Department of Chemistry CHEM1010 General Chemistry *********************************************** Instructor: Dr. Hong Zhang Foster Hall, Room 221 Tel: 931-6325 Email: [email protected]
Jul 15, 2015
Department of Chemistry
CHEM1010 General Chemistry***********************************************
Instructor: Dr. Hong ZhangFoster Hall, Room 221
Tel: 931-6325Email: [email protected]
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
NO3-: nitrate, an anion
SO42-: 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 6.0221367×1023
But, 6.02×1023 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, 6.02×1023.
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 O2 = 16.0u×2 = 32.0u
Formula mass of SO2 = 32.1u + 16.0u×2 = 64.1u
Formula mass of CO2 = 12.0u + 16.0u×2 = 44.0u
Chapter 6. (L22)-Chemical Accounting
• Molar massDefinition: 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 Namass of 1 mole CO2 = 44.0 g CO2
mass of 1 mole O2 = 32.0 g O2
mass of 1 mole CO32- = 60.0 g CO3
2-
Chapter 6. (L22)-Chemical Accounting
• Introduction to mole and mass relationships in chemical reaction equations
The need for mole calculations in chemistry:
2H2 + O2 = 2H2O
? mole ? mole 4 mole0.5 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:
2H2 + O2 = 2H2O
? 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 MassH2 2u 2g
O2 32u 32g
H2O 18u 18g
2H2 + O2 = 2H2O
2 molecules 1 molecule 2 molecules2 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 MassH2 2u 2g
O2 32u 32g
H2O 18u 18g
2H2 + O2 = 2H2O
2 molecules 1 molecule 2 molecules2 moles 1 mole 2 moles4g 32g 36g
Chapter 6. (L22)-Chemical Accounting
• Molar relationships in chemical equationsCalculation example:2H2 + O2 = 2H2O
2 moles 1 moles 2 moles? moles ? mole 1.5 moles
? mole H2 = 1.5 mole H2O×2 mole H2/2 mole H2O
= 1.5 mole H2
? mole O2 = 1.5 mole H2O×1 mole O2/2 mole H2O
= 0.75 mole O2
Chapter 6. (L22)-Chemical Accounting
• Molar relationships in chemical equationsCalculation example:2H2 + O2 = 2H2O
2 moles 1 mole 2 moles? moles 2.5 moles ? Moles
? mole H2 = 2.5 mole O2×2 mole H2/1 mole O2
= 5.0 mole H2
? mole H2O= 2.5 mole O2×2 mole H2O/1 mole O2
= 5.0 mole H2O
Chapter 6. (L22)-Chemical Accounting
• Mass relationships in chemical equationsCalculation example:2H2 + O2 = 2H2O
4.0g 32g 36g?g ?g 9.0g
? g H2 = 9.0g H2O×4g H2/36 g H2O
= 1.0 g H2
? g O2 = 9.0 H2O×32g O2/36g H2O
= 8.0 mole O2
Chapter 6. (L22)-Chemical Accounting
• Mass relationships in chemical equationsCalculation example:2H2 + O2 = 2H2O
4.0g 32g 36g?g 4.0g ?g
? g H2 = 4.0g O2×4.0g H2/32g O2
= 0.5g H2
? g H2O= 4.0g O2×36g H2O/32g O2
= 4.5g H2O
Chapter 6. (L22)-Chemical Accounting
• Mass relationships in chemical equationsCalculation example:C + O2 = CO2
12.0g 32g 44g10.0g ?g ?g
? g O2 = 10.0g C×32g O2/12g C
= 26.7g O2
? g CO2 = 10.0g C×44g CO2/12g C
= 36.7g CO2
Chapter 6. (L22)-Chemical Accounting
• Mass relationships in chemical equationsCalculation example:Molar mass of NaN3: 65 g
Molar mass of N2: 28 g
2NaN3 = 2Na + 3N2
130.0g 46.0g 84.0g60.0g ?g
? g N2 = 60.0g NaN3×84g N2/130g NaN3
= 38.8g N2
Chapter 6. (L22)-Chemical Accounting
Quiz Time
How many grams of water can be produced out of 8 g H2 gas from the reaction:
2H2 + O2 = 2H2O
(a) 8.0 g H2O;
(b) 18.0 g H2O;
(c) 72.0 g H2O;
(d) 36.0 g H2O.
Chapter 6. (L22)-Chemical Accounting
Quiz Time
How many moles of water can be produced out of 8 moles of CH4 gas from the reaction:
CH4 + 2O2 = 2H2O + CO2
(a) 8.0 mole H2O;
(b) 32.0 mole H2O;
(c) 4.0 mole H2O;
(d) 16.0 mole H2O.
Chapter 6. (L22)-Chemical Accounting
Quiz Time
How many grams of water can be produced out of 8 g of CH4 gas from the reaction:
CH4 + 2O2 = 2H2O + CO2
(a) 8.0 g H2O;
(b) 32.0 g H2O;
(c) 16.0 g H2O;
(d) 18.0 g H2O.
Chapter 6. (L22)-Chemical Accounting
Quiz Time
How many grams of ammonia can be produced out of 20.0 g of N2 from the reaction:
3H2 + N2 = 2NH3
(a) 29.4 g N2;
(b) 14.9 g N2;
(c) 24.3 g N2;
(d) 29.1 g N2.