Transcript File - IUST Dentistry

International University
for Science & Technology
College of Pharmacy
General Chemistry
(Students of Dentistry)
Prof. Dr. M. H. Al-Samman
CHEMISTRY
Chapter 3
Covalent Bond and Molecular Compounds
The Covalent Bond:
A single covalent bond: it is the bond formed by a shared pair
of electrons between two atoms, one electron of each.
H – H , Cl – Cl ,F – F
A double covalent bond: it is the bond formed by sharing two
pairs of electron between two atoms, two electron of each.
O=O ,
A triple covalent bond : it is the bond formed by sharing three
pairs of electrons between two atoms, three electrons of each.
Covalent Bond and Molecular Compounds
Notice:
-Triple bond is shorter and stronger than double bond, double
bond is shorter and stronger than single bond.
-We call the compounds with covalent bonds Molecular
Compounds.
-Coordinate Bond: it is the bond formed by a shared pair of
electrons between two atoms, one of them donates the lone pair
of the electrons while the other atom provides the empty
orbital.
Molecules and Formulas
A molecular formula gives the number of each
kind of atom in a molecule.
An empirical formula simply gives the (whole
number) ratio of atoms of elements in a
compound.
Compound
Molecular
formula
Empirical
formula
Hydrogen peroxide H2O2
HO
Octane
????
C8H18
5
Molecular Compounds
Ball-and-stick model vs. Space-filling
model
EOS
6
Empirical and Molecular
Formulas
Empirical formula: the simplest whole
number ratio of elements in a compound
Example:
Molecular formula of glucose – C6H12O6
The elemental ratio C:H:O is 1:2:1, so the
empirical formula is CH2O
EOS
7
Structural Formulas
Shows how atoms are attached to one
another.
EOS
8
Mass Percent Composition
from Chemical Formulas
The mass percent composition of a
compound refers to the proportion of the
constituent elements, expressed as the number
of grams of each element per 100 grams of the
compound.
-Each g amount of the compound contains g amount of
the element
-Each 100 g of compound contains x g of the element .
Percentage Composition of Butane
Relating Molecular Formulas
to Empirical Formulas
A molecular formula is a simple integer multiple
of the empirical formula.
That is, an empirical formula of CH2 means that
the molecular formula is CH2, or C2H4, or
C3H6, or C4H8, etc.
So: we find the molecular formula by:
molecular formula mass
= integer (nearly)
empirical formula mass
We then multiply each subscript in the empirical formula by the integer.
Mass Percent Composition
and Chemical Formulas
Example :
Phenol, a general disinfectant, has the composition 76.57% C,
6.43% H, and 17.00% O by mass. Determine its empirical formula.
Answer:
Moles of C percent :76.57/12.011 = 6.374 %
Moles of H percent : 6.43/1.0079 = 6.379 %
Moles of O percent :17.00/15.998=1.062 %
- We divide each figure by the smallest figure to obtain the ratio
between the elements in the formula:
The number of O atoms = 1.062/1.062 = 1 atom
The number of C atoms = 6.347/ 1.062 = 6 atom
The number of H atoms = 6.379/1.062 = 6 atom
The empirical formula of phenol is : C₆H₆O
Mass Percent Composition
and Chemical Formulas
Example:
Diethylene glycol, used in antifreeze, as a softening
agent for textile fibers and some leathers, and as a
moistening agent for glues and paper, has the
composition 38.70% C, 9.67% H, and 51.61% O by
mass.
1-Determine its empirical formula.
2-Determine its molecular formula if the molecular mass
of the compound is 62 u
Answer:
Moles of C % = 38.70/12.011 = 3.225 mol
Moles of H % = 9.67/1.0079 = 9.594 mol
Moles of O % = 51.61/15.998 = 3.226 mol
Mass Percent Composition
and Chemical Formulas
Example: cont.
-Number of O atoms = 1 atom
-Number of C atoms = 1 atom
-Number of H atoms = 2.97=3 atom
-The empirical formula is : CH₃O
The empirical formula mass: 12 + ( 3x1) + 16 = 31 uThe integer number : 62/31 =2The molecular formula : C₂H₆O₂-
Mass Percent Composition
and Chemical Formulas
Example:
The empirical formula of hydroquinone, a
chemical used in photography, is C3H3O, and
its molecular mass is 110 u. What is its
molecular formula?
