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# Chemistry- JIB Topic 4 Stoichiometry

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### Chemistry- JIB Topic 4 Stoichiometry

1. 1. STOICHIOMETRY
2. 2. <ul><li>Molar Mass: g/mol </li></ul><ul><li>Molecular Mass: sum of atomic masses </li></ul><ul><li>Mass of element Moles of element atoms of element 1 mole 6.02 x 10 23 atoms PT mass 1 mole OR PT mass 1 mole 1 mole 6.02 x 10 23 atoms </li></ul>
3. 3. Percent Composition <ul><li>% mass of each element in a compound </li></ul><ul><li>Example: H 3 PO 4 = 98.00 g/mol </li></ul><ul><ul><li>H = 3.03 g / 98.00 g = 3.09% </li></ul></ul><ul><ul><li>P = 30.97 g / 98.00 g = 31.60% </li></ul></ul><ul><ul><li>O = 64.00 g / 98.00 g = 65.31% </li></ul></ul><ul><ul><ul><li>Putting this in reverse, you can calculate the empirical formula </li></ul></ul></ul>
4. 4. Empirical Formula <ul><li>H = 3.09 g x (1 mol/1.01 g) = 3.06 mol H </li></ul><ul><li>P = 31.60 g x (1 mol/30.97 g) = 1.020 mol P </li></ul><ul><li>O = 65.31 g x (1 mol/16.00 g) = 4.082 mol O Divide all mol #s by smallest mol # </li></ul><ul><li>H = 3.06 / 1.020 = 3.00 </li></ul><ul><li>P = 1.020 / 1.020 = 1.000 </li></ul><ul><li>O = 4.082 / 1.020 = 4.002 </li></ul><ul><ul><li>These become the subscripts – H 3 PO 4 </li></ul></ul><ul><ul><li>Remember, you can only round 0.1 and 0.9. Otherwise, multiply by a factor to get whole number integers </li></ul></ul>
5. 5. <ul><li>Molecular Formula: how many atoms of each element are present in a compound </li></ul><ul><ul><li>1. Calculate molecular mass of empirical formula </li></ul></ul><ul><ul><li>2. Divide given molecular mass by the calculated molecular mass of empirical formula. (Whole #) </li></ul></ul><ul><ul><li>Example: The molecular mass of ascorbic acid is 176 g/mol. The empirical formula is C 3 H 4 O 3 = 88.07 g/mol </li></ul></ul><ul><ul><ul><li>176 g/mol / 88.07 = 2 Double subscripts to get C 6 H 8 O 6 </li></ul></ul></ul>
6. 6. Chemical Equations <ul><li>Formulas for substances </li></ul><ul><li>Balance equations </li></ul><ul><li>Add state symbols (s, l, g, aq) </li></ul><ul><li>If you know </li></ul><ul><ul><li>1. moles of substance present </li></ul></ul><ul><ul><li>2. balanced equation </li></ul></ul><ul><ul><ul><li>You can calculate moles of another substance in equation </li></ul></ul></ul>
7. 7. <ul><li>How to calculate moles </li></ul><ul><ul><li>1. write equation </li></ul></ul><ul><ul><li>2. moles present </li></ul></ul><ul><ul><li>3. use coefficients to find reacting ratio </li></ul></ul>
8. 8. <ul><li>Volumetric Analysis (Molarity and Molality) </li></ul><ul><li>M = mol Molarity L </li></ul><ul><li>m = mol Molality kg </li></ul>
9. 9. Dilutions <ul><li>Dilution – solutions prepared by adding water to more concentrated ones (stock solutions) </li></ul><ul><li>M 1 V 1 = M 2 V 2 (molarity)(volume) = (molarity)(volume) </li></ul><ul><li>The volume of water that must be added to the concentrated solution is the difference between the volume of the final (diluted) solution and the volume of the concentrated solution </li></ul><ul><li>Example – What volume of water must be added to prepare 2L of 3M KOH from an 8M KOH solution? </li></ul>
10. 10. Spectroscopy and Beer-Lambert Law <ul><li>Spectroscopy is the study of the interaction of electromagnetic radiation and matter (most common are UV, Visible, and Infrared) </li></ul><ul><li>Beer-Lambert Law </li></ul><ul><ul><li>Used to relate the concentrations of colored solutions to the amount of light they absorb </li></ul></ul><ul><ul><li>The amount of absorbance is calculated using the formula: A = Σ bc </li></ul></ul><ul><ul><ul><li>A = absorbance </li></ul></ul></ul><ul><ul><ul><li>Σ = molar absorptivity (constant that depends on nature of material during absorption) </li></ul></ul></ul><ul><ul><ul><li>b = path length (length of sample that light passes through) </li></ul></ul></ul><ul><ul><ul><li>c = concentration </li></ul></ul></ul>
