Transcript Unit #5 - Wikispaces

Unit #5
Polyatomic Ions / Mole
/ Molarity /
Electrochemistry
This unit builds upon previous
topics:
Electron Configuration
Orbital Notation
Electron Dot Notation
Octet Rule
Remember:
Ion-
an atom or group of atoms that has either lost
or gained electron(s) and as a result has an
electric charge.
Cation-
is a positively charged ion.
Anion-
is a negatively charged ion.
Recall how we represented atoms with the
electron dot notation.
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We will represent ions the same way with two
exceptions:
1)
If the atom has at least 5, but less than 8 valence
electrons; we will draw as many additional electrons as
needed to acquire a total of 8 valence electrons. (This will
give the ion an Octet of valence electrons.)
1st exception con’t:
After the "atomic symbol" of the ion, in the upper
right corner of the symbol, we will write a negative
sign (or digit, if more than one) for each additional
electron we added to achieve the nearest noble gas
core. An octet of valence electrons.
1st exception con’t:
Example #1:
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1st exception con’t:
Example #2: Bromine
1st exception con’t:
Example #3: Oxygen
We will represent ions the same way with two
exceptions (con’t.):
2)
If the atom has less than 4 valence electrons; we
will remove as many electrons as needed to move back to
the nearest noble gas core. A total of 8 valence electrons.
(This will give the ion an Octet of valence electrons.)
2nd exception con’t:
After the "atomic symbol" of the ion, in the upper
right corner of the symbol, we will write a positive
sign (or digit, if more than one) for each additional
electron we subtracted to achieve the nearest noble
gas core. An octet of valence electrons.
2nd exception con’t:
Example #1: Sodium
2nd exception con’t:
Example #2: Aluminum
2nd exception con’t:
Example #3: Barium
Polyatomic ionA type of ion made of more than 2 atoms.
These will also be classified as either an anion
or a cation.
4 steps to drawing polyatomic ions:
Step 1) Determine and write down the "4 key
numbers".
These are:
p
e
c
v
"4 key numbers” (con’t.)
“p”
# of protons (add up the atomic number of all
involved atoms).
“e”
# of total electrons (add up the atomic numbers
(which equals the number of total electrons in a
neutral atom) of all atoms in the ion.
"4 key numbers” (con’t.)
“c”
# of core electrons (atomic number minus the
number of "s" and "p" electrons in the highest
energy level).
“v”
# of valence electrons (determined by
subtracting the # of core electrons from the #
total electrons).
of
Step 2)
Draw the "skeleton" of the ion
Step 3)
Give all atoms eight electrons (except
Hydrogen, it only wants two), (remember
each bond (represented by a dash)
counts as 2 electrons).
Step 4)
Count the number of electrons drawn and
compare this to the calculated number ("Key
number #4” the “v” ).
Adjust as necessary by adding double or triple
bonds to reduce the number of electrons while
still ensuring all atoms still have an octet of
valence electrons.
Draw brackets and valence charge to finish.
Examples:
Carbonate
Cyanide
Ammonium
The Mole Concept.
Mole The amount of substance that
contains Avagadro's number of particles
of that substance.
Abbreviated: mol.
NOT m, M, or M.
Avagadro's number This is equal to 6.02 * 1023 .
It is specifically the number of C12 atoms in 12 grams
(the chart mass in grams) of this isotope.
Think of this word like the word "Dozen", which
always means twelve. Well this number always means
6.02 * 1023 particles.
Atomic weight The weighted average of the masses of the
isotopes of an element.
Based on C12 , 1 atom = 12 AMUs (atomic mass units).
AMU -
arbitrary unit equal to 1/12 the mass
of a C12 atom. (Atomic Mass Unit)
AMU continued:
Imagine if we had to determine the "weight" of
everyone in the classroom, but, we didn't have a
scale. We could select someone and then assign to
that individual an arbitrary "weight" unit. We might
say he/she weight 10 CRUs (Chem. Room Units).
Everyone in the room will now be weighed relative to
that student.
The atomic weight of all elements is relative to the
"weight" of carbon 12.
