Download Chemical Quantities(mole).

Chemical Quantities: Standard C-4.4
What is a mole?
The Mole:
Is the SI unit for the amount of substance and is
abbreviated “mol”. One mole contains 6.02 x 1023
representative particles.
For instance, one mole = 6.02 x 1023 atoms
one mole = 6.02 x 1023 ions
one mole = 6.02 x 1023 molecules
one mole = 6.02 x 1023 ion pairs.
Avogadro’s Number, NA (or constant) = 6.02 x 1023 atoms.
Compare:
One dozen contains 12 objects.
1 dozen eggs = 12 eggs
=
1 dozen cookies = 12 cookies =
1 mole Na = 6.02 x 1023 atoms Na
1 mole Pb = 6.02 x 1023 atoms Pb
600 g
200 g
= 22.99 g
= 207.2 g
Atomic Mass: the weighted average of the masses of the
isotopes of an element. Ex: 1 mole of Na = 22.99 g.
How many atoms are there in 0.500 moles of Na?
EE or Exp
0.500 mol Na x 6.02 x 1023 atoms Na = 3.01 x 1023 atoms Na
1 mole of Na
0.250 mol Na x 6.02 x 1023 atoms Na = 1.51 x 1023 atoms Na
1 mole of Na
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How many moles are there in 2.50 x 1023 atoms of Na?
2.50 x 1023 atoms x
1 mole
= 0.415 moles of Na
23
6.02 x 10 atoms Na
(note: Number on bottom means that you have to divide by that
number)
Molar Mass
(tp p.409, cp ch.7)
Molecular Mass of a covalent compound is the (atomic)
mass of one molecule.
H2O = H
O
2 x 1.01 = 2.02 g
+ 1 x 15.99 = 15.99 g
18.01 g/mol
Calculate the molar (molecular) mass of CH4 (methane)
CH4 = 16.05(u or g) u is units, g is grams
Formula Mass of an ionic compound is the (atomic)
mass of one formula unit.
NaCl = 22.99 + 35.45 = 58.44 (u or g)
MnCl2 = 54.94 + 70.90 = 125.84 g/mol
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How do you determine the amount of moles in a
compound?
1. If you have 9.0 grams of water, you have ____moles of
water.
Water - H2O = 18.01 g/mole (gmm)
9.0 g H2O x 1 mole of H2O = 0.50 mol of H2O
18.01 g H2O
1.b 13.0g H2O x 1 mole of H2O
18.01 g H2O
= 0.722 mol of H2O
2. If you weigh out 5.00g of Formaldehyde, how many
moles is that? 60.0 g of formaldehyde?
CH2O – Formaldehyde 30.02 g/mol (gmm)
a. 5.00g CH2O x 1 mole CH2O = 0.167mol CH2O
30.02 g CH2O
b. 60.0g CH2O x 1 mole CH2O = 2.00 mol CH2O
30.02 g CH2O
Independent Practice:
I. Calculate the molar masses of the following
compounds.
1. CO2
2. NaOH
3. HCl
4. N2O5
5. MgSO4
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II. Determine the amount of moles and particles from
the given mass amounts of the following compounds.
1. 50.0 g of CO2
2. 50.0 g of NaOH
3. 18.0 g of HCl
4. 100. g of N2O5
5. 100. g of MgSO4
III. Determine the amount of grams from the given
moles of compounds and then determine the amount of
particles.
1. 1.50 moles of CO2
2. 2.00 moles of HCl
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Stoichiometry: (CP:Ch 9)
Stoichiometry is that portion of chemistry dealing with
numerical relationships in chemical reactions; the
calculations of quantities of substances involved in
chemical equations.
Example: S’mores
2 graham crackers, 1 lg marshmallow, 2 sect. of
chocolate
2 Gc + 1Mm + 2 Ch  1 S’m
Try:
How many Gc, Mm, Ch do you need to make 15 S’m?
30 Gc + 15 Mm + 30 Ch
How many graham crackers do you need if need to make
45 S’mores?
45 Sm x
2 Gc =
1 Sm
90 Gc
Chemical equations are like recipes.
What do the equations mean? (Figure 9.4)
The following can be determined from an equation.
1. Particles
2. Moles
3. Mass
4. Volume
5. physical state
Stoichiometry continued
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For example:
Problem:
2H2 (g) + O2 (g)  2H2O
Atoms:
4
2
6
Molecules: 2
1
2
Moles:
2
1
2
Mass (g):
4x1
2x16
2x18
Volume (L): 44.8
22.4
44.8
What else can be determined from an equation?
Ratio (relationship):
2 mol H2
1 mol O2
1 mol O2
2 mol H2O
2 mol H2
2 mol H2O
1 mol O2
2 mol H2
2 mol H2O
1 mol O2
2 mol H2O
2 mol H2
Why are the ratios important?
Mole to mole calculations, mass to mass calculations, to
determine L.R., to determine the amount of product
produced.
a. Mole to Moles Calculations:
x mol G x b mol W = xb mol W
a mol G
a
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(cp:9b,p.244)
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b. Mass to Mass Calculations (convert mass of reactant to
mass of product OR reactant):
3 Conversion Steps : (see Figure 9.9 cp)
1. Mass to mole
2. Mole to mole (ratio)
3. Mole to mass
c. Limiting Reagent (or reactant):
The reagent that limits or determines the amount of
substance (product) that is formed in a reaction.
