U3-LM2B-WS Molar Mass and Conversions



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U3-LM2B-WS Molar Mass and Conversions Name: KEY 1. The molar mass of chlorine is: 2 x 35.45 g/mol Cl = 70.90 g/mol Cl 2 (Remember that chlorine exists as a diatomic molecule in nature) 2. The molar mass of carbon dioxide is: 12.01 g/mol C + (2 x 16.00 g/mol O) = 44.01 g/mol CO 2 3. The molar mass of aluminum carbonate, Al 2 (CO 3 ) 3, is: (2 x 26.98 g/mol Al) + (3 x 12.01 g/mol C) + (9 x 16 g/mol O) = 234.0 g/mol Al 2 (CO 3 ) 3 4. The molar mass of ascorbic acid (Vitamin C), C 6 H 8 O 6 is: (6 x 12.01 g/mol C) + (8 x 1.01 g/mol H) + (6 x 16 g/mol H) = 176.14 g/mol C 6 H 8 O 6 5. A 4.0g/mol represents the molar mass of the element. helium 6. A 2.0 g/mol represents the molar mass of the element. hydrogen 7. A 40.0 g sample of sodium is 1.74 moles of sodium and 1.05 x 10 24 atoms of sodium. 40.0 g Na 1 mol Na = 1.74 moles Na 22.99 g Na 1.74 moles Na 6.022 x 10 23 atoms Na = 1.05 x 10 24 atoms Na 1 mol Na 8. One mole of elemental diatomic chlorine is 70.9 grams of chlorine and contains atoms 1.20 x 10 24 of chlorine. 1 mol Cl 2 70.9 g Cl = 70.9 g Cl 1 mol Cl 2 1 mol Cl 2 2 mol Cl 6.022 x 10 23 atoms Cl = 1.20 x 10 24 atoms Cl 1 mol Cl 2 1 mol Cl 1

9. If 2 moles of magnetite, Fe 3 O 4, are needed, one needs to weigh 463.1 grams of the substance. This amount corresponds to two formula units and it contains 3.61 x 10 24 ions of iron and ions 4.82 x 10 24 of oxygen. 2 mol Fe 3 O 4 6.022 x 10 23 formula units Fe 3 O 4 = 1.20 x 10 24 formula units Fe 3 O 4 1 mol Fe 3 O 4 1.20 x 10 24 formula units Fe 3 O 4 3 ions Fe = 3.61 x 10 24 ions Fe 1formula unit Fe 3 O 4 1.20 x 10 24 formula units Fe 3 O 4 4 ions O = 4.82 x 10 24 ions O 1 formula unit Fe 3 O 4 10. A sample that is 36.0 grams of water represents 2 moles of water. It contains 4 grams of hydrogen and 32 grams of oxygen. It also contains 4 moles of H atom and moles 2 of O atoms. This sample also represents 1.20 x 10 24 molecules of water, 2.41 x 10 24 atoms of hydrogen and 1.20 x 10 24 atoms of oxygen. 36.0 g H 2 O = 2 mol H 2 O 18 g H 2 O 2 mol H 2 O 2 mol H = 4 mol H 1 g H = 4 g H 1 mol H 2 mol H 2 O 1 mol O = 2 mol O 16 g H = 32 g H 1 mol H 2 mol H 2 O 6.022 x 10 23 molecules H 2 O = 1.20 x 10 24 molecules H 2 O 2 mol H 2 O 1 mol O 6.022 x 10 23 atoms O = 1.20 x 10 24 atoms O 2 mol H 2 O 6.022 x 10 23 molecules H 2 O = 1.20 x 10 24 molecules H 2 O 11. a-calculate the mass in grams of 2.5 moles of calcium. 2.5 mol Ca 40.08 g Ca = 1.0 x 10 2 g Ca 1 mol Ca b-how many atoms are there is 20.0 grams of calcium? 20.0 g Ca 1 mol Ca 6.022 x 10 23 atoms Ca = 3.00 x 10 23 atoms Ca 40.08 g Ca 1 mol Ca 2

