Homework #2 Chapter 3 Stoichiometry. Find the atomic weight on the periodic table that matches amu. Ti titanium

Transcription

1 Homework #2 Chapter 3 Stoichiometry 23. Need to find averae ar mass of X (M X M X = (Fracrion of 46 X (M46 X + (Fracrion of 47 X (M47 X + (Fracrion of 48 X (M48 X + (Fracrion of 49 X (M49 X + (Fracrion of 50 X (M50 X M X = (0.0800( amu + (0.0730( amu + (0.7380( amu + (0.0550( amu + (0.0540( amu = amu Find the atomic weiht on the periodic table that matches amu. Ti titanium 26. Need to find the mass present of 151 Eu and 153 Eu Know M Eu = amu fraction of 151 Eu = x m151 = amu Eu fraction of 153 Eu = 1-x m153 = amu Eu M Eu = (Fracrion of Eu (M151 Eu + (Fracrion of Eu (M Eu amu = x( amu + (1 x( amu 0.96 amu = ( amux x = 0.48 Therefore 48% 151 Eu and 52% 153 Eu 27. Need to find the ar mass of 185 Ru(M185 Ru M Ru = (Fracrion of Ru (M187 Ru + (Fracrion of Ru (M Ru amu = (0.6260( amu + ( M185 Ru M185 = amu Ru Need to find the averae ar mass of X (M X M X = (Fracrion of a X (M a X + (Fracrion of b X (M b X + (Fracrion of c X (M X + (Fracrion of X d (M X d M X = (0.0140( amu + (0.2410( amu + (0.2210( amu + (0.5240( amu = amu Find the atomic weiht on the periodic table that matches amu. Lead (Pb c 1

2 34. Mass of Total Atoms in Moles of Sample Molecules in Sample Sample Sample a 4.24 C 6H ec atoms b H 2O ec atoms c ec CO atoms d ec atoms in CH 3OH Note: Grey information is iven a Moles of Sample 4.24 C 6 H 6 ( 1 C 6H C 6 H 6 = C 6 H 6 Molecules in Sample C 6 H 6 ( ecules C 6 H 6 = ecules C 1 C 6 H 6 H 6 6 Atoms in Sample 12 atoms ecules C 6 H 6 ( = ecule C 6 H 1023 atoms 6 b Mass of Sample H 2 O ( H 2O = 4.04 H 1 H 2 O 2O Molecules in Sample H 2 O ( ecules H 2 O = ecules H 1 H 2 O 2 O Total Atoms in Sample 3 atoms ecules H 2 O ( = ecule H 2 O 1023 atoms c Moles of Sample 1 CO ecules CO 2 ( = CO ecule CO 2 2 Mass of Sample CO 2 ( CO 2 = 1.98 CO 1 CO 2 2 Total Atoms in Sample 3 atoms ecules CO 2 ( = ecule CO 1022 atoms 2 d Molecules in Sample atoms CH 3 OH ( 1 ecule CH 3OH 6 atoms = ecules CH 3 OH Moles of Sample 1 CH 3 OH ecules CH 3 OH ( ecule CH 3 OH = CH 3 OH Mass of Sample CH 3 OH ( CH 3OH = CH 1 CH 3 OH 3OH 35. a 20.0 m C 8 H 10 N 4 O 2 ( m ( 1 C 8H10N4O C8H10N4O2 = C 8 H 10 N 4 O 2 b C2H ecules C 2 H 5 OH ( 5 OH = C ecule CH3OH 2H 5 OH c 1.50 CO 2 ( 1 CO CO2 = CO 2 2

