COPYRIGHTED MATERIAL. Gravimetrics. Key Terms. Key Concepts. Words that can be used as topics in essays: accuracy atomic theory

Size: px

Start display at page:

Download "COPYRIGHTED MATERIAL. Gravimetrics. Key Terms. Key Concepts. Words that can be used as topics in essays: accuracy atomic theory"

Jewel Wendy Wheeler
6 years ago
Views:

1 8684-X Ch02.F 2/9/01 7:49 AM Page 23 Gravimetris Key Terms Words that an be used as topis in essays: auray atomi theory density empirial formula extensive property frational rystallization heterogeneous homogeneous intensive property isotopes law of onservation of mass and energy Key Conepts Equations and relationships that you need to know: 1 nm = 1 l0 9 m = 10 Å 1 m 3 = 1 ml F = 1.8( C) + 32 K = C density= mass volume Avogadro s number = = 1 mole number of moles = mass moleular weight moleular weight of ideal gas = density molar volume atual yield % yield = theoretial yield 100% mass of element in ompound % omposition = mass of ompound 100% observed value - expeted value % error = expeted value 100% law of definite proportions (law of onstant omposition) limiting reatant mixture moleular formula perentage yield preision random error systemati error theoretial yield unertainty COPYRIGHTED MATERIAL 23

2 8684-X Ch02.F 2/9/01 7:49 AM Page 24 Part II: Speifi Topis Measurement Terms SI (International System) Multipliers Multiple Prefix Symbol tera T 10 9 giga G 10 6 mega M 10 3 kilo k 10 2 heto h 10 1 deka da 10 1 dei d 10 2 enti 10 3 milli m 10 6 miro µ 10 9 nano n pio p femto f atto a SI Base Units meter kilogram seond ampere kelvin mole andela m kg s (se) A K mol d SI Derived Units bequerel Bq 1 disintegration/se oulomb C A se farad F A se/v = A 2 se 4 kg 1 m 2 gray Gy J/kg henry H Wb/A hertz Hz se 1 (yle/se) joule J kg m 2 se 2 = 10 7 ergs 24

3 8684-X Ch02.F 2/9/01 7:49 AM Page 25 Gravimetris SI Derived Units (ontinued) lumen lm d sr lux lx lm/m 2 newton N kg m se 2 pasal Pa N/m 2 = kg m 1 se 2 ohm Ω V/A = kg m 2 se 3 A 2 siemens S Ω 1 = A V 1 tesla T Wb/m 2 volt V J A 1 se 1 = kg m 2 se 3 A 1 watt W J/se = kg m 2 se 3 weber Wb V se Non-SI Units angstrom Å 10 8 m atmosphere atm 101,325 N/m 2 or 760 mm Hg or kpa or 760 torr bar bar 10 5 N/m 2 alorie al J dyne dyn 10 5 N= 1 g m se 2 erg erg 10 7 J inh in 2.54 m millimeter of mm Hg N/m 2 merury pound lb kg torr torr N/m 2 = al m 1 25

4 8684-X Ch02.F 2/9/01 7:49 AM Page 26 Part II: Speifi Topis Samples: Multiple-Choie Questions 1. A popular Bourbon whiskey is listed as being 92 Proof. The liquor industry defines Proof as being twie the volume perentage of alohol in a blend. Ethanol (drinking alohol) has the strutural formula CH 3 CH 2 OH (MW = 46 g/mol). The density of ethanol is 0.79 g/ml. How many liters of whiskey must one have in order to have 50. moles of arbon? A liters B. 1.6 liters C. 3.2 liters D. 4.0 liters E. 6.4 liters Answer: C Step 1: Write down an equals sign (=). Step 2: To the right of the equals sign, write down the units you want the answer to be in. Examination of the problem reveals that you want your answers in liters of whiskey, so you have = liters of whiskey Step 3: Begin the problem with the item you are limited to. In this ase, it is 50. moles of arbon no more, no less. Plae the 50. moles of arbon over moles of arbon 1 = liters of whiskey Step 4: Get rid of the units moles of arbon by plaing them in the denominator of the next fator. What do you know about moles of arbon? There are 2 moles of arbon in eah mole of ethanol. 50. moles of arbon mole ethanol moles arbon = liters of whiskey Step 5: Continue in this fashion, getting rid of unwanted units until you are left in the units you desire. At that point, stop and do the alulations. 50. moles arbon 1 mole ethanol 46 grams ethanol 1 2 moles arbon 1 mole ethanol 1 ml ethanol 200 ml whiskey 1 L whiskey 079. g ethanol 92 ml ethanol 3 = 32. L whiskey 10 ml whiskey 26

