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Mass Relationships

Mass Relationships. Lab Data All masses are relative to one another. Cu:O = 4:1 O:H = 16:1 Cu:H= 64:1 Cu:O:H 64:16:1 Must choose a standard, but which one? C = 12 O = 16 F = 19 H = 1 H hard to work with, variable results

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Mass Relationships

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  1. Mass Relationships • Lab Data • All masses are relative to one another. • Cu:O = 4:1 O:H = 16:1 Cu:H= 64:1 • Cu:O:H 64:16:1 • Must choose a standard, but which one? • C = 12 O = 16 F = 19 H = 1 • H hard to work with, variable results • F gives best results, very hard to duplicate • O,C used most often Eventually settled on C

  2. Chlorine Problem • Two naturally occuring isotopes • Cl mass number 35 Actual relative mass = 34.97 amu • Cl mass number 37 Actual relative mass = 36.97 amu • Cl 35 - 75.53% abundance in nature • Cl 37 - 24.47% abundance in nature • 34.97 x .7553 = 26.41 • 36.97 x .2447 = 9.04 • 35.45 amu average atomic mass • This is the number found on the periodic tables

  3. Practice • Sr relative mass 83.9 - 0.50 % abundance • Sr relative mass 85.9 - 9.9 % abundance • Sr relative mass 86.9 - 7.0 % abundance • Sr relative mass 87.9 - 82.6 % abundance • Average atomic mass = ?

  4. Practice • Sr relative mass 83.9 0.50 % abundance • Sr relative mass 85.9 9.9 % abundance • Sr relative mass 86.9 7.0 % abundance • Sr relative mass 87.9 82.6 % abundance • Average atomic mass = 87.61amu

  5. Discovery Activity

  6. Practice • Practice: molar amounts of elements • 1 mole K atoms = ? grams • 2 moles Mg atoms = ? grams • .3 moles Al atoms = ? grams

  7. Chemical Formulas • tells: * atom ratio within the molecule • * mole ratio within the substance as well • formula weight(molar mass) • sum total of masses of all atoms in the formula • ex. CaCO3 (Ca) 40.08 x 1 = 40.08 • (C) 12.01 x 1 = 12.01 • (O) 16.00 x 3 = 48.00 • 100.09

  8. Formula Weight Practice • NaCl = ? • K2I = ? • Ca(OH)2 = ?

  9. % Composition

  10. % Composition • in a sample of a compound, • total mass of one element x 100 = % • total mass of compound sample • ex. compound containing K, Cr, O • 25.0 gram sample contains 6.64 g K, 8.84 g Cr, 9.52 g O • 6.64 g K x 100 = 26.7 % K 8.84 g Cr x 100 = 35.4 % Cr • 25.0 g 25.0 g • 9.52 g O x 100 = 38.1 % O • 25.0 g

  11. Empirical Formula • simplest ratio of elements in a compound • may not be actual, just simplest • total mass of one element = element quotient • formula weight • ratio of element quotients = empirical formula • 6.64 g K = 0.170 8.84 g Cr = 0.170 9.52 g O = 0.595 • 39.10 52.00 16.00 • if ratio is not obvious, divide all quotients by the smallest • 0.170 = 1 0.595 = 3.5 • 0.170 0.170 • No half atoms, so actual ratio must be 2:2:7 • empirical formula = K2Cr2O7

  12. Empirical vs Actual • simplest ratio of elements in a compound CH2O • empirical F.W. = 30 • actual F.W. = 180 • actual formula must be 6 times the empirical or C6H12O6

  13. Reaction Types • First 5 types - most common • A) Composition • A + B → AB • B) Decomposition • AB → A + B • C) Single Replacement • X + AB → A + XB • D) Double Replacement • XA + ZB → ZA + XB • E) Combustion • CxHz + O2→ CO2 + H2O

  14. Terms • XA + ZB →ZA + XB • reactants reaction symbol products • reaction = equation • specifically: • reactions happens in the lab • equations symbolize reactions on paper.

  15. Writing Double Replacement Reactions • 1. Write the names of the reactants. • 2. Switch the first names of the reactants. • 3. Write these names on the right side of the arrow. • 4. Equation needs to be written all on one line. • 5. Starting on the far right, write the formulas. • 6. Moving to the left, continue to write formulas.

  16. Example • calcium + sodium→ • chloride carbonate • calcium + sodium → sodium + calcium • chloride carbonate chloride carbonate • CaCl2+ Na2CO3→ NaCl + CaCO3 • CaCl2+ Na2CO3→ NaCl + CaCO3 • CaCl2+ Na2CO3→ NaCl + CaCO3 • CaCl2+ Na2CO3→ NaCl + CaCO3 • 2. Switch the first names of the reactants. • 3. Write these names on the right side of the arrow. • 4. Starting on the far right, write the formulas. • 5. Moving to the left, continue to write formulas.

