Download CHAPTER 3: Molecules, Compounds and Chemical Equatio and

Type II: Metal forms more than one cation. These metals present in
the middle of the periodic table.
2+
3+
Example. Fe can form Fe and Fe
CHAPTER 3: Molecules, Compounds and Chemical Equation
COMPOUNDS
Two or more different chemical elements join together by chemical force
(chemical bond) to form compounds.
Eg.
+
Na + Cl
-
+
NaCl
-
NH4 + NO3
NH4NO3
Classification of compounds
Ga
3+
1. Covalent compounds are formed by sharing electrons between
non-metals.
Example:
H + H → H2
Consist of positive (metal cation) and
2. Ionic compounds negative (non-metal anion) charged ions were attracted together
because of opposite charge. Compounds made in this manner are
also called as ionic solids or salts.
Example:
+
-
Na + Cl → NaCl
NAMING SIMPLE COMPOUNDS
Classification of Ions (depending on number of atoms):
Ions are classified into monoatomic ions and polyatomic ions depend on the
number of atoms present.
Monoatomic ion: An atom which has net positive or negative
charge.
+
+
Example: H , Na , Cl
5 million chemical compounds are currently known. In this chapter, we are
going to study the naming of only simple compounds, mostly binary
compounds (compounds
compounds composed of two elements).
elements
Before naming a simple compound, we need to check the properties of
elements (which are joined together in the compound).
i)
Polyatomic ion: Group of atoms (more than 1 atom) which have net
positive or negative charge
+
Example: NH4 , NO3
If one element is metal (in the left side or middle of the periodic table)
and another is non-metal
metal (right side of the periodic table) – it is ionic
compound.
-
Check the number of atoms in anion and cation.
cation
If one atom – Monoatomic ion
If more than one atom – Polyatomic ion
-
For monoatomic ion,, check whether the cation it is type I (left
side of periodic table) or type II (middle of periodic table)
Classification of Ions (depending on number of atoms):
Ions also classified based on the capacity of the metal to form one or more
cations,
Type I: Metals forms only one cation. These metals present in the
+
2+
2+
3+
3+
left side of the periodic table (and Ag , Zn , Cd , Al , Ga ).
+
2+
3+
Example. Na form only Na (not Na or Na ).
ii)
If both the elements are non-metals
metals (both of them
t
in the right side of
periodic table) – It is covalent compound.
compound
General Chemistry - Handout_Ch-3
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Example.
Cl = Chlorine (Root name = Chlor)
-
Cl (anion) = Chloride
Chlor
Common monoatomic cations and anions
Cation
Name
Anion
+
Hydrogen
H
+
LIthium
F
H
Li
+
K
2+
Mg
Al
Potassium
Br
Magnesium
I
Iodide
2+
Barium
S
+
Ag
Bromide
-
O
3+
Fluoride
-
Calcium
Ba
Hydride
-
2+
Ca
Name
-
2-
oxide
2-
Sulfide
3-
Nitride
3-
Phosphide
Aluminum
N
Silver
P
TYPE OF SIMPLE COMPOUNDS
1) Naming type I ionic compounds
1. Type I ionic compounds
2. Type II ionic compounds
3. Type III covalent compounds
Compound containing monoatomic metal cation (Type I: metal form only one
type of cation) and non-metal anion.
4. Ionic compounds with polyatomic ions
Elements of the compound present in
5. Acids
Metal (Left side of periodic table) + Non-metal
Non
(right side of periodic table)
Before learning about naming of compounds, you need to know about
naming of ions.
Name of monoatomic cations (positive charged ions) is same as that of
the element
Example.
+
Na = Sodium, Na (cation) also sodium
Naming
Name of cation (element name) + Name of anion (root name + -ide)
Example:
+
-
NaCl contains Na (cation) + Cl (anion)
Na = Sodium, Na+ = Sodium
Name of monoatomic anions (negative charged ions) is ‘Root name’ of
the element + (-ide).
