TOPIC 1 TOPIC 1 STOICHIOMETRIC STOICHIOMETRIC RELATIONSHIPSRELATIONSHIPS

1.31.3

REACTING MASSES AND VOLUMESREACTING MASSES AND VOLUMES

ESSENTIAL IDEAESSENTIAL IDEA

Mole ratios in chemical equations Mole ratios in chemical equations can be used to calculate reacting can be used to calculate reacting ratios by mass and gas volume.ratios by mass and gas volume.

NATURE OF SCIENCE (1.8) NATURE OF SCIENCE (1.8)

Making careful observations and obtaining Making careful observations and obtaining evidence for scientific theories – Avogadro’s evidence for scientific theories – Avogadro’s

initial hypothesis.initial hypothesis.

STOICHIOMETRYSTOICHIOMETRY

For any stoichiometry problem, you must For any stoichiometry problem, you must know what unit you are given and what know what unit you are given and what unit you are looking for.unit you are looking for.

Step one is to ALWAYS convert to moles Step one is to ALWAYS convert to moles unless you are already in moles.unless you are already in moles.

Step two is to multiply by the mole ratio Step two is to multiply by the mole ratio with the unknown on top.with the unknown on top.

Step three is to convert to the unit you are Step three is to convert to the unit you are looking for using the mole conversion looking for using the mole conversion chart.chart.

Stoich Problems – Mole to MoleStoich Problems – Mole to Mole Given moles – looking for molesGiven moles – looking for moles

Step 1: Label your equation

How many moles of H2O are produced from 6.0 moles of oxygen?

Step 2: Write down the given and multiply by the mole ratio with the unknown on top.

2H2H22 + O + O22 2H 2H22OO

MOLE TO MOLE EXAMPLESMOLE TO MOLE EXAMPLES

1. N2 + 3H2 2 NH3

How many moles of N2 are needed to make 12.2 moles of NH3?

2. 2C2H2 + 5O2 4CO2 + 2H2O

How many moles of CO2 are produced from .80 moles of O2?

3. 2H2S + 3O2 2SO2 + 2H2OHow many moles of H2S must react with .68 moles O2?

PRACTICE - MOLE TO MOLEPRACTICE - MOLE TO MOLE 2C2C22HH22 + 5O + 5O22 4CO 4CO22 + 2H + 2H22OO 1. How many moles of CO1. How many moles of CO22 are produced are produced

when 14.3 moles of Owhen 14.3 moles of O22 are burned? are burned?

2. How many moles of water are produced 2. How many moles of water are produced when .8432 moles of COwhen .8432 moles of CO22 are produced? are produced?

3. How many moles of oxygen are needed 3. How many moles of oxygen are needed to react with 11.44 moles of Cto react with 11.44 moles of C22HH22??

Stoich Problems – Mole to MassStoich Problems – Mole to Mass Given moles – looking for gramsGiven moles – looking for grams

2H2H22 + O + O22 2H 2H22OOHow many grams of water are formed from 6.0 moles of H2?

Step 1: Label the equation correctly.

Step 2: Write down the given then multiply by the mole ratio – unknown on top.

Step 3: Multiply by molar mass of unknown to convert from moles to grams.

MOLE TO MASS EXAMPLESMOLE TO MASS EXAMPLES

2H2H22 + O + O22 2H 2H22OO 1. How many grams of water are 1. How many grams of water are

produced when 4.84 mol hydrogen produced when 4.84 mol hydrogen reacts?reacts?

2. How many grams of oxygen are need 2. How many grams of oxygen are need to react with 54.2 mol of hydrogen?to react with 54.2 mol of hydrogen?

3. How many grams of hydrogen are 3. How many grams of hydrogen are needed to produce 1.634 mol of water?needed to produce 1.634 mol of water?

PRACTICE – MOLE TO MASSPRACTICE – MOLE TO MASS CaCa33(PO(PO44))22 + 3SiO + 3SiO22 + 5C 3CaSiO + 5C 3CaSiO33 + +

5CO + 2P5CO + 2P 1. How many grams of SiO1. How many grams of SiO22 are needed are needed

to produce 10.0 mol CO?to produce 10.0 mol CO?