Answer:
The empirical formula mass : (12x3)+ (3x1) + 16 = 55 u
The integer number : 110/55 = 2
The molecular formula : C₆H₆O₂
Mass Percent Composition
and Chemical Formulas
Example :
Burning a 0.1000-g sample of a carbon–hydrogen–oxygen
compound in oxygen yields 0.1953 g CO2 and 0.1000 g H2O. A
separate experiment shows that the molecular mass of the
compound is 90 u. Determine (a) the mass percent composition,
(b) the empirical formula, and (c) the molecular formula of the
compound.
Mass of carbon=0.1935x12/44=0.052779
Mass of hydrogen=0.1000x2/18=0.11111
Percent c=(0.052779/0.1000)x100=52.77%
Percent H=(0.0555/0.1000)x100=11.11%
Percent of O=100 – 63.88 = 36.12%
C mol%=52.77/12=4.3975 mol
H mol %=5.55/1=5.55 mol
O mole %=36.12/16=2.25mol
Mass Percent Composition
and Chemical Formulas
Then by dividing each molar percent by the smallest figure which is 2.25
we get:
-Number of O atoms=1 atom
-Number of C atoms =2 atom
-Number of H atom=5 atom
-The empirical formula : C₂H₅O
-Empirical formula mass : 24 + 5+ 16 = 45 u
Integer number: 90/45 = 2 -
The molecular formula : C₄H₁₀O₂-
Balancing
Chemical Equations
Example :
Balance the equation
(not balanced)
Fe + O2  Fe2O3
2Fe + 3/2 O₂  Fe2O3
4Fe + 3 O₂  2Fe2O3
Example :
Balance the equation
C2H6 + O2  CO2 + H2O
C2H6 + 7/2 O2  2CO2 + 3 H2O
2C2H6 + 7 O2  4 CO2 + 6 H2O
Example :
Balance the equation
H3PO4 + NaCN  HCN + Na3PO4
H3PO4 + 3 NaCN  3 HCN + Na3PO4
Balancing
Chemical Equations
Example:
When 0.105 mol propane is burned in an
excess of oxygen, how many moles of oxygen
are consumed? The reaction is
C3H8 + 5 O2  3 CO2 + 4 H2O
1mole ----5mole
0.105------?
Number of O moles = 0.105 x 5 /1 = 0.525 mol
Balancing
Chemical Equations
QWhen 11.6 g of butane is burned in an excess
of oxygen.
how many grams of oxygen are consumed?(a
The reaction is(b
C₄H₁₀ + 13/2 O2  4 CO2 + 5 H2O
58
(13/2)x32
11.6g
x
The mass of consumed oxygen: 41.6 g
Limiting Reactants
If the reactants are not present in stoichiometric amounts, at
end of reaction some reactants are still present (in
excess).
Limiting Reactant: it is the substance which is
completely consumed during the reaction and limits
the reaction.
Limiting Reactants
Limiting Reactants
The amount of product predicted from stoichiometry taking into
account limiting reagents is called the theoretical yield.
Theoretical yield: it is the yield which we predict by
mathematical calculations.
The actual yield: it is the yield which we actually ( really )
obtain.
The percent yield :relates the actual yield (amount of
material recovered in the laboratory) to the theoretical
yield:
Determining Limiting Reagents
Practice Problem
Example:
During a process we burned 11.60 g of butane C₄H₁₀
with 80 g of oxygen , we obtained 30 g of carbon
dioxide .
1- write the equation of the reaction.
2- balance the equation.
3- assign the limiting reactant in the reaction.
4- calculate the theoretical yield of carbon dioxide.
5- calculate the percentage yield of the reaction.
Determining Limiting Reagents
Practice Problem
Answer:
1- C₄H₁₀ + O₂ → CO₂ + H₂O
2- C₄H₁₀ + O₂ → 4 CO₂ + 5 H₂O
C₄H₁₀ + 13/2 O₂ → 4 CO₂ + 5 H₂O
3- to assign the limiting reagent, we write:
C₄H₁₀ + 13/2 O₂ → 4 CO₂ + 5 H₂O
58 g
208 g
11.6 g
xg
The needed amount of oxygen= (11.6x208)/58 = 41.6g
So, Butane is the limiting reactant .