Cl = Chlorine, Cl- = Chloride
Chlor
NaCl = Sodium Chloride
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Some more examples
Common type II cations
MgS = Magnesium sulfide
Ion
KBr = Potassium bromide
Fe
3+
Fe
2+
2+
K2S = Potassium sulfide (Note: both K and K2 named as potassium)
Cu
LiBr = Lithium bromide
+
Cu
Name
Ion
Iron(III)
Co
3+
Cobalt(III)
Iron(II)
Co
2+
Cobalt(II)
Copper(II)
Pb
4-
Lead(IV)
Pb
2+
Lead(II)
Copper(I)
Name
Some more examples
NaI = Sodium iodide
CaBr2 = Calcium bromide
Fe2O3 = Iron(III) oxide
CuS = Copper(II) sulfide
NaBr = Sodium bromide
CuCl2 = Copper(II) chloride
CuBr = Copper(I) bromide
CoO = Cobalt(II) oxide
PbCl2 = Lead(II) chloride
3) Naming covalent compounds (Type III)
2) Naming type II ionic compounds (Type II)
Compound containing monoatomic metal cation (Type II: metal form more
than one type of cation) and non-metal anion.
Compound contain 2 non-metals.
Naming
[
1) Name of first element + Name of second element (written as anion)
Elements of the compound present in
2) Prefixes are used to denote the number of atoms present.
Metal (Middle of periodic table) + Non-metal (right side of periodic table)
Number
Naming
1
mono-
2
di-
3
Tri-
4
tetra-
5
penta-
6
hexa-
7
hepta-
Cl = Chloride
8
octa-
FeCl2 = Iron(II) chloride
9
nona-
10
-
deca-
Name of cation (Indicate charge) + Name of anion
Example:
Prefix
FeCl2 contains Fe
2+
and 2 Cl
2+
Fe = Iron
Charge = 2+ (write as II)
-
-
Mono is never used for the first element.
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Example
5) Naming acids
+
N2O = Dinitrogen monoxide
Acids are molecules with one or more H ions attached to an anion.
NO = Nitrogen monoxide
2 types of acids: 1) Anion does not contain oxygen
NO2 = Nitrogen dioxide
2) Anion contains oxygen
N2O5 = Dinitrogen Pentoxide
i) Acid: Anion does not contain oxygen
NF3 = Nitrogen trifluoride
Naming
SO2 = Sulfur dioxide
Prefix: hydro + Root name of anion + Suffix: -ic
4) Naming compounds with polyatomic ions
Example:
Compound contain polyatomic metal cation or/and polyatomic non-metal
anion.
HCl (Hydrogen + Chlorine) – Hydrochloric acid
Naming
[
ii) Acids: Anion contains oxygen
Name of cation + Name of anion
Naming
Common polyatomic ions (memorize the table)
Ion
Name
2-
CO3
-
HCO3
Ion
Sulfate
Sulfite
-
Nitrate
-
Nitrite
NO2
PO4
Phosphate
-
Bicarbonate
2SO3
NO3
3-
Carbonate
2SO4
Name
C2H3O2
OH
-
Acetate
Hydroxide
Root name of anion + Suffix –ic or –ous
If the anion name ends in –ate, the suffix –ic is added.
Example:
-
H2SO4 (contain sulfate anion, SO4 ) – Sulfuric acid
-
HNO3 (contains nitrate anion, NO3 ) – Nitric acid
+
NH4
Ammonium
-
CH3CO2H (contains CH3CO2 ) - Acetic acid
Example
+
-
NH4NO3 = NH4 , NO3 = Ammonium nitrate
+
If the anion name ends in –ite, the suffix –ous is added.
-
NaOH = Na , OH = Sodium hydroxide
2+
Fe (NO3)2 = Fe ,
2NO3
= iron(II) nitrate
Example:
2-
H2SO3 (contain sulfite anion, SO3 ) – Sulfurous acid
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FORMULA OF A COMPOUND
PERCENT COMPOSITION
Formula of a compound = Atomic (elemental) composition
Percent composition of an element =
2 types of formula:
1. Molecular formula
Mass of element
x 100%
Total mass of compound
2. Empirical formula
Percentage composition is also called as mass percentage.
Molecular formula = Total amount of each atom in a compound.
Example:
Empirical formula = Relative amounts (ratio) of each atom, but NOT the
.
total amount of each atom in a compound.
What is the percent composition of C and O in CO2?
