General Chemistry 101 - University of Bahrain - PDF

UNIVERSITY OF BAHARAIN DEPARTMENT OF CHEMISTRY GENERAL CHEMISTRY 101 Textbook CHEMISTRY, Steven S. Zumdahl, Susan A. Zumdahl, and Donald J. Decoste, 10th Edition, CENGAGE Learning COURSE INSTRUCTOR Section 3.2 Atomic Masses Modern System of Atomic Masses...

UNIVERSITY OF BAHARAIN DEPARTMENT OF CHEMISTRY GENERAL CHEMISTRY 101 Textbook CHEMISTRY, Steven S. Zumdahl, Susan A. Zumdahl, and Donald J. Decoste, 10th Edition, CENGAGE Learning COURSE INSTRUCTOR Section 3.2 Atomic Masses Modern System of Atomic Masses  Instituted in 1961  Standard taken is 12C  12C is assigned a mass of exactly 12 atomic mass units (amu or u) Masses of all other atoms are given relative to this standard. Copyright ©2018 Cengage Learning. All Rights Reserved. Section 3.2 Atomic Masses Mass Spectrometer  Helps to accurately compare the masses of atoms Copyright ©2018 Cengage Learning. All Rights Reserved. Atom Atomic mass Atomic mass unit(amu) Relative atomic mass Mass spectrometer Abundance Average atomic mass Average Atomic mass = ∑%abundance X atomic mass 100 Abundance of isotope is expressed in % Sum of natural abundance of all the isotopes of an element =100% If there are two isotopes, abundance of one of the isotope = X % abundance of other isotope = 100- X Q1) Carbon has two isotopes, C-12 has atomic mass 12 amu, C-13 has atomic mass 13.0035amu, and the % abundance of these isotopes is 98.90% and 1.10 % respectively. Calculate average atomic mass. isotopes %abundance Atomic mass(amu) Q1) Carbon has two isotopes, C-12 has atomic mass 12 amu, C-13 has atomic mass 13.0035amu, and the % abundance of these isotopes is 98.90% and 1.10 % respectively. Calculate average atomic mass. isotopes %abundance Atomic mass(amu)  Q1) Carbon has two isotopes, C-12 has atomic mass 12 amu, C-13 has atomic mass 13.0035amu, and the % abundance of these isotopes is 98.90% and 1.10 % respectively. Calculate average atomic mass? Isotopes % abundance At.mass(amu) C-12 98.90 12 C-13 1.10 13.0035 ∑%abundance X atomic mass Average Atomic mass = 100 = 98.90 𝑋12+1.10 𝑋13.0035 =12.01 amu 100 Q2) Calculate average atomic mass from the following data isotopes Atomic mass (amu) % abundance 10 B 10.0129 19.80 11 B 11.0093 80.20 Q2) Calculate average atomic mass from the following data isotopes Atomic mass (amu) % abundance 10 B 10.0129 19.80 11 B 11.0093 80.20 Q2) Calculate average atomic mass from the following data isotopes Atomic mass (amu) % abundance 10 B 10.0129 19.80 11 B 11.0093 80.20 ∑%abundance X atomic mass Average Atomic mass = 100 19.80 𝑋10.0129+80.20 𝑋11.0093 =10.81 amu 100 Q3) Calculate the average atomic mass isotopes Atomic mass % abundance 35 Cl 34.968 75.53 37 Cl 36.956 24.47 Q3) Calculate the average atomic mass isotopes Atomic mass % abundance 35 Cl 34.968 75.53 37 Cl 36.956 24.47  Q3) Calculate average atomic mass from the following data isotopes Atomic mass % abundance 35 Cl 34.968 75.53 37 Cl 36.956 24.47 ∑%abundance X atomic mass Average Atomic mass = 100 75.53 𝑋34.968+24.47𝑋36.956 = 35.45 amu 100  Q4)Calculate average atomic mass from the following data isotopes Atomic mass(amu) % abundance 63 Cu 62.9296 69.17 65 Cu 64.9278 30.83 ∑%abundance X atomic mass Average Atomic mass = 100 (69.17 𝑋62.9296)+(30.83 𝑋64.9278)= 63.5 amu 100 Q5) Nitrogen has two isotopes N-14 and N-15 and the atomic masses of these isotopes are 14.0031 amu and 15.0001amu respectively. The average atomic mass of these isotopes is 14.0067 amu.Calculate % abundance of heavier isotopes Answer Isotopes Atomic mass(amu) % abundance Q5) Nitrogen has