Chemical Calculations Percents Percent means parts of 100

Chemical Calculations

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

Percents • If you get 24 questions correct on a 30 question exam, what is your percent? 24/30 x 100 = 80% • 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? 40 x 95/100 = 38 correct

Percent Error • Percent error = |accepted value – experimental value| X 100 percent accepted 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 %.

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/cm 3 - Mass = Density X Volume - Volume – Mass / Density

Density D=M/V • Calculate the density of a piece of metal with a volume of 18. 9 cm 3 and a mass of 201. 0 g. D= 201. 0 g / 18. 9 cm 3 = 10. 6 g/cm 3 • The density of CCl 4 is 1. 58 g/m. L. What is the mass of 95. 7 m. L of CCl 4? 1. 58 g/m. L = X / 95. 7 m. L X = 1. 58 g/m. L X 95. 7 m. L X = 151 g • What is the volume of 227 g of olive oil if its density is 0. 92 g/m. L? X = 227 g / 0. 92 g/m. L = 227 g / X X= 247 or 2. 5 X 102 m. L

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? 5. 0 cm X 10. 0 cm X 2. 0 cm = 100 cm 3 D = 800. 0 g / 100 cm 3 = 8. 0 g/cm 3 • 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/cm 3? Is this acceptable? Explain. 8. 0 – 8. 5 / 8. 5 X 100 = 5. 9% No, the % error is greater than 2 %

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 g solution – Used to express small concentrations Pg. 460

Molarity • What is the molarity of a potassium chloride solution that has a volume of 400. 0 m. L and contains 85. 0 g KCl? • Gather Info Volume of solution = 400. 0 m. L Mass of solute = 85. 0 g KCl Molarity of KCl solution = ? • Plan Work – Calculate the mass of KCl into moles using molar mass: 1 mol 85. 0 g KCl = 1. 14 mol KCl 74. 55 g KCl – Convert the volume in milliliters into volume in liters 400. 0 m. L 1 L = 0. 4000 L 1000 m. L • Calculate – Molarity is moles of solute divided by volume of solution 1. 14 mol KCl = 2. 85 mol / L = 2. 85 M KCl 0. 4000 L Pg. 465

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 1 g 2. 2 mg = 2. 2 X 10 -3 g 1000 mg - Divide this by 500 g to get the amount of lead in 1 g water, then multiple by 1, 000 to get the amount of lead in 1, 000 g water. • Calculate 0. 0022 g Pb 500 g H 2 O 1, 000 parts 1 million = 4. 4 ppm Pb ie: 4. 4 parts Pb per million parts H 2 O Pg. 461

Specific Heat • Specific Heat – the quantity of energy that must be transferred as heat to raise the temperature of 1 g of a substance by 1 K. • The quantity of energy transferred as heat depends on: 1. The nature of the material 2. The mass of the material 3. The size of temperature change • Ex: 1 g of Fe 100°C to 50°C transfers 22. 5 J of energy. 1 g of Ag 100°C to 50°C transfers 11. 8 J of energy. à Fe has a larger specific heat than Ag à Meaning that more energy as heat can be transferred to the iron than to the silver

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.

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) Cp = ∆T = difference btwn. initial and final temperatures (K) (Final Temp – Initial Temp) Q = Cp (m X ΔT) Mass = (Cp)(ΔT) / Q q___ m X ∆T

Specific Heat Example (pg. 61) • A 4. 0 g sample of glass was heated from 274 K to 314 K and was found to absorb 32 J of energy as heat. Calculate the specific heat of this glass. 1. Gather Info A. Mass (m) of sample = 4. 0 g B. Initial Temp = 274 K C. Final Temp = 314 K D. Amt. of Energy absorbed (q) = 32 J 2. 3. - Plan Work Cp = q___ m X ∆T Calculate Fill in formula 32 J Cp = _______ = 4. 0 g X 40 K 2 SD 32 J = _____ 160 g • K 2 SD 0. 20 J/g • K

Practice Problems Page 61 1 -4

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

Specific Heat #2 • The temperature of a piece of copper with a mass of 95. 4 g increases from 298. 0 K to 321. 1 K when the metal absorbs 849 J of energy as heat. What is the specific heat of copper? Use Sig Figs.

Specific Heat #3 • If 980 k. J of energy as heat are transferred to 6. 2 L of H 2 O at 291 K, what will the final temp of H 2 O be? The specific heat of water is 4. 18 J/g • K. Assume that 1. 0 m. L of H 2 O equals 1. 0 g or H 2 O. Use Sig Figs.

Specific Heat #4 • How much energy as heat must be transferred to raise them temperature of a 55 g sample of Al from 22. 4°C to 94. 6°C? The specific heat of Al is 0. 897 J/g • K. Note that a temperature change of 1°C is the same as a temperature change of 1 K because the sizes of the degree divisions on both scales are equal. Use Sig Figs.

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)

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

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 H 2 O(s) is 37. 4 J/K • mol 1. Gather Info Initial Temp = -5. 4°C = 267. 6 K Final Temp = -0. 2°C = 272. 8 K C = 37. 4 J/K • mol 2. Plan Work ∆H = C∆T 3. Calculate ∆H = 37. 4 J/K • mol (272. 8 K – 267. 6 K) = (37. 4 J/K • mol)(5. 2 K) = 194. 48 J/mol = 194 J/mol Pg. 346

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. 1. Gather Info Initial Temp = 19. 2°C = 292 K Final Temp = 4. 00°C = 277 K C = 24. 2 J/K • mol 2. Plan Work ∆H = C∆T 3. Calculate ∆H = (24. 2 J/K • mol)(277 K – 292 K) (24. 2 J/K • mol)(-15 K) = -363 J/mol Pg. 347

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

Page 346 #1 • Calculate the molar enthalpy change of H 2 O(l) when liquid water is heated from 41. 7°C to 76. 2°C.

Page 346 #2 • Calculate the ∆H of Na. Cl when it is heated from 0. 0°C to 100. 0°C.

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.

Page 347 #2 • Lead has a molar heat capacity of 26. 4 J/K • mol. What molar enthalpy change occurs when lead is cooled from 302°C to 275°C.

Simple Conversions • If you had 6. 0 mops, how many pops would you have? 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 6. 0 mops 6 jips 1 dip 2 nips 2 kops 1 fop 3 gops 1 pop 3 mops 6 jips 7 dips 4 nips 3 kops 1 fop 3 gops = 0. 0952 pops = 9. 5 X 10 -2 pops