1 mol of CO2 contain 1 mol C and 2 mol O
Example:
1 mol C = 12.01 g
Molecular formula of glucose = C6H12O6
-
1 molecule of glucose contains 6 C atoms, 12 H atoms
and 6 O atoms
2 mol O = 2 x 16 = 32 g
1 mol CO2 = 12.01 + 32 = 44.01 grams
Empirical formula of glucose = C1H2O1
-
Percent composition of C (in CO2) = (12.01 / 44.01) x 100% = 27.3%
Glucose contains C, H and O atoms in the ratio of 1:2:1
Percent composition of O (in CO2) = (32/44.01) x 100% = 72.7%
Solution 1: Which one of the following is not an empirical formula?
a) CHO
b) CH2O
c) C2H4O
At last check the total percentage is 100% (not more or less than 100)
d) C2H4O2
-
C2H4O2 is not an empirical formula, because the number of atoms can be
divided by 2 to get ratio.
Solution: Determine the mass percentage of carbon in C2H5OH?
1 mol of C2H5OH contains 2 mol C, 6 mol H and 1 mol O.
Correct empirical formula for C2H4O2 is CH2O.
2 mol C = 2 x 12.01 = 24.02 g
Remember:
6 mol H = 6 x 1.008 = 6.048
Moles = Mass / Molar mass
1 mol O = 16 g
1 mol C2H5OH = 24.02 + 6.048 + 16 = 46.068 g
FORMULA MASS:
st
st
Formula Mass = (number of atoms of 1 element x atomic mass of 1
nd
nd
element) + (number of atoms of 2 element x atomic mass of 2 element) +
4
Mass percentage of C = (24.02 / 46.068) x 100% = 52.14%
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DETERMINING THE FORMULA OF A COMPOUND
2. Determining empirical formula from % composition
The empirical or molecular formula can be determined in 3 ways.
1. Determining molecular formula from molecular mass and %
composition
•
•
•
Determine the mass of each element (using molar mass and %
composition).
Determine the moles of each element present in the compound
(1 mole = atomic mass in grams).
Write the molecular formula (using the elements and their moles).
•
Determine the mass of each element (by assuming that the molar
mass is 100 gram).
•
Convert grams to moles (1 mole = atomic mass in grams).
•
Write the empirical formula (using the elements and their ratio)
Find the element with the smallest number of moles, and then divide
the number of moles for each element by that smallest number.
If you do not have whole numbers, multiply all the numbers in the
equation by integers until you get whole numbers.
Example
Example
What is the molecular formula for Ribose, with a Molar Mass of
150.13 g, and a % composition of 40.00% C, 6.71% H and 53.29%
O?
Hydrazine contains 87.42% nitrogen and 12.58% hydrogen. What is
its empirical formula?
• Mass of each element (Assume that the molar mass is 100 g)
Ribose contains C, H and O.
Nitrogen (87.42%) = 87.42 g N
• Mass of each element
Hydrogen (12.58%) = 12.58 g H
Mass of C = 150.13 g x (40.00/100) = 60.05 g
• Moles of each element
Mass of H = 150.13 g x (6.71 /100) = 10.07 g
Moles of nitrogen = 87.42 g x (1 mol / 14.01 g) = 6.24 mol
Mass of O = 150.13 g x (53.29/100) = 80.00 g
Moles of hydrogen = 12.58 g x (1 mol / 1.008 g) = 12.48 mol
• Moles of each element
Moles of C = 60.05g X (1 mole/12g) = 5 moles
• Empirical formula
Ratio: 12.48 moles H/ 6.24 moles of N = 2:1 H:N
Moles of H = 10.07 g x (1 mole/1.008 g) = 10 moles
Moles of O = 80.00g x (1 mole/16g) = 5 moles
Empirical formula = N1H2 (or) NH2
• Molecular formula
One mole of ribose = 5 moles of C + 10 moles of H + 5 moles of O.
Molecular formula of ribose = C5H10O5
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3. Determining empirical formula by thermal decomposition (burning)
•
•
Weight O = Weight of the compound – Weight of C and H
= 100 g – ((3.33 x 12.01) + (6.66 x 1.008))
Thermal decomposition (burning) is a simple analysis to determine
the empirical formula of the compound.
When the compound is burned, you can determine how much H2O,
CO2, N2 etc. came out from the compound, and from this you can
find out the empirical formula.