3 36. a 5.00 C 2 H 5 O 2 N ( 1 C 2H 5 O2N C2H 5 O2N ( 1 N 1 C2H 5 O2N ( atoms 1 N = N atoms b 5.00 M 3 N 2 ( 1 M 3N2 2 N ( M3N2 1 M3N2 ( atoms = N atoms 1 N c 5.00 Ca(NO 3 2 ( 1 Ca(NO 3 2 ( 2 N Ca(NO3 2 1 Ca(NO3 2 ( atoms 1 N = N atoms d 5.00 N 2 O 4 ( 1 N 2O4 2 N ( N2O4 1 N2O4 ( atoms = N atoms 1 N 38. a M C14 H 18 N 2 O 5 = b 10.0 C 14 H 18 N 2 O 5 ( 1 C 14H18N2O 5 = C C14H18N2O 5 14 H 18 N 2 O 5 c 1.56 C 14 H 18 N 2 O 5 ( C 14H18N2O 5 1 C14H18N2O 5 = 459 C 14 H 18 N 2 O 5 d 5.0 m C 14 H 18 N 2 O 5 ( m ( 1 C 14H18N2O C14H18N2O 5 ( ecules 1 C14H18N2O 5 = ecules e 1.2 C 14 H 18 N 2 O 5 ( 1 C 14H18N2O 5 2 N ( ( atoms C14H18N2O 5 1 C14H18N2O 5 = atoms 1 N f ecules C 14 H 18 N 2 O 5 ( 1 C 14H18N2O ecules ( C 14H18N2O 5 1 C14H18N2O 5 = C 14 H 18 N 2 O 5 1 ecules C 14 H 18 N 2 O 5 ( 1 C 14H18N2O ecules ( C 14H18N2O 5 1 C14H18N2O 5 = C 14 H 18 N 2 O Assume 1 of C 20H 29FO 3 1 C 20 H 29 FO 3 ( C 20H29FO3 = C 1 20H C20H29FO3 29 FO 3 Mass % C 20 C C 1 C 20 H 29 FO 3 ( ( = C 1 C 20 H 29 FO 3 1 C mass % C = ( 100% = 71.38% Mass % H 29 H 1.01 H 1 H 1 C 20 H 29 FO 3 ( ( = 29.3 H 1 C 20 H 29 FO mass % H = ( 100% = 8.71% Mass % F 1 F F 1 C 20 H 29 FO 3 ( ( = F 1 C 20 H 29 FO 3 1 F mass % F = ( 100% = 5.647% Mass % O 3 O O 1 C 20 H 29 FO 3 ( ( = O 1 C 20 H 29 FO 3 1 O mass % O = ( 100% = 14.26%

4 44. C 8H 10N 4O 2 Assume 1 C 8H 10N 4O 2 1 C 8 H 10 N 4 O 2 ( C 8H 10 N 4 O 2 = C 1 C 8 H 10 N 4 O 8 H 10 N 4 O C C 1 C 1 C 8 H 10 N 4 O 2 ( ( = C 1 C 8 H 10 N 4 O mass % C = ( 100% = 49.47% C 12H 22O 11 Assume 1 C 12H 22O 11 1 C 12 H 22 O 11 ( C 12H 22 O 11 = C 1 C 12 H 22 O 12 H 22 O C C 1 C 12 H 22 O 11 ( ( = C 1 C 12 H 22 O 11 1 C mass % C = ( 100% = 42.10% C 2H 5OH Assume 1 C 2H 5OH 1 C 2 H 5 OH ( C 2H 5 OH = C 1 C 2 H 5 OH 2H 5 OH 1 C 2 H 5 OH ( 2 C C ( C 2 H 5 OH 1 C mass % C = ( 100% = 52.13% In order of increase mass % of carbon C 12H 22O 11 < C 8H 10N 4O 2 < C 2H 5OH or sucrose < caffeine < ethanol = C 47. Assume 1 of funal laccase In 1 funal laccase ecule there are 4 Cu atoms Therefore, in 1 of funal laccase there are 4 es of Cu atoms Find the mass of 4 Cu. 4 Cu ( Cu 1 Cu Find mass of funal laccase = Cu mass % Cu = ( m Cu 100% m total m Cu m total = ( 100% = ( % = 65,200 mass % Cu 0.390% The mass total was for 1 of the funal laccase, therefore, in order to et the ar mass one needs to divide by 1. M funal laccase = 65, Compound 1 Heatin of compound 1 leaves H m O = m total m H = = Determine es of H H ( 1 H H = H 4

5 Determine es of O O ( 1 O O = O Divide throuh by smallest number of es ( Mercury Oxyen = = Empirical Formula HO Compound 2 Heatin of compound 2 results in O bein produced (mass of O is the mass lost m H = m total m O = = Determine es of H H ( Determine of O O ( 1 H H 1 O O = H = O Divide throuh by smallest number of es ( Mercury Oxyen = = Empirical Formula H 2O 54. Assume 100 sample therefore C, 6.56 H, O, and 8.28 N Calculate the number of es of C, H, O, and N Carbon Hydroen Oxyen Nitroen C ( 6.56 H ( O ( 8.28 N ( 1 C C 1 H 1.01 H 1 O O 1 N N = C = 6.50 H = O = H Divide throuh by smallest amount (0.591 Carbon Hydroen = = Oxyen Nitroen = = Empirical Formula C 8H 11O 3N 5

