Transcript platinum and mixing table

General Chemistry
101 Chem
Dr. Fahad Ahmed Alharthi
Room: 2A104, Building No. 5
Email: [email protected]
‫(المراجع( ‪References‬‬
‫‪Chemistry the General Science 11E‬‬
‫‪T. L. Brown, H. E. LeMay, B. E.‬‬
‫‪Bursten and C. J. Murphy.‬‬
‫الكيمياء العامة‪.‬‬
‫د‪.‬أحمد العويس‪ ،‬د‪.‬سليمان الخويطر‪،‬‬
‫د‪.‬عبدالعزيز الواصل‪ ،‬د‪.‬عبدالعزيز‬
‫السحيباني‪.‬‬
MATTER
We define matter as anything that has mass and
takes up space.
Three States of Matter
A Gas has neither a definite shape nor
definite volume:
it adopts the volume and shape of the
vessel containing it.
A Liquid has definite volume but not shape:
it adopts the shape of the vessel containing
it.
A Solid has definite volume and shape:
which is independent of the vessel containing
it.
Changes of States
Vaporization
(heat or reduce pressure)
Condensation
(cool or increase pressure)
LIQUID
Melting
(heat)
Freezing
(cool)
SOLID
Liberates Energy
Requires Energy
GAS
COMPOSITION
Composition is the type and amount of substance that
make up a sample of matter.
• Atoms are the building blocks of matter.
• Each element is made of the same kind of atom.
• A compound is made of two or more different kinds of
elements.
PROPERTIES
Properties are the characteristics that give each substance
a unique identity.
Physical Properties
those which the substance
shows by itself without
interacting with another
substance, such as: colour,
melting point, boiling
point, density
Chemical Properties
those which the substance
shows as it interacts with,
or transforms into, other
substances, such as:
flammability, corrosiveness
PROBLEM: Decide whether each of the following
process is primarily a physical or a chemical change,
and explain briefly.
(a) Frost forms as the temperature drops on a humid winter
night.
(b) Dynamite explodes to form a mixture of gases.
(c) Dissolving sugar and water.
(d) A silver fork tarnishes in air.
Criteria: “Does the substance change composition or
just change form?”
SOLUTION:
(a) physical change
(b) chemical change
(c) physical change
(d) chemical change
Units of Measurement
Table 1.1 SI Base (or Fundamental) Units
Base Quantity
Name of Unit
Symbol
Length
meter
m
Mass
kilogram
kg
Time
second
s
Current
ampere
A
Temperature
kelvin
K
Amount of substance
mole
mol
Luminous intensity
candela
cd
• Système International d’Unités
• A different base unit is used for each quantity.
Metric System
Table 1.2 Prefixes Used with SI Units
Prefix
Symbol
TeraT
Meaning
1012
GigaMegaKilo-
G
M
k
109
106
103
DeciCentiMilli-
d
c
m
10-1
10-2
10-3
MicroNanoPico-
m
n
p
10-6
10-9
10-12
Volume
• The most commonly used metric
units for volume are the liter (L)
and the milliliter (mL).
– A liter is a cube 1 dm long on each
side.
– A milliliter is a cube 1 cm long on
each side.
1 cm3 = (1 x 10-2 m)3 = 1 x 10-6 m3
1 dm3 = (1 x 10-1 m)3 = 1 x 10-3 m3
1 L = 1000 mL = 1000 cm3 = 1 dm3
1 mL = 1 cm3
1 cm3 = (1 x 10-2 m)3 = 1 x 10-6 m3
1 dm3 = (1 x 10-1 m)3 = 1 x 10-3 m3
1 L = 1000 mL = 1000 cm3 = 1 dm3
1 mL = 1 cm3
How many mL are in 1.63 L?