5 8684-X Ch02.F 2/9/01 7:49 AM Page 27 Gravimetris 2. A sample of a pure ompound was found to ontain grams of arbon, grams of hydrogen, and grams of hlorine. What is the empirial formula for the ompound? A. CHCl 3 B. CH 2 Cl C. CH 2 Cl 2 D. CH 3 Cl E. C 2 H 2 Cl 4 Answer: C First, hange the grams of eah element to moles. You end up with mole of arbon, mole of hydrogen, and mole of hlorine. This represents a 1 arbon : 2 hydrogen : 2 hlorine molar ratio. 3. Balane the following equation using the lowest possible whole-number oeffiients: NH 3+ CuO " Cu + N 2+ H 2O The sum of the oeffiients is A. 9 B. 10 C. 11 D. 12 E. 13 Answer: D Step 1: Begin balaning equations by trying suitable oeffiients that will give the same number of atoms of eah element on both sides of the equation. Remember to hange oeffiients, not subsripts. 2NH 3" 1N 2 Step 2: Look for elements that appear only one on eah side of the equation and with equal numbers of atoms on eah side. The formulas ontaining these elements must have the same oeffiients. CuO " Cu 27

6 8684-X Ch02.F 2/9/01 7:49 AM Page 28 Part II: Speifi Topis Step 3: Look for elements that appear only one on eah side of the equation but in unequal numbers of atoms. Balane these elements. 2NH 3" 3H 2O Step 4: Balane elements that appear in two or more formulas on the same side of the equation. Step 5: Double hek your balaned equation and be sure the oeffiients are the lowest possible whole numbers. 2NH 3+ 3CuO " 3Cu + N 2+ 3H 2O = 12 (Be sure to inlude the unwritten 1 that is in front of N 2.) 4. When mole of BaCl 2(aq) is mixed with mole of K 3 AsO 4(aq), what is the maximum number of moles of solid Ba 3 (AsO 4 ) 2 that ould be formed? A mole B mole C mole D mole E mole Answer: A Begin by writing a balaned equation: 3BaCl 2 + 2K 5 AsO 4 " Ba 3( AsO 4) 2 + 6KCl ( aq) ( aq) ( s) ( aq) You may want to write the net-ioni equation: Ba + 2AsO " Ba ^AsO h ( aq ) 4 ( aq ) s ] g Next, realize that this problem is a limiting-reatant problem. That is, one of the two reatants will run out first, and when that happens, the reation will stop. You need to determine whih one of the reatants will run out first. To do this, you need to be able to ompare them on a 1:1 basis. But their oeffiients are different, so you need to relate both reatants to a ommon produt, say Ba 3 (AsO 4 ) 2. Set the problem up like this: mole BaC12 1 mole Ba 3( AsO 4 ) moles BaC1 = mole Ba ( AsO ) mole K 3 AsO 4 1 mole Ba 3( AsO 4) moles K AsO = mole Ba ( AsO )

7 8684-X Ch02.F 2/9/01 7:49 AM Page 29 Gravimetris Given the two amounts of starting materials, you disover that you an make a maximum of mole of Ba 3 (AsO 4 ) 2, beause at that point you will have exhausted your supply of K 3 AsO A test tube ontaining CaCO 3 is heated until all of the ompound deomposes. If the test tube plus alium arbonate originally weighed grams and the loss of mass during the experiment was grams, what was the mass of the empty test tube? A g B g C g D g E g Answer: A Begin by writing a balaned equation. Remember that all Group II arbonates deompose to yield the metalli oxide plus arbon dioxide gas. CaCO 3^sh" CaO^ sh+ CO 2_ gi Aording to your balaned equation, any loss of mass during the experiment would have to have ome from the arbon dioxide gas leaving the test tube grams of CO 2 gas orrespond to mole. Beause all of the alium arbonate deomposed, and the alium arbonate and arbon dioxide gas are in a 1:1 molar ratio, you must originally have had mole of alium arbonate, or grams. The alium arbonate and test tube weighed grams, so if you get rid of the alium arbonate, you are left with grams for the empty test tube grams of oxygen gas, 32.0 grams of methane gas, and 32.0 grams of sulfur dioxide gas are mixed. What is the mole fration of the oxygen gas? A B C D E Answer: B 29

