to water. You add 1 drop of phenolphthalein and the solution turns pink. This observation tells you Na 2

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1 1A End of Lab Questions Notes 9/11/18 Lab 4 1 You add 100 g of to water You add 1 drop of phenolphthalein and the solution turns pink This observation tells you is a base 2 You add vinegar (09 M acetic acid) to the CO 3 solution from Question 1 Calculate the volume in ml of vinegar that reacts with Balanced chemical equation: + 2 HC 2 NaC 2 + H 2 O + C (g) 100 g /(1 mole /106 g )(2 moles HC 2 /1 mole ) = 0018 moles HC 2 O 2 Molarity of HC 2 = moles of HC 2 /Volume so Volume = moles of HC 2 /Molarity = 0018 moles HC 2 / 09 moles/lter = 0020 liters 0020 liters (1000 ml/1 liter) = 20 ml of 09 M acetic acid 3 You want the solution to stay pink Should you use more or less of the volume of vinegar you calculated in Question 2? + 2 HC 2 NaC 2 + H 2 O + C (g) Solution is pink because it is a base so you want (base) to be the excess reactant and vinegar to be the limiting reactant So use less vinegar (less than 20 ml) to keep the solution pink 4 Which reactant is the limiting reactant in Question 3? 5 a For your Lab 4 demonstration, white solid A is Choose a base that soluble in water, eg, KOH, NaH, Mg(OH) 2 and Ca are bases but both are insoluble in water NaCl, Mg(NO 3, and Ca(NO 3 are neutral b For your Lab 4 demonstration, colorless liquid B is Liquid B is an acid that reacts with a base that produces C gas Liquid B is either HCl and HC 2 The base that reacts with this acid that produces C gas must contain H ion or ion, eg, NaH, c For your Lab 4 demonstration, colorless liquid C is Liquid C reacts with the pink solution after Liquid B reacts with Solution A to form a white solid The pink solution is a base and contains excess base, either H ion or ion For Liquid C, identify a cation that combines with H ion or ion to form an insoluble solid - see Lab 4 Materials List and the solubility table H compounds are soluble in water So, NaH cannot be used for Solid A compounds are insoluble in water (with several exceptions) So, KOH and NaCl and NaH will not work because K + and Na + 2- will not combine with to form a precipitate (solid) Ethanol (C 2 H 5 OH) will not work because it is a molecular compound and does not have ions that can combine with CO 3 Vinegar (HC 2 ) will not work because it is a weak acid and only has a few ions in solution (H + and C 2 ions will not combine with to form a precipitate (solid))

2 HCl will not work Although it is a strong acid, the H + and Cl ions will not combine with to form a precipitate (solid)) Ca has the Ca 2+ ion that could combine with to form a precipitate (solid) but Ca is not soluble in water and cannot be used as Liquid C Ca(NO 3 has the Ca 2+ ion that could combine with to form a Ca precipitate (solid) Ca(NO 3 is soluble in water and can be used as Liquid C Mg(NO 3 has the Mg 2+ ion that could combine with to form a Mg precipitate (solid) Mg(NO 3 is soluble in water and can be used as Liquid C 9/6/18 Lab 3, Part 2 1 a Which antacid did you make? b Actual yield in g of your antacid Give a number only; do not include text The actual yield of your antacid is the mass of dry antacid that you made in this experiment c % yield of your antacid Give a number only; do not include text % yield = (actual yield in g/theoretical yield in g) x 100 d If the % yield of antacid is not 100%, identify a specific step in your experiment that led to this error If the % yield is less than 100%, antacid is usually lost during the filtration step or not all of the reactants reacted If the % yield is greater than 100%, the antacid is probably not dry (still wet, not all of the water was removed) 2 NaOH (aq) reacts with AlCl 3 (aq) to form a solid: NaOH + AlCl 3 ---> a Write a balanced chemical equation for this reaction 3 NaOH + AlCl 3 ---> 3 NaCl + Al(OH) 3 b What is the chemical formula of the solid that is formed in this reaction? Solid is Al(OH) 3 See solubility table Al(OH) 3 is insoluble in water NaCl is soluble in water c Write a net ionic equation for this reaction 3 NaOH + AlCl 3 ---> 3 NaCl + Al(OH) 3 See solubility table The soluble compounds dissociate into ions See table of common ions to check formula and charge Ionic equation: 3 Na OH + Al Cl ---> 3 Na + + Cl + Al(OH) 3 Net ionic equation: 3 OH + Al > Al(OH) 3 Spectator ions are Na + and Cl 3 a 540 g of NaOH reacts enough AlCl 3 so all of the NaOH reacts Calculate the theoretical yield of Al(OH) 3 Convert mass of NaOH to moles of NaOH Use molar mass of NaOH for your conversion factor Convert moles of NaOH to moles of Al(OH) 3 Use coefficients in balanced chemical equation for your conversion factor