two isotopes N-14 and N-15 and the atomic masses of these isotopes are 14.0031 amu and 15.0001amu respectively. The average atomic mass of these isotopes is 14.0067 amu.Calculate % abundance of heavier isotopes Isotopes Atomic mass(amu) % abundance Answer  Q6) Bromine has two isotopes Br-79 and Br-81 and the atomic masses of these isotopes are 78.92 amu and 80.92amu respectively. The average atomic mass of these isotopes is 79.90 amu.Calculate % abundance of both isotopes. Isotopes Atomic mass(amu) % abundance  Q6) Bromine has two isotopes Br-79 and Br-81 and the atomic masses of these isotopes are 78.92 amu and 80.92amu respectively. The average atomic mass of these isotopes is 79.90 amu.Calculate % abundance of both isotopes. Isotopes Atomic mass(amu) % abundance Q6)Silver has two isotopes Ag-1 and Ag-2 and the atomic mass of Ag -1 isotope is 106.90509 amu and abundance is 51.84%. The average atomic mass of these isotopes is 107.90 amu.Calculate atomic mass of Ag-2 isotope? isotopes Atomic mass % abundance Average atomic Mass(amu) 1Ag 106.90509 51.84 107.90 2Ag Silver has two isotopes Ag-1 and Ag-2 and the atomic mass of Ag -1 isotope is 106.90509 amu and abundance is 51.84%. The average atomic mass of these isotopes is 107.90 amu.Calculate atomic mass of Ag-2 isotope? isotopes Atomic mass % abundance Average atomic Mass(amu) 1Ag 106.90509 51.84 107.90 2Ag Q7) Rubidium has two isotopes.85Rb has mass 84.9117amu and 87Rb has mass 86.9085 amu. If the average atomic mass is 85.4678amu ,what is the abundance of each isotope? isotopes Atomic mass % abundance 85 Rb 87 Rb Q7) Rubidium has two isotopes.85Rb has mass 84.9117amu and 87Rb has mass 86.9085 amu. If the average atomic mass is 85.4678amu ,what is the abundance of each isotope? isotopes Atomic mass % abundance Average atomic Mass(amu) 85 Rb 84.9117 X =? 85.4678 72.2% 87 Rb 86.9085 100-X =? 100-72.2=27.8% ∑%abundance X atomic mass Average Atomic mass = 100 85.4678= 84.9117𝑥𝑋 + 100 − 𝑋 𝑥 86.9085 100 X= 72.2 % Q.8) The atomic mass of 6Li and 7Li are 6.0151amuand 7.0610amu respectively. Calculate the natural abundances of these isotopes. isotopes Atomic mass % abundance Q8)The atomic mass of 6Li and 7Li are 6.0151amuand 7.0610amu respectively. Calculate the natural abundances of these isotopes. isotopes Atomic mass % abundance Average atomic Mass(amu) 6 Li 6.0151 X =? 11.4 % 6.941 (FROM 7 Li 100-11.4=88.6% PERIODIC TABLE) 7.0610 100-X =? ∑%abundance X atomic mass Average Atomic mass = 100 6.941= 6.0151𝑥𝑋 + 100 − 𝑋 𝑥 7.0610 100 X= 11.4 % One pair = One dozen= One mole= Avodadro number= Mole and molar mass Mass of one mole is molar mass; unit is g/mol Atomic mass- mass of one atom; unit-u Molecular mass- mass of one molecule; unit-u Molar mass- mass of one mole; unit-g/mol Atom Atomic mass Molarmass C O N Cl Atom Atomic mass Molarmass C O N Cl Molecule Molecular mass Molarmass Cl2 H2O2 C12H22O11 CuSO4.5H2O H2C2O4.2H2O Molecule Molecular mass Molarmass Cl2 H2O2 C12H22O11 CuSO4.5H2O H2C2O4.2H2O Mass of one mole(Avogadro number) of atoms =Molar mass Mass of one atom in grams = 𝑚𝑜𝑙𝑎𝑟 𝑚 𝑎 𝑠 𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑎 𝑡 𝑜 𝑚 𝐴𝑣𝑜𝑔𝑎𝑑𝑟𝑜 𝑛𝑢𝑚𝑏𝑒𝑟 Mass of one molecule in gram = 𝑚𝑜𝑙𝑎𝑟𝐴𝑣𝑜𝑔𝑎𝑑𝑟𝑜 𝑚 𝑎𝑠 𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 Q1) Find the mass of one atom of chlorine in grams Q2) Find the mass of 200 atoms of chlorine in grams Q3) Find the mass of a molecule of chlorine in grams Mass of one mole(Avogadro number) of atoms =Molar mass Mass of one atom in grams = 𝑚𝑜𝑙𝑎𝑟 𝑚 𝑎 𝑠 𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑎 𝑡 𝑜 𝑚 𝐴𝑣𝑜𝑔𝑎𝑑𝑟𝑜 