2. How many grams of carbon are 2. How many grams of carbon are needed to react with 3.0 mol of needed to react with 3.0 mol of CaCa33(PO(PO44))22??

3. 42.0 mol of C produces how many 3. 42.0 mol of C produces how many grams of CaSiOgrams of CaSiO33??

Stoich Problems – Mass to MoleStoich Problems – Mass to Mole Given grams – looking for moles Given grams – looking for moles

2H 2H22 + O + O22 2H 2H22OO

How many moles of water are formed from 16.0 grams of H2?

Step 1: Label the equation correctly.

Step 2: Put down the given and divide by molar mass of the given to

get moles.

Step 3: Multiply by mole ratio – unknown on top – to get moles.

MASS TO MOLE EXAMPLESMASS TO MOLE EXAMPLES

2H2H22 + O + O22 2H 2H22OO 1. How many moles of water are 1. How many moles of water are

produced when 4.84 grams of hydrogen produced when 4.84 grams of hydrogen reacts?reacts?

2. How many moles of oxygen are need 2. How many moles of oxygen are need to react with 54.2 grams of hydrogen?to react with 54.2 grams of hydrogen?

3. How many moles of hydrogen are 3. How many moles of hydrogen are needed to produce 1.634 grams of water?needed to produce 1.634 grams of water?

PRACTICE – MASS TO MOLEPRACTICE – MASS TO MOLE CaCa33(PO(PO44))22 + 3SiO + 3SiO22 + 5C 3CaSiO + 5C 3CaSiO33 + +

5CO + 2P5CO + 2P 1. How many moles of SiO1. How many moles of SiO22 are needed are needed

to produce 420 g of CO?to produce 420 g of CO?

2. How many moles of CaSiO2. How many moles of CaSiO33 are are produced when 100.0 g of Caproduced when 100.0 g of Ca33(PO(PO44))2 2 are are reacted?reacted?

3. How many moles of Ca3. How many moles of Ca33(PO(PO44))2 2 are are needed to produce 280.0 g of CO?needed to produce 280.0 g of CO?

Stoich Problems – Mass to Stoich Problems – Mass to MassMass

Given grams – looking for gramsGiven grams – looking for grams

2H2H22 + O + O2 2 2H 2H22OO

How many grams of O2 are needed to react with 20.0 g of H2 ?

Step 1: Label the equation.

Step 2: Write down the given and divide by the molar mass of the given.

Step 3: Multiply by the mole ratio to find moles of unknown.

Step 4: Multiply by molar mass of unknown to find grams of unknown.

MASS TO MASS EXAMPLESMASS TO MASS EXAMPLES

2H2H22 + O + O22 2H 2H22OO 1. How many grams of water are 1. How many grams of water are

produced when 4.84 grams of hydrogen produced when 4.84 grams of hydrogen reacts?reacts?

2. How many grams of oxygen are need 2. How many grams of oxygen are need to react with 54.2 grams of hydrogen?to react with 54.2 grams of hydrogen?

3. How many grams of hydrogen are 3. How many grams of hydrogen are needed to produce 1.634 grams of water?needed to produce 1.634 grams of water?

PRACTICE – MASS TO MASSPRACTICE – MASS TO MASS NN22 + 3H + 3H2 2 2NH 2NH33

1. How many grams of ammonia are 1. How many grams of ammonia are formed from 15.0 g Nformed from 15.0 g N22??

2. How many grams of H2. How many grams of H22 are needed to are needed to react with 48.3 g Nreact with 48.3 g N22??

3. How many grams of H3. How many grams of H22 are needed to are needed to produce .914 g NHproduce .914 g NH33??

UNDERSTANDINGS/KEY IDEAUNDERSTANDINGS/KEY IDEA1.3.A1.3.A

Reactants can be either limiting or Reactants can be either limiting or excess.excess.

APPLICATION/SKILLSAPPLICATION/SKILLS

Be able to solve problems relating Be able to solve problems relating to reacting quantities, limiting and to reacting quantities, limiting and excess reactants, theoretical, excess reactants, theoretical, experimental and percentage experimental and percentage yields.yields.

Limiting ReactantLimiting Reactant

The reactant that will run out first in a The reactant that will run out first in a reaction.reaction.

