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Chemical Calculations

Chemical Calculations. Percents. Percent means “parts of 100” or “parts per 100 parts” The formula: Part Whole. x 100. Percent =. Percents. If you get 24 questions correct on a 30 question exam, what is your percent?

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Chemical Calculations

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  1. Chemical Calculations

  2. Percents • Percent means “parts of 100” or “parts per 100 parts” • The formula: Part Whole x 100 Percent =

  3. Percents • If you get 24 questions correct on a 30 question exam, what is your percent? • A percent can also be used as a RATIO • A friend tells you she got a grade of 95% on a 40 question exam. How many questions did she answer correctly? 24/30 x 100 = 80% 40 x 95/100 = 38 correct

  4. Percent Error X 100 percent accepted value • Percent error = |accepted value – experimental value| • Percent error is used to find out the degree of error you have in an experiment. • There will always be some error, scientists like to keep error below 2 %.

  5. Density • Density- the ratio of the mass of a substance to the volume of the substance. - Expressed as: Liquids & solids= grams/cubic centimeters Gasses= grams/liters • Density = Mass/Volume = g/cm3 • Mass = Density X Volume • Volume – Mass / Density

  6. Density D = M / V D= 201.0 g / 18.9 cm3 = 10.6g/cm3 • Calculate the density of a piece of metal with a volume of 18.9 cm3 and a mass of 201.0 g. • The density of CCl4 is 1.58 g/mL. What is the mass of 95.7 mL of CCl4? • What is the volume of 227 g of olive oil if its density is 0.92 g/mL? 1.58 g/mL = X / 95.7 mL X = 1.58 g/mL X 95.7 mL X = 151 g X = 227 g / 0.92 g/mL 0.92 g/mL = 227 g / X X= 247 or 2.5 X 102 mL

  7. Density and % Error Practice • If you were given an object that had a length of 5.0 cm, a width or 10.0 cm and a height of 2.0 cm, what would the density of this object be if you weighed it and found that it had a mass of 800.0 g? • What would the % error be for your measurments if I told you the accepted value for the density of this object is 8.50 g/cm3? Is this acceptable? Explain. 5.0 cm X 10.0 cm X 2.0 cm = 100 cm3 D = 800.0 g / 100 cm3 = 8.0 g/cm3 8.0 – 8.5 / 8.5 X 100 = 5.9% No, the % error is greater than 2 %

  8. Concentration Measurements • Molarity: M • Molarity = mol solute / L solution • Use in solution stoichiometry calculations • Mole solute = Molarity X Liters solution • Liters solution = moles solution / Molarity • Molality: m • mol solute / kg solvent • Used with calculation properties such as boiling point elevation and freezing point depression • Parts per Million: ppm • g solute / 1 000 000 g solution • Used to express small concentrations Pg. 460

  9. Molarity • What is the molarity of a potassium chloride solution that has a volume of 400.0 mL and contains 85.0 g KCl? • Gather Info Volume of solution = 400.0 mL Mass of solute = 85.0 g KCl Molarity of KCl solution = ? • Plan Work • Calculate the mass of KCl into moles using molar mass: 85.0 g KCl • Convert the volume in milliliters into volume in liters 400.0 mL • Calculate • Molarity is moles of solute divided by volume of solution 1 mol = 1.14 mol KCl 74.55 g KCl 1 L = 0.4000 L 1000 mL 1.14 mol KCl = 2.85 mol / L = 2.85 M KCl Pg. 465 0.4000 L

  10. Parts Per Million • A chemical analysis shows that there are 2.2 mg of lead in exactly 500 g of water. Convert this measurement to parts per million. • Gather Info Mass of Solute = 2.2 mg Mass of Solvent = 500 g Parts per Million = ? • Plan Work • First change 2.2 mg to grams 2.2 mg - Divide this by 500 g to get the amount of lead in 1 g water, then multiple by 1,000,000 to get the amount of lead in 1,000,000 g water. • Calculate 0.0022 g Pb 1 g = 2.2 X 10-3 g 1000 mg 1,000,000 parts = 4.4 ppm Pb 1 million 500 g H2O ie: 4.4 parts Pb per million parts H2O Pg. 461

  11. Specific Heat • Specific Heat – the quantity of energy that must be transferred as heat to raise the temperature of 1g of a substance by 1K. • The quantity of energy transferred as heat depends on: • The nature of the material • The mass of the material • The size of temperature change • Ex: 1g of Fe 100°C to 50°C transfers 22.5J of energy. 1g of Ag 100°C to 50°C transfers 11.8J of energy. • Fe has a larger specific heat than Ag • Meaning that more energy as heat can be transferred to the iron than to the silver

  12. Explain Specific Heat in My Terms • Metals = Low Specific Heat = little energy must be transferred as heat to increase temperature. • Water = High Specific Heat (Highest of most common substances) = can absorb a large quantity of energy before temperature increases.