3 Convert moles of Al(OH) 3 to mass of Al(OH) 3 Use molar mass of Al(OH) 3 for your conversion factor 540 g NaOH (1 mole NaOH/40 g NaOH)(1 mole Al(OH) 3 /3 mole NaOH)(78 g Al(OH) 3 /mole Al(OH) 3 ) = 351 g Al(OH) 3 b You react 540 g of NaOH with 1000 g of AlCl 3 You collect 325 g of solid Al(OH) 3 Calculate the % yield of Al(OH) 3 % yield = (actual yield Al(OH) 3 in g/theoretical yield Al(OH) 3 in g) x 100 % yield = (325 g/351 g) x 100 = 926% 4 Baking soda (NaH ) is used as an antacid Explain why you cannot synthesize baking soda using a double replacement reaction like you did with Ca and Mg(OH Baking soda is soluble in water so you won t be able to do a double replacement precipitation reaction Double replacement precipitation reaction: AB + CD AD + CB + NaH + A = Na + -, D = H See solubility table: all Na + compounds are soluble in water All H (bicarbonate) compounds are soluble in water So when the soluble Na + compound (AB) reacts with the soluble H (bicarbonate) compound (CD), each product will be soluble in water (NaH (AD) and CB 9/4/18 Lab 3, Part 1 + Ca(NO 3 ---> NaNO 3 + CaCO 3 1 Write a balanced chemical equation for this reaction + Ca(NO 3 ---> 2 NaNO 3 + CaCO 3 e Write a net ionic equation for this reaction Use solubility table to determine solubility of each compound The soluble compounds dissociate (break up) into ions The insoluble compounds do not dissociate into ions = soluble so it dissociates into 2 Na + and Ca(NO 3 = soluble so it dissociates into Ca 2+ and NO 3 NaNO 3 = soluble so it dissociates into Na + and NO 3 Ca = insoluble so keep it as a compound Ionic equation: 2 Na Ca 2+ + NO 3 ---> 2 Na NO 3 + Ca Note Na + and NO 3 are on each side of the equation so they cancel out These are spectator ions that do not participate in the reaction Net ionic equation: + Ca > Ca 2 Calculate the masses of and Ca(NO 3 to make 1 g of Ca a molar mass of Ca in g/mole

4 molar mass of Ca = 40 g/mole molar mass of C = 12 g/mole molar mass of O = 16 g/mole So the molar mass of Ca = (16) = 100 g/mole b moles of CO 3 Given mass of Ca = 10 g, calculate moles of CaCO 3 10 g Ca / 100 g/mole = 001 moles CaCO moles Ca x (1 mole of /1 mole of Ca ) = 001 moles CO 3 c mass of in g 001 moles x (106 g /1 mole ) = 106 g CO 3 d moles of Ca(NO 3 ) 2 Given mass of Ca = 10 g, calculate moles of CaCO 3 10 g Ca / 100 g/mole = 001 moles CaCO moles Ca x (1 mole of Ca(NO 3 /1 mole of Ca ) = 001 moles Ca(NO 3 ) 2 e mass of Ca(NO 3 in g 001 moles Ca(NO 3 x (164 g Ca(NO 3 /1 mole Ca(NO 3 ) = 164 g Ca(NO 3 ) 2 8/30/18 Lab 2, Part 3 1 Report the % ethanol before you started your distillation See your Table 4 Ethanol Distillation Data 2 a Report the density of ethanol in g/ml from your 1st distillation b Report the % ethanol from your 1st distillation c Were you able to separate ethanol from water in the first distillation? Give reasons Compare the % ethanol before you started the distillation to the % ethanol from your first distillation If the % ethanol increased, then you were able to separate ethanol from water 3 a Report the density of ethanol in g/ml from your 2nd distillation b Report the % ethanol from your 2nd distillation c Were you able to separate ethanol from water in the second distillation? Give reasons Compare the % ethanol from your first distillation to the % ethanol from your second distillation If the % ethanol increased, then you were able to separate ethanol from water 4 If you did a 3rd distillation, would you expect the % ethanol to increase, decrease, or stay the same? Give reasons based on your answer to Questions 2b and 3b Compare the % ethanol before you started the distillation to the % ethanol from your first distillation to the % ethanol from your second distillation If the % ethanol increased from before distillation to the 1st distillation to the 2nd distillation, then you would expect the % ethanol to increase for a 3rd distillation These successive simple distillations are like doing a fractional distillation