11. 11. <ul><li>When absorbance measurements are made at a fixed wavelength in a cell of constant path length, Σ and b are constant and A will be directly proportional to c. </li></ul><ul><li>If a solution of a compound obeys the Beer-Lambert Law, a plot of aborbance (y-axis) vs. concentration (x-axis) gives a straight line with a slope of Σ b. </li></ul><ul><li>The y-intercept is zero (the line will pass through the origin of the graph) </li></ul><ul><li>One can use the graph to read corresponding concentrations and absorption values </li></ul>
12. 12. <ul><li>A plot of absorption against wavelength can be used to determine the exact color of a solution </li></ul><ul><li>The point at which the greatest absorption is observed can be used to determine, via a color wheel, which wavelength is being reflected and, therefore, the color of the solution. </li></ul><ul><li>The color that is observed is due to the wavelengths of light that the sample did not absorb </li></ul><ul><ul><li>For example, if a sample absorbs light in the orange region of the visible spectrum, then the solution will appear blue. </li></ul></ul>
13. 13. <ul><li>A color wheel can be used to relate absorbed and transmitted colors……the transmitted color being the complement color of the absorbed light. </li></ul><ul><li>Although, theoretically the spectrophotometer can be used at a number of different wavelengths, because of the limitation of electronics, the optimal wavelength is one where the absorbance is highest </li></ul>
14. 14. <ul><li>Performing an experiment at the point of highest absorbance offers 2 advantages: </li></ul><ul><ul><li>1. Beer’s Law linear relationship between concentration and absorbance is most likely to hold around this point </li></ul></ul><ul><ul><li>2. When diluting the solution in order to investigate other lower concentrations, it is likely that if one starts at a point of maximum absorbance, that the absorbance will still remain significant and, therefore, detectable at the lower concentrations. </li></ul></ul>
15. 15. Basic Spectrophotometer
16. 16. Spectrophotometer Diagram 2
17. 17. Typical Beer’s Law Graph
18. 18. Color Wheel
19. 19. Analysis of Hydrates <ul><li>Hydrates are formula units with water associated with them </li></ul><ul><li>CuSO 4 • 5H 2 O Copper (II) sulfate pentahydrate </li></ul><ul><li>Strong heating evaporates the water </li></ul><ul><li>When water is removed, the salt is called anhydrous </li></ul><ul><ul><li>Example: 8.69 g of CuSO 4 • xH 2 O is heated and left a residue of 5.56 g. What is the formula? </li></ul></ul>
20. 20. Limiting Reactant <ul><li>Determines if a certain reactant will be completely used up during a reaction </li></ul><ul><li>Example: Phosphorus reacts with chlorine according to the equation P 4 (s) + 6 Cl 2 (g) 4 PCl 3 (l) </li></ul><ul><ul><li>If 215 g of P 4 are allowed to react with 725 g of Cl 2 , how much PCl 3 will be formed and what is the limiting reactant? </li></ul></ul>
21. 21. Percent Yield <ul><li>% Yield = actual x 100 theoretical </li></ul><ul><li>Yield is less than 100% because </li></ul><ul><ul><li>1. reactants are not pure </li></ul></ul><ul><ul><li>2. some product is lost during purification </li></ul></ul><ul><ul><li>3. side reactions take place and give by-products </li></ul></ul>