Download CHAPTER 4: CHEMICAL QUANTITIES and AQUEOUS REACTIONS

CHAPTER 4: CHEMICAL QUANTITIES and AQUEOUS REACTIONS
REACTION STOICHIOMETRY: Amounts of Reactants & Products
Chemical equation represents the number of molecules involved in a
reaction, not the mass of the molecule.
However, the mass reactants should be equal to the mass of products. Thus,
if you know the mass or reactants, you can calculate the mass of products.
Mass to Mass Conversion:
Steps involved in stoichiometric calculation
4. Write the equation
5. Balance the chemical equation
Mole to Mole Conversion:
6. Find out the moles of desired reactants and products (using the
ratios derived from the balanced equation)
Steps involved in stoichiometric calculation
7. Convert moles to grams
1. Write the equation
2. Balance the chemical equation
3. Find out the moles of desired reactants or products (using the ratios
derived from the balanced equation)
Example: How many moles of ammonia are produced when 0.6 moles of N2
react with H2?
1. Write the equation
N2 + H2 → NH3
2. Balance the chemical equation
N2 + 3H2 → 2NH3
3. From the balanced equation, find out the moles of desired reactants
and products.
1 moles of N2 gives 2 moles of NH3
4. Find the number of moles of Product formed.
0.6 moles N2 x 2 mole NH3
1 mole N2
8. Calculate the amount of desired reactant or product
Example
209 g of methanol (CH3OH) burns in air to give CO2 and H2O. What is
the mass of water?
5. Write the equation
CH3OH + O2 → CO2 + H2O
6. Balance the chemical equation
2CH3OH + 3O2 → 2CO2 + 4H2O
7. From the balanced equation, find out the moles of desired reactants
and products.
2 moles of CH3OH gives 4 moles of H2O.
8. Convert moles to grams.
Molar mass of CH3OH = (1 x Mass of C) + (4 x Mass of H) +
.
(1xMass of O) in g
= (1 x 12.01) + (4 x 1.008) + (1 x 16) g
= 32 g
Molar mass of H2O = (2 x Mass of H) + (1xMass of O) in g
= 1.2 moles of NH3 produce
= (2 x 1.008) + (1 x 16) g
= 18 g
General Chemistry – Handout (Page:
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2 moles of CH3OH produce 4 moles of H2O.
Therefore,
Solving limiting reactant problems
1. Write the equation.
2 x 32 g of CH3OH produce 4 x 18 g of H2O
2. Balance the chemical equation.
64 g of CH3OH produce 72 g of H2O
3. Convert grams of reactants to moles
9. Calculate the amount of desired reactant or product.
64 g of CH3OH give 72 g of H2O
Amount of H2O produced from 209 g of CH3OH
= 209 g CH3OH x 72 g of H2O
64 g CH3OH
= 235.12 g of H2O
235.12 g of H2O is produced from 209 g of methanol.
LIMITING REACTANT
4. Find out the moles of product by each reactant. The reactant giving
the least no. of product will be limiting reactant.
5. Calculate the mass of the product from the Limiting reactant. That
will be the theoretical yield.
Example: Methanol (CH3OH) is produced form 1.00 g H2 and 6.44 g CO.
Which is the limiting reactant? What is the Limiting reactant and theoretical
yield?
Reaction:
H2 + CO → CH3OH
Balanced equation:
2H2 + CO → CH3OH
Convert grams to moles:
The limiting reactant is the reactant that is consumed first, limiting the
amounts of products formed.
Molar mass of H2 = 2 x 1.008 g = 2 g/mole
Moles of H2 = 1.00 g x (1 mol /2 g) = 0.5 mol
Molar mass of CO = 12.01 g + 16 g = 28 g
Moles of CO = 6.44 g x (1 mol /28 g) = 0.23 mol
Limiting reactant:
Moles of CH3OH from H2:
0.5 mol. H2 x 1 mol CH3OH
2 mol H2
=
0.25 mol CH3OH
Moles of CH3OH from CO:
6 moles of H + 12 moles of Cl → 6 moles of HCl
Cl – Limiting reactant
0.23 mol. CO x 1 mol CH3OH = 0.23 mol CH3OH
1 mol CO
As CO is giving less amount (i.e. 0.23mol) of CH3OH therefore, it is the
limiting reactant and H2 is in Excess.
