Q3U3 Molarity, Solubility, Equilibrium, and Determining Experimental Yield.
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Transcript of Q3U3 Molarity, Solubility, Equilibrium, and Determining Experimental Yield.
Q3U3 Molarity, Solubility, Equilibrium, and Determining Experimental Yield
Chemical Reactions
Solutions consist of 2 components1. solute: a substance that is dissolved2. Solvent: the medium it is dissolved in
Solutions are homogeneous mixtures: constant composition throughout
Solutes do not have to be solids, They can be gas
CO2 dissolves in water, as does O2 (this is what fish breathe)
They can be liquid Ethylene glycol (antifreeze)
The most common solvent is water! The strength of a solution is important and can be
described in a quantitative way
Solutions and Molarity
The Mole is used to describe the concentration of a solutionOr the “strength” of a solution
A 1 Molar solution contains 1 mole of a solute dissolved in enough solvent (usually water) to make 1 liter of solutionMolarity (M) = moles of solute liter of solutionOne mole of NaCl is 58.5g, if 2 moles of NaCl are dissolved in water to make 1 liter of solution, the solution is a 2M solution This number is very consistent and uniform in all 2 molar solutions of NaCl
Molarity
Calculate the Molarity of 1.50 L of solution that contains 200.0g of MgCl2
Molarity= (200.0g MgCl2) X (1 molMgCl2) = 1.40 mol/L
( 1.50 L ) (95.3 g MgCl) =1.40 M
Example
Calculate the molarity of a solution that contains 10.0g of sodium hydroxide in 5.00 X 102 mL of solution
Molarity = 10.0 g NaOH X 1 mol NaOH X 1000mL
5.0 X 102 mL 40.0 g NaOH 1L
= 0.500 mol/L = .500 M
Example 2
Pg 109 #’s 1-3
Home Work
1. a. 1.00 M b. 0.25 M c. 1.21 M 2 a. 41.6 g b. 421 g c. 71.0 g3 a. 1.00 L b. 4.00 L c. 0.737 L
Home Work Answers
Another way to express concentration of solutions
Defined as the moles of a substance divided by the moles of solution (sum of moles of solute and moles of solvent)
X = moles solute (moles solute + moles solvent)
The sum of all of the mole fractions in a solution equals 1
Mole Fraction (X)
What are the mole fractions of glucose and water in a solution made of 7.59g of glucose, dissolved in water?
Molecular mass of glucose is 180.0g, find the moles of glucose.
mol glucose= (7.59g glucose) (1 mol glucose)= 0.0422 mol
( 1 ) (180.0 g glucose)The molecular mass of water is 18.0g, find the moles of
water.mol water = (125 g water) (1 mol water) = 6.94 mol (18.0 g water)
Mole Fraction Example
Now find the mole fraction of glucose
Mole fraction = mole solute/(mole solute + mol solvent)
=0.0422mol/ (0.0422+6.94)= 0.00605
Now find the mole fraction of water
Mol fraction water = 6.94/6.98 =0.994
If you multiply the mole fraction by 100, you get the mole percent!
Mole Fraction Example continued
Calculate the mole fraction of each component in the following
1. 22.5g CH3CH2OH in 1.0 x 102 g H2O
H2O = 0.919CH3CH2OH = 0.0807
2. 39.5 g C6H5CH3 toluene, in 1.5 X 102 g C6H6, benzene
C6H5CH3 = 0.183C6H6 = 0.817
Now you try these
Calculate the molarity of the following solution 1. 500.0mL with 82.0g Ca(NO3)2
Calculate the mass of solute in the following solutions2. 250.0mL Na2SO4. 7H2O solution that is 2.0 M
3. 1.50 L of KH2PO4 solution that is 0.24 M
How many liters of solution can be made from the following?
