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12.1 Test Review

12.1 Test Review. What is percent composition?. the percent (by mass) of all the elements in a compound. What is Avagadros’ constant – 6.02 X 10 23. the number of particles contained in 1 mole of a substance. What is the SI unit of chemical quantity?. the mole. What is molar mass?.

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12.1 Test Review

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  1. 12.1 Test Review

  2. What is percent composition? • the percent (by mass) of all the elements in a compound

  3. What is Avagadros’ constant – 6.02 X 1023 • the number of particles contained in 1 mole of a substance

  4. What is the SI unit of chemical quantity? • the mole

  5. What is molar mass? • the mass of 1 mole of a substance

  6. classify the following substance as:a. formula unit b. atom c. molecule d. ion • Mg • atom

  7. classify the following substance as:a. formula unit b. atom c. molecule d. ion • C2O4-2 • ion

  8. classify the following substance as:a. formula unit b. atom c. molecule d. ion • Na(NO3)2 • formula unit

  9. classify the following substance as:a. formula unit b. atom c. molecule d. ion • C6H12O6 • molecule

  10. determine the molar mass for the following substance: • Br • 80

  11. determine the molar mass for the following substance: • C6H12O6 • 180

  12. determine the molar mass for the following substance: • Na(NO3)2 • 147

  13. Simplified mole conversion

  14. How many moles are in 32 grams of SCl2? 32g SCl2 X 1 1 mole SCl2 102 g SCl2 = 0.31 mol SCl2

  15. Calculate the number of atoms in 3.2 moles of gold (Au) 3.2g mol Au X 1 6.02 X 10 23 atoms Au 1 mol Au = 1.9 X 1024 atoms Au

  16. What is the volume in liters of 12.4 mol of CO2 gas? 12.4 mol CO2 X 1 22.4 L CO2 1 mol CO2 = 277.8 L CO2

  17. What is the volume in liters of 21.3g of N2 gas? 1 mol N2 28 g N2 21.3 g N X 1 X 22.4 L N2 1 mol N2 = 17.04 L N2

  18. How many molecules are in 72.3 L of O2? 72.3 L O2 X 1 1 mol O2 22.4 L O2 X 6.02 X 1023 O2 1 mol O2 = 1.9 X1024 molecules O2

  19. What is the mass in grams of 2.1 X 1021 formula units of MgCl2? X 94 g MgCl2 1 mol MgCl2 2.1 X 1021 fu MgCl2 1 X 1 mol MgCl2 6.02 X 1023 fu MgCl2 = 0.33 g MgCl2

  20. Determine the percent composition for the following compound:CaCl2 • Ca = 40.078 g • Cl2 = 2 x 35.453 g = + 70.906 g • CaCl2 = 110.984 g Ca = 40.1g X 100 = 36.1% Ca 110.9 Cl = 70.9g X 100 = 63.9% Cl 110.9g

  21. empirical formula of a compound with percent composition of 38.4 % Mn, 16.8 % C, 44.7 % O 38.4 g Mn X 1mol Mn = 0.699 mole Mn 1 55 g Mn 16.8 g C X 1mol C = 1.339 mole C 1 12 g C 45.7 g O X 1mol O = 2.798 mole O 1 16 g O

  22. 0.699 mole Mn = 1 mol Mn 0.699 1.339 mole C = 2 mol C 0.699 2.798 mole O = 4 mol O 0.699 So formula is MnC2O4

  23. empirical and molecular formulas of a compoundwith percent composition of 71.0 % Cl, 25.0% C, 4.0 % H and mass 98.96 g/mol 71.0 g Cl X 1mol Cl = 2.0285 mole Cl 1 35 g Cl 25.0 g C X 1mol C = 2.0833 mole C 1 12 g C 4.0 g H X 1mol H = 4.0 mole H 1 1 g H

  24. 2.0285 mole Cl = 1 mol Cl 2.0285 2.0833 mole C = 1.02 mol C 2.0285 4.0 mole H = 1.97 mol H 2.0285 So formula is ClCH2

  25. molecular mass of compound 98.96g = 2.01 molar mass of ClCH2 49 g 2 X (ClCH2) = Cl2C2H4

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