8 8684-X Ch02.F 2/9/01 7:49 AM Page 30 Part II: Speifi Topis First hange all the grams to moles: 32.0 grams of O 2 = 1.00 mole 32.0 grams of CH 4 = 2.00 moles 32.0 grams of SO 2 = mole Mole fration of oxygen gas: Mole fration of oxygen gas: mole O total moles = mole fration 7. Element X is found in two forms: 90.0% is an isotope that has a mass of 20.0, and 10.0% is an isotope that has a mass of What is the atomi mass of element X? A B C D E Answer: B To solve this problem, multiply the perentage of eah isotope by its atomi mass and add those produts. ( ) + ( ) = 20.2 atomi mass of element X 8. What is the formula of a ompound formed by ombining 50. grams of element X (atomi weight = 100.) and 32 grams of oxygen gas? A. XO 2 B. XO 4 C. X 4 O D. X 2 O E. XO Answer: B Aording to the information given, you have 0.50 mole of element X (50. g/100. g mole 1 = 0.50 mole). For the oxygen, remember that you will use 16 g/mole for the atomi weight, giving you 2.0 moles of oxygen atoms. A 0.50:2.0 molar ratio is the same as a 1:4 molar ratio, so the answer is XO 4. 30

9 8684-X Ch02.F 2/9/01 7:49 AM Page 31 Gravimetris 9. An oxide is known to have the formula X 2 O 7 and to ontain 76.8% X by mass. Whih of the following would you use to determine the atomi mass of X? A. B. C. D. E m 2 7 m m 7 m m 7 m m 2 7 m m 2 7 m Answer: C From the information provided, you know that the oxide X 2 O 7 ontains 76.8% X and 23.2% oxygen by weight. If you had g of the oxide, you would have 76.8 g of X and 23.2 g of O (or 23.2/16.0 moles of O atoms). Beause for eah mole of O in the oxide you have 2/7 mole of X, you have in effet (23.2/16.0) 2/7 mole of X. Sine the units of atomi mass are g/mole, the setup is: m 7 m Finding the solution is unneessary for seleting an answer; however, here is the solution for your information gX g/ mol go mole X g O/ mole O 7 2 = 186 e o e mole O o (The element is rhenium.) 31

10 8684-X Ch02.F 2/9/01 7:49 AM Page 32 Part II: Speifi Topis 10. A freshman hemist analyzed a sample of opper(ii) sulfate pentahydrate for water of hydration by weighing the hydrate, heating it to onvert it to anhydrous opper(ii) sulfate, and then weighing the anhydride. The % H 2 O was determined to be 30%. The theoretial value is 33%. Whih of the following hoies is definitely NOT the ause of the error? A. After the student weighed the hydrate, a piee of rust fell from the tongs into the ruible. B. Moisture driven from the hydrate ondensed on the inside of the ruible over before the student weighed the anhydride. C. All the weighings were made on a balane that was high by 10%. D. The original sample ontained some anhydrous opper(ii) sulfate. E. The original sample was wet. Answer: E 30% H 2 O in the hydrate sample represents mass of hydrate - mass of anhydride mass of hydrate 100% In a problem like this, I like to make up some easy fititious numbers that I an use to fit the senarios and see how the various hanges affet the final outome. Let s say the mass of the hydrate is 10 g and the mass of the anhydride is 7 g. This would translate as 10 g- 7 g 10 g 100% = 30% H2 O In examining hoie A, the original mass of the hydrate would not hange; however, beause rust will not evaporate, the final mass of the anhydride would be higher than expeted let s say 8 g. Substituting this value into the formula for % water would give 10 g- 8 g 10 g 100% = 20% H2 O whih is less than the theoretial value of 30%. This is in the diretion of the student s experimental results, and sine we are looking for the hoie that is NOT the ause, we an rule out A as an answer. In hoie B, the mass of the hydrate would not hange, but the mass of the anhydride would be higher than it should be. Let s estimate the anhydride at 8 g again. 10 g- 8 g 10 g 100% = 20% H2 O 32