The excess reactant is the reactant that The excess reactant is the reactant that is not used up completely in reaction.is not used up completely in reaction.

Limiting ReactantLimiting Reactant

You will recognize a limiting reactant You will recognize a limiting reactant problem because there will be 2 “givens” problem because there will be 2 “givens” in the problem.in the problem.

2H2H22 + O + O2 2 2H 2H22OO

If 4 moles of H2 react with 8 moles of O2 , how much water will be formed?

Limiting Reactant - CalculationsLimiting Reactant - Calculations

Step 1: Write down and convert both givens to moles (if not already in moles).

Step 2: Set up two stoichiometric problems with the opposite reactant as the mole ratio.

Step 3: Interpret the equations by crossing and comparing to determine the limiting reactant.

4 mols H2 given X = 2 mols O2needed

8 mols O2 given X = 16 mols H2needed

1 mol O2

2 mol H2

2 mol H2

1 mol O2

You need 2 moles of O2 and you have 8 mols O2 given

You need 16 moles of H2, but you were only given 4 moles H2.

Since you do not have enough H2 given, it will run out first. Therefore, H2 is the limiting reactant.

Step 4: Use the limiting reactant, calculate the answer.

2H2H2(g)2(g) + O + O2(g) 2(g) 2H 2H22OO(l)(l)

How many grams of water are formed when 12.0 g of How many grams of water are formed when 12.0 g of hydrogen reacts with 17.0 g of oxygen?hydrogen reacts with 17.0 g of oxygen?

First of all, I recognize that I am given an amount of hydrogen First of all, I recognize that I am given an amount of hydrogen and an amount of oxygen so I need to figure out which one is and an amount of oxygen so I need to figure out which one is going to run out first and stop the reaction. This one is called the going to run out first and stop the reaction. This one is called the limiting reactant. The other reactant will be the excess reactant.limiting reactant. The other reactant will be the excess reactant.

12.0 g H12.0 g H22 x x mol mol = 5.94 mol H= 5.94 mol H22 given given

2.02g2.02g 17.0 g O17.0 g O22 x x molmol = .531 mol O = .531 mol O22 given given

32.0g32.0g Notice that my first step was to convert to moles and to actually Notice that my first step was to convert to moles and to actually

stop after the conversion to moles and write the word “given”.stop after the conversion to moles and write the word “given”.

2H2H2(g)2(g) + O + O2(g) 2(g) 2H 2H22OO(l)(l)

The next step is to find out how much of each reactant is needed to The next step is to find out how much of each reactant is needed to react with each other. You do this by multiplying each reactant by the react with each other. You do this by multiplying each reactant by the mole ratio with each other.mole ratio with each other.

12.0 g H12.0 g H22 x x mol mol = 5.94 mol H= 5.94 mol H22 given x given x 1 mol O1 mol O22 = 2.97 mol O = 2.97 mol O22 needed needed

2.02g 2 mol H2.02g 2 mol H22

17.0 g O17.0 g O22 x x molmol = .531 mol O = .531 mol O22 given x given x 2 mol H2 mol H22 = 1.06 mol H= 1.06 mol H22 needed needed

32.0g 1 mol O32.0g 1 mol O22

Now cross compare to find the reactant that you do not have enough Now cross compare to find the reactant that you do not have enough of. You need 1.06 mol Hof. You need 1.06 mol H22 and you are given 5.94 mol H and you are given 5.94 mol H22 so you have so you have

more than enough. You need 2.97 mol Omore than enough. You need 2.97 mol O22 and are only given .531 mol and are only given .531 mol

OO22 so you do not have nearly enough. This means that oxygen is your so you do not have nearly enough. This means that oxygen is your

limiting reactant; therefore, hydrogen is your excess reactant.limiting reactant; therefore, hydrogen is your excess reactant.

2H2H2(g)2(g) + O + O2(g) 2(g) 2H 2H22OO(l)(l)

To solve the problem, you use the limiting reactant. (Important! Be sure to use the given moles, not the needed moles.)

The problem asked you for the mass of water produced. You now have a mole – mass problem.

.531 mol O2 x 2 mol H2O x 18.02g = 19.1 g H2O

1 mol O2 mol The answer is your theoretical yield. If you were asked how much excess reactant remained,

simply subtract the moles H2 needed by the moles given. 5.94 mol H2 given – 1.06 mol H2 needed = 4.88 mol H2 in

excess.

