Chapter 3 -Chapter 4

Chapter 3 -Chapter 4 - General Chemistry I Fall 2007 Joann...

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Unformatted text preview: General Chemistry I Fall 2007 Joann S. Monko Chemistry 9th ed. Raymond Chang Quantitative Analysis of a Mixture A substance, (? amount), reacts with a known quantity of another chemical. Stoichiometric ratio is known from equation. ? amount can be determined. 0 A substance, (? composition), can be converted into one or more substances of known composition. Stoichiometric ratio is known from equation. ? composition can be determined. Kotz &Treichel Chemistry & Chemical Reactivity 5th ed. Kotz &Treichel Chemistry & Chemical Reactivity 0 5th ed. Quantitative Analysis of Al The aluminum in a 0.764 g sample of an unknown material was precipitated as aluminum hydroxide, Al(OH)3, which was then converted to aluminum oxide, Al2O3, by heating strongly. If 0.127 g of Al2O3 is obtained from the 0.764 g sample, what is the mass percent of aluminum in the sample? (Al + stuff) Al(OH)3 Al2O3 0.764 g 0.127 g g Al2O3 moles Al2O3 moles Al g Al De te rm ining Fo rm ula s In chemical analysis we determine the % by weight of each element in a given amount of pure compound and derive the EMP IR IC AL or S IMP LES T Formula. P R O BLEM: A c o m p o und o f B a nd H is 8 1 .1 0 % B. Wh a t is its e m p iric a l fo rm ula ? Kotz &Treichel Chemistry & Chemical Reactivity 5th ed. Empirical Formulas 1. Convert weight % to mass Convert mass to moles % A g A x mole A % B g B y mole B 2. Find mole ratio 3. Ratio gives formula x mole A y mole B A#B# A c o m p o und o f B a nd H is 8 1 .1 0 % B. Wh a t is its e m p iric a l fo rm ula ? Because it contains only B and H, it must contain 18.90% H. In 100.0 g of the compound there are 81.10 g of B and 18.90 g of H. Calculate the number of moles of each constituent. Kotz &Treichel Chemistry & Chemical Reactivity 5th ed. A c o m p o und o f B a nd H is 8 1 .1 0 % B. Wh a t is its e m p iric a l fo rm ula ? Calculate the number of moles of each element in 100.0 g of sample. 1 mol 81.10 g B = 7.502 mol B 10.81 g 1 mol 18.90 g H = 18.75 mol H 1.008 g Kotz &Treichel Chemistry & Chemical Reactivity 5th ed. A compound of B and H is 81.10% B. What is its empirical formula? Take the ratio of moles of B and H. Always divide by the smaller number. 18.75 mol H 2.499 mol H 2.5 mol H = = 7.502 mol B 1.000 mol B 1.0 mol B **need a whole number ratio: = 5 mol H to 2 mol B EMP IR IC AL FO R MULA = B 2 H 5 Kotz &Treichel Chemistry & Chemical Reactivity 5th ed. Is the molecular formula B H , B H , B2H6 Its empirical formula is B H . What is its Molecular Formula ? Kotz &Treichel Chemistry & Chemical Reactivity 5th ed. B H is one example of this class 2 compounds. of 5 4 10 A compound of B and H is 81.10% B. Its empirical formula is B2H5. What is its molecular formula? Do an EXPERIMENT to find the MOLAR MASS. The experiment gives 53.3 g/mol Compare with the mass of B H = 26.66 g/unit 53.3 g/mol 2 units of B2H5 = 26.66 g/unit of B2H5 1 mol Molecular formula = B4H10 Kotz &Treichel Chemistry & Chemical Reactivity 5th ed. Clicker Question Cumene is a compound that contains only carbon and hydrogen. It is 89.94% carbon and has a molar mass of 120.2 g/ mol. What is the empirical formula for Cumene? a. CH3 b. C2H6 c. C3H4 d. CH e. C4H3 Formula from Mass Data One reagent will be used completely Limiting Reagent One reagent will be in excess Same as before! 1. 2. 3. 4. Determine masses that reacted. Convert to moles. Find the mole ratio. Mole ratio determines formula. Determine the formula of a compound of Sn and I using the following data. Reaction of excess Sn and I . Mass of iodine (I ) used = 1.947 g 2 Kotz &Treichel Chemistry & Chemical Reactivity 5th ed. Tin and Iodine Compound Find the mass of Sn that combined with 1.947 g I . __ 1 mol -3 mol Sn 0.455 g Sn = 3.83 x 10 2 118.7 g Tin and Iodine Compound Now find the number of moles of I2 . Mass of I used was 1.947 g. 1 mol 1.947 g I2 = 7.671 x 10 -3 mol I2 253.81 g How many mol of iodine atoms? -3 mol I 2 mol I atoms 7.671 x 10 2 1 mol I2 Kotz &Treichel Chemistry & Chemical Reactivity 5th ed. 2 = 1.534 x 10-2 mol I atoms Tin and Iodine Compound Next, find the ratio of number of moles of moles of I and Sn that combined. 