11 8684-X Ch02.F 2/9/01 7:50 AM Page 33 Gravimetris In hoie C, beause all masses are being measured on a onsistently wrong balane, the faultiness does not matter in the final answer g g g 100% = 30% H2 O In hoie D, the original mass of the hydrate would remain unhanged. However, the mass of the anhydride would be higher than expeted beause the sample would lose less water than if it had been a pure hydrate. This fits the senario of 10 g- 8 g 10 g 100% = 20% H2 O with the error being onsistent with the diretion of the student s results. Therefore, D is not the orret answer. In hoie E, the original sample is wet. The freshman hemist weighed out 10 g of the hydrate, but more weight is lost in the heating proess than expeted, making the final mass of the anhydride lower than expeted, say 6 g. Using the equation for % H 2 O shows 10 g- 6 g 10 g 100% = 40% H2 O whih is higher than the theoretial value of 30% and in line with the reasoning that this ould NOT have aused the error. 11. When 100 grams of butane gas (C 4 H 10, MW = 58.14) is burned in exess oxygen gas, the theoretial yield of H 2 O is: A B. C / % D. E Answer: E Begin with a balaned equation: C4H / 2 O2 " 4CO2 + 5H2O Next, set up the equation in fator-label fashion: 100 gc H mole C 4 H10 5 mole H 2O gH2O gc H 1mole C H 1mole H O = gh 2 O

12 8684-X Ch02.F 2/9/01 7:50 AM Page 34 Part II: Speifi Topis 12. Element Q ours in ompounds X, Y, and Z. The mass of element Q in 1 mole of eah ompound is as follows: Compound Grams of Q in Compound X Y Z Element Q is most likely: A. N B. O C. F D. Ir E. Cs Answer: C All of the numbers are multiples of (fluorine). Use the law of multiple proportions. 13. Whih one of the following represents an intensive property? A. temperature B. mass C. volume D. length E. heat apaity Answer: A The measured value of an intensive property does NOT depend on how muh matter is being onsidered. The formula for heat apaity C is C = m C p, where m = mass and C p = speifi heat. 34

13 8684-X Ch02.F 2/9/01 7:50 AM Page 35 Gravimetris 14. Whih of the following would have an answer with three signifiant figures? A B. ( )( ) / ( ) C. ( ) / ( ) D. ( ) / ( ) E. ( ) / (3.2 l0 1 ) Answer: D ( ) = = = - 6 = 3 sf.. ` j

14 8684-X Ch02.F 2/9/01 7:50 AM Page 36 Part II: Speifi Topis Samples: Free-Response Questions 1. A student performed the following experiment in the laboratory: She suspended a lean piee of silver metal in an evauated test tube. The empty test tube weighed grams. The silver weighed grams. Next, she introdued a stream of hlorine gas into the test tube and allowed it to reat with the silver. After a few minutes, a white ompound was found to have formed on the silver strip, oating it uniformly. She then opened the apparatus, weighed the oated strip, and found it to weigh grams. Finally, she washed the oated strip with distilled water, removing all of the white ompound from the silver strip, and then dried the ompound and the strip and reweighed. She disovered that the silver strip weighed grams. (a) (b) () Show how she would determine (1) the number of moles of hlorine gas that reated (2) the number of moles of silver that reated Show how she ould determine the simplest formula for the silver hloride. Show how her results would have been affeted if (1) some of the white ompound had been washed down the sink before it was dried and reweighed (2) the silver strip was not thoroughly dried when it was reweighed Answer Step 1: Do a restatement of the general experiment. In this ase, I would draw a sketh of the apparatus before and after the reation, labeling everything. This will get rid of all the words and enable you to visualize the experiment. Cl 2 Initial: Empty test tube = g Ag strip = g Final: Coated strip = g Ag strip = g Ag 36

