Last week…


Review of ionic and net ionic equations.
Review of water solubilities.
Table of Water Solubilities
Being able to use this table is essential for both writing
ionic equations and ranking the solubility of
compounds.
bromide
carbonate
chloride
hydroxide
iodide
nitrate
oxide
phosphate
sulfate
sulfide
acetate
sodium
s
s
s
s
s
s
s
d
s
s
S
silver
ss
i
i
i
n
i
s
i
i
ss
i
i= nearly insoluble
ss=slightly soluble
s=soluble
d=decomposes
n=not isolated
Ionic and Net Ionic Equations
Last week we mixed solution #2 (NaCl) with solution
#4 (AgNO3).
This equation is written as:
NaCl + AgNO3  NaNO3 + AgCl
bromide
carbonate
chloride
hydroxide
iodide
nitrate
oxide
phosphate
sulfate
sulfide
acetate
sodium
s
s
s
s
s
s
s
d
s
s
S
silver
ss
i
i
i
n
i
s
i
i
ss
i
i= nearly insoluble
ss=slightly soluble
s=soluble
d=decomposes
n=not isolated
Ionic and Net Ionic Equations
To write an ionic equation from the given equation:
NaCl + AgNO3  NaNO3 + AgCl
We first need to check the table to see which
compounds disassociate in water…
bromide
carbonate
chloride
hydroxide
iodide
nitrate
oxide
phosphate
sulfate
sulfide
acetate
sodium
s
s
s
s
s
s
s
d
s
s
S
silver
ss
i
i
i
n
i
s
i
i
ss
i
i= nearly insoluble
ss=slightly soluble
s=soluble
d=decomposes
n=not isolated
Ionic and Net Ionic Equations
Since the compounds in red disassociate, we need to split them up:
NaCl + AgNO3  NaNO3 + AgCl
becomes
Na+ + Cl- + Ag+ + NO3-  Na+ + NO3- + AgCl
Notice the insoluble AgCl stays together.
bromide
carbonate
chloride
hydroxide
iodide
nitrate
oxide
phosphate
sulfate
sulfide
acetate
sodium
s
s
s
s
s
s
s
d
s
s
S
silver
ss
i
i
i
n
i
s
i
i
ss
i
i= nearly insoluble
ss=slightly soluble
s=soluble
d=decomposes
n=not isolated
Ionic and Net Ionic Equations
To write a net ionic equation from the newly created ionic equation:
Na+ + Cl- + Ag+ + NO3-  Na+ + NO3- + AgCl
Simply eliminate any ion that appears on both the reactant and
product side of the equation (shown in orange).
This leaves us with: Ag+ + Cl-  AgCl
bromide
carbonate
chloride
hydroxide
iodide
nitrate
oxide
phosphate
sulfate
sulfide
acetate
sodium
s
s
s
s
s
s
s
d
s
s
S
silver
ss
i
i
i
n
i
s
i
i
ss
i
i= nearly insoluble
ss=slightly soluble
s=soluble
d=decomposes
n=not isolated
Table of Water Solubilities
To rank solubility of compounds, simply use the table.
For example, in order of increasing solubility:
silver chloride (i)  silver sulfate (ss)  silver nitrate (s)
bromide
carbonate
chloride
hydroxide
iodide
nitrate
oxide
phosphate
sulfate
sulfide
acetate
sodium
s
s
s
s
s
s
s
d
s
s
S
silver
ss
i
i
i
n
i
s
i
i
ss
i
i= nearly insoluble
ss=slightly soluble
s=soluble
d=decomposes
n=not isolated
Questions?

If anything from the last two labs is still unclear,
be sure to see me today BEFORE you go take
the quiz!
Lab #6
Stoichiometry
Chemistry 108
Instructor:
Kristine Cooper
Stoichiometry

Chemical math

The branch of chemistry that quantifies the substances
in a chemical reaction

The art of figuring how much stuff you'll make in a
chemical reaction from the amount of each reagent you
start with
Atomic weight

Atomic weight of an element (found on the
periodic table) is given in AMUs.

For example, the atomic weight of carbon is 12
amu.

This is the mass of ONE atom of carbon.
Gram atomic weight

The gram atomic weight of an element is the
same NUMBER as the element’s atomic
weight, the UNITS change.

For example, the gram atomic weight of carbon
is 12 g.

This is the mass of 6.02x1023 atoms of carbon
(an Avagardo’s number or one mole).
Formula weight / Molecular weight


Formula weight is the combined weight of all
atoms in a compound.
Example:
–
Propanol CH3(CH2)2OH
What is the formula weight?
Count the number of each atom, multiply by it’s
atomic weight of that atom, then total.
Percent composition


Calculated by determining the percent (by
mass) of each individual atom of the formula
weight of the compound.
Example:
–
Propanol CH3(CH2)2OH
We determined the formula weight earlier to be 60 amu.
Now determine what % of 60 amu the total mass of
each atom comprises.
Use the total mass of each type of atom
determined before, then divide by the formula
weight.
Simplest empirical formula
Example:
A compound was found to contain 88.8% oxygen and 11.2%
hydrogen, what is the empirical formula?
1.
Set the amount of the compound as 100g.
In a 100g sample, there would be 88.8g oxygen and 11.2g
hydrogen.
2.
Determine the moles of each kind of atom.
88.8 g O / 16 g/mol=5.55 mol O
11.2 g H / 1 g/mol = 11.1 mol H
3.
Set these numbers as subscripts, then simplify the ratio
(divide by the smallest subscript present).
H11.1O5.5  H2O
True molecular formula



Simple empirical formulas only tell us the ratio
of atoms present in a compound.
True molecular formulas tell us the true number
of each atom present.
THESE ARE ALWAYS EVEN MULTIPLES OF
THE EMPIRICAL FORMULA!
True molecular formula


For example, we determined water’s empirical
formula to be H2O, which has a formula weight
of 18g for each water molecule.
What is the true molecular formula for 90g of
water?
True molecular formula

First, divide 90g by the molecular weight of
water (18g). 90g/18g= 5

Next, multiply each subscript in the empirical
formula (H2O) by this number (5).

H=2x5=10 and O=1x5=5, so the true molecular
formula is H10O5.
Law of Definite Composition

The chemical composition of a substance is
fixed, it never varies with sample size.

Simply stated, water is still H2O whether you
have a mL vs. 5L, etc.
Stoichiometry practice
Calculate the molecular weight
of methane (CH4).
Stoichiometry practice
12 moles of a certain
compound weighs 702 g.
What is the molecular weight
of the compound?
Stoichiometry practice
Calculate the percent
composition of each element
in hydrogen peroxide (H2O2).
Stoichiometry practice
Calculate the mass (in grams)
of 1.15 moles of vandium
sulfide (VS2).
Stoichiometry practice
You want to manufacture sodium
monoxide following the process:
2Na + O2 → 2Na2O.
If you begin with 11.5g Na, how much
O2 will be necessary to ensure all the
Na has reacted?
Today in Lab



Given an unknown oxygen-containing
potassium salt, we will determine its chemical
formula.
By removing the oxygen from the compound
and then calculating the difference in weight
with and without the oxygen, we can determine
the original composition.
Check out the demo compounds, compare
volumes.