  17. Example • calcium + sodium→ • chloride carbonate • calcium + sodium → sodium + calcium • chloride carbonate chloride carbonate • CaCl2+ Na2CO3→ NaCl + CaCO3 • CaCl2+ Na2CO3→ NaCl + CaCO3 • CaCl2+ Na2CO3→ NaCl + CaCO3 • CaCl2+ Na2CO3→ NaCl + CaCO3 • 2. Switch the first names of the reactants. • 3. Write these names on the right side of the arrow. • 4. Starting on the far right, write the formulas. • 5. Moving to the left, continue to write formulas.

  18. Example • calcium + sodium→ • chloride carbonate • calcium + sodium → sodium + calcium • chloride carbonate chloride carbonate • CaCl2+ Na2CO3→ NaCl + CaCO3 • CaCl2+ Na2CO3→ NaCl + CaCO3 • CaCl2+ Na2CO3→ NaCl + CaCO3 • CaCl2+ Na2CO3→ NaCl + CaCO3 • 2. Switch the first names of the reactants. • 3. Write these names on the right side of the arrow. • 4. Starting on the far right, write the formulas. • 5. Moving to the left, continue to write formulas.

  19. Example • calcium + sodium→ • chloride carbonate • calcium + sodium → sodium + calcium • chloride carbonate chloride carbonate • CaCl2+ Na2CO3→ NaCl + CaCO3 • CaCl2+ Na2CO3→ NaCl + CaCO3 • CaCl2+ Na2CO3→ NaCl + CaCO3 • CaCl2+ Na2CO3→ NaCl + CaCO3 • 2. Switch the first names of the reactants. • 3. Write these names on the right side of the arrow. • 4. Starting on the far right, write the formulas. • 5. Moving to the left, continue to write formulas.

  20. Balancing Equations • CaCl2+ Na2CO3→ NaCl + CaCO3 • Must be the same number and kind of atoms on each side of the arrow. • CaCl2+ Na2CO3→ NaCl + CaCO3 • 1 Ca 2 Na1 Na 1 Ca • 2 Cl 1 CO3 1 Cl 1 CO3 • Use coefficients to multiply the number of molecules to achieve balance. • CaCl2+ Na2CO3→ 2 NaCl + CaCO3 • 1 Ca 2 Na 2 Na 1 Ca • 2 Cl 1 CO3 2 Cl 1 CO3

  21. Balancing Equations • CaCl2+ Na2CO3→ NaCl + CaCO3 • Must be the same number and kind of atoms on each side of the arrow. • CaCl2+ Na2CO3→ NaCl + CaCO3 • 1 Ca 2 Na1 Na 1 Ca • 2 Cl 1 CO3 1 Cl 1 CO3 • Use coefficients to multiply the number of molecules to achieve balance. • CaCl2+ Na2CO3→ 2 NaCl + CaCO3 • 1 Ca 2 Na 2 Na 1 Ca • 2 Cl 1 CO3 2 Cl 1 CO3

  22. Balancing Equations • CaCl2+ Na2CO3→ NaCl + CaCO3 • Must be the same number and kind of atoms on each side of the arrow. • CaCl2+ Na2CO3→ NaCl + CaCO3 • 1 Ca 2 Na1 Na 1 Ca • 2 Cl 1 CO3 1 Cl 1 CO3 • Use coefficients to multiply the number of molecules to achieve balance. • CaCl2+ Na2CO3→2 NaCl + CaCO3 • 1 Ca 2 Na2 Na 1 Ca • 2 Cl 1 CO3 2 Cl 1 CO3

  23. Stoichiometry stoy key om e treenn • The numerical relationship between the molecules in a reaction. • CaCl2+ Na2CO3→2 NaCl + CaCO3 • basic 1 molecule 1 molecule 2 molecules 1 molecule • ratio • 367 molecules 367 molecule 734 molecules 367 molecule • 1 mole 1 mole 2 moles 1 mole • 7 mole 7 mole 14 moles 7 mole • .09 moles .09 moles .18 moles .09 moles

  24. Stoichiometry 2nn • CaCl2+ Na2CO3→2 NaCl + CaCO3 • Since moles and coefficients both are related to the number of molecules, the coefficients in the equation, and the theoretical/actual moles in the reaction MUST have exactly the same ratio. • coefficient CaCl2 = moles CaCl2 = 1 • coefficient Na2CO3 moles Na2CO3 1 • coefficient CaCl2 = moles CaCl2 = 1 • coefficient NaCl moles NaCl 2 • coefficient CaCl2 = moles CaCl2 = 1 • coefficient CaCO3 moles CaCO3 1

