# Stoichiometry

Document Sample

```					Solutions are homogenous mixtures.
When a substance is mixed with a liquid and it
disintegrates into sub microscopic particles such as
molecules atoms or ions, the mixture is a homogenous
mixture and therefore a solution.

Solute
Solutions                          The substance dissolved
in the other.

Solvent
The substance present in
excess that dissolves the
solute.
Why is water called the
universal solvent?

because it can dissolve most ionic and
polar compounds.

Example of the dissolving process of an ionic
compound in water
http://www.northland.cc.mn.us/biology/Biology1111/animations/dissolve.html
Vocabulary
Solubility: amount of substance that can be dissolved in a
certain solvent.
Diluted: Contains small amount of solute for the given
amount of solution.
Concentrated: Contains large amounts of solute for the
given amount of solution.

Concentrated
solution                                 Diluted solution
Saturated solution: It cannot contain more solute at the
given temperature.

Supersaturated solution: Contains more solute that the
amount it can contain at the given temperature.

This is an unstable solution and the excess of solute will
eventually separate, with any disturbance, or with the use of
a seed crystal, from the solution forming a precipitate.

This process is called crystallization
http://genchem.chem.wisc.edu/demonstrations/Gen_Chem_Pages/11solutionspag
e/crystallization_from_super.htm
Concentration:

Is the amount of substance
contained within a given
volume of solution.

c = Amount of substance
Solution volume in dm3
MOLARITY

Molarity is the concentration of
a solution in moles of solute per
liter of solution.

M= n       or
M= n
VL             Vdm3
•Other forms of concentration used
are: % composition (parts per
hundred) and ppm (parts per million)

•ppm is used when the concentration
is very small (like the addition of
fluorine to tap water)

% by mass =    g solute    X   100
g solution

ppm = mg dm-3
•Sample Problems

Calculate the molarity of a solution prepared by dissolving
11.5g of solid NaOH in enough water to make 1.50 L of solution.

What do we need to find first?

The moles of NaOH…

1 mol NaOH
11.5g NaOH X                      = 0.288 mol NaOH
40.00 g NaOH
Molar mass
Na: 22.99
O: 16.00        M=n=       0.288 mol NaOH = 0.192 mol L-1
H: 1.01           V(L)      1.50L solution
40.00g/mol
•Sample Problems

What is the concentration in ppm of the solution in
the previous problem?.

What do we need to find first?

The mg of NaOH…

1,000 mg NaOH
11.5g NaOH X                = 11,500 mg NaOH
1 g NaOH

ppm = mg =      11,500 NaOH         = 7,670 ppm
V(L)      1.50L solution
Solve the problems in your notes
Dilution
When a solution is prepared
dissolving a concentrated solution
the dilution formula can be used.

Formula:   VcMc = VdMd

Remember that you can       only use
this formula for dilution
Stoichiometry

Stoichiometry refers to the use of a
balanced equation to calculate the
mass of products and/or reactants.

Example: How many moles and what mass
of carbon dioxide and water can be obtained
when 5.00g of ethane (C2H6) react with enough
oxygen (Combustion reaction).
Steps   1. Write and balance the equation for the
reaction
2. Convert the known (given) mass of the
reactant or product to moles of the
substance
3. Use the balanced equation to set the
appropriate mole ratio
4. Use the ratio to calculate the number of
moles of the desired reactant or product
5. Convert from moles back to grams if
required by the problem.

Solve the problems in your notes
Limiting and excess
reactants
Limiting reactant is the reactant that is totally
consumed during the reaction and therefore
controls the amount of products that can be
formed.

Excess reactant is the reactant that doesn’t
react completely during the reaction and
therefore there is an excess of it remaining after
the reaction.

Mg(s) + FeSO4(aq) ----- MgSO4 (aq) + Fe(s)
2.00g   2.00g     Which one will react completely?
Example
Mg(s) + FeSO4(aq) ----- MgSO4 (aq) + Fe(s)
2.00g Mg X 1 mol Mg = 0.0823 mole Mg
24.31g Mg
0.0823 mole Mg X 1 mole FeSO4 = 0.0823 mol FeSO4
1 mol Mg    Moles of FeSO4 needed to
react completely with
0.0823 moles Mg
2.00g FeSO4 X    1mole FeSO4 = 0.0132 moles FeSO4
151.92 g FeSO4 Moles of FeSO4 that we have

We need 0.0832 moles FeSO4 to react with 2.00 g of Mg
and we have only 0.0132 moles of FeSO4

Which reactant is going to be totally consumed in the reaction?
FeSO4 is the limiting reactant
It will be totally consumed during this reaction
It determines the amount of products that can
be obtained
How much iron (Fe) will be obtained?    (explain)
Answer: 0.737g Fe

Mg is the reactant in excess. This means that at the end
of the reaction there will be some magnesium left.

How much Mg will be left after the reaction?        (explain)

Answer: 0.320g Mg

Do prob.15 p 177. Students do 16
Molarity (review)

Molarity is the concentration of a
solution in moles of solute per liter
of solvent.
M= n
Vin dm3
M= Molarity
N= number of moles of solute
VL= Volume in liters of solution.
Do prob.17and 18 p 179.
Percent YIELD
The real amounts of products formed during a
chemical reaction are usually different from
the expected amounts.
The real amount of one product obtained
(experimental) is called the ACTUAL YIELD
The expected amount, the amount of products calculated
through the stoichiometry of the reaction, from the
limiting reactant, is the THEORETICAL YIELD
Actual yield    X 100 %
Percent Yield =
Theoretical Yield

Do prob.19and 20 p 182.
combustion reaction to
find the Empirical formula
A 0.496 g sample of an unknown hydrocarbon was
completely burned in oxygen. The sample produced
1.56 g of carbon dioxide and 0.638 g of water.
a) How many moles of carbon dioxide were
formed?
b) How many moles of water were formed?
c) What is the empirical formula of the
hydrocarbon?
d) A 26.0 g sample of another unknown
compound containing only carbon and hydrogen
was burned in excess oxygen and 88g of CO2
were produced. What is the possible molecular
formula of this compound?

```
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
 views: 21 posted: 3/26/2012 language: English pages: 18