= 100 – (40 + 6.67) = 53.33 g
Moles O = 53.33/16 = 3.33 mole
• Divide by lowest moles
Lowest moles is with C or O
Example
A compound that contains C, H, and O. Thermal decomposition
(burning) 1.5 g of the compound gives 2.2 g of CO2 and 0.9 g of H2O.
What is the empirical formula?
• Convert the release of CO2 & H2O for 100 g of compound.
C/C = 3.33/3.33 = 1
H/C = 6.66/3.33 = 2
O/C = 3.33/3.33 = 1
Empirical formula = CH2O
CO2 = 2.2 g CO2 x 100 g = 146.7 g
1.5 g
H2O = 0.9 g H2O x 100 g = 60 g
1.5 g
• Convert grams to moles
(1 mole of CO2 = 44.01 g, 1 mole of H2O = 18.01 g)
Moles of CO2 = 146.7 g X 1 mole CO2
44.01 g
Solution: A compound contains 1.40 g S and 2.10 g O, and has a molar
mass of 160 g/mol. What is the empirical and molecular formula of the
compound?
Molar mass of S = 32 g
Molar mass of O = 16 g
Moles of S = 1.40 g x (1 mol / 32 g) = 0.04375 mol
= 3.33 moles of CO2
1 mole of CO2 contains 1 mole of C.
3.33 moles of CO2 contain 3.33 moles of C
Moles of H2O = 60 g X 1 mole H2O
18.01 g
= 3.33 moles of H2O
1 mole of H2O contains 2 mole of H.
3.33 moles of H2O contain (2 x 3.33 =) 6.66 moles of H
Moles of O = 2.10 g x (1 mol / 16 g) = 0.13125 mol
Empirical formula = S0.04375O0.13125
To get whole number, divide both moles by 0.04375.
= SO3
Molar formula = Empirical formula x Molar mass of the compound
Molar mass of the elements
= SO3 x (160 / 80) = (SO3) x 2 = S2O6
• Find the missing O
General Chemistry - Handout_Ch-3
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CHEMICAL REACTION
A chemical change involves reorganization of the atoms in one or more
substances.
In the reaction, reactants are turned into products.
• Gas - (g)
• Dissolved in water (aqueous solution) - (aq)
2. Relative numbers of reactants and products
(atomic/molecular/mole ratios).
Example
•
In a chemical reaction, atoms are neither created nor destroyed.
All atoms present in the reactants must be present in the
product, in the same number.
No new atoms may appear in the products that were not
present in the reactants
Example
Chemical reaction between HCl and NaOH (reactants) gives NaCl
and H2O (products).
CHEMICAL EQUATION
•
Reactants are listed on the left hand side.
•
Products are listed on the right hand side.
•
Balanced equation: The number of atoms in left and right hand side
should be same.
Example
Reactants
2 H, 1 Na, 1 Cl, 1 O atoms
Meaning of the equation is
1. HCl, NaOH and NaCl are aqueous solutions, H2O is a liquid.
2. 1 molecule of HCl reacts with one molecule of NaOH to give 1
molecule of NaCl and one molecule of H2O.
(or)
1 mole of HCl reacts with 1 mole of NaOH to give 1 mole of NaCl
and 1 mole of H2O.
BALANCING CHEMICAL EQUATIONS
Chemical equation is the symbolic representation of a chemical reaction.
HCl (aq) + NaOH (aq)
HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)
→
NaCl (aq) + H2O (l)
Products
2 H, 1 Na, 1 Cl, 1 O atoms
Balance the chemical reaction by inspection, starting with the most
complicated molecules.
Change only the number of molecules. Do not change the formula or any of
the reactant or product.
a) Write the reaction
Hydrogen + oxygen → water
b) Write the unbalanced equation
H2 + O2 → H2O
c) Balance the equation (by HIT and TRIAL).
Equalize the number of atoms in left and right sides of the equation.
Meaning of a Chemical Equation
2H2 + O2
Chemical equation gives 2 important type of information.
4 H atoms
2 O atoms
1. Physical States of reactants and products
• Solid - (s)
• Liquid - (l)
→ 2H2O
4 H atom
2 O atom
d) Include phase information
2H2 (g) + O2 (g) → 2H2O (l)
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