5 If the % ethanol from before distilling to the first to the second distillation did not show a pattern, then you cannot draw a conclusion about the % ethanol for a 3rd distillation 8/28/18 Lab 2, Part 2 1 a Soda brand that you used in Lab 2 Example: Coca-Cola 1 b What is the % sugar according to the soda label from Lab 2? Give a number only; do not include text Example: 12 oz (355 ml) can of Coca-Cola contains 39 g of sugar % sugar = (mass of sugar/volume of soda) x 100 = (39 g sugar/355 ml soda) x 100 = 11% sugar 1 c What is the experimental % sugar from Lab 2? Give a number only; do not include text Measure known volume of soda into an empty beaker Heat soda to evaporate and separate the sugar from the rest of the soda Mass of sugar = mass of soda after heating - mass of beaker % sugar = (mass of sugar/volume of soda) x d What is the experimental % sugar from Lab 1? Give a number only; do not include text Calculate % sugar from mass of sugar per ml of soda from Lab 1 mass of sugar per ml of soda from Lab 1 = density of regular soda - density of diet soda This number tells you the mass of sugar in 1 ml of soda % sugar = (mass of sugar/volume of soda) x 100 Example: 105 g/ml density of regular soda g/ml density of diet soda = 010 g/ml This difference in density means there is 010 g of sugar in 1 ml of regular soda % sugar = (01 g sugar/1 ml of soda) x 100 = 10% sugar 1 e Calculate the % error in the % sugar in soda from Lab 1 % error = (true % sugar - experimental % sugar from Lab 1)/(true % sugar) x f Calculate the % error in the % sugar in soda from Lab 2 % error = (true % sugar - experimental % sugar from Lab 2)/(true % sugar) x g Which experiment, Lab 1 or Lab 2, was the better method to determine the mass of sugar in soda? Give reasons Compare the % error in % sugar for Lab 1 and Lab 2 The experiment that gave the smaller % error is closer (more accurate) to the true % error and is probably the better method 2 You measure the mass of an empty 250 ml beaker to be 1142 g You pour 100 ml of soda into this beaker; the mass of the beaker and soda is 2042 g You heat the soda and beaker in a 100 o C oven for 30 minutes The mass of the beaker and residue is 1234 g

6 a What is the mass of sugar in g? Give a number only; do not include text Mass of sugar = mass of beaker and residue - mass of beaker = 1234 g g = 92 g sugar b What is the % sugar in the soda? % sugar = (mass of sugar/volume of soda) x 100 = (92 g sugar/100 ml soda) x 100 = 92% sugar 8/23/18 Lab 2, Part 1 Problem 1: You find a fluid leak under your car The fluid is a colorless liquid Liquid A vinegar Liquid B NaCl (aq) Liquid C water Liquid D NaOH (aq) Problem 2: You are watching your favorite TV police drama The detective sees a white powdery substance, tastes it, Don't taste, Test! Solid 1 NaHCO 3 Solid 2 CaCO 3 Solid 3 NaCl Solid 4 sugar 8/21/18 Lab 1, Part 2 1 a What brand of soda did you use? b The experimental density of regular soda is g/ml Example: density of regular soda = 105 g/ml c The experimental density of diet soda is g/ml Example: density of diet soda = 095 g/ml d The volume of soda in your soda can or bottle is ml Examples: volume of soda can = 355 ml (12 oz) volume of soda bottle = 591 ml (20 oz) 2 Sugar is the ingredient that causes the difference in density a Calculate the difference in density between regular and diet soda in g/ml Difference in density = density of regular soda - density of diet soda Example: 105 g/ml g/ml = 010 g/ml This difference in density means there is 010 g of sugar in 1 ml of regular soda