Gram Atomic Weight The mass in grams of 1 mole of a substance.
Example 1: Hydrogen.
1 atom weighs ?
1.0079 amu
1 mole of hydrogen atoms weighs ? 1.0079 g.
However what is special about Hydrogen ? Its diatomic
What is the formula for a molecule of Hydrogen?
H2
1 molecule of Hydrogen weighs ?
2.0158 amu
1 mole of Hydrogen gas weighs ?
2.0158 g.
Example #2: CO2
1 molecule of Carbon dioxide weighs ?
44.0098 amu
1 mole of Carbon dioxide weighs ?
44.0098 g.
Make sure you are aware of what you
are working with:
Atoms or Molecules
AMUs or Grams
Molarity Is a concentration unit in:
#moles / liter. (The liter is usually water)
Note that the numerator is # of moles.
We will learn how to convert this to
grams in the next topic.
Abbreviated: M
NOT mol, M, m, or mol.
DIMO An acronym that means:
Divide In Multiply Out.
It is a drawing that you have to be able to
reproduce and use.
# of grams
# of particles
# of moles
22.4
# of liters of a gas
4 Steps to use the DIMO chart.
1)
Determine the chart mass of the
substance you're working with.
2)
Deal with the concentration of that substance.
3)
Deal with the volume of the solution you're
working with.
4)
Use "DIMO" to solve.
Example 1:
STEP 1:
How many grams of NaOH are in 1 liter of a
1M solution of the NaOH?
Determine the chart mass
Example 1 (con’t.):
STEP 1: 40 g/mol
STEP 2:
Deal with the concentration
Example 1 (con’t.):
STEP 1: 40 g/mol
STEP 2: 1 mol/L
STEP 3: Deal with the volume
Example 1 (con’t.):
STEP 1: 40 g/mol
STEP 2: 1 mol/L
STEP 3: 1 mol of NaOH in sample
STEP 4: Use "DIMO" to solve
Solution to Example 1:
How many grams of NaOH are in 1 liter of a
1M solution of the NaOH?
There are 40 g of NaOH in 1 liter of a 1M
solution of NaOH.
Example 2:
STEP 1:
0.75 liters of a 0.5M solution contains how
many grams of CaCl2?
Determine the chart mass
Example 2 (con’t.):
STEP 1: 111 g/mol
STEP 2:
Deal with the concentration
Example 2 (con’t.):
STEP 1: 111 g/mol
STEP 2: 0.5 mol/L
STEP 3: Deal with the volume
Example 2 (con’t.):
STEP 1: 111 g/mol
STEP 2: 0.5 mol/L
STEP 3: 0.375 mol of CaCl2 in sample
STEP 4: Use "DIMO" to solve
Solution to Example 2:
How many grams of CaCl2 are in 0.75 liter of a
0.5M solution of the CaCl2?
There are 41.625 g of CaCl2 in 0.75 liter of a
0.5M solution of CaCl2.
Notice in the previous examples we
calculated the number of grams of a
substance.
In the next example, we will determine
how many moles of a substance are in a
given volume of a solution.
Example 3:
STEP 1:
What is the Molarity of a solution created by
dissolving 150 grams of NaI into 250 mL of
distilled water?
Determine the chart mass
Example 3 (con’t.):
STEP 1: 150 g/mol
STEP 2:
Deal with the concentration
Example 3 (con’t.):
STEP 1: 150 g/mol
STEP 2: 150 g / 0.25 L
STEP 3: Deal with the volume
Example 3 (con’t.):
STEP 1: 150 g/mol
STEP 2: 150 g / 0.25 L
STEP 3: 600 g / 1 L
STEP 4: Use "DIMO" to solve
Solution to Example 3:
What is the Molarity of a solution created by
dissolving 150 grams of NaI into 250 mL of
distilled water?
600 g of NaI is 4 moles
There are 4 moles of NaI in 1 liter of the
solution of NaI.
Therefore the molarity is ____
4M
Finding the number of particles (atoms or
molecules) is done the same way: except you would
divide or multiply in or out by "Avagadro's number".
You will get practice doing this on your study guide.