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Chemical Equations

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Moles: A New Conversion Factor. 1 mol = 6.022 1023 parts. 1 mol atoms = atomic weight in grams ... atom is equal to its molar mass in grams (conversion factor) ... – PowerPoint PPT presentation

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Title: Chemical Equations


1
Chemical Equations
  • Law of Conservation of Mass
  • Matter can not be created or destroyed
  • Balanced chemical equations reflect this

2
Balancing Chemical Equations
Reactants
Products
Not balanced
MgCl2 (aq) AgNO3 (aq) ? AgCl (s)
Mg(NO3)2 (aq)
Magnesium chloride
Silver(I) chloride
Magnesium nitrate
Silver(I) nitrate
  • Numbers and identities of atoms must be the same
    on both sides of a reaction (rxn) arrow
  • Total charge must be the same on both sides of
    the rxn arrow

3
Balanced Equations and Coefficients
MgCl2 (aq) AgNO3 (aq) ? AgCl (s)
Mg(NO3)2 (aq)
  • Coefficients go in front of chemical and indicate
    how many
  • Coefficients should be the smallest possible
    whole number

4
Reaction Amounts
A B ? C
  • How much A do we mix with B to get a certain
    amount of C?
  • Depends on relationship between A, B, and C
  • Coefficients tell us relationship
  • parts 1 part A to 1 part B makes 1 part C

5
Parts in Chemistry equals Moles
  • Avogadros number and Moles
  • 6.022 ? 1023 of anything equals 1 mole
  • In Chemistry 6.022 ? 1023 atoms or molecules
    equal 1 mole
  • 2 parts red to 2 parts green to 1 part yellow
    equals 1 Rah
  • (microscopic world)
  • 2 moles red to 2 moles green to 1 mole yellow
    equals
  • 1 mole Rah
  • (macroscopic world)

1 yellow 2 red 2 Green ? 1 rah C 2 Cl
2 H ? CH2Cl2
6
Moles and Grams
  • Avogadros number
  • 6.022 ? 1023 atoms or molecules equal 1 mole
  • 1 mole of atoms equals the atomic weight in
    grams
  • ie. 1 mole of O atoms (6.022 ? 1023 atoms)
    16.00 g
  • 1 atom of O 16.00 amu, how many atoms weigh
    16.00 g?
  • 16.00 g O ? 6.022 ? 1023 amu ? 1 atom O 6.022
    ? 1023 atoms O
  • 1 g 16.00 amu

7
Moles A New Conversion Factor
  • 1 mol 6.022 ? 1023 parts
  • 1 mol atoms atomic weight in grams
  • 1 mol of molecules molar mass in grams
  • Molar Mass, Molecular Weight, Formula Weight
  • Sum of all atomic weights of the atoms in a
    formula
  • Can be determined using a mass spectrometer

8
Converting Between Grams, Molecules, and Moles
  • If you have 20.0 g of CO2, how many moles of CO2
    do you have? How many molecules of CO2 do you
    have?

9
Formulas, moles, and atoms
Moles talk to moles
  • Subscripts indicate the relative molar ratio of
    elements
  • ie. CO2 has 1 mol of C for every 1 mol of CO2
    and
  • 2 mol of O for every 1 mol of CO2 and 2 mol of O
    for every 1 mol of C

10
Percent Composition (By Mass)
mass of x in substance total mass of substance
Mass Percent of x
? 100
Percent Composition in a compound
  • Determine the molar mass of the compound (total
    mass of 1 mole of substance)
  • Determine the mass of each element in 1 mole of
    compound (mass of x in substance)
  • Use the above formula
  • The mass percent of all elements in a compound
    should add up to 100

11
Working backwards from compositions to formulas
  • Start with mass of elements (i.e. empirical
    data) and calculate a formula
  • Mass often determined using combustion analysis

12
Stoichiometric Amounts of Products and Reactants
  • Stoichiometric amounts represent the ideal or
    perfect amount of reactants and products
  • Stoichiometric amounts come from the mole to mole
    ratios of reactants and products (coefficients in
    balanced chemical equation)

13
Mole Relationships and Chemical Equations
Pb(NO3)2 (aq) KI (aq) ?
KNO3 (aq) PbI2 (s)
2
2
2.54 g
  • 1. Start with balanced chemical equation

2. Coefficients in front of compounds tell you
the mole relationships
1 mole of Pb(NO3)2 plus 2 moles of KI makes 2
moles of KNO3 and 1 mole of PbI2
14
Mole Relationships and Grams
  • Determine the molar mass of compounds of interest
  • 1 mole of a compound or atom is equal to its
    molar mass in grams (conversion factor)
  • Calculate the moles of each starting material (if
    moles are given in the beginning start here)
  • Convert to moles of compound of interest
  • Convert to grams of compound of interest

15
Experimentally Determine Amount of Product
  • Mix preweighed starting materials together
  • Filter final solution into pre-weighed piece of
    filter paper
  • Let filter paper and product dry
  • Weigh product and filter paper
  • Subtract the weight of the filter paper from the
    weight in step 3.
  • Mass of Product also called experimental yield of
    product or actual yield of product.

16
Percent Yield
  • Relates experimental findings to theoretical
    predictions
  • Percent Yield Experimental yield ? 100
  • Theoretical yield

17
Limiting Reagent
Pb(NO3)2 (aq) 2 KI (aq) ?
2 KNO3 (aq) PbI2 (s)
2.54 g
3.00 g
  • How much PbI2 (s) should we get from mixing the
    amounts of reactants shown above?
  • One reactant runs out before the other, ie. it
    limits the amount of product(s) that can be
    formed
  • The other product is in excess, ie. some is left
    over at the end of the reaction
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