4.01 mol I = -3 mol Sn 1.00 mol Sn 3.83 x 10 1.534 x 10 -2 mol I Empirical formula is SnI4 Kotz &Treichel Chemistry & Chemical Reactivity 5th ed. Formula from Mass Data Example #2 Phosphorus, P, is combined with fluorine, F , to give a gaseous Kotz &Treichel Chemistry & Chemical Reactivity 5th ed. Concentration The amount of solute in a solution is given by its concentration. concentration Molarity moles solute (M) = liters of solution Concentration (M) = [ ...] The concentration of a calcium chloride solution could be written as: 2.5 M CaCl2 or 2.5 mol CaCl2 L Kotz &Treichel Chemistry & Chemical Reactivity 5th ed. CuCl Solution Ion Concentrations 2 2 CuCl (aq) --> Cu2+(aq) + 2 Cl-(aq) If [CuCl2] = 0.30 M, Then [Cu2+] = 0.30 M [Cl-] = 2 x 0.30 M = 0.60 M Kotz &Treichel Chemistry & Chemical Reactivity 5th ed. Chemistry, 8th ed. Chang Making Solutions: Molarity = moles of solute Liters of solution 1.0 L of water was used to make 1.0 L of solution. Water is left over?? Kotz &Treichel Chemistry & Chemical Reactivity 5th ed. PROBLEM: Dissolve 5.00 g of NiCl26 H2O in enough water to make 250. mL of solution. Calculate molarity. Step 1: Calculate moles of 1 NiCl26H2O mol 5.00 g = 0.0210 mol 237.7 g Step 2: Calculate molarity 0.0210 mol = 0.0841 M 0.250 L [NiCl26 H2O ] = 0.0841 M Kotz &Treichel Chemistry & Chemical Reactivity 5th ed. Clicker Question Prepare a 500.0 mL solution with 7.82 g of NaCl. What is the molarity of the solution? (Molar Mass NaCl = 58.44 g/mol) US ING MO LAR IT Y What mass of oxalic acid, H2C2O4, is required to make 250. mL of a 0.0500 M solution? Because Conc (M) = moles volume this means = mol v m o le s = C V Kotz &Treichel Chemistry & Chemical Reactivity 5th ed. US ING MO LAR IT Y What mass of oxalic acid, H2C2O4, is required to make 250. mL of a 0.0500 M solution? m o le s = C V Step 1: Calculate moles of acid required. (0.0500 mol)(0.250 L) = 0.0125 mol L Step 2: Calculate mass of acid required. (0.0125 mol )(90.00 g ) = 1.13 g mol Kotz &Treichel Chemistry & Chemical Reactivity 5th ed. Clicker Question How would you prepare 1.50 L of 0.125 M KNO3? (Molar Mass of KNO3 = 101.11 g/mol) P re p a ring S o lutio ns Weigh out a solid solute and dissolve in a given quantity of solvent. Dilute a concentrated solution to give one that is less concentrated. Kotz &Treichel Chemistry & Chemical Reactivity 5th ed. P R O BLEM: Y o u h a ve 5 0 .0 m L o f 3 .0 M Na O H a nd y o u wa nt 0 .5 0 M Na O H. Wh a t d o y o u d o ? Add water to the 3.0 M solution to lower its concentration to 0.50 M H2 O s o lutio n! Kotz &Treichel Chemistry & Chemical Reactivity 5th ed. -Dilute th e 3.0 M NaOH Concentrated 0.50 M NaOH Dilute P R O BLEM: Y o u h a ve 5 0 .0 m L o f 3 .0 M Na O H a nd y o u wa nt 0 .5 0 M Na O H. Wh a t d o y o u d o ? How much water is added? The important point is that ---> m o le s o f Na O H in O R IG INAL s o lutio n = m o le s o f Na O H in FINAL s o lutio n Kotz &Treichel Chemistry & Chemical Reactivity 5th ed. P R O BLEM: Y o u h a ve 5 0 .0 m L o f 3 .0 M Na O H a nd yo u wa nt 0 .5 0 M Na O H. Wh a t d o yo u d o ? Amount of NaOH in original solution = CV = (3.0 mol/L)(0.050 L) = 0.15 mol NaOH Amount of NaOH in final solution must also = 0.15 mol NaOH Volume of final solution = (0.15 mol NaOH)(1 L/0.50 mol) = 0.30 L or Kotz &Treichel Chemistry & Chemical Reactivity 5th ed. 300 mL P R O BLEM: Y o u h a ve 5 0 .0 m L o f 3 .0 M Na O H a nd yo u wa nt 0 .5 0 M Na O H. Wh a t d o y o u d o ? Conclusion: H2 O add enough water to 50.0 mL of 3.0 M NaOH to make 300 mL of 0.50 3.0 M NaOH Concentrated 0.50 M NaOH Dilute M NaOH. Kotz &Treichel Chemistry & Chemical Reactivity 5th ed. P re p a ring S o lutio ns b y Dilutio n A shortcut Cinitial Vinitial = Cfinal Vfinal Or... C1V1 = C2V2 P R O BLEM: Y o u h a ve 5 0 .0 m L o f 3 .0 0 M Na O H a nd yo u wa nt 0 .5 0 0 M Na O H. Wh a t d o y o u d o ? C1 = 3.00 M V1 = 50.0 mL C2 = 0.500 M V2 = ? (3.00 M) (50.0 mL) = (0.500 M) (V2) 300. mL = V2 Kotz &Treichel Chemistry & Chemical Reactivity 5th ed. Dilutions Cinitial Vinitial = Cfinal C1V1 = C2V2 VfinalO BLEM: Y o u wa nt to m a ke 1 .5 L o f 1 .3 M PR C1 = 5.0 M V1 = ? C2 = 1.3 M V2 = 1.5 L Na O H fro m a s to c k s o lutio n o f 5 .0 M Na O H. Wh a t d o yo u d o ? (5.0 M) ( ? L V1) = (1.3 M) (1.5 L) 0.39 L = V1 Put 0.39 L (or 390 mL) of the 5.0 M stock solution into a 1.5 L volumetric flask and dilute with water to the mark on the neck. Clicker Question A student prepared 1.5 L of a 0.080 M NaCl solution. It took 75 mL of stock solution. What was the concentration of the stock solution? ...
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This note was uploaded on 04/08/2008 for the course CHEM 100 taught by Professor Monko during the Winter '08 term at Kutztown.

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