1 L = 1000 mL
1.63 L x 1000 mL= 1630 mL
1L
Density
Density – SI derived unit for density is kg/m3
1 g/cm3 = 1 g/mL = 1000 kg/m3
mass
Density = volume
m
d= V
A piece of platinum metal with a density of 21.5 g/cm3
has a volume of 4.49 cm3. What is its mass?
m
d= V
Temperature
Comparison of the three temperature scales
K = 0C + 273.15
273 K = 0 0C
373 K = 100 0C
0F
= 9 x 0C + 32
5
32 0F = 0 0C
212 0F = 100 0C
Convert 172.9 0F to degrees Celsius and Kelvin
= 9 x 0C + 32
5
0F – 32 = 9 x 0C
5
0F
5 x (0F – 32) = 0C
9
0C = 5 x (0F – 32)
9
5 x (172.9 – 32) =
0C =
9
K=
Atomic structure
Subatomic Particles
(Table 1.3)
Particle
Mass
(g)
Charge
(Coulombs)
Charge
(units)
Electron (e-) 9.1 x 10-28 -1.6 x 10-19
1.67 x 10
-27
Neutron (n) 1.67 x 10
-27
Proton (p)
+1.6 x 10
-19
0
mass p = mass n = 1840 x mass e-
-1
+1
0
Atomic number (Z) = number of protons in nucleus
Mass number(A) = number of protons + number of neutrons
= atomic number (Z) + number of neutrons
Isotopes are atoms of the same element (X) with different numbers of
neutrons in the nucleus
Mass Number
A
ZX
Atomic Number
1
1H
235
92
2
1H
U
Element Symbol
3
1H
(D)
238
92
U
(T)
Cation is an ion with a positive charge
If a neutral atom losses one or more electrons
it becomes a cation.
Na
11 protons
11 electrons
Na+
11 protons
10 electrons
Anion is an ion with a negative charge
If a neutral atom gains one or more electrons
it becomes an anion.
Cl
17 protons
17 electrons
Cl-
17 protons
18 electrons
Monatomic ion contains only one atom
Na+, Cl-, Ca2+, O2-, Al3+, N3-
Polyatomic ion (molecular ion) contains more than
one atom
OH-, CN-, NH4+, NO3-
Examples
3+
How many protons and electrons are in
27
13
Al
How many protons and electrons are in
78
34
Se 2- ?
?
Chemical Equations
Chemical equations are concise representations
of chemical reactions.
CH4 (g) + 2 O2 (g)
CO2 (g) + 2 H2O (g)
Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g)
CO2 (g) + 2 H2O (g)
Reactants appear on the left side of the equation.
CH4 (g) + 2 O2 (g)
CO2 (g) + 2 H2O (g)
Products appear on the right side of the equation.
CH4 (g) + 2 O2 (g)
CO2 (g) + 2 H2O (g)
The states of the reactants and products are written
in parentheses to the right of each compound.
CH4 (g) + 2 O2 (g)
CO2 (g) + 2 H2O (g)
Coefficients are inserted to balance the equation.
Subscripts and Coefficients Give
Different Information
• Subscripts tell the number of atoms of each element in
a molecule.
Subscripts and Coefficients Give
Different Information
• Coefficients tell the number of molecules.
Examples of balancing chemical
equation
2 HCl + Zn
ZnCl2 + H2
C 3H8 + 5 O 2
3 CO2 + 4 H2O
Zn + 2 HNO3
Zn(NO3)2 + H2
Different Types of Chemical Reactions
Examples
-Combination reactions
- Decomposition reactions
- Combustion in Air
Combination Reactions
• Examples:
• In this type
of reaction
two or more
substances
react to form
one product.
– 2 Mg (s) + O2 (g)  2 MgO (s)
– N2 (g) + 3 H2 (g)  2 NH3 (g)
– C3H6 (g) + Br2 (l)  C3H6Br2 (l)
Decomposition Reactions
• In a decomposition one
substance breaks down
into two or more
substances.
• Examples:
– CaCO3 (s)  CaO (s) + CO2 (g)
– 2 KClO3 (s)  2 KCl (s) + O2 (g)
– 2 NaN3 (s)  2 Na (s) + 3 N2 (g)
Combustion Reactions
• Examples:
• These are generally
rapid reactions that
produce a flame.