General Chemistry – Handout (Page:
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Theoretical Yield (mass of product):
Molar mass of CH3OH:
12
1
16
= (1 x Mass of C) + (4 x Mass of H) + (1 x Mass of O) in grams
= (1 x 12) + (4 x 1) + (1 x 16) = 32 grams/mol.
Molar mass of aspirin (C9H8O4) =
12
1
16
(9 x Mass of C) + (8 x Mass of H) + (4 x Mass of O) in grams
= (9 x 12) + (8 x 1) + (4 x 16)
= 180 grams
3. Calculate the amount of desired reactant or product
Mass of CH3OH:
0.23 mol. CH3OH x 32 g CH3OH
1 mol CH3OH
=
7.36 g CH3OH
180 g of aspirin produced from 138 g of salicylic acid (limiting
reactant)
Amount of aspirin formed from 14.4 grams of salicylic acid (S.A)
= 14.4 g S.A x (180 g aspirin / 138 g S.A)
= 18.78 g aspirin
PERCENT YIELD
Theoretical Yield and actual yield is required to calculate percent yield.
Theoretical Yield is the amount of product that would result if the
entire limiting reagent reacted. Its amount is calculated using the
balanced equation.
4. Calculate the percent yield
Theoretical yield = 18.78 g
Actual yield = 6.26 g
Percent yield =
Actual Yield is the amount of product actually obtained from a
reaction. It is usually given.
Percent yield =
Actual Yield
Theoretical Yield
= 33.33%
SOLUTION
14.4 g of salicylic acid (limiting reactant) gives actual yield of 6.26 g
of aspirin. Calculate the percent yield of aspirin using the following
equation.
Salicylic acid
+
C4H6O3 (l)
Acetic anhydride
→ C9H8O4 (s)
Aspirin
x 100%
= (6.26/18.78) x 100%
x 100%
Example
C7H6O3 (s)
Actual Yield
Theoretical Yield
+
CH3CO2H (l)
A homogeneous mixture of two or more substances, which may be solids,
liquids, gases.
Solvent: Used to dissolve (mix homogenously) solid, liquid or gas. So it is
called dissolving medium. Mostly liquids are used as solvent.
-
In general, solvent is present in greater amount.
Acetic acid
Solute: Solids, liquids or gases that are dissolved in the solvent.
1. 1 mole of salicylic acid (limiting reactant) produce 1 mole of aspirin
2. Convert moles to grams.
Molar mass of salicylic acid (C7H6O3) =
(7 x Mass of 12C) + (6 x Mass of 1H) + (3 x Mass of 16O) in grams
= (7 x 12) + (6 x 1) + (3 x 16) = 138 grams
-
In general, solute is present in lesser amount.
Solid (or) Liquid (or) Gas + Liquid
Solute
→
Solvent
Liquid
Solution
General Chemistry – Handout (Page:
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Moles of solute = Molarity x Liter of solution
Example
Sugar (solid solute) + Water (liquid solvent) → Sugar solution (liquid)
= 4 moles/L
Amount of solute in grams
x 2 L = 8 moles
= Mole x Molar mass
= 8 x 40 g (molar mass of NaOH)
= 320 g
Exercise 1. Calculate the molarity of a solution prepared by dissolving 11.5 g
of solid NaOH in enough water to make 1.5 L of a solution.
SOLUTION CONCENTRATION AND SOLUTION STOICHIOMETRY
Gram → Moles → Molarity
In solutions, the solute is dissolved in a solvent. The amount of solute in a
solution is measured using a ‘concentration unit’ called molarity.
Gram to Mole conversion:
Molarity is defined as moles of solute per liter of solution.
Molarity (M) =
Moles
Liters of solution
1 M = 1 mole of solute dissolved in 1 L of solution
Example
1 moles of NaOH (40 gram) is present in 2 liters of solution. What is the
molarity?