LIMITING REACTANT LIMITING REACTANT EXAMPLEEXAMPLE

If you have 6.70 mol Na reacting with 3.20 If you have 6.70 mol Na reacting with 3.20 mol Clmol Cl22, what is your limiting reactant, how , what is your limiting reactant, how many moles of product will be formed and many moles of product will be formed and how much excess reactant remains?how much excess reactant remains?

2Na + Cl2Na + Cl22 2NaCl 2NaCl

UNDERSTANDINGS/KEY IDEAUNDERSTANDINGS/KEY IDEA1.3.B1.3.B

The experimental yield can be The experimental yield can be different from the theoretical yield.different from the theoretical yield.

Theoretical Yield – Theoretical Yield – the maximum amount of product that can the maximum amount of product that can be produced from a given amount of be produced from a given amount of reactant (found by using stoichiometry).reactant (found by using stoichiometry).

Experimental Yield – Experimental Yield – the measured amount of product the measured amount of product obtained from a reaction (what you got in obtained from a reaction (what you got in the lab)the lab)

Percent YieldPercent Yield

You will recognize this type of You will recognize this type of problem when they ask you to find the problem when they ask you to find the percent yield.percent yield.

The experimental value will be given The experimental value will be given in the problem and you have to in the problem and you have to calculate the theoretical yield by calculate the theoretical yield by using stoichiometry.using stoichiometry.

EXAMPLE

2H2H2(g)2(g) + O + O2(g) 2(g) 2H 2H22OO(l)(l)

If 17.0 g of oxygen reacts to form 18.7 g of If 17.0 g of oxygen reacts to form 18.7 g of water, what is the percent yield?water, what is the percent yield? First of all, you recognize that this is a percent yield First of all, you recognize that this is a percent yield

problem. Circle 18.7 g and label it “experimental”. Next problem. Circle 18.7 g and label it “experimental”. Next work the problem to solve for the theoretical yield.work the problem to solve for the theoretical yield.

17.0 g O17.0 g O22 x x mol mol x x 2 mol H2 mol H22OO x x 18.02 g18.02 g = = 19.1 g 19.1 g

32.0 g 1 mol O32.0 g 1 mol O22 mol mol % yield = % yield = expexp x 100% = x 100% = 18.7g18.7g x 100% = 97.9% x 100% = 97.9% theo 19.1gtheo 19.1g

UNDERSTANDINGS/KEY IDEAUNDERSTANDINGS/KEY IDEA1.3.C1.3.C

Avogadro’s law enables the mole Avogadro’s law enables the mole ratio of reacting gases to be ratio of reacting gases to be determined from volumes of the determined from volumes of the gases.gases.

Avogadro’s hypothesis states that Avogadro’s hypothesis states that equal volumes of different gases equal volumes of different gases contain equal numbers of particles at contain equal numbers of particles at the same temperature and pressure.the same temperature and pressure.

This means that if you are given gas This means that if you are given gas volumes and asked for gas volumes at volumes and asked for gas volumes at the same conditions, you can find your the same conditions, you can find your answer using the mole ratios without answer using the mole ratios without doing any conversions.doing any conversions.

APPLICATION/SKILLSAPPLICATION/SKILLS

Be able to calculate reacting Be able to calculate reacting volumes of gases using Avogadro’s volumes of gases using Avogadro’s law.law.

Example ProblemExample Problem

40 cm40 cm33 of carbon monoxide is reacted with 40 of carbon monoxide is reacted with 40 cmcm33 of oxygen in the following reaction. of oxygen in the following reaction.

2CO + O2CO + O22 → 2CO → 2CO22

What volume of carbon dioxide is produced?What volume of carbon dioxide is produced?

This is actually a limiting reactant “volumes of This is actually a limiting reactant “volumes of gases” problem and is quite often assessed on gases” problem and is quite often assessed on the IB exam.the IB exam.

40 cm40 cm33 CO given x CO given x 1 mol O1 mol O22 = 20 cm = 20 cm33 O O22 needed needed

2 mol CO2 mol CO

40 cm40 cm33 O O22 given x given x 2 mol CO2 mol CO = 80 cm= 80 cm33 CO needed CO needed

1 mol O1 mol O22

Notice you can use the mole ratio directly Notice you can use the mole ratio directly without converting cmwithout converting cm33 to moles first. to moles first.

