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


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


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


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.


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


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


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


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 %


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

Q3U3 Molarity, Solubility, Equilibrium, and Determining Experimental Yield.

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