6 59. Assume 100 sample therefore 26.7 P, 12.1 N, and 61.2 Cl Calculate the number of es of P, N, and Cl Phosphorus Nitroen Chlorine 26.7 P ( 12.1 N ( 62.2 Cl ( 1 P P 1 N N 1 Cl Cl = P = N = 1.75 Cl Divide throuh by smallest amount (0.862 Phosphorus Nitroen Hydroen = = Empirical Formula PNCl 2 M PNCl2 = Find ecular formula P 5N 5Cl 10 = 5.09 Multiply empirical formula by CxHyOz + O2 CO2 + H2O (Combustion Reaction All of the carbon in the compound oes into formin CO2. Calculate n C m CO 2 ( m ( 1 CO 2 1 C = 2.03 ( = C CO 2 1 CO 2 All of the hydroen in the compound oes into formin H2O. Calculate n H 6.45 m H 2 O ( 1 ( 1 H 2O 2 H ( = m H 2 O 1 H 2 O 10 4 H To find the mass of O, find the mass of C and H and subtract them from the overall mass of the compound. Calculate m C m C = C ( C 1 C = C Calculate m H m H = H H ( = H 10 4 H Calculate m O m O = m Cx H y O z m C m H m Cx H y O z = m ( m = m O = = Calculate n O 1 O ( = O 10 4 O 6

7 Divide throuh by smallest e amount ( Carbon Hydroen Oxyen = = = Multiple throuh by 2 to et whole numbers Empirical Formula C4H3O2 Find ecular formula M = m n = = 166 M C4 H 3 O 2 = = 2.00 Multiply empirical formula by 2: C8H6O4 61. CxHyOz + O2 CO2 + H2O (Combustion Reaction All of the C in the compound oes into formin CO2. Therefore, n CO2 = n C m CO 2 ( 1 ( 1 CO 2 1 C ( = C 1000 m CO 2 1 CO 2 All of the H in the compound oes into formin H2O. Therefore, n H2 O = 2n H m H 2 O ( 1 ( 1 H 2O 2 H ( = m H 2 O 1 H 2 O 10 4 H To find the mass of O, find the mass of C and H, and subtract them from the overall mass of the compound. Calculate m C m C = C C ( = C 1 C Calculate m H m H = H H ( = H 10 4 H Calculate m O m O = m Cx H y O z m C m H m Cx H y O z = m ( m = m O = = Calculate n O 1 O ( = O 10 4 O Divide throuh by smallest e amount ( Carbon Hydroen Oxyen = = = Multiple throuh by 3 to et whole numbers Empirical Formula C3H4O3 M C3 H 4 O 8 = Find ecular formula = Multiply empirical formula by 2 C6H8O6 63. X+2Y XY 2 7

8 65. a C 6H 12O 6(s + 6O 2( 6CO 2( + 6H 2O( b Fe 2S 3(s + 6HCl( 2FeCl 3(s + 3H 2S( c CS 2(l + 2NH 3( H 2S( + NH 4SCN(s 68. a C 2H 5OH(l + 3O 2( 2CO 2( + 3H 2O( b 3Pb(NO 3 2(aq + 2Na 3PO 4(aq Pb 3(PO 4 2(s + 6NaNO 3(aq 69. a 16Cr(s + 3S 8(s 8Cr 2S 3(s b 2NaHCO 3(s heat Na 2CO 3(s + CO 2( + H 2O( c 2KClO 3(s heat 2KCl(s + 3O 2( d 2Eu(s + 6HF( 2EuF 3(s + 3H 2( k Al ( k ( 1 Al Al (3 NH 4ClO4 ( NH 4ClO4 = 4,000 3 Al 1 NH4ClO4 74. Use the mass of HNO 3 to calculate the es of HNO 3 needed k HNO 3 ( ( 1 HNO 3 = k HNO3 107 HNO 3 Use es of HNO 3 to calculate the es of NH 3 needed 2 NO HNO 3 ( 3 NO 2 ( ( 4 NH 3 = HNO 3 2 NO 2 4 NO 107 NH 3 Use the es of NH 3 to calculate the kilorams of NH 3 produces Fe ( 15.0 Fe ( 15.0 Fe ( NH 3 ( NH 3 1 NH3 ( 1 k 1000 = k NH NO( + O 2( 2NO 2( The limitin reaent is NO. 1 Fe Al Al (2 ( Fe 2 Fe 1 Al 1 Fe Fe (2 Fe 2O 3 2 Fe ( Fe 2O 3 1 Fe Fe (2 Al 2O 3 2 Fe ( Al 2O 3 = 7.25 Al 1 Fe 2 O 3 = 21.4 Fe 2 O 3 1 Al 2 O 3 = 13.7 Al 2 O NH 3( + 5O 2( 4NO( + 6H 2O( NH 4 O 2 (L.R. NO H 2O Initial 10 ec 10 ec 0 0 Chane in terms of x -4x -8-5x -5x=10 x=2 Final 2 ec 0 8 ec 12 ec The Reaction will o until either the NH 4 or O 2 is used up, therefore, x=2 and oxyen is the limitin reaent ICF tables can be done in ecules or es. 22 total eucles are present in the container once the reaction oes to completion. +4x +8 +6x +12 8