Molarity (M) = Moles = 1 mole = 0.5 M
L. of soln.
2L
Using molarity (M), you can calculate the moles of solute (amount of solute)
present in the solution.
1 mole of NaOH (Molar mass) = 23 + 16 + 1 = 40 g
11.5 g NaOH x
1 mole
= 0.288 mol
40 g NaOH
Molarity = Moles of solute = 0.288 mol = 0.192 M
Liters of solution
1.5 L
Exercise 2. How many mL of a 0.155 M KCl solution contains 2.55 g KCl?
Molarity = number of Moles of solute
Liters of solution
⇒ Liters of Solution = no. of moles of solute
Molarity
19
17
Moles of solute = Molarity x Liter of solution (Volume)
Molar mass of KCl = (1 x Mass of K) + (1 x Mass of Cl) in grams
= (1 x 39.10) + (1 x 35.45)
= 74.55 g/mol
Mole to gram conversion:
no. of moles of solute
Amount of solute (in grams) = Moles of solute x Molar mass
Example
What is amount of NaOH present in 2 liters of 4M solution?
Molarity → Moles → Amount of solute
= mass in g / molar mass
= 2.55 g / 74.55 g/mol. = 0.0342 mol
⇒ Liters of Solution = no. of moles of solute
Molarity
3
= 0.0342 mol. = 2.206 L = 2.206 x 10 mL
0.155 M
General Chemistry – Handout (Page:
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DILUTION
Exercise
Dilution means reducing
educing the concentration of a solution (reducing the
molarity of a solution).
-
The concentration of a solution can be reduced by adding
solvent into the solution.
-
Dilution changes only the concentration of the solution, but does
not change the number of moles present.
What is the volume of 16 M H2SO4 must be used to prepare 1.5 L of
a 0.10 M H2SO4 solution?
Concentrated solution: M1 = 16 M, V1 = ?
Diluted solution: M2 = 0.10 M, V2 = 1.5 L
M1 V1 = M2V2
Moles of solute before dilution = Moles of solute after dilution
16 x V1 = 0.10 x 1.5
In the lab, concentrated solution is called as stock solution and diluted
solution is called as working solution.
Formula for dilution related problems:
M1 V1 = M2V2
V1 = (0.10 x 1.5) / 16 = 0.0094 L = 9.4 ml
Example: What volume (in L) of 2.50 M NaCl solution is required to
completely react with 0.150 L of a 0.150 M AgNO3 solution according to the
reaction:
NaCl(aq) + AgNO3(aq) AgCl(ss) + NaNO3(aq)
M1 , V1 = Molarity and volume
lume of concentrated solution
(before dilution)
Strategy:
M2 , V2 = Molarity and volume of diluted solution (after dilution)
L of AgNO3 soln mol AgNO3 mol NaCl L of NaCl soln
Example
2 M solution (2 moles of solute/1 L solution)
Dilution
0.150 mol AgNO3
1 L solution
M [AgNO3 ] =
1 mol NaCl
1 mol AgNO3
1 L solution
0.250 mol NaCl
0.150 mol AgNO3
1 L solution
Add 1 L solvent
1 M solution (2 moles of solute / 2 L solution)
1 mol AgNO3 = 1 mol NaCl
M [NaCl] =
Concentration: Changed from 2 M to 1 M
Number of moles: Does not change (2 moles)
M1V1 = 2 (M) x 1 (L) = 2
M2V2 = 1 (M) x 2 (L) = 2
= 0.150 L ×
0.250 mol NaCl
1 L solution
0.150 mol AgNO3
1 L
×
1 mol NaCl
1L
×
1 mol AgNO3 0.250 mol NaCl
= 0.0900 L of NaCl
Therefore, M1 V1 = M2V2
General Chemistry – Handout (Page:
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+
Aqueous solution
The meaning of ‘Aqua’ is water. If water is the solvent, the solution is called
aqueous solution (solute can be solid, liquid or gas).
-
Finally positive (Na ) and negative (Cl ) ions were separated from the
molecule (NaCl), and surrounded by water. This process is called
hydration.
The ions surrounded by water are called hydrated ions.