4. A 0.10 M solution using 117.0g NaCl
5. A 1.00 M solution using 50.0 g CuSO4 5H2O
6. 0.20 M solution using 200.0g Na2S
Quiz
1. Molarity of ---500.0mL with 82.0g Ca(NO3)2(82.0 g/500.mL) (1mol/164.0g) (1000mL/1L) =1.o M2. Find mass --250.0mL Na2SO4. 7H2O solution that is 2.0
M(250.0mL/1) (2.0 mol/1L) (268g/1mol) (1L/1000mL) =134g3. Find mass of -- 1.50 L of KH2PO4 solution that is 0.24 M(1.50 L/1) (0.24 moles/1L) ( 136g/1mol) = 48.96 g KH2PO44. liters solution ? -- a 0.10 M solution using 117.0g NaCl(117.0g/1) (1 mol/58.5 g) (1L/0.1 mol)= 20.0L5. liters?- 1.00 M solution using 50.0 g CuSO4 5H2O(50.0g/1) (1 mol/249.5 g) (1L/1 mol) =0.20 L6. liters?- 0.20 M solution using 200.0g Na2S
(200.g/1) (1 mol/78.0 g) (1 L/0.2 mol) = 12.8 L
Quiz problem solutions
Previously we have assumed that all reactions have gone to completion ( react until one of the reactants is used up, then the reaction stops)
Reactions go to completion because the elements from the initial reaction are removed in one of the following 3 ways:1. Formation of a gas2. Formation of a slightly ionized substance3. Formation of a precipitateExperimentation has proved that, in some cases,
not all of the reactants are converted to the product, no matter how much time is given.
These reactions are reversible!
Equilibrium Reactions
Equilibrium reactions do not go to completionAs the reaction proceeds, the reactant
concentrations decrease as the product is formed. However at a certain point, the concentrations level
off. There is no additional product formed, or additional
reactant consumedThe concentration of reactants and products no longer
change, A state of Chemical equilibrium has been established.
Ex. N2 (g) + 3H2 (g) <----> 2 NH3 (g) + energyA reversible chemical reaction is in equilibrium
when :the rates of the opposing reactions are equal and overall concentrations remain constant
Equilibrium Reactions
Le Chatelier’s Principle: If a system in equilibrium is subjected to a disturbance, the equilibrium will shift in an attempt to reduce the disturbance and regain equilibrium Disturbances or outside influences can consist
of the following:Concentration changePressure changeTemperature change
ALSO> If more reactant is added to the system in equilibrium, the reaction shifts right and more product is formed
Le Chatelier’s Principle and Reactants
Concentration Change Example: Ammonia equilibrium equation
N2 (g) + 3H2 (g) ----- 2NH3 (g) + energy
Adding N2 (g) disturbs the system, the reaction shifts right This can be corrected through the consumption
of N2, so the reaction shifts right and more NH3 is made.
If H2 (reactant) is removed, the reaction shifts leftThe disturbance is relieved by producing more
H2 and N2, consuming NH3
Le Chatelier’s Principle and Reactants
Scale /gumdrops demo
Pressure Change Example: Ammonia equilibrium equation
N2 (g) + 3H2 (g) ----- 2NH3 (g) + energyNOTE! Pressure Change Only Affects Gaseous
equilibrium systems!If Pressure on reactant gas is increasing,
this causes the reaction to shift towards the side with the least volume.
In the above reaction, the shift would be to the right because this would make 4 particles (N2(g) + 3H2) combine to make only 2 particles (NH3)This would reduce particle collisions, this reduces the
pressure, thereby relieving the disturbance
Le Chatelier’s Principle and Reactants
Pressure Demo-Gumdrops
Temperature Change Example: Ammonia equilibrium equation
N2 (g) + 3H2 (g) ----- 2NH3 (g) + energy
A temperature increase will cause the reaction to shift in the direction of the endothermic reaction
In the example, the left to right reaction is exothermic produces heatA rise in temp will therefore shift the reaction
toward the left (endothermic reaction)!Using up the heat, corrects the disturbance
Le Chatelier’s Principle and Reactants
The Following disturbances do not cause equilibrium shiftsCatalysts, which speed up the reaction by
lowering the activation energy, DO NOT CAUSE AN EQUILIBRIUM SHIFT!They speed up the reaction but do not increase the
amount of product producedInhibitors, which slow down reactions, also do not
cause a shift.They cause the reaction to take longer, but do not
reduced the amount of product produced.