• Most often involve
hydrocarbons
reacting with
oxygen in the air.
– CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (g)
– C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (g)
Formula Weights(FW)
• A formula weight is the sum of the atomic
weights for the atoms in a chemical
formula.
• So, the formula weight of calcium chloride,
CaCl2, would be
Ca: 1(40.1 amu*)
+ Cl: 2(35.5 amu)
111.1 amu
• Formula weights are generally reported for
ionic compounds.
*atomic mass unit
Molecular Weight (MW)
• A molecular weight is the sum of the
atomic weights of the atoms in a molecule.
• For the molecule ethane, C2H6, the
molecular weight would be
C: 2(12.0 amu)
+ H: 6(1.0 amu)
30.0 amu
Percent Composition
One can find the percentage of the mass of
a compound that comes from each of the
elements in the compound by using this
equation:
(number of atoms)(atomic weight)
% element =
(FW of the compound)
x 100
Example
The percentage of carbon in ethane (C2H6)
is:
(2)(12.0 amu)
%C =
=
(30.0 amu)
24.0 amu
30.0 amu
= 80.0%
x 100
Avogadro’s Number
Avogadro’s number is 6.02 x 1023
The mole is a unit measurement express the amounts of
a chemical substance.
One mole of chemical substance contains Avogadro’s
number of elementary entities (e.g., atoms,
molecules, ions, electrons).
Mass (m)
Number of mole (n)=
Molecular weight (Mwt)
Example
1 mole of H2O has
a mass of 18g.
Molar Mass
• By definition, a molar mass is the mass of 1
mol of a substance (i.e., g/mol).
– The molar mass of an element is the mass
number for the element that we find on the
periodic table.
– The formula weight (in amu’s) will be the same
number as the molar mass (in g/mol).
Mole Relationships
• One mole of atoms, ions, or molecules contains
Avogadro’s number of those particles.
• One mole of molecules or formula units contains
Avogadro’s number times the number of atoms or ions
of each element in the compound.
Examples
1- Calculate how many atoms there are in 0.200 moles
of copper.
The number of atoms in one mole of Cu is equal to the
Avogadro number = 6.02 x 1023 .
Number of atoms in 0.200 moles of Cu = (0.200 mol) x
(6.02x1023 mol-1 ) = 1.20 x 1023 .
2- Calculate how molecules of H2O there are in 12.10
moles of water.
Number of water molecules = (12.10 mol)x(6.02 x 1023)
= 7.287 x 1024
3- Calculate the number of moles in 5.380 g of glucose
(C6H12O6) .
Moles of C6H12O6 =
5.380 g
180.0
gmol-1
= 0.02989 mol.
4- Calculate the mass, in grams, of 0.433 mol of
Ca(NO3)2.
Mass = o.433 mol x 164.1 g/mol = 71.1 g.
5- How many molecules are in 5.23 g of glucose
(C6H12O6)? How many oxygen atoms are in this
sample?
Molecules of C6H12O6 =
5.23 g
180.0 gmol-1
x (6.02x1023)
= 1.75 x 1022 molecules
Atoms O = 1.75 x 1022 x 6 = 1.05 x 1023
Calculating Empirical Formulas
A molecular formula shows the exact number of atoms of
each element in the smallest unit of a substance
An empirical formula shows the simplest whole-number
ratio of the atoms in a substance
molecular
empirical
H2O
H2O
C6H12O6
CH2O
O3
O
N2H4
NH2
Example
The compound para-aminobenzoic acid (PABA) is
composed of carbon (61.31%), hydrogen (5.14%),
nitrogen (10.21%), and oxygen (23.33%).
Find the empirical formula of PABA.