Example: 1) Sugar + Water → Sugar solution (or) aqueous solution of sugar
2) NaCl (salt) + Water → Salt solution (or) aqueous solution of salt
H2O
WATER: The common solvent
-
Structure of Water (H2O)
Cl surrounded
by H2O
NaCl
Two hydrogen atoms share electrons
unequally with one oxygen atom to form
two polar covalent bonds.
+
Na surrounded
by H2O
Because of the unequal sharing of
electron (‘O’ has more electrons than ‘H’),
Example 2: NH4NO3
+
Hydrogen has partial positive charge (Delta positive; δ ).
Oxygen has partial negative charge (δ ).
NH4NO3 dissolves in water to produces hydrated ions (NH4+ and NO3-).
-
(Please note that some ionic compounds are not soluble in water).
Thus, water is a polar molecule.
Water dissolves Polar Molecules
Water dissolves Ionic molecules
H2O also dissolve polar (non-ionic) molecules which has partial positive (δ+)
and partial negative (δ+) charge (like water).
Ionic molecules (salt) have positive ions and negative ions.
Example 1: NaCl
Example: Ethanol (C2H5OH) is a polar molecule.
+
-
When NaCl dissolved in water, it produces Na and Cl ions
Negative end (δ ) of polar solute molecule (ethanol) is attracted to
+
water's hydrogen (δ ).
Positive end (δ ) of polar solute molecules (ethanol) is attracted to
water's oxygen (δ ).
H2O (l)
NaCl (s) →→→→
Na+ (aq) + Cl- (aq)
+
-
Positive ions (Na ) attracted to the oxygen end (δ ) of water.
Negative ions (Cl ) attracted to the hydrogen end (δ ) of water.
-
-
+
+
General Chemistry – Handout (Page:
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1. Strong electrolytes
All the solutes (100%) are dissolved in water to produce separate ions. The
solution strongly conducts electric current.
Example:
H2O
i.
Ionic compounds: NaCl
H2O (l)
+
NaCl (s) →→→→
Na (aq) + Cl (aq)
SOLUBILITY
ii.
Strong acids: HCl, H2SO4, HNO3
Solubility is the property of a solute (solid, liquid, or gas) to dissolve in a
solvent (liquid).
H2O (l) +
HCl →→→→
H (aq)+ Cl (aq)
"Like Dissolves Like" Rule
H2O (l)
+
H2SO4 →→→→
H (aq)+ HSO4 (aq)
1. Polar solvents dissolve polar and ionic compounds.
Example:
Ethanol (polar) and NaCl (ionic) dissolve in water (polar).
H2O (l) H+ (aq)+ NO - (aq)
HNO3 →→→→
3
iii.
Strong base: NaOH, KOH
H2O (l) Na+ (aq) + OH- (aq)
NaOH →→→→
2. Non-polar solvents dissolves non-polar compounds.
Example:
Fat (non-polar) dissolve in benzene (non-polar).
H2O (l) K+ (aq) + OH- (aq)
KOH →→→→
2. Weak electrolytes
ELECTROLYTES
Solute dissolved in water solvent to produce aqueous solution.
Solutes are dissolved in water, but only a small percentage of dissolved
molecules produce ions. The solution weakly conducts electric current.
Solute (solid, liquid or gas) + Water (solvent) → Aqueous solution.
If the aqueous solution conducts electric current, the solute is called as
electrolytes. Electrolytes are classified into 3 types.
Example:
i.
1. Strong electrolyte
Weak acids: Acetic acid (HC2H3O2)
H2O (l) H O+ (aq) + C H O - (aq) + HC H O (aq)
HC2H3O2 →→→→
3
2 3 2
2 3 2
2. Weak electrolyte
3. Non-electrolyte
Ions present in the solution (+ve and –ve ions produced form the solute) are
responsible to conduct electric current.
ii.
Weak base: Ammonia (NH3)
H2O (l)
+
NH3 →→→→
NH4 (aq) + OH (aq) + NH3 (aq)
3. Non-electrolytes
General Chemistry – Handout (Page:
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+
Solutes are dissolved in water, but do not produce any ions. The solution
does not conduct an electric current.
