CHEM1010-Fall_04-22.ppt

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


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.)


• 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


• 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.


• 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


• 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-


• 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


• 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


• 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


• 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


• 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


• 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


• 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


• 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


• 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


• 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


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.


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.


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.


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.


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.

Technology

CHEM1010-Fall_04-22.ppt