Excess reagent (reactant): any substance (reactant) that is left
over.
How to determine a L.R.:
1. Write one of the reactant’s given quantities.
2. Set a math problem comparing the reactants by using
the mole to mole ratio obtained from the balanced
equation.
3. Perform the math functions to determine the amount
needed.
4. Compare the answer of this with the given quantities.
If the needed quantity is less than the given, then this is
the excess reagent (E.R.).
If the needed quantity is more than the given, then this is
the L.R.
Needed < Given  E.R.
Needed > Given  L.R.
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Example: p. 254
Given 6.70 moles of Na & 3.20 moles of Cl2, which
reactant is the L.R.?
2 Na (s) + Cl2 (g)  2 NaCl(s)
6.70 moles 3.20 moles
6.70 mol Na x 1mol Cl2
2 mol Na
(cp: p. 254, sample prb 9-8)
= 3.35 mol Cl2 required
(needed)
(Is there at least 3.35 moles?  NO. )
Needed > Given  L.R. is Cl2
d. How to determine the amount of product(s)
produced: (This is considered theoretical yield.)
1. Write the given quantity of the L.R. down.
2. Set a math problem comparing the L.R. to the desired
product by using the mole to mole ratio obtained from the
balanced equation.
3. Perform the math functions to determine the amount
produced.
How many moles of NaCl (product) produced?
3.20 moles Cl2 X 2 mol NaCl = 6.40 mol NaCl
1mole Cl2
Student tries:
2 Na (s) + Cl2 (g)  2 NaCl(s)
How many moles of NaCl (product) produced from the
6.70 moles of Na?
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6.70 moles Na X ? mol NaCl = ???? mol NaCl
?mol Na
2 mol NaCl = 6.70 mol NaCl
2 mol Na
Other important Math concepts:
Actual yield: the amount of product actually formed.
Theoretical yield: the maximum amount of product that
could actually be formed with the given reactants
Percent yield: the ratio of actual yield to the theoretical
yield expressed as a percent.
% yield = Actual yield
Theoretical yield
x 100%
Ex: In a lab experiment, you produce 15.0 g of NaCl. The
theoretical yield for the experiment is 25.0 g. What is the
percent yield for this experiment?
15.0 g
25.0 g
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x 100% = 60.0 % Yield
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Percent Composition of Representative Compound:
(cp: p. 188)
% mass of element E =
grams of E
x 100
grams of compound
Ex: What is the percent composition of Mg and O in
MgO?
24.31 g Mg + 16.00g O = 40.31 g MgO
24.31g Mg x100 = 60.3 % Mg
40.31 g MgO
16.00 g O x100 = 39.7 % O
40.31 g MgO
Independent Practice: CP: p. 189, #29
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What is Molar Volume?
It is the volume of a gas that a mole of a gas occupies at a
pressure of one atmosphere (101 kPa) and a temperature
of 0.00°C. Under these conditions (STP), the volume of
one mole is 22.3 L.
Air bags:
(tp: p.417); (cp: p.260)
2NaN3(s)  3N2(g) + 2Na(s)
6Na(s) + Fe2O3 
3Na2O(s) + 2Fe (s)
Na2O(s) + 2CO2 (g) + H2O(g)  2NaHCO3
Ideal gas Law: PV = nRT
P = pressure (kPa)
V = volume (Liters)
n = 1 mole
R = constant value (8.31 kPa ∙ L)
mol ∙ K
T = temperature (Kelvin)
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Empirical Formulas
The formula of a compound having the smallest wholenumber ratio of atoms in the compound is called the
empirical formula.
Chemical formulas for ionic compounds are the same as
their empirical formulas.
For covalent compounds they are not the same. For
example, many covalent compounds have the same
empirical formula CH2O.
CH2O – Formaldehyde
C2H4O2 – Acetic Acid
C6H12O6 – Glucose
How can you determine the ratio of moles from
percentages of elements present in a covalent
compound?
If you have a 100g of a compound with 40.0% Carbon,
6.70% Hydrogen, and 53.3% Oxygen. 1st: calc. g of each
100 x 40% = 40.0 g C
100 x 6.70% = 6.70g H
100 x 53.3% = 53.3g O
2nd: calculate moles
40.0g C X 1mol C
= 3.33 mol C
12.0 g C
6.70g H X 1 mol H =
6.63 mol H
1.01 g H
53.3g O X 1 mol O = 3.33 mol O
16.0 g O
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3rd Divide all mole numbers by the smallest one.
3.33 mol C = 1.00 mol C
3.33
6.63 mol H = 1.99 mol H
3.33
3.33 mol O = 1.00 mol O
3.33
Write as a chemical formula: C1.00H1.99O1.00
Round to whole numbers: CH2O
The molar mass of CH2O is 30.0 g/mol. What if you have
a compound with the same ratio of C to H to O, but the
molar mass is 60.0 g/mol. What is the chemical formula?
Unknown
CH2O
60.0 g = 2 times
30.0 g
Unknown has 2 times the molar mass (g).
Unknown is equal to 2 x (CH2O) = C2H4O2 . This is
Acetic Acid.
What about 180 g/mol?
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C6H12O6 – Glucose
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