You need 80 cmYou need 80 cm33 CO and only have 40cm CO and only have 40cm33 so CO is the limiting reactant.so CO is the limiting reactant.

40 cm40 cm33 CO given x CO given x 2 mol CO2 mol CO22 = 40 cm = 40 cm33 CO CO22 produced produced

2 mol CO2 mol CO

Oxygen is in excess by 20 cmOxygen is in excess by 20 cm33..

UNDERSTANDINGS/KEY IDEAUNDERSTANDINGS/KEY IDEA1.3.D1.3.D

The molar volume of an ideal gas is The molar volume of an ideal gas is a constant at specified temperature a constant at specified temperature and pressure.and pressure.

Standard Temperature and Pressure STP Standard Temperature and Pressure STP = 273 K (0= 273 K (0°C) and 101.3 kPa (1atm)°C) and 101.3 kPa (1atm)

(Use 100 kPa when you can’t use a calculator.)(Use 100 kPa when you can’t use a calculator.)

Room Temperature and PressureRoom Temperature and Pressure

RTP = 298 K (25°C) and 101.3 kPaRTP = 298 K (25°C) and 101.3 kPa The molar volume of a gas at STP is 22.4 The molar volume of a gas at STP is 22.4

dmdm33/mol./mol. The molar volume of a gas at RTP is 24 The molar volume of a gas at RTP is 24

dmdm33/mol./mol.

APPLICATION/SKILLSAPPLICATION/SKILLS

Be able to solve problems and Be able to solve problems and analyze graphs involving the analyze graphs involving the relationship between temperature, relationship between temperature, pressure and volume for a fixed pressure and volume for a fixed mass of an ideal gas.mass of an ideal gas.

KINETIC THEORY OF MATTER

1. The volume of the gas particles is assumed to be zero.

2. The gas particles are in constant motion. 3. The collisions of the gas particles with the

sides of the container cause pressure. 4. The particles exert no forces on each other. 5. The average kinetic energy is directly

proportional to the Kelvin temperature of the gas.

PROPERTIES OF GASES

No definite shape No definite volume Very easily compressed High rate of diffusion Gas particles exert pressure on their

surroundings.

It is important to remember that you ALWAYS use Kelvin temperatures when working gas law problems.

K = oC + 273

Zero K (-273 oC) is called absolute zero and this is the temperature when all motion theoretically ceases to exist.

BOYLE’S LAW Robert Boyle, an Irish chemist, discovered

that the volume of a gas was inversely proportional to the pressure applied. In other words, as pressure is increased, volume decreases.

P1V1 = P2V2

Graph of pressure vs volume

pressure 1/volumepr

essu

re

volu

me

Boyle’s Law: The pressure of a gas is inversely proportional to the volume.

CHARLES’S LAW

Jacques Charles, a French physicist, discovered the relationship between temperature and volume at constant pressure. He was the first person to make a solo balloon flight and it was on his flight that he discovered that the volume of a gas was dependent upon the temperature.

He discovered that the volume of a gas is directly proportional to the Kelvin temperature. In other words, if you increase the temperature, the volume increases.

T1V2 = T2V1

Graph of volume vs temperature

Temperature (K)vo

lum

e

Charles’ Law – the volume of a gas is proportional to the Kelvin temperature.

GAY LUSSAC’S LAW

Joseph Gay Lussac, a French physicist and chemist, another avid balloonist discovered the relationship between temperature and pressure.

He discovered that at constant volume, if you increase the temperature, the pressure increases.

He also invented many types of chemical glassware that is still in use today.

T1P2 = P1T2

Graph of pressure vs temperature

Temperature (°C) Temperature (K)pr

essu

re

pres

sure

Gay Lussac’s Law – the pressure of a gas is directly proportional to the Kelvin temperature.

Effect of temperature on gas volume

1. If you double the temperature and keep volume constant, the pressure doubles.

2. If you then double the volume and keep the temperature constant, the pressure is halved.

3. Therefore, the net overall effect of doubling the temperature and doubling the volume is that there is no overall change in pressure.

