chem1010-fall_04-22.ppt

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Department of Chemistry CHEM1010 General Chemistry *********************************************** Instructor: Dr. Hong Zhang Foster Hall, Room 221 Tel: 931-6325 Email: [email protected]

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Page 1: CHEM1010-Fall_04-22.ppt

Department of Chemistry

CHEM1010 General Chemistry***********************************************

Instructor: Dr. Hong ZhangFoster Hall, Room 221

Tel: 931-6325

Email: [email protected]

Page 2: CHEM1010-Fall_04-22.ppt

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

Page 3: CHEM1010-Fall_04-22.ppt

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.)

Page 4: CHEM1010-Fall_04-22.ppt

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

Page 5: CHEM1010-Fall_04-22.ppt

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.

Page 6: CHEM1010-Fall_04-22.ppt

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

Page 7: CHEM1010-Fall_04-22.ppt

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 CO2 = 44.0 g CO2

mass of 1 mole O2 = 32.0 g O2

mass of 1 mole CO32- = 60.0 g CO3

2-

Page 8: CHEM1010-Fall_04-22.ppt

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 mole

0.5 mole ? mole

2 moles ? mole

Page 9: CHEM1010-Fall_04-22.ppt

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

Page 10: CHEM1010-Fall_04-22.ppt

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

H2 2u 2g

O2 32u 32g

H2O 18u 18g

2H2 + O2 = 2H2O2 molecules 1 molecule 2 molecules2 moles 1 mole 2 moles

Page 11: CHEM1010-Fall_04-22.ppt

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

H2 2u 2g

O2 32u 32g

H2O 18u 18g

2H2 + O2 = 2H2O2 molecules 1 molecule 2 molecules2 moles 1 mole 2 moles4g 32g 36g

Page 12: CHEM1010-Fall_04-22.ppt

Chapter 6. (L22)-Chemical Accounting

• Molar relationships in chemical equationsCalculation example:

2H2 + O2 = 2H2O2 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

Page 13: CHEM1010-Fall_04-22.ppt

Chapter 6. (L22)-Chemical Accounting

• Molar relationships in chemical equationsCalculation example:

2H2 + O2 = 2H2O2 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

Page 14: CHEM1010-Fall_04-22.ppt

Chapter 6. (L22)-Chemical Accounting

• Mass relationships in chemical equationsCalculation example:

2H2 + O2 = 2H2O4.0g 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

Page 15: CHEM1010-Fall_04-22.ppt

Chapter 6. (L22)-Chemical Accounting

• Mass relationships in chemical equationsCalculation example:

2H2 + O2 = 2H2O4.0g 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

Page 16: CHEM1010-Fall_04-22.ppt

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

Page 17: CHEM1010-Fall_04-22.ppt

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

Page 18: CHEM1010-Fall_04-22.ppt

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.

Page 19: CHEM1010-Fall_04-22.ppt

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.

Page 20: CHEM1010-Fall_04-22.ppt

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.

Page 21: CHEM1010-Fall_04-22.ppt

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.

Page 22: CHEM1010-Fall_04-22.ppt

Chapter 6. (L22)-Chemical Accounting

Quiz Time

How many moles or grams of N2 gas would be needed, respectively, to produce 1.25 moles of NH3 from the reaction:

3H2 + N2 = 2NH3

(a) 0.625 mole or 6.5 g N2;

(b) 1.0 mole or 1.0 g N2;

(c) 0.625 mole or 1.75 g N2;

(d) 0.625 mole or 17.5 g N2.