Determining Limiting Reagents
Practice Problem
Answer:
4- To calculate the theoretical yield, we write:
C₄H₁₀ + 13/2 O₂ → 4 CO₂ + 5 H₂O
58 g
4x 44 g
11.6 g
yg
The theoretical yield= (11.6 x 176)/ 58= 35.2 g
5- to calculate the percentage yield, we write:
The percentage yield=(actual yield/ theoretical yield)x100
The percentage yield = (30/35.2)x100 = 85.22%
Determining Limiting Reagents
Guided Practice Problem
Example:
Part of the SO2 that is introduced into the atmosphere
ends up being converted to sulfuric acid, H2SO4.
The net reaction is:
2SO2(g) + O2(g) + 2H2O(l)  2H2SO4(aq)
• Answer:
How much H2SO4 can be formed from 5.0 mol of SO2,
1.0 mol O2, and an unlimited quantity of H2O?
Assign the limiting reagent of the reaction
Determining Limiting Reagents
Guided Practice Problem
2SO2(g) + O2(g) + 2H2O(l)  2H2SO4(aq)
2mol
1 mol
2 mol
1 mol
2mol
- The amount of H2SO4 in grams= 2x98 = 196 g
- Oxygen is the limiting reagent in the reaction.
Determining Limiting Reagents
Guided Practice Problem
Consider the following reaction:
2Na3PO4(aq) + 3Ba(NO3)2(aq)  Ba3(PO4)2(s) + 6NaNO3(aq)
Suppose that a solution containing 3.50 g of Na3PO4 is mixed
with a solution containing 6.40 g of Ba(NO3)2. How many
grams of Ba3(PO4)2 can be formed? What is the % yield, if
experimentally, only 4.70 g were obtained from the reaction?
Solution
At first, it must determine which of reactants is completely
consumed and is therefore the limiting reactant. The
quantity of this reactant, in turn, will determine the quantity of
Barium Phosphate Ba3(PO4)2
• It is needed a grams-to-moles conversion factor to convert from the
given reactant masses and moles-to-grams factor to convert to the
desired product mass. The quantity of excess reactant can be calculated
as the difference between the given mass of this reactant and the mass
consumed (reacted) in the reaction
• The balanced equation is given as following:
2 Na3PO4(aq) + 3Ba(NO3)2(aq)  Ba3(PO4)2(s) + 6NaNO3(aq)
• It can identify the limiting reactant by finding the number of moles
Barium Phosphate Ba3(PO4)2 produced by assuming first one reactant,
and then the other is as the limiting reactant.
No. moles Na3PO4 = 0.021 moles (MW of Na3PO4 = 164)
No. moles Ba(NO3)2 = 0.025 moles (MW of Ba(NO3)2 = 261)
- 0.021 moles Na3PO4 produce 0.011 moles of Ba3(PO4)2
- 0.025 moles Ba(NO3)2 produce 0.008 moles of Ba3(PO4)2
Because the amount of product in the second calculation [0.008 mol
Ba3(PO4)2] is smaller than (0.021 mol Ba3(PO4)2), thus the Barium
Nitrate is the limiting reactant. So thus when 0.008 mol of Ba3(PO4)2
has been formed, a quantity 0.025 mol of Ba(NO3)2is completely
Consumed and the reaction stops, producing a specific mass of
Ba3(PO4)2
Determining Limiting Reagents
Guided Practice Problem
• Having found that the amount of product is 0.0083 mol Ba3(PO4)2, thus
the mass of Ba3(PO4)2 is 3.7 grams.
Number of moles of reacted Na3PO4 is 0.017 moles. The mass of
Na3PO4 is 2.8 grams the unreacted of Na3PO4 is 0.71 gr.
• The theoretical yield = 4.99 gr. Ba3(PO4)2
• [MW of Ba3(PO4)2= 601]
• The experimental yield = 4.70 gr. Ba3(PO4)2
• The percentage yield = (4.70 / 4.99)x100 = 94.2 %
The End of Chapter 3
The test will cover Chapters 1-3, Scheduled
Homework: 3.9, 3.11, 3.15, 3.17, 3.19, 3.21, 3.25,
3.27, 3.31, 3.33, 3.43, 3.47