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    CHEM1010-Fall_04-22.ppt CHEM1010-Fall_04-22.ppt Presentation Transcript

    • Department of Chemistry CHEM1010 General Chemistry *********************************************** Instructor: Dr. Hong Zhang Foster Hall, Room 221 Tel: 931-6325 Email: hzhang@tntech.edu
    • CHEM1010/General Chemistry _________________________________________ Chapter 6. (L22)-Chemical Accounting
      • Today’s Outline
      • ..Review of Avogadro’s number, formula mass, mole, molar mass, and molar volume
      • ..Introduction to mole and mass relationships in chemical reaction equations
      • ..Molar relationships in chemical equations
      • ..Mass relationships in chemical equations
    • Chapter 6. (L22)-Chemical Accounting
      • Building Your
      • Chemical Vocabulary
      • NO 3 - : nitrate, an anion
      • SO 4 2- : sulfate, an anion
      • Both anions are the major anionic components of acid rain (acid precipitation, including acid rain, acid snow, acid fog, etc.)
    • Chapter 6. (L22)-Chemical Accounting
      • One of the most important numbers in chemistry: Avogadro’s number
      • Avogadro’s number:
      • The number of atoms in a 12-g sample of carbon-12 is called Avogadro’s number
      • Avogadro’s number has been experimentally determined to be 6.0221367 ×10 23
      • But, 6.02 ×10 23 is sufficiently enough for our purpose in general chemistry
    • Chapter 6. (L22)-Chemical Accounting
      • Counting molecules, the unit of mole (like dozen in chemistry)
      • Definition: 1 mole is an amount of substance that contains the same number of elementary units (molecules, or atoms, or ions) as there are atoms in exactly 12 g of carbon-12.
      • By the definition of Avogadro’s number, we know that
      • 1 mole is the amount of substance that contains the same number of elementary units (molecules, or atoms, or ions) exactly as the Avogadro’s number, that is, 6.02 ×10 23 .
    • Chapter 6. (L22)-Chemical Accounting
      • Formula masses
      • Definition: Formula mass of a molecule or ion is the sum of the masses of each of the atoms represented by the formula
      • Example:
      • Formula mass of O 2 = 16.0u ×2 = 32.0u
      • Formula mass of SO 2 = 32.1u + 16.0u×2 = 64.1u
      • Formula mass of CO 2 = 12.0u + 16.0u×2 = 44.0u
    • Chapter 6. (L22)-Chemical Accounting
      • Molar mass
      • Definition: The molar mass of a substance is the mass of 1 mole of that substance in the unit of gram. It is numerically equivalent to the atomic mass of the atom or the formula mass of the molecule of concern.
      • Examples:
      • mass of 1 mole Na = 23.0 g Na
      • mass of 1 mole CO 2 = 44.0 g CO 2
      • mass of 1 mole O 2 = 32.0 g O 2
      • mass of 1 mole CO 3 2- = 60.0 g CO 3 2-
    • Chapter 6. (L22)-Chemical Accounting
      • Introduction to mole and mass relationships in chemical reaction equations
      • The need for mole calculations in chemistry:
      • 2H 2 + O 2 = 2H 2 O
      • ? mole ? mole 4 mole
      • 0.5 mole ? mole
      • 2 moles ? mole
    • Chapter 6. (L22)-Chemical Accounting
      • Introduction to mole and mass relationships in chemical reaction equations
      • The need for mass calculations in chemistry:
      • 2H 2 + O 2 = 2H 2 O
      • ? g ? g 72 g
      • 4 g ? g
      • 64 g ? g
    • Chapter 6. (L22)-Chemical Accounting
      • Calculations of quantities of mole and mass in chemical equations: The principle of mole ratio
      • Molecule Formula Mass Molar Mass
      • H 2 2u 2g
      • O 2 32u 32g
      • H 2 O 18u 18g
      • 2H 2 + O 2 = 2H 2 O
      • 2 molecules 1 molecule 2 molecules
      • 2 moles 1 mole 2 moles
    • Chapter 6. (L22)-Chemical Accounting
      • Calculations of quantities of mole and mass in chemical equations: The principle of mass ratio