  13. Specific Heat Formula Cp = specific heat at a given pressure (J/g•K) q = energy transferred as heat (J) m = mass of the substance (g) ∆T = difference btwn. initial and final temperatures (K) (Final Temp – Initial Temp) Q = Cp (m X ΔT) Mass = (Cp)(ΔT) / Q Cp = q___ m X ∆T

  14. Specific Heat Example (pg.61) • A 4.0g sample of glass was heated from 274K to 314K and was found to absorb 32J of energy as heat. Calculate the specific heat of this glass. • Gather Info A. Mass (m) of sample = 4.0g B. Initial Temp = 274K C. Final Temp = 314K D. Amt. of Energy absorbed (q) = 32J 2.Plan Work Cp = q___ m X ∆T 3. Calculate • Fill in formula Cp = _______ = _____ = X 2 SD 32 J 32 J 0.20 J/g•K 4.0 g 40 K 160 g•K 2 SD

  15. Practice Problems Page 61 1-4

  16. Specific Heat #1 • Calculate the specific heat of a substance if a 35g sample absorbs 48J as the temperature is raised from 293K to 313K. Be sure to use the correct number of sig. figs. in your answer.

  17. Specific Heat #2 • The temperature of a piece of copper with a mass of 95.4g increases from 298.0K to 321.1K when the metal absorbs 849J of energy as heat. What is the specific heat of copper? Use Sig Figs.

  18. Specific Heat #3 • If 980kJ of energy as heat are transferred to 6.2L of H2O at 291K, what will the final temp of H2O be? The specific heat of water is 4.18J/g•K. Assume that 1.0mL of H2O equals 1.0g or H2O. Use Sig Figs.

  19. Specific Heat #4 • How much energy as heat must be transferred to raise them temperature of a 55g sample of Al from 22.4°C to 94.6°C? The specific heat of Al is 0.897J/g•K. Note that a temperature change of 1°C is the same as a temperature change of 1K because the sizes of the degree divisions on both scales are equal. Use Sig Figs.

  20. Enthalpy • Enthalpy- the sum of the internal energy of a system plus the product of the system’s volume multiplied by the pressure that the system exerts on its surroundings. (heat content, total energy of the system) • When calculating enthalpy if the change in enthalpy is positive, it means that heating the sample requires energy making it an endothermic process.(run up the hill) • When the change is negative, the sample has been cooled, meaning that the sample has released energy making it an exothermic process.(fall down the hill)

  21. Molar Enthalpy Formula ∆H = molar enthalpy (J/mol) C = molar heat capacity (J/K•mol) ∆T = change in temperature (K) C = ΔH / ΔT Note: A mole is the amount of a substance ∆H = C∆T

  22. Molar Enthalpy Heating • How much does the molar enthalpy change when ice warms from -5.4°C to -0.2°C? The molar heat capacity of H2O(s) is 37.4J/K•mol • Gather Info Initial Temp = -5.4°C Final Temp = -0.2°C C = 37.4J/K•mol • Plan Work ∆H = C∆T • Calculate ∆H = 37.4J/K•mol (272.8K – 267.6K) = (37.4J/K•mol)(5.2K) = 194.48 J/mol = 194 J/mol = 267.6 K = 272.8 K Pg. 346

  23. Molar Enthalpy Cooling • Calculate the molar enthalpy change with an aluminum can that as a temperature of 19.2°C is cooled to a temperature of 4.00°C. The molar heat capacity for Al is 24.2 J/K•mol. • Gather Info Initial Temp = 19.2°C Final Temp = 4.00°C C = 24.2 J/K•mol • Plan Work ∆H = C∆T • Calculate ∆H = = 292 K = 277 K (24.2 J/K•mol) (277 K – 292 K) (24.2 J/K•mol) (-15 K) = -363 J/mol Pg. 347

  24. Enthalpy Practice Page 346 1 & 2 Page 347 1 & 2

  25. Page 346 #1 • Calculate the molar enthalpy change of H2O(l) when liquid water is heated from 41.7°C to 76.2°C.

  26. Page 346 #2 • Calculate the ∆H of NaCl when it is heated from 0.0°C to 100.0°C.

  27. Page 347 #1 • The molar heat capacity of Al(s) is 24.2 J/K•mol. Calculate the molar enthalpy change when Al(s) is cooled from 128.5°C to 22.6°C.

  28. Page 347 #2 • Lead has a molar heat capacity of 26.4J/K•mol. What molar enthalpy change occurs when lead is cooled from 302°C to 275°C.

  29. Simple Conversions 2 kops = 4 nips 1 dip = 6 jips 1 fop = 3 gops 1 pop = 3 gops 3 mops = 6 jips 7 dips = 2 nips 3 kops = 1 fop • If you had 6.0 mops, how many pops would you have? 6 jips 6.0 mops 1 dip 2 nips 2 kops 1 fop 3 gops 1 pop 4 nips 3 gops 6 jips 7 dips 3 kops 1 fop 3 mops = 0.0952 pops = 9.5 X 10-2 pops

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