9 84. Calculate the es of A 2O and AC 10H 9N 4SO A 2 O ( 1 A 2O A 2 O = A 2O 50.0 C 10 H 10 N 4 SO 2 ( 1 C 10H 10 N 4 SO C 10 H 10 N 4 SO 2 = C 10H 10 N 4 SO 2 To completely react all the A 2O how many es of C 10H 9N 4SO 2 would you need? A 2 O ( 2 C 10H 10 N 4 SO 2 = C 1 A 2 O 10 H 10 N 4 SO 2 Since we only have of C 10H 10N 4SO 2 the C 10H 10N 4SO 2 is the limitin reaent. A 2O C 10H 10N 4SO 2 (L.R. AC 10H 9N 4SO 2 H 20 Initial Chane -x x -2x= x= x x Final How much AC 10H 9N 4SO 2 could you produce? AC 10 H 9 N 4 SO 2 ( AC 10H 9 N 4 SO 2 1 AC 10 H 9 N 4 SO 2 = 71.4 AC 10 H 9 N 4 SO a Calculate the es of the chlorobenzene and chloral present at the beinnin of the reaction C 6 H 5 Cl ( 1 C 6H 5 Cl C 6 H 5 Cl 6H 5 Cl 485 C 2 HOCl 3 ( 1 C 2HOCl C 2 HOCl 3 = 3.29 C 2 HOCl 3 Calculate the es of C 2HOCl 3 needed to fully react 10.1 es of C 6H 5Cl C 6 H 5 Cl ( 1 C 2HOCl 3 = C 2 C 6 H 5 Cl 2HOCl 3 Since we only have 3.29 C 2HOCl 3, it is our limitin reaent C 6H 5Cl C 2HOCl 3 (L.R. C 14H 9Cl 5 H 20 Initial Chane -2x x -x=-3.29 x=3.29 +x x Final Calculate the mass ofc 14H 9Cl 5 formed 3.29 C 14 H 9 Cl 5 ( C 14H 9 Cl 5 = C 1 C 14 H 9 Cl 14 H 9 Cl 5 5 b The limitin reaent is C 2HOCl 3 and C 6H 5Cl is present in excess. See part a. c Calculate the amount of chlorobenzene used 3.29 C 2 HOCl 3 ( 2 C 6H 6 Cl ( C 6H 6 Cl = 741 C 1 C 2 HOCl 3 1 C 6 H 6 Cl 6H 6 Cl = 401 actual yield d % yield = ( 100% = ( % = 17.1% theoretical yield They are askin you to determine the chemical formula of salicylic acid. What you know Salicylic acid + C4H6O3 aspirin + C2H4O2 9

10 The problem tells you that the coefficients in front of each of the species are 1 Determine the ecular formula of aspirin CxHyOz + O2 CO2 + H2O (Combustion Reaction All of the carbon in the compound oes into formin CO2. Therefore, n CO2 = n C. 1 C 2.20 CO 2 ( 1 CO 2 ( = C CO 2 1 CO 2 All of the hydroen in the compound oes into formin H2O. Therefore, n H2 O = 2n H H 2 O ( 1 H 2O 2 H ( = H H 2 O 1 H 2 O To find the mass of O, find the mass of C and H, and subtract them from the overall weiht of the compound. Calculate m C C m C = C ( = C Calculate m H m H = H ( 1 C 1.01 H 1 H = H Calculate m O m O = m Cx H y O z m C m H m O = = 0.35 Calculate n O 1 O 0.35 ( = 0.22 O O Divide throuh by smallest amount (0.22 Carbon Hydroen Oxyen = = = Multiple throuh by 4 to et whole numbers Empirical Formula C9H8O4 Since the ecular formula weiht is between 170 M C9 H 8 O 4 = ecular formula are the same. Determine the formula or salicylic acid Salicylic acid + C4H6O3 aspirin (C9H8O4 + C2H4O2 Salicylic acid: C7H6O3 and 190 the empirical and 137. You need to determine the amount of NO( produced. You know that 2.00 of NH 3 are reacted with of O 2 and 6.75 es of O 2 remains after the reaction oes to completion. This indicates that NH 3 is the limitin reaent Determine the es of O 2 reacted n O2 (reacted = n O2 (oriinal n O2 (unreacted = = 3.25 NH 3 reacts by one of the followin equations 4NH 3( + 5O 2( 4NO( + 6H 2O( 4NH 3( + 7O 2( 4NO 2( + 6H 2O( Therefore for equation 1 n O2 = 5 4 n NO and for equation 2 n O2 = 7 4 n NO 2 The total number of es of O 2 used was 3.25 ivin 5 4 n NO n NO 2 = 3.25 Similarly for equation 1 n NH3 = n NO and for equation 2 n NH3 = n NO2 10

11 The total number of es of NH 3 used was 2.00 ivin n NO + n NO2 = 2.00 Combine equations and solve for e of NO n NO2 = 2.00 n NO 5 4 n NO (2.00 n NO = n NO n NO = 13 2n NO = 1 n NO =