  25. Stoichiometry 3nn • 2 AlCl3+ 3 CaCO3→ 3 CaCl2 + Al2(CO3)3 • Since moles and coefficients both are related to the number of molecules, the coefficients in the equation, and the theoretical/actual moles in the reaction MUST have exactly the same ratio. • coefficient AlCl3 = moles AlCl3 = 2 • coefficient CaCO3 moles CaCO3 3 • coefficient AlCl3 = moles AlCl3 = 2 • coefficient CaCl2 moles CaCl2 3 • coefficient AlCl3 = moles AlCl3 = 2 • coefficient Al2(CO3)3 moles Al2(CO3)3 1

  26. Practicenn • 2 AlCl3+ 3 CaCO3→ 3 CaCl2 + Al2(CO3)3 • If 1.6 moles of AlCl3 is used to react with CaCO3, how many moles of CaCl2 are produced, how many moles of Al2(CO3)3 are produced, and how many moles of CaCO3 are needed? • coefficient AlCl3 = moles AlCl3 = 2 = 1.6 • coefficient CaCO3 moles CaCO3 3 ? • coefficient AlCl3 = moles AlCl3 = 2 = 1.6 • coefficient CaCl2 moles CaCl2 3 ? • coefficient AlCl3 = moles AlCl3 = 2 = 1.6 • coefficient Al2(CO3)3 moles Al2(CO3)3 1 ?

  27. Practicenn • 2 AlCl3+ 3 CaCO3→ 3 CaCl2 + Al2(CO3)3 • If 1.6 moles of AlCl3 is used to react with CaCO3, how many moles of CaCl2 are produced, how many moles of Al2(CO3)3 are produced, and how many moles of CaCO3 are needed? • coefficient AlCl3 = moles AlCl3 = 2 = 1.6 • coefficient CaCO3 moles CaCO3 3 ? • coefficient AlCl3 = moles AlCl3 = 2 = 1.6 • coefficient CaCl2 moles CaCl2 3 ? • coefficient AlCl3 = moles AlCl3 = 2 = 1.6 • coefficient Al2(CO3)3 moles Al2(CO3)3 1 ?

  28. Datann • 2 HCl + Mg → H2 + MgCl2 • Trial Mass HCl(g) Mass Mg(g) ml H2 produced • 1 .178 .020 21 • 1 .178 .040 42 • 1 .178 .080 82 • 1 .178 .120 122 • 1 .178 .160 122 • 1 .178 .200 122 • Patterns?? • amount of H2 levels out

  29. Datann • 2 HCl + Mg → MgCl2 + H2 • Trial Mass HCl(g) Mass Mg(g) ml H2 produced • 1 .178 .020 21 • 1 .178 .040 42 • 1 .178 .080 82 • 1 .178 SEP .120 SEP 122 • 1 .178 .160 122 • 1 .178 .200 122 • Patterns?? • amount of H2 levels out • called ‘ stoichiometric equivalence point’

  30. Datann • 2 HCl + Mg → MgCl2 + H2 • Trial Mass HCl(g) Mass Mg(g) ml H2 produced • 1 .178 excess .020 21 • 1 .178 excess .040 42 • 1 .178 excess .080 82 • 1 .178SEP .120SEP 122 • 1 .178 .160 excess 122 • 1 .178 .200 excess 122 • Patterns?? • amount of H2 levels out • called ‘ stoichiometric equivalence point’ • reactant that determines product amount called limiting

  31. Datann • 2 HCl + Mg → MgCl2 + H2 • Trial Mass HCl(g) Mass Mg(g) ml H2 produced • 1 .178 excess .020 limiting 21 • 1 .178 excess .040 limiting 42 • 1 .178 excess .080 limiting 82 • 1 .178SEP .120SEP 122 • 1 .178 limiting .160 excess 122 • 1 .178 limiting .200 excess 122 • Patterns?? • amount of H2 levels out • called ‘ stoichiometric equivalence point’ • reactant that determines product amount called limiting • reactant that determines product amount called limiting

  32. Practicenn • 2 Sb + 3 I2→ 2 SbI3 • a) if 1.20 moles of Sb react, how much SbI3 is produced? • b) if 2.40 moles of I2 react, how much SbI3 is produced? • c) if 1.20 moles of Sb react with 2.40 moles of I2, how • much SbI3 is produced? • d) if 1.00 moles of SbI3 are produced, what per cent yield • of final product is produced? • actual yield x 100 = % yield • predicted yield

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