Le Chatelier’s Principle and Reactants
For the following gaseous equilibrium reactions, indicate what happens to the equilibrium position when the given disturbance or condition change occurs. (shift left, right or no shift)
a. Remove NH3 gas b. Decrease pressure
N2 + 3H2 2NH3 + energyc. Decrease temperature d. add a
catalyst CO2 + H2 + energy CO + H2O e. Increase SO2 concentration f. increase
temperature 2SO2 + O2 2 SO3 + energyg. Increase temperature h. increase CO
concentration CO2 + C + energy 2CO
Practice Equilibrium Problems
For the following gaseous equilibrium reactions, indicate what happens to the equilibrium position when the given disturbance or condition change occurs. (shift left, right or no shift)
i. Decrease pressure j. remove N2O4 N2O4 + energy 2 NO2
k. Increase H2 concentration l. increase pressure
H2 + Cl2 2HCl + energy
m. Decrease O2 concentration n. add catalyst N2 + O2 + energy 2NO
Equilibrium Problems
A. rightB. left , gas will fill spaceC. left, produces heat (exothermic)D. no change, increases speed but does not change
volumeE. rightF. left, toward endothermicG. right, toward endothermicH. left, consume excessI. right, increase volumeJ. leftK. right L. no change, particle number is not changed either wayM. leftN. no change
Answers
The submicroscopic interactions that occur between water molecules and various solute particles determine the extent to which water is able to dissolve a solute.
Ionic compoundsLike salt in water
Dissociation: the process by which changed particles in an ionic solid separate from one another
Produces ions
Covalent compoundsSugar
No dissociation, just separation (neutral particles)
Solubility
Both cases: interparticle attractive forces between the solvent and the solute particles, overcome attractive forces between the solute particles
Like dissolves like: dissolving occurs when similarities occur between solvent and solutePolar and polarNon-polar and non-polar
Solubility
Saturated solution: a solution that holds the maximum amount of solute per solution, the solution can not dissolve any more solute
Unsaturated solution: The amount of solute dissolved is less than the amount of solute than can be dissolved, the solvent can dissolve more solute
Supersaturated solution: Unstable solutions which contain more dissolved solute than they can usually hold. Conditions were manipulated to get additional solute dissolved, temporarily. Ex fudge making, more sugar is dissolved by
heating the mixture, then slow cooling with constant stirring keeps the sugar trapped in small crystals in the fudge
Solution Saturation
Temperature effects solubility!The solubility of most SOLID
solutes generally increases with increasing temperature
The solubility of GASES always decreases with increased temperature
The amount of increase or decrease depends on the individual solutes
Solubility
Many reactants continue until one of the reactants is consumed, the reactant used up first is called the limiting reactant.
This is used only in non-reversible reactions!Limiting reactant problems are solved by comparing
the moles of each reactant. Steps
1. Write a balanced equation2. Change given to moles3. Determine moles produced by given quantity, for each4. Determine predicted quantities produced using the
lesser amount produced (limiting reaction)
Limiting Reactants
Example: If 40.0g H3PO4 react with 60.0g MgCO3. calculate the volume of CO2 produced at STP.
2H3PO4 + 3MgCO3 Mg3(PO4)2 + 3CO2 +3H2O
Step 2: Change Grams to molesMol H3PO4 = 40g H3PO4 1mol H3PO4 =
0.408 mol 1 98.0 g H3PO4Mol MgCO3 = 60.0 g MgCO3 1mol MgCO3 =
0.712 mol 1 84.3g MgCO3
Problem using limiting reactants
Step 3 :determine moles of CO2 which will be produced by each reactant(molar ratio from balanced equation)
Mol CO2 = 0.408 mol H3PO4 3 mol CO2 = 0.612 mol
1 2 mol H3PO4
Mol CO2 = 0.712 mol MgCO3 3 mol CO2 =0.712 mol
1 3 mol MgCO3The limiting reactant produces the lesser amount of
product- H3PO4
Ex Problem, limiting reactants continued
Step 4 :Use the limiting reactant to complete the problem (remember molar volume is 22.4L/mole at STP)
Volume CO2 = 0.612 mol CO2 22.4 L = 13.7 L at STP
1 1 mol CO2
4og of H3Po4 will produce 13.7 L of CO2 at STP
Example Problem, limiting reactants continued again!
If 20.og of NaOH react with 30.0g of H2SO4 to produce Na2SO4, which reactant is limiting?
2NaOH + H2SO4 Na2SO4 +2H2O30 g H2SO4 1 mol H2SO4 = 0.30 mol
H2SO4 1 98 g H2SO420g NaOH 1mol NaOH = 0.5 mol NaOH 1 40.0g NaOH0.30 mol H2SO4 1 mol Na2SO4 = 0.3 mol
Na2SO4 prod 1 1 mol H2SO40.5 mol NaOH 1 mol Na2SO4 = 0.25 mol
Na2SO4 prod 1 2 mol NaOHNaOH is limiting reactant, 0.25 mol Na2SO4
prod
Now you try one
If 5.00g of copper metal react with a solution containing 20.0g of AgNO3 to produce silver metal, which reactant is limiting?