Example:
i.
-
-
+
AgCl (s) + K (aq) + NO3 (aq)
3) Net ionic equation (show only components that actually react)
Ethanol: C2H5OH
+
-
Ag (aq) + Cl (aq) → AgCl (s)
H2O (l)
C2H5OH →→→→
C2H5OH (aq)
ii.
+
-
Ag (aq) + NO3 (aq) + K (aq) + Cl (aq) →
-
+
K and NO3 are present in both sides of the reaction (see
reaction 2). They are called spectator ions.
Sugar
1) PRECIPITATION REACTION
TYPES OF CHEMICAL REACTIONS
Solution + Solution → Insoluble solid (precipitate)
Chemical reaction is a change in which substances (reactants) are changed
into one or more new substances (products).
Solution reaction: If the reactants are solutions, the reaction is called
solution reaction.
Important solution reactions
1) Precipitation reactions
Solution + Solution → Insoluble solid (precipitate)
Description of the Acid-Base reaction:
1) Formula equation
AgNO3 (aq) + NaCl (aq) → AgCl (s) + NaNO3 (aq)
2) Complete ionic equation
+
-
+
-
Ag (aq) + NO3 (aq) + Na (aq) + Cl (aq) →
-
+
2) Acid-base reactions or Neutralization reactions
Transfer of proton from one to another.
Acid solution + Base solution → Salt solution + Water
AgCl (s) + Na (aq) + NO3 (aq)
3) Net ionic equation
+
3) Oxidation-reduction reactions
Solution + Solution → Transfer of electron from one to another
-
Ag (aq) + Cl (aq) → AgCl (s)
Simple rules for the solubility of salts in water
-
Describing solution reactions
We can describe the solution reaction in 3 ways.
1) Formula equation (reactants and products are represented in
molecular formula)
1. Most nitrate (NO3 ) salts are soluble.
+
+
+
+
2. Most salts containing the alkali metals (Group 1A: Li , Na , K , Cs ,
+
+
Rb ) and ammonium (NH4 ) ions are soluble.
-
Ionic compounds separated into +ve and –ve ions in the solution.
-
+
2+
2+
4. Most sulfate salts are soluble (NOT BaSO4, PbSO4, HgSO4, CaSO4)
AgNO3 (aq) + KCl (aq) → AgCl (s) + KNO3 (aq)
2) Complete ionic equation (all strong electrolytes shown as ions)
-
3. Most Cl , Br , and I salts are soluble (NOT Ag , Pb , Hg2 )
-
5. Most OH salts are not soluble (NaOH, KOH are soluble, Ba(OH)2,
Ca(OH)2 are marginally soluble)
2-
2-
2-
3-
6. Most S , CO3 , CrO4 , PO4 salts are not soluble.
General Chemistry – Handout (Page:
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Description of the Acid-Base reaction:
Exercise 1: Calculate the mass of PbSO4 formed when 1.25 L of 0.0500 M
Pb(NO3)2 and 2.00 L of 0.0250 M Na2SO4 are mixed. The reaction is
1) Formula equation
HCl (aq) + NaOH (aq) → NaCl (s) + H2O (aq)
Na2SO4 (aq) + Pb(NO3)2 (aq) → PbSO4 (s) + 2NaNO3 (aq)
2) Complete ionic equation
+
-
+
-
+
-
H (aq) + Cl (aq) + Na (aq) + OH (aq) → Na (aq) + Cl (aq) + H2O (l)
Moles of Pb(NO3)2 = 1.25 L X 0.0500 moles = 0.0625 moles
1L
3) Net ionic equation
+
Moles of Na2SO4 = 2.00 L X 0.0250 moles = 0.0500 moles
1L
-
H (aq) + OH (aq) → H2O (l)
Na2SO4 is the limiting reactant.
Acid-Base Titrations
Therefore, 0.0500 mol PbSO4 is formed.
+
-
In the neutralization reaction (H + OH → H2O):
+
Exercise 2: On the basis of solubility rules, predict whether aluminum nitrate
and iron (III) phosphate are soluble in water or not.