15 8684-X Ch02.F 2/9/01 7:50 AM Page 37 Gravimetris Step 2: Write a balaned hemial equation that desribes the reation. silver + hlorine gas yields silver hloride. 2Ag^ sh + Cl 2 _ gi " 2AgCl^sh Step 3: Begin to answer the questions asked. Remember to give a brief restatement for eah question, to label eah speifi question so that the grader knows whih question you are answering, and to underline the onlusion(s) where neessary. 1. (a) (1) Restatement: Number of moles of hlorine gas that reated. mass of hlorine that reated = (mass of silver strip + ompound) mass of original silver strip g g = g of hlorine atoms moles of hlorine atoms that reated = mass of hlorine atoms/atomi mass of hlorine g / g mole 1 = mole of hlorine atoms (a) (2) Restatement: Moles of silver that reated. (b) _ mass of original silver strip - mass of dry strip after washingi atomi mass of silver g g = g g / g mole 1 = mole of silver atoms Restatement: Empirial formula for silver hloride. empirial formula = moles of silver atoms / moles of hlorine atoms mole of silver mole of hlorine = " AgCl () (1) Restatement: Effet on the empirial formula if some of the white ompound had been washed down the sink before the oated strip was dried and reweighed. The white produt was silver hloride. Had she lost some before she weighed it, the mass of silver hloride would have been less than what it should have been. This would have made the number of grams of hlorine appear too low, whih in turn would have made the number of moles of hlorine appear too low. Thus, in the ratio of moles of silver to moles of hlorine, the denominator would have been lower than expeted and the ratio would have been larger. Beause the mass of the ompound does not enter into the alulations for the moles of silver that reated, the moles of silver would not have been affeted. 37

16 8684-X Ch02.F 2/9/01 7:50 AM Page 38 Part II: Speifi Topis () (2) Restatement: Effet on the empirial formula if the silver strip had not been dried thoroughly after being washed free of the silver hloride. Beause the strip has been washed free of the ompound (silver hloride), you assume that any silver missing from the strip went into the making of the silver hloride. If the strip had been wet when you weighed it, you would have been led to think that the strip was heavier than expeted, and therefore that less silver had gone into the making of the silver hloride. Thinking that less silver had been involved in the reation, you would have alulated fewer moles of silver. The alulated ratio of moles of silver to moles of hlorine would have been less than expeted. 2. Three ompounds, D, E, and F, all ontain element G. The perent (by weight) of element G in eah of the ompounds was determined by analysis. The experimental data are presented in the following hart. Compound % by Weight of Element G Moleular Weight D E F (a) Determine the mass of element G ontained in 1.00 mole of eah of ompounds D, E, and F. (b) What is the most likely value for the atomi weight of element G? () Answer Compound F ontains arbon, hydrogen, and element G. When 2.19 g of ompound F is ompletely burned in oxygen gas, 3.88 g of arbon dioxide gas and 0.80 g of water are produed. What is the most likely formula for ompound F? 2. Given: Compounds D, E, and F with % (by weight of element G) and their respetive MW s (a) Restatement: Calulate the mass of element G in 1.00 mole of ompounds D, E, and F g/mole = 71.0 g G/mole D g/mole = 107 g G/mole E g/mole = 35.5 g G/mole F (b) Restatement: Most likely atomi weight of G. Aording to the law of multiple proportions, the ratios of the mass of element G to the masses of ompounds D, E, and F must be small, whole numbers. The largest ommon denominator of 71.0, 106.5, and 35.5 is 35.5, so our best estimate is that the atomi weight of G is 35.5 (hlorine). 38

17 8684-X Ch02.F 2/9/01 7:50 AM Page 39 Gravimetris () Restatement: Compound F = C x H y G z or C x H y Cl z? C xh ycl z+ O 2_ gi" CO 2_ gi+ H 2O^, h + Cl 2_ gi g+? g" g g+? gcl2_ gi moles of arbon = moles of CO gco g : mole - 1 = moles of hydrogen = 2 moles of H 2 O 080. gh2o g : mole - 1 = (2.19 grams of F)(1 mole F / 74.5 g F) = mole of ompound F This means that eah mole of F ontains 3 moles of C ( / ) and 3 moles of H (0.088), or 39 grams of CH. This leaves = 36 grams, orresponding to 1 mole of element G (Cl). Therefore, the empirial formula is C 3 H 3 Cl. 39

18 8684-X Ch02.F 2/9/01 7:50 AM Page 40

General Equilibrium. What happens to cause a reaction to come to equilibrium?

General Equilibrium. What happens to cause a reaction to come to equilibrium? General Equilibrium Chemial Equilibrium Most hemial reations that are enountered are reversible. In other words, they go fairly easily in either the forward or reverse diretions. The thing to remember