Assume that 100.00 g of para-aminobenzoic acid:
C:
61.31 g
12.01 g.mol -1
= 5.105 mol C
H:
5.14g
1.01 g.mol
= 5.09 mol H
N:
10.21 g
14.01 g.mol -1
= 0.7288 mol N
O:
61.31 g
12.01 g.mol -1
= 1.456 mol O
-1
Calculate the mole ratio by dividing by the smallest
number of moles:
C:
5.105 mol
0.7288 mol
= 7.005  7
H:
5.09 mol
0.7288 mol
= 6.984  7
N:
0.7288 mol
0.7288 mol
= 1.000
O:
1.458 mol
0.7288 mol
= 2.001  2
The empirical formula for para-aminobenzoic
acid is:
C7H7NO2
Combustion Analysis
• Compounds containing C, H and O are routinely analyzed
through combustion in a chamber like this.
– C is determined from the mass of CO2 produced.
– H is determined from the mass of H2O produced.
– O is determined by difference after the C and H have
been determined.
Determining Empirical Formula by
Combustion Analysis
Example:
Combustion of 0.255 g of isopropyl alcohol produces
0.561 g of CO2 and 0.306 g of H2O. Determine the
empirical formula of isopropyl alcohol.
0.561 g CO2
Grams of C =
x 1 x 12 gmol-1 = 0.153 g
-1
44 gmol
0.306 g H2O
Grams of H =
x 2 x 1 gmol-1 = 0.0343 g
-1
18 gmol
Grams of O = mass of sample – (mass of C + mass of H)
= 0.255 g – ( 0.153 g + 0.0343 g) = 0.068 g
0.153 g
Moles of C =
= 0.0128 mol
-1
12 gmol
0.0343 g
Moles of H =
1 gmol-1
= 0.0343 mol
0.068 g
Moles of O =
16 gmol-1
= 0.0043 mol
Calculate the mole ratio by dividing by the smallest
number of moles:
C:H:O
2.98:7.91:1.00
C3H8O
Stoichiometric Calculations
The coefficients in the balanced equation give the
ratio of moles of reactants and products.
Starting with the mass of Substance A you can use the
ratio of the coefficients of A and B to calculate the mass
of Substance B formed (if it’s a product) or used (if it’s a
reactant).
C6H12O6 + 6 O2  6 CO2 + 6 H2O
Starting with 1.00 g of C6H12O6…
calculate the moles of C6H12O6…
use the coefficients to find the moles of H2O…
and then turn the moles of water to grams.
Limiting Reactants
• The reactant that is completely consumed in a
reaction is called the limiting reactant (limiting
reagent).
– In other words, it’s the reactant that run out first (in this
case, the H2).
• In example below, O2 would be the excess reactant
(excess reagent).
Theoretical (calculated) Yield
• The theoretical yield is the maximum
amount of product that can be made.
• The amount of product actually obtained in
a reaction is called the actual yield.
• However, the actual yield is always less than
the theoretical yield.
Why?
• Part of the reactants may not react.
• Side reaction.
• Difficult recovery.
Percent yield calculation
• The percent yield of a reaction relates to
the actual yield to the theoretical
(calculated) yield.
Percent Yield =
Actual Yield
Theoretical Yield
x 100
Examples
Fe2O3(s) + 3 CO(g)
2 Fe(s) + 3 CO2(g)
If we start with 150 g of Fe2O3 as the limiting
reactant, and found actual yield of Fe was
87.9 g, what is the percent yield?
Actual Yield
The percent yield =
x 100
Theoretical Yield
Fe2O3(s) + 3 CO(g)
2 Fe(s) + 3 CO2(g)
150 g Fe2O3
105 g Fe
150 g
159 g mol-1
0.943 mol Fe2O3
1.887 mol x 55.85 g mol-1
X
2 mol Fe
1 mol
Fe2O3
Theoretical yield = 105 g.
87.9 g
The percent yield =
x 100 = 83.7 %
105 g
1.887 mol Fe
Solutions
• Solutions are defined as
homogeneous mixtures of
two or more pure
substances.
• The solvent is present in
greatest abundance.
• All other substances are
solutes.
Molarity
• Molarity is one way to measure the
concentration of a solution.
• The molarity of a solution is calculated by
taking the moles of solute and dividing by
the liters of solution.
