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Stoichiometry: Mathematics of chemical formulas and equations. mole = (mol) 602000000000000000000000. Mole. A mole is just a number pair = 2 trio = 3 quartet = 4 dozen = 12 baker’s dozen = 13 gross = 144. Avogadro’s Number. 6.02x10 23. How BIG is a mole?.

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stoichiometry mathematics of chemical formulas and equations
Stoichiometry:

Mathematics of chemical formulas and equations

slide2

mole = (mol) 602000000000000000000000

Mole
  • A mole is just a number

pair = 2

trio = 3

quartet = 4

dozen = 12

baker’s dozen = 13

gross = 144

how big is a mole
How BIG is a mole?

There are ~ 6.6 billion people on Earth

How many Earths would it take to equal the population of 1 mole?

9.12 x 1013

slide5
If you had a mole of cats . . .

They would create a sphere

larger than Earth!

slide6
If you had a mole of $$$$$ and you spent $800 billion dollars a day how many years would it take to spend a MOLEion dollars?

2.06 x 109 years

slide8
If you had a mole of H2O could you swim in it?

NO!

Water molecules are so small

that a mole of H2O = 18ml

how small are atoms
How small are atoms?
  • There are more atoms in one gram of salt than grains of sand on all the beaches of all the oceans in all the world.
slide10
Just one granule of sugar contains 1 x 1017 molecules
  • Each time you take a breath of air, you inhale about 2 x 1022 molecules of nitrogen and 5 x 1021 molecules of oxygen.
slide11
In chemistry we don’t work with individual atoms or molecules because they are too small to be weighed or measured
  • We have to work with LOTS of atoms in order to measure them

THAT’s WHERE THE

MOLE COMES IN!

gram atomic mass

of the

in

Gram ATOMic Mass
  • mass in grams of 1 mole of atoms of an element
  • In other words……

1 mol C atoms = 6.02 x 10 23 C atoms = 12g C

gram formula mass gram molecular mass molar mass

of the

in

Gram Formula MassGram Molecular Mass Molar Mass
  • mass in grams of 1 mole of a substance
  • In other words . . . Add it all up!

1 mole of NaCl = 58g = 6.02x1023 particles of NaCl

1 mole of H2O = 18g = 6.02x1023 molecules of H20

slide17
Now…..
  • Use the gram formula mass and the gram atomic mass to determine
    • how many moles or atoms of an element are found in some mass of a substance
    • how much mass that element contributes to the mass of the entire substance

Use factor label & follow the units!

stoich iometry of chemical formulas18
Stoichiometry of Chemical Formulas
  • If you have 1 molecule of (NH4)2SO4
    • How many atoms of N are there?
    • How many atoms of H?

SO…..

  • If you have 1 mole of of (NH4)2SO4
    • How many moles of N are there?
    • How many moles of H?
molar volume

Gases ONLY

@STP

Molar Volume
  • 1 mole of ANYgas

O2 (g)

NH3 (g)

He (g)

contains 6.02 x 1023 molecules and

occupies a volume of 22.4L

slide21
STP

standardtemperature & pressure

0oC or 273K

101.3kPa or 1 atm

remember
Remember . . .
  • Gases also have mass

1 mole of O2(g) = 32g

1 mole of NH3(g) = 17g

 we can calculate Density!

d m v
D = m/v
  • A sample of oxygen contains 3 moles of particles at STP what is its density?
  • 2 steps
    • Convert moles to mass AND volume
    • Calculate density
  • If mass or volume is given, use it and convert the other……. THEN calculate D!
slide24

Remember . . .

1 mole = 6.02x1023atoms or molecules =gfm = 22.4L (g)

stoich iometry of chemical equations

“it’s a simple matter

of weight ratios . . .”

Stoichiometry of Chemical Equations
  • The study of quantitative relationships that can be derived from chemical equations.
stoichiometry cookies
Stoichiometry cookies
  • If you look at chemical equations as recipes it may be easier to understand that
    • changing the amount of a reactant will change the amount of the product IN THE SAME RATIO!
examining molar relationships in balanced equations
Examining Molar Relationships in Balanced Equations

6CO2 + 12 H2O + 2804kJ  6O2 + C6H12O6 +6H20

Balanced equations

  • Law of conservation of mass / matter
    • ATOMS are not createdor destroyed during a chemical reaction, they are only rearranged to form new substances.
    • # atoms on reactant side = # atoms on product side
  • Law of conservation of E
    • E on the reactant side = E on the product side
synthesis reaction
SYNTHESIS Reaction
  • 2 or more reactants combine to build a single product

General Formula:

A + B  AB

3H2 + N2 2NH3

decomposition reaction
DECOMPOSITION reaction
  • A compound is broken down into 2 or more simpler substances

General Formula:

AB  A + B

2H2O  2H2 + O2

single replacement reaction
SINGLE REPLACEMENT reaction
  • One of the reactants is a single element. It becomes part of a compound as a product.

General Formula:

A + BX  B + AX

Mg + CaBr2 Ca + MgBr2

double replacement reaction
DOUBLE REPLACEMENT reaction
  • TWO elements switch places during the process of the reaction.

General Formula:

AB + CD AD + CB

LiCl + KBr LiBr + KCl

combustion
Combustion
  • Oxygen is always a reactant
  • CO2 and H2O are always products
  • _______ + O2 CO2 + H2O
practice
Practice
  • 2Al + 3CuSO4 Al2(SO4)3 + 3Cu
  • 2H2 + O2  2H2O
  • C12H22O11  11H2O + 12C
  • KCl + AgNO3  KNO3 + AgCl
  • CH4 + 2O2  CO2 + 2H2O