      • Molecule Formula Mass Molar Mass
      • H 2 2u 2g
      • O 2 32u 32g
      • H 2 O 18u 18g
      • 2H 2 + O 2 = 2H 2 O
      • 2 molecules 1 molecule 2 molecules
      • 2 moles 1 mole 2 moles
      • 4g 32g 36g
    • Chapter 6. (L22)-Chemical Accounting
      • Molar relationships in chemical equations
      • Calculation example:
      • 2H 2 + O 2 = 2H 2 O
      • 2 moles 1 moles 2 moles
      • ? moles ? mole 1.5 moles
      • ? mole H 2 = 1.5 mole H 2 O ×2 mole H 2 /2 mole H 2 O
      • = 1.5 mole H 2
      • ? mole O 2 = 1.5 mole H 2 O ×1 mole O 2 /2 mole H 2 O
      • = 0.75 mole O 2
    • Chapter 6. (L22)-Chemical Accounting
      • Molar relationships in chemical equations
      • Calculation example:
      • 2H 2 + O 2 = 2H 2 O
      • 2 moles 1 mole 2 moles
      • ? moles 2.5 moles ? Moles
      • ? mole H 2 = 2.5 mole O 2 ×2 mole H 2 /1 mole O 2
      • = 5.0 mole H 2
      • ? mole H 2 O= 2.5 mole O 2 ×2 mole H 2 O/1 mole O 2
      • = 5.0 mole H 2 O
    • Chapter 6. (L22)-Chemical Accounting
      • Mass relationships in chemical equations
      • Calculation example:
      • 2H 2 + O 2 = 2H 2 O
      • 4.0g 32g 36g
      • ?g ?g 9.0g
      • ? g H 2 = 9.0g H 2 O ×4g H 2 /36 g H 2 O
      • = 1.0 g H 2
      • ? g O 2 = 9.0 H 2 O ×32g O 2 /36g H 2 O
      • = 8.0 mole O 2
    • Chapter 6. (L22)-Chemical Accounting
      • Mass relationships in chemical equations
      • Calculation example:
      • 2H 2 + O 2 = 2H 2 O
      • 4.0g 32g 36g
      • ?g 4.0g ?g
      • ? g H 2 = 4.0g O 2 ×4.0g H 2 /32g O 2
      • = 0.5g H 2
      • ? g H 2 O= 4.0g O 2 ×36g H 2 O/32g O 2
      • = 4.5g H 2 O
    • Chapter 6. (L22)-Chemical Accounting
      • Mass relationships in chemical equations
      • Calculation example:
      • C + O 2 = CO 2
      • 12.0g 32g 44g
      • 10.0g ?g ?g
      • ? g O 2 = 10.0g C ×32g O 2 /12g C
      • = 26.7g O 2
      • ? g CO 2 = 10.0g C ×44g CO 2 /12g C
      • = 36.7g CO 2
    • Chapter 6. (L22)-Chemical Accounting
      • Mass relationships in chemical equations
      • Calculation example:
      • Molar mass of NaN 3 : 65 g
      • Molar mass of N 2 : 28 g
      • 2NaN 3 = 2Na + 3N 2
      • 130.0g 46.0g 84.0g
      • 60.0g ?g
      • ? g N 2 = 60.0g NaN 3 ×84g N 2 /130g NaN 3
      • = 38.8g N 2
    • Chapter 6. (L22)-Chemical Accounting
      • Quiz Time
      • How many grams of water can be produced out of 8 g H 2 gas from the reaction:
      • 2H 2 + O 2 = 2H 2 O
      • (a) 8.0 g H 2 O;
      • (b) 18.0 g H 2 O;
      • (c) 72.0 g H 2 O;
      • (d) 36.0 g H 2 O.
    • Chapter 6. (L22)-Chemical Accounting
      • Quiz Time
      • How many moles of water can be produced out of 8 moles of CH 4 gas from the reaction:
      • CH 4 + 2O 2 = 2H 2 O + CO 2
      • (a) 8.0 mole H 2 O;
      • (b) 32.0 mole H 2 O;
      • (c) 4.0 mole H 2 O;
      • (d) 16.0 mole H 2 O.
    • Chapter 6. (L22)-Chemical Accounting
      • Quiz Time
      • How many grams of water can be produced out of 8 g of CH 4 gas from the reaction:
      • CH 4 + 2O 2 = 2H 2 O + CO 2
      • (a) 8.0 g H 2 O;
      • (b) 32.0 g H 2 O;
      • (c) 16.0 g H 2 O;
      • (d) 18.0 g H 2 O.
    • Chapter 6. (L22)-Chemical Accounting
      • Quiz Time
      • How many grams of ammonia can be produced out of 20.0 g of N 2 from the reaction:
      • 3H 2 + N 2 = 2NH 3
      • (a) 29.4 g N 2 ;
      • (b) 14.9 g N 2 ;
      • (c) 24.3 g N 2 ;
      • (d) 29.1 g N 2 .
    • Chapter 6. (L22)-Chemical Accounting
      • Quiz Time
      • How many moles or grams of N 2 gas would be needed, respectively, to produce 1.25 moles of NH 3 from the reaction:
      • 3H 2 + N 2 = 2NH 3
      • (a) 0.625 mole or 6.5 g N 2 ;
      • (b) 1.0 mole or 1.0 g N 2 ;
      • (c) 0.625 mole or 1.75 g N 2 ;
      • (d) 0.625 mole or 17.5 g N 2 .