AVOGADRO’S LAW

Amadeo Avogadro, an Italian chemist, postulated that equal volumes of gases at the same temperature and pressure contained the same number of particles.

So for a gas at constant temperature and pressure, the volume is directly proportional to the number of moles of a gas.

V1n2 = n1V2

COMBINED GAS LAW

You can combine the first three gas laws into one equation. Remember to always use Kelvin temperature.

In this equation, you will be given 5 knowns and solve for 1 unknown.

P1V1T2 = P2V2T1

APPLICATION/SKILLSAPPLICATION/SKILLS

Be able to solve problems relating Be able to solve problems relating to the ideal gas equation.to the ideal gas equation.

IDEAL GAS LAW Use the ideal gas law when given moles or mass

in a gas law problem or if you are asked to solve for moles or mass.

PV = nRT Where P is pressure, V is volume, n is moles, T

is Kelvin temp, and R is the universal gas law constant.

R = 8.31 J K-1 mol-1 or 8.31 m3 Pa / K mol

(A J is a N-m and a pascal is a Nm-2) You must use the correct units. P has to be in

Pa and V must be in m3.

APPLICATION/SKILLSAPPLICATION/SKILLS

Be able to explain the deviation of Be able to explain the deviation of real gases from ideal behavior at real gases from ideal behavior at low temperature and high pressure.low temperature and high pressure.

REAL GASES

An ideal gas exactly obeys the gas laws. Real gases actually have attractive forces

between them. Real gases actually take up some volume. A gas most behaves like an ideal gas at

high temperatures and low pressures.

APPLICATION/SKILLSAPPLICATION/SKILLS

Be able to obtain and use Be able to obtain and use experimental values to calculate experimental values to calculate the molar mass of a gas from the the molar mass of a gas from the ideal gas equation.ideal gas equation.

MOLAR MASS OF A GAS

Using density and the ideal gas law, the molar mass can be derived.

M = dRT or M = mRT

P V P

where d is density and M is molar mass.

UNDERSTANDINGS/KEY IDEAUNDERSTANDINGS/KEY IDEA1.3.E1.3.E

The molar concentration of a The molar concentration of a solution is determined by the solution is determined by the amount of solute and the volume of amount of solute and the volume of solution.solution.

The The concentrationconcentration is the composition of the is the composition of the solution expressed in moles or grams solute solution expressed in moles or grams solute over volume of the total solution. over volume of the total solution.

When no more solute can be dissolved in a When no more solute can be dissolved in a solution, the solution is saturated.solution, the solution is saturated.

Square brackets [ ] are used to represent Square brackets [ ] are used to represent concentration or molarity.concentration or molarity.

Concentration [ ] = moles of solute/volume of Concentration [ ] = moles of solute/volume of solutionsolution

The unit of volume in [conc] is usually dmThe unit of volume in [conc] is usually dm33.. ***To solve for moles, multiply the concentration ***To solve for moles, multiply the concentration

by the volume given. This is a very common by the volume given. This is a very common calculation.***calculation.***

If you are given a volume of cmIf you are given a volume of cm33 and and asked to find moles with a concentration asked to find moles with a concentration using mol/dmusing mol/dm33, you must divide by 1000., you must divide by 1000.

1 dm1 dm33 = 1000 cm = 1000 cm33

GUIDANCEGUIDANCE

Units of concentration include:Units of concentration include:

g/dmg/dm33, mol/dm, mol/dm33 and parts per and parts per

million (ppm).million (ppm).

GUIDANCEGUIDANCE

Use square brackets to denote Use square brackets to denote molar concentration.molar concentration.

UNDERSTANDINGS/KEY IDEAUNDERSTANDINGS/KEY IDEA1.3.F1.3.F

A standard solution is one of known A standard solution is one of known concentration.concentration.

UNDERSTANDINGS/KEY IDEAUNDERSTANDINGS/KEY IDEA

Strong electrolytes are assumed to Strong electrolytes are assumed to completely break down into ions in completely break down into ions in solution.solution.

ELECTROLYTESELECTROLYTES

Electrolytes are substances that when in Electrolytes are substances that when in solution break into ions and conduct solution break into ions and conduct electricity due to the presence of said ions.electricity due to the presence of said ions.

