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Chemical Calculations

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

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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

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Percent Error 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 %. accepted value X 100 percent

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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

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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. The density of CCl 4 is 1.58 g/mL. What is the mass of 95.7 mL of CCl 4 ? What is the volume of 227 g of olive oil if its density is 0.92 g/mL? D= 201.0 g / 18.9 cm 3 = 10.6 g/cm 3 1.58 g/mL = X / 95.7 mLX = 1.58 g/mL X 95.7 mL X = 151 g 0.92 g/mL = 227 g / X X = 227 g / 0.92 g/mL X= 247 or 2.5 X 10 2 mL

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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/cm 3 ? Is this acceptable? Explain. 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 8.0 – 8.5 / 8.5 X 100 = 5.9% No, the % error is greater than 2 %

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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

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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 –C–Calculate the mass of KCl into moles using molar mass: 85.0 g KCl –C–Convert the volume in milliliters into volume in liters 400.0 mL Calculate –M–Molarity is moles of solute divided by volume of solution 1 mol 74.55 g KCl = 1.14 mol KCl 1 L 1000 mL = 0.4000 L 1.14 mol KCl 0.4000 L = 2.85 mol / L = 2.85 M KCl Pg. 465

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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 –F–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 1000 mg = 2.2 X 10 -3 g 1,000,000 parts 500 g H 2 O 1 million = 4.4 ppm Pb ie: 4.4 parts Pb per million parts H 2 O Pg. 461

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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: 1.The nature of the material 2.The mass of the material 3.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

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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.

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Specific Heat Formula C p = specific heat at a given pressure (J/gK) 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 C p = q___ m X ∆T

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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. 1.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 C p = q___ m X ∆T 3. Calculate -Fill in formula C p = _______ = _____ = X 2 SD 32 J 4.0 g40 K160 gK 32 J 0.20 J/gK

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Practice Problems Page 61 1-4

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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.

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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.

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Specific Heat #3 If 980kJ of energy as heat are transferred to 6.2L of H 2 O at 291K, what will the final temp of H 2 O be? The specific heat of water is 4.18J/gK. Assume that 1.0mL of H 2 O equals 1.0g or H 2 O. Use Sig Figs.

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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/gK. 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.

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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)

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

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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 37.4J/Kmol 1.Gather Info Initial Temp = -5.4°C Final Temp = -0.2°C C = 37.4J/Kmol 2.Plan Work ∆H = C∆T 3.Calculate ∆H = 37.4J/Kmol (272.8K – 267.6K) = ( 37.4J/Kmol)(5.2K) = 194.48 J/mol = 194 J/mol = 267.6 K = 272.8 K Pg. 346

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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/Kmol. 1.Gather Info Initial Temp = 19.2°C Final Temp = 4.00°C C = 24.2 J/Kmol 2.Plan Work ∆H = C∆T 3.Calculate ∆H = = 292 K = 277 K (24.2 J/Kmol)(277 K – 292 K) (24.2 J/Kmol)(-15 K) =-363 J/mol Pg. 347

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

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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.

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

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

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

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Simple Conversions If you had 6.0 mops, how many pops would you have? 2 kops = 4 nips1 dip = 6 jips1 fop = 3 gops 1 pop = 3 gops3 mops = 6 jips7 dips = 2 nips 3 kops = 1 fop 6.0 mops 3 mops 6 jips 1 dip 7 dips 2 nips 4 nips 2 kops 3 kops 1 fop 3 gops 1 pop = 0.0952 pops = 9.5 X 10 -2 pops

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