𝑛𝑢𝑚𝑏𝑒𝑟 Mass of one molecule in gram = 𝑚𝑜𝑙𝑎𝑟𝐴𝑣𝑜𝑔𝑎𝑑𝑟𝑜 𝑚 𝑎𝑠 𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 Q1) Find the mass of one atom of chlorine in grams Q2) Find the mass of 200 atoms of chlorine in grams Q3) Find the mass of a molecule of chlorine in grams Q4) Find the mass of a molecule of C6H12O6 in grams Q5) Find the mass of 100 molecules of C6H12O6 in grams Q4) Find the mass of a molecule of C6H12O6 in grams Q5) Find the mass of 100 molecules of C6H12O6 in grams  Q6) Find the mass of a molecule of C60 in grams Q7)Mass of an atom of an element is 5.3156x10-23. Identify the atom. g  Q6) Find the mass of a molecule of C60 in grams Q7) Mass of an atom of an element is 5.3156x10-23 g. Identify the atom. n=number of moles MM =Molar mass in g/mol Mass in grams mass n MM Find the no. of moles in a) 9 g H2O b)0.36 kg C6H12O6 c) 6.0 g of He d) 87.3 g of Mg Find the no. of moles in a) 9 g H2O b)0.36kg C6H12O6 c) 6.0 g of He d) 87.3 g of Mg Find the mass of : a. 10 moles of H2O a. 2 moles of C6H12O6 Mass of a 10 moles of a compound is 3600 g. Find the molar mass of the compound Find the mass of : a. 10 moles of H2O a. 2 moles of C6H12O6 Mass of a 10 moles of a compound is 3600 g. Find the molar mass of the compound  n=no. of moles NA=Avogadro number Number of particles n NA Find the number molecules of NaOH in 1.5 moles of NaOH Find the number molecules of NaOH in 80g of NaOH Find the number molecules of NaOH in 1.5 moles of NaOH Find the number molecules of NaOH in 80g of NaOH How many molecules are present in 90g of glucose(MM=180g/mol)? How many moles are present in 1.204x1024 molecules of water? How many atoms of O,H&C are present in 100 molecules of glucose ? How many molecules are present in 90g of glucose(MM=180g/mol)? How many moles are present in 1.204x1024 molecules of water? How many atoms of O,H&C are present in 100 molecules of glucose ? Q 1 How many grams of Zn are in 0.356 mol of Zn? Q 2 Calculate the number of grams of lead (Pb) in 12.4 moles of lead. 𝑚𝑎𝑠𝑠 Q.1 n = 𝑀.. 𝑀 Mass = M.M x n. = 65.39 X 0.356 = 23.3 g Q 2 n = 𝑚𝑎𝑠𝑠 𝑀.. 𝑀 Mass =n * M.M =12.4 X 207.2 =2569.28 g Q. How many moles of rubidium (Rb) are there in 3.75 × 1024 Rb atoms? n= 𝑁 = 3.75 × 1024 =6.229 moles 𝑁 𝐴 6.02 𝑋1023 N n A.N Q. What is the mass in grams of 1.68 moles of vanadium (V)? Mass = n XM.M mass = 1.68 X 50.94= 85.57 g n M.M Q. What is the mass of 10 moles of H2SO4? Mass = n * M.M = 10mol * 98g/mol = 980 g Q. Calculate No.of molecules in 10 moles of H2O No.of molecules = n* A.N No.of molecules = 10 X 6.022 X1023 =6.022 X1024 Q. Calculate Moles of H2SO4 in 6.06 x10 23 molecules of H2SO4 n ==6.06 𝑥 10 23 =1.006 6.022 𝑋1023 Q. Find the no.of molecules of H2O in 72g H2O n= 𝑚𝑎𝑠𝑠 = 72. =4 mole 𝑀.𝑀 18 No.of molecules N= n * A.N = 4 X 6.02 x1023=2.408 X10 24 Q. Complete the following table for C6H12O6 Mass (N2H4CO) No.of moles of N2H4CO No.of molecules No.of atoms of H Of N2H4CO 30 g 2 mol 1.806x1024 9.632x1024 Mass (N2H4CO) No.of moles of N2H4CO No.of molecules No.of atoms of H Of N2H4CO 30 g 2 mol 1.806x1024 9.632x1024 Mass of C6H12O6 Moles of C6H12O6 Molecules of Atoms of H (g) Moles C6H12O6 90 2 1.806 x1024 2.889 x1025 Mass of C6H12O6 Moles of C6H12O6 Molecules of Atoms of H (g) Moles C6H12O6 90 90 =0.499 0.499 x 6.02 X10 23 = 12 X 3.004 x10 23 = 180.156 =3.004 x 10 23 3.60 X 10 24 2 = 12 * 1.204 X1024 2X 180.156 2 X 6.02 X 10 23 =1.44 x 10 24 =360.312 =1.204 X10 24 1.806 x1024 =12 * 1.806X10 24 =2.17 x 10 25 3 x180.156=540.5 1.806 𝑋 10 24 =3 6.02 𝑋10 23 2.41 𝑋 10 24 =2.889 𝑥 10 25 2.889 x1025 = 4 12 6.02 𝑋 10 23 4 x 180.156 =2.41 x 10 24 = 720.6