho$ and the absorption coefficient, $\alpha$ multiplied by the intensity, I shown in the 1st order differential equation: $$ -\frac{\partial{I}}{\partial{x}} = I \alpha \rho $$
Combine like-terms to each side of the equation:
$$\int_{I_{0}}^{I} \frac{\partial{I}}{I} = -\int_{0}^{x} \alpha \rho \partial{x} $$We know that $\int \frac{1}{x}dx = ln(x)$, so
$$ln(\frac{I}{I_{0}}) = - \alpha \rho x, $$To get the general solution of the D.E we can take the exponential of both sides
$$\frac{I}{I_{0}} = e^{-\alpha \rho x} $$General Solution to the D.E:
$$I (x) = I_{0} e^{-\alpha \rho x}$$Otherwise, to continue deriving Beer's Law we can use the property of logarithms:
$$-ln(\frac{I}{I_{0}}) = ln(\frac{I_{0}}{I}) = \alpha \rho x, $$and since we know the following
$$log_{10}(x) = \frac{ln(x)}{ln(10)},$$then we can say
$$ log_{10}(\frac{I_{0}}{I}) = \frac{\alpha \rho x}{ln(10)}$$Finally, we can say that $\rho \propto c$. We can also simplify further by saying $\epsilon =\frac{\alpha}{ln(10)}$, which has units of $M^{-1}cm^{-1}$ and $x = b$, where b is in cm.
$$ A = ln(\frac{I_{0}}{I}) = \epsilon b c$$Theory - a brief Introduction to Quantum Mechanics - particle in a box (PIB)¶
$$\Delta E = E_{S_{1}} − E_{S_{0}}$$$$E_{n} =\frac{h^{2}n^{2}}{8ma^{2}}$$$$ \Delta E = \frac{h^{2}(N+1)^{2}}{8ma^{2}} = h\nu = \lambda $$n = 1, 2, 3..., h = 6.62608 × 10-34 J·s (Planck’s constant), m = 9.10939 × 10-31 kg (mass of an electron), and a = length of the box (in meters if using SI units).
Solving for the box length, a, gives:
$$a = \sqrt{\frac{h\lambda (N+1)}{8mc} } $$The box length, a, may also be related to the polyene structure with the empirical formula:
$$a = (P · l) + E$$where $P$ is the number of C atoms in the conjugated chain, $l$ is the C-C bond length in the chain, and $E$ is the effective size of the end-groups.
Where & how can this be applied?¶
$\beta$-carotin is the chemical that gives carrots their characteristic orange color. Absorbance is around 450 nm (violet-blue), so we see orange, which is the light transmitted.
Color Wheel with corresponding wavelengths:¶
Jump to table of contents.¶
Appendix:¶
Some of the Main Types of Reactions:¶
1) Combination (synthesis)¶
$$ \mathrm{A}+\mathrm{B} \rightarrow \mathrm{C} $$$$ {2H_{2}}_{(g)} + {O_{2}}_{(g)} \rightarrow {2H_{2}O}_{(g)}\tag{1} $$2) Decomposition¶
$$ \mathrm{C} \rightarrow \mathrm{A}+\mathrm{B} $$$$ {2H_{2}O}_{(g)} \rightarrow {2H_{2}}_{(g)} + {O_{2}}_{(g)} \tag{1}$$3) Single Displacement¶
$$ \mathrm{AB}+\mathrm{C} \rightarrow \mathrm{CB}+\mathrm{A} $$$${{CuSO}_{4}}_{(aq)} + Zn_{(s)} \rightarrow {Zn{SO}_{4}}_{(aq)}+Cu_{(s)}\tag{1}$$4) Double Displacement¶
$$ \mathrm{AB}+\mathrm{CD} \rightarrow \mathrm{AD}+\mathrm{BC} $$$$ \mathrm{AgNO}_{3}+\mathrm{NaCl} \rightarrow \mathrm{AgCl}+\mathrm{NaNO}_{3}\tag{1}$$$$ {NaHCO_{3}}_{(aq)} + {CH_{3}COOH}_{(aq)} \rightarrow {H_{2}CO_{3}}_{(aq)} + {NaCH_{3}COO}_{(s)} \rightarrow {H_{2}O}_{(l)} + {CO_{2}}_{(g)} + {NaCH_{3}COO}_{(s)}\tag{2}$$5) Neutralization¶
Double displacement reactions between strong acids and strong bases results in salt and water:
$$\mathrm{HCl}_{(aq)}+\mathrm{NaOH}_{(aq)} \rightarrow \mathrm{NaCl}_{(aq)}+\mathrm{H}_{2} \mathrm{O}_{(l)}\tag{1}$$Double displacement reactions between acids and bases (water as solvent) results in salt :
$${NaHCO_{3}}_{(aq)} + {CH_{3}COOH}_{(aq)} \rightarrow {H_{2}CO_{3}}_{(aq)} + {NaCH_{3}COO}_{(s)}\tag{2}$$6) Combustion¶
These reactions always result in a change in enthalpy, $\Delta H_{combust} < 0$ and forms $CO_{2}$ and $H_{2}O$.
$$CH_{4} + 2O_{2} \rightleftharpoons CO_{2} + 2H_{2}O\hspace{0.2cm};\hspace{0.2cm} \Delta H < 0 \tag{1}$$Suppose wood has the chemical formula $\mathbf{C}_{6} \mathbf{H}_{12} \mathbf{O}_{6}$. Then,
$$\mathbf{C}_{6} \mathbf{H}_{12} \mathbf{O}_{6} \quad+6 \mathbf{O}_{2}\rightleftharpoons\quad 6 \mathrm{CO}_{2} \quad+\quad 6 \mathbf{H}_{2} \mathbf{O}\hspace{0.2cm};\hspace{0.2cm} \Delta H <0 \tag{2}$$7) Redox¶
$${{CuSO}_{4}}_{(aq)} + Zn_{(s)} \rightarrow {Zn{SO}_{4}}_{(aq)}+Cu_{(s)}\tag{1}$$Jump to table of contents.¶
External Links:
Wolfram Alpha - Computational engine based on a vast collection of built-in data, algorithms and methods - Free version offers limited user interface.
Sympygamma - Similar to Wolfram alpha - Free step-by-step solutions of mathematical operations (requires a numerical pythonic input - easy)
Scale of the Universe - Applet you should check out.