Molarity (M) =
moles of solute
volume of solution in liters
mol
M=
L
Mixing a Solution
• To create a solution of a
known molarity, one weighs
out a known mass (and,
therefore, number of moles)
of the solute.
• The solute is added to a
volumetric flask, and solvent
is added to the line on the
neck of the flask.
Examples
Calculate the molarity of a solution made by
dissolving 0.750 g of sodium sulfate (Na2SO4)
in 850 mL of water.
Moles of Na2SO4 =
0.750 g
= 0.0053 mol
-1
142 g mol
0.0053 mol
Molarity =
= 0.0062 M
0.850 L
How many moles of KMnO4 are present in 250
mL of a 0.0475 M solution?
mol
M=
L
mol
Moles of KMnO4 = 0.0475
x 0.25 L
L
= 0.012 mol
How many milliliters of 11.6 M HCl solution
are needed to obtain 0.250 mol of HCl?
Molarity (M) =
moles of solute
volume of solution in liters
0.250 mol
11.6 M =
volume in liters
volume in liters = 0.022 L = 22 mL.
What are the molar concentrations of each of
the ions present in a 0.025 M aqueous
solution of Ca(NO3)2?
In aqueous solution:
Ca(NO3)2
Ca2+ + 2 NO3Ca2+ = 0.025 M
NO3- = 0.025 x 2 = 0.05 M
Dilution
• One can also dilute a more concentrated solution
by
– Using a pipet to deliver a volume of the solution to a
new volumetric flask, and
– Adding solvent to the line on the neck of the new flask.
- The molarity of the new solution can be determined
from the equation
Mc  Vc = Md  Vd
where Mc and Md are the molarity of the concentrated and dilute
solutions, respectively, and Vc and Vd are the volumes of the two
solutions.
-
Mc  Vc = Md  Vd
Moles solute before dilution = moles solute after dilution
Examples
How many milliliters of 3.0 M H2SO4 are
needed to make 450 mL of 0.10 M H2SO4 ?
Mc  Vc = Md  VV
3.0 M x Vc = 0.10 M x 450 mL
Vc = 15 mL
Ways of Expressing Concentrations of
Solutions
• Mass Percentage, ppm, and ppb
 Mass Percentage
mass of A in solution
Mass % of A =
total mass of solution
 100
Note: use the same unit in both mass of A in solution and
total mass of solution
Example:
36% HCl by mass contains 36 g of HCl for
each 100 g of solution (64 g H2O)
 Parts per Million (ppm)
mass of A in solution
ppm =
total mass of solution
 106
Parts per Billion (ppb)
mass of A in solution
ppb =
total mass of solution
 109
Note: use the same unit in both mass of A in solution and
total mass of solution
Examples
Calculate the mass percentage of Na2SO4 in
a solution containing 10.6 g Na2SO4 in 483 g
water.
10.6 g
 100
Mass % of Na2SO4 =
(483 + 10.6) g
= 2.15 %
An ore contains 2.86 g of silver per ton of
ore. What is the concentration of silver in
ppm?
ppm =
2.86 g
 106
106 g
= 2.86 ppm
• Mole Fraction, Molarity, and Molality
 Mole Fraction (X)
XA =
moles of A
total moles in solution
 Molarity (M)
Molarity (M) =
moles of solute
volume of solution in liters
Since volume is temperature-dependent,
molarity can change with temperature.
 Molality (m)
moles of solute
m=
Kilograms of solvent
Since both moles and mass do not change
with temperature, molality (unlike molarity)
is not temperature-dependent.
Examples
An aqueous solution of hydrochloric acid
contains 36 % HCl by mass. (a) Calculate the
mole fraction of HCl in the solution. (b)
Calculate the molality of HCl in the solution.
Moles HCl =
36 g
= 0.99 mol
36.5 g mol-1
Moles H2O =
64 g
= 3.6 mol
18 g mol-1
moles HCl
XHCl =
moles H2O + moles HCl
0.99
=
3.6 + 0.99
= 0.22
0.99 mol HCl
Molality of HCl =
= 15.5 m
0.064 kg H2O