c-what is the mass of 1.40x10 20 atoms of calcium? 1.40 x 10 20 atoms Ca 1 mol Ca 40.08 g Ca=.00692 g Ca 6.022 x 10 23 atoms Ca 1 mol Ca d-calculate the mass in grams of one calcium atom. 1atom Ca 1 mol Ca 40.08 g Ca= 6.7 x 10-23 g Ca 6.022 x 10 23 atoms Ca 1 mol Ca 12. a-how many atoms are contained in 28.0 grams of nitrogen? 28.0 g N 2 1 mol N 2 2 mol N 6.022 x 10 23 atoms N = 1.20 x 10 24 atoms N 28.0 g N 2 1 mol N 2 1 mol N b-how many moles of N atoms are represented in 5.0 x 10 30 atoms of nitrogen? 5.0 x 10 30 atoms N 1 mol N = 8.3 x 10 6 moles N 6.022 x 10 23 atoms N c-calculate the mass in grams of one molecule of nitrogen. 1 molecule N 2 1 mol N 2 28.0 g N 2 = 4.6 x 10-23 g N 2 6.022 x 10 23 molecules N 2 1 mol N 2 d-calculate the mass in grams of one atom of nitrogen. 1atom N 1 mol N 14.0 g N= 2.3 x 10-23 g N 6.022 x 10 23 atoms N 1 mol N 13. a-what masses of each element are presented in 5.60 moles of acetic acid, CH 3 COOH? 5.6 mol CH 3 COOH 2 mol C 12.01 g C = 135 g C 1 mol CH 3 COOH 1 mol C 5.6 mol CH 3 COOH 2 mol O 16 g O = 179 g O 1 mol CH 3 COOH 1 mol O 5.6 mol CH 3 COOH 4 mol H 12.01 g H = 22.6 g H 1 mol CH 3 COOH 1 mol H 3

b- How many moles of H atoms and how many atoms of H does the above sample contain? 5.6 mol CH 3 COOH 4 mol H = 22.4 mol H 1 mol CH 3 COOH 22.4 mol H 6.022x10 23 atom H = 1.35 x 10 25 atoms H 1 mol H c-what is the mass of acetic acid that contains 4.0 g of hydrogen? 4.0 g H 1 mol H 1 mol CH 3 COOH 60.06 g CH 3 COOH = 60.06 g CH 3 COOH 1 g H 4 mol H 1 mol CH 3 COOH d-what is mass of acetic acid that contains 32.0 g of oxygen? 32.0 g O 1 mol O 1 mol CH 3 COOH 60.06 g CH 3 COOH = 60.06 g CH 3 COOH 16 g O 2 mol O 1 mol CH 3 COOH e-what mass of acetic acid contains 48.0 grams of carbon? 48.0 g C 1 mol C 1 mol CH 3 COOH 60.06 g CH 3 COOH = 120.12 g CH 3 COOH 12 g C 2 mol C 1 mol CH 3 COOH 14. There are 737 g of sodium chloride in a can of salt. a-how many moles of sodium chloride does the can of salt contain? 737 g NaCl 1 mol NaCl = 12.6 mol NaCl 58.44 g NaCl b-how many formula units of salt does the can of salt contain? 12.6 mol NaCl 6.022 x 10 23 formula unit NaCl = 7.59 x 10 24 formula unit NaCl 1 mol NaCl c-calculate the mass in grams of one formula unit of sodium chloride. 1 formula unit NaCl 1 mol NaCl 58.44 g NaCl= 9.7 x 10-23 g NaCl 6.022 x 10 23 formula unit NaCl 1 mol NaCl d-how many ions of sodium does the can of salt contain? 12.6 mol NaCl 1 mol Na + 6.022 x 10 23 ions Na + = 7.59 x 10 24 ions Na + 1 mol NaCl 1 mol Na + 4

e-how many moles of chloride ions does the can contain? 12.6 mol NaCl 1 mol Cl - = 12.6 mol Cl - 1 mol NaCl f-how many grams of sodium does the can of salt contain? 12.6 mol NaCl 1 mol Na + 22.99 g Na + = 289.67 g Na + 1 mol NaCl 1 mol Na + 15. a-how many moles of ammonium sulfate are in 32.0 g of ammonium sulfate? 32.0 g (NH 4 ) 2 SO 4 1 mol (NH 4 ) 2 SO 4 =.242 mol (NH 4 ) 2 SO 4 132.17 g (NH 4 ) 2 SO 4 b-how many formula units of ammonium sulfate are there in 32.0 g of ammonium sulfate? 32.0 g (NH 4 ) 2 SO 4 1 mol NH 4 ) 2 SO 4 6.022 x 10 23 formula units (NH 4 ) 2 SO 4 = 1.46 f.u.s (NH 4 ) 2 SO 4 132.17 g (NH 4 ) 2 SO 4 1 mol (NH 4 ) 2 SO 4 * f.u = formula unit c-how many atoms of H are found in 32.0 g of ammonium sulfate? 32.0 g (NH 4 ) 2 SO 4 1 mol NH 4 ) 2 SO 4 8 mols H 6.022 x 10 23 atoms H = 1.17 X 10 24 atoms H 132.17 g (NH 4 ) 2 SO 4 1 mol (NH 4 ) 2 SO 4 1 mol H d-how many grams of hydrogen are found in 32.0 g of ammonium sulfate? 32.0 g (NH 4 ) 2 SO 4 1 mol NH 4 ) 2 SO 4 8 mols H 1.01 g H = 1.94 g H 132.17 g (NH 4 ) 2 SO 4 1 mol (NH 4 ) 2 SO 4 1 mol H 5

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