Cu + 2AgNO3 Cu(NO3)2 + 2Ag
Answer: AgNO3, 0.118 mol Ag5.0 g copper moles20.0 g AgNO3 molesMoles and mole ratio # of moles produced1 mol Cu : 2 mol Ag2 mol AgNO3 : 2 mol Ag Smallest mole amount produced is limiting reactant
Try another one!
RememberThe amount of product produced by a chemical
reaction, as predicted by stoichiometry is called Theoretical Yield
when actual yield, produced via experimentation, is different than theoretical yield, the efficiency of the reaction is expressed by determining percent yield.
Percent yield = (actual yield X 100 (Theoretical yield) This yield can be affected by many different
things
Theoretical and Experimental Yield
Example 1. A student conducts a single displacement
reaction that produces 2.755 grams of copper.
Mathematically he determines that 3.150 grams of copper should have been produced.
Calculate the student's percentage yield.
Percent Yield Problems
Solve: actual amount of product: 2.755 g expected amount of product: 3.150 g
actual amount of productpercentage yield = --------------------------------------- x 100 expected amount of product
2.755g
percentage yield = --------------- x 100 3.150g
percentage yield = 87.4603174 % = 87.46 %
Example 1 Solution
Example 2. A student completely reacts 5.00g of
magnesium with an excess of oxygen to produce magnesium oxide.
Analysis reveals 8.10 g of magnesium oxide.
What is the student's percentage yield?
Example 2
In this problem, you are given the actual amount of the product, but you are not given the expected amount of the product.
The second mass shown, 5.00g, is the mass of one of the reactants.
In order to determine the expected amount of the product in this problem, you must begin with a mass-mass problem.
Solution to Example 2
First, write a balanced chemical equation for the reaction:
2Mg(s) + O2(g) ----> 2MgO(s)
Now, label the given and the unknown: given unknown
2Mg(s) + O2(g) ----> 2MgO(s)
5.00 g X g
Solution to Example 2 continued
Change the mass given to moles by dividing by the molar mass of Mg.
mass given: 5.00g molar mass of Mg: 24.3 g
5.00gNumber of moles of Mg = ----------- 24.3 g/mole
Number of moles of Mg = 0.206 moles
Solution to Example 2 continued
Compare the molar ratio between the given and the unknown to determine the number of moles produced.
Coefficient of Mg; 2 Coefficient of MgO:2
# of moles of Mg: = 0.206 moles Number of moles of MgO: = ?
number of moles of given number of moles of unknown-------------------------------- = -------------------------------------- coefficient of given coefficient of unknown
0.206 moles X moles -------------- = ------------- 2 2
Number of moles of MgO produced = 0.206 moles
Solution to Example 2 continued
Now, change the number of moles of MgO produced to the mass by multiplying by the molar mass of MgO.
# of moles of MgO = 0.206 molesMolar mass of MgO = 40.3g/mole
mass = # of moles x molar mass mass of MgO = 0.206 moles x 40.3
g/mole mass of MgO = 8.30 g
Solution to Example 2 continued
Now, you are ready to solve the percentage yield problem.
actual mass of MgO produced = 8.10 g expected mass of MgO = 8.30 g
actual amount of product
percentage yield = expected amount of product X 100
8.10 g percentage yield = ------------ x 100 8.30 g
percentage yield = 97.6 %
Solution to Example 2 continued
A student completely reacts 6.00g of magnesium with an excess of oxygen to produce magnesium oxide.
Analysis reveals 8.50 g of magnesium oxide.
What is the student's percentage yield?
Your Turn; Try This One
2Mg(s) + O2(g) ----> 2MgO(s)
Change the mass given to moles mass given: 6.00g
molar mass of Mg: 24.3 g 6.00g
Number of moles of Mg = ----------- 24.3 g/mole
Number of moles of Mg = 0.247 molesCompare the molar ratio:
number of moles of given number of moles of unknown-------------------------------- = -------------------------------------- coefficient of given coefficient of unknown
0.247 moles X moles -------------- = ------------- 2 2
Number of moles of MgO produced = 0.247 moles mass = # of moles x molar mass mass of MgO = 0.247 moles x 40.3 g/mole mass of MgO = 9.95 g 8.50 g
percentage yield = ------------ x 100 9.95 g
percentage yield = 85.4 %
Answer