7 b Calculate the mass of sugar in g in one can or bottle of soda based on the difference in density Mass of sugar = (density of regular soda - density of diet soda) x volume of soda Examples: In one 355 ml can of soda, the mass of sugar = 010 g/ml x 355 ml = 355 g In one 591 ml bottle of soda, the mass of sugar = 010 g/ml x 591 ml = 591 g c Report the mass of sugar in g in one can or bottle of soda from the label 12 oz (355 ml) can of Coca-Cola contains 39 g of sugar 20 oz (591 ml) bottle of Coca-Cola contains 65 g of sugar d Calculate the % error in the mass of sugar % error = (experimental value - true value)/true value x 100 Examples: 12 oz can of soda % error = (355 g - 39 g)/39 g x 100 = 9% error 20 oz bottle of soda % error = (591 g - 65 g)/65 g x 100 = 9% error 8/16/18 Lab 1, Part 1 1 a Based on class data, which volume measuring device is the most accurate? b Report the % error of each device to support your answer The most accurate volume measuring device shows the lowest % error % error = (experimental value - true value)/true value x 100 For 8-11 lab, 10 ml graduated cylinder data showed a 07% error and is the most accurate 50 ml beaker data showed a 11% error and is the 2nd most accurate 50 ml graduated cylinder data showed a 12% error and is the least accurate For 11-2 lab, 10 ml graduated cylinder data showed a 05% error and is the most accurate 50 ml beaker data showed a 1% error and is the 2nd most accurate 50 ml graduated cylinder data showed a 10% error and is the least accurate If you gave numbers only, you did not earn credit Match the device to the % error Eg, For the 50ml beaker, the %error is 1125%, for the 50ml GC, %error is 12%, and for the 10ml GC, %error is -07% c Based on class data, which volume measuring device is the most precise? d Report the % difference of each device to support your answer The most precise volume measuring device shows the lowest % difference % difference= (high value - low value)/average x 100 For 8-11 lab, 10 ml graduated cylinder data showed a 22% difference and is the most precise 50 ml graduated cylinder data showed a 71% error and is the 2nd most precise 50 ml beaker data showed a 20% error and is the least precise For 11-2 lab, 50 ml beaker and 50 ml graduated cylinder data showed a 0% difference and are the most precise 10 ml graduated cylinder data showed a 2% difference and is the least precise

8 If you gave numbers only, you did not earn credit Match the device to the % difference Eg, The % difference for the 50ml beaker is 20%, for the 50ml GC, % difference is 71%, and for the 10ml GC, % difference is 218% 2 a You calibrate a thermometer using boiling water as your calibration substance You record the temperature of boiling water as 971 degrees C When you use this thermometer to measure temperature, how many degrees should you add or subtract to your temperature reading? = 29 o C b You measure the temperature of water as 255 degrees C using the thermometer from Question 2a What is the actual temperature of the water in degrees C? From Question 2a, the thermometer reads 29 o C low So you want to add 29 o C to the measured temperature reading to give the actual temperature: = 284 o C 8/14/18, Lab 1, Day 1 1 a What does each line in ml on a 10 ml graduated cylinder represent? 01 ml b What is the uncertainty in volume when you use a 10 ml graduated cylinder? +/- 001 ml c You add enough water to the 5 ml mark on your 10 ml graduated cylinder You want to report the volume as ml 500 ml 2 Each line on a 100 ml graduated cylinder represents 1 ml What is the uncertainty in volume when you use a 100 ml graduated cylinder to measure volume? +/- 01 ml