Aluminum nitrate – Soluble (rule 1)
Iron (III) phosphate – Not soluble (rule 6)
-
Moles of Acid (H ) = Moles of Base (OH )
Therefore, if the concentration of one reactant is known, we can find out the
concentration another reactant required for complete neutralization. This can
be measured by ‘titration’ (with the use of a chemical indicator). The point at
which the indicator changes color is called ‘end point’.
3) OXIDATION-REDUCTION REACTIONS
Exercise 3: When Al(NO3)3 and Ba(OH)3 solutions are mixed, what
precipitate (if any) will form?
Oxidation = Lose of electrons
Reduction = Gain of electrons
Possible reaction: Al(NO3)3 + Ba(OH)3 → Ba(NO3)3 + Al(OH)3
As per solubility rule 5, Al(OH)3 is not soluble. So it form precipitate.
2Mg (s) + O2 (g) → 2MgO (s)
Oxidation: 2Mg → 2Mg
-
2) ACID-BASE REACTION (OR) NEUTRALIZATION REACTION
+
2+
Reduction: O2 + 4e → 2O
-
An acid produces H when dissolved in water, and a base produced OH .
Brønsted -Lowry definition: ACID is a proton donor, and BASE is a proton
acceptor
-
+ 4e
2-
2Mg2+ + 2O2- → 2MgO
In this reaction, Mg is oxidized and O2 is reduced.
Mg is reducing agent and O2 is oxidizing agent.
Acid solution + Base solution → Salt solution + Water
General Chemistry – Handout (Page:
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Oxidation state
Oxidation number of the compound = 0
Oxidation state (or oxidation number) is an indicator of the degree of
oxidation of an atom in a chemical compound.
Therefore, (2 x +1) + (2 x oxidation number of Cr) + (7 x -2) = 0
+ 2 + Oxidation number of 2Cr -14= 0
S.No
Oxidation state (or oxidation number)
1
The oxidation number of the atom of a free
element is zero
Element
(H, Na, Cl) = 0
2
In ionic compounds: The oxidation number
of a monatomic ion equals its charge
Na in Nacl = +1
In covalent compounds: The oxidation
number of the atoms depends on sharing of
electrons.
O in H2O = -2
3
4
The sum of the oxidation states for an
electrically neutral compound must be zero
Example
Oxidation number of 2Cr = -2+14 = +12
Oxidation number of Cr = +6
+
-
Cl in NaCl = -1
H in H2O = +1
Examples: Identify the oxidizing agent and reducing agent in the
following reactions:
1. 2Li (s) + Cl2 (g) → 2LiCl (s)
F in HF = -1
NaCl = +1-1 = 0
H2O = (2 x +1) -2 = 0
Example 1
2 Li0 (s) + Cl20 (g) → 2 Li+1Cl-1
Li0 (s) → Li+1 (Loss of electron – Oxidation – Li is Reducing Agent)
0
-1
Cl2 (g) → Cl (Gain of electron – Reduction – Cl2 is Oxidizing agent)
In IF7 (compound), the oxidation number of F is -1. What is the
oxidation number of I?
2. 2Al (s) + 3 Sn
2+
3+
(aq) → 2 Al
(aq) + 3 Sn (s)
Oxidation number of F = -1
2Al0 (s) + 3 Sn2+ (aq) → 2 Al3+ (aq) + 3 Sn0 (s)
The oxidation number of the compound = 0
Al → Al (Loss of electrons – Oxidation – Al is Reducing Agent)
Therefore,
Sn
0
2+
+3
0
2+
→ Sn (Gain of electrons – Reduction – Sn
is Oxidizing agent)
Oxidation number of I + Oxidation number of F7 = 0
Oxidation number of I + (-7) = 0
Combustion Reaction:
Oxidation number of I = +7
These are the redox reaction in which substance reacts with oxygen (O2) to
form oxygen-containing compound.
Example 2
In K2Cr2O7 (compound), the oxidation number of O is -2 & K = +1.
What is the oxidation number of Cr?
O = -2, K = +1
General Chemistry – Handout (Page:
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