All ionic compounds are electrolytes All ionic compounds are electrolytes meaning that when in solution, they meaning that when in solution, they dissociate into the ions making up the dissociate into the ions making up the compound.compound.

Strong acids and strong bases dissociate Strong acids and strong bases dissociate completely into their ions.completely into their ions.

APPLICATION/SKILLSAPPLICATION/SKILLS

Be able to determine the ions Be able to determine the ions produced by an electrolyte in produced by an electrolyte in aqueous solution and calculate the aqueous solution and calculate the concentration of those ions.concentration of those ions.

Sample ProblemsSample Problems

1.0 mol/dm1.0 mol/dm33 HCl breaks into 1.0 mol/dm HCl breaks into 1.0 mol/dm33 H H++ ions and 1.0 mol/dmions and 1.0 mol/dm33 Cl Cl-- ions ions

2.0 mol/dm2.0 mol/dm33 Ba(OH) Ba(OH)22 breaks into 2.0 mol/dm breaks into 2.0 mol/dm33

BaBa2+2+ ions and 4.0 mol/dm ions and 4.0 mol/dm33 OH OH-- ions ions

You should be able to break any ionic You should be able to break any ionic compound into its ions and be able to compound into its ions and be able to determine the concentration of said ions by determine the concentration of said ions by the original concentration of the ionic cmpd.the original concentration of the ionic cmpd.

APPLICATION/SKILLSAPPLICATION/SKILLS

Be able to solve problems involving Be able to solve problems involving molar concentration, amount of molar concentration, amount of solute and volume of solution.solute and volume of solution.

A A solutionsolution is a homogeneous mixture of the is a homogeneous mixture of the solvent with the solute.solvent with the solute. The The solutesolute is the substance being dissolved in the is the substance being dissolved in the

solvent and it is usually the less abundant component.solvent and it is usually the less abundant component. The The solventsolvent is the substance doing the dissolving. It is the substance doing the dissolving. It

is usually a liquid and is the more abundant is usually a liquid and is the more abundant component.component.

An aqueous solution is a solution with water as An aqueous solution is a solution with water as the solvent.the solvent.

Solutes can be solids, liquids or gases, but the Solutes can be solids, liquids or gases, but the solvent is generally a liquid.solvent is generally a liquid.

EXAMPLE

Calculate the mass of copper II sulfate pentahydrate, CuSO4

.5H20, required to prepare 500 cm3 of a 0.400 mol/dm3 solution.

First of all, they are asking for mass so you know you need to solve for moles.

You can solve for moles by multiplying volume by the concentration – watch for units.

n = 500 cm3 x 1 dm3/1000cm3 x 0.400 mol/dm3 = .200 mol To solve for mass when you have moles, multiply by the molar

mass. Mass = .200 mol x 249.61g/mol = 49.9 g

APPLICATION/SKILLSAPPLICATION/SKILLS

Be able to use the experimental Be able to use the experimental method of titration to calculate the method of titration to calculate the concentration of a solution by concentration of a solution by reference to a standard solution.reference to a standard solution.

Standard solutions are used to find Standard solutions are used to find concentrations of other solutions.concentrations of other solutions.

The volumetric technique “The volumetric technique “titrationtitration” is most ” is most commonly used to find the concentration commonly used to find the concentration of an unknown solution.of an unknown solution.

EXAMPLE What volume of 2.00 mol/dm3 hydrochloric acid would have to be added to

25.0 cm3 of a 0.500 mol/dm3 sodium carbonate solution to produce a neutral solution of sodium chloride?

First you need a balanced equation:

2HCl + Na2CO3 2NaCl + H2O + CO2

Next find moles of sodium carbonate: 25.0cm3 x 1dm3/1000cm3 x 0.500mol/dm3 = .0125mol Use the mole ratio to find moles of HCl needed:

.0125 mol Na2CO3 x 2 mol HCl = .0250 mol HCl

1 mol Na2CO3

Use the concentration equation to solve for volume since you have moles and concentration.

[Conc] = mol/volume so volume = mol/[conc] volume = .0250mol/2.00mol/dm3 = .0125dm3 x 1000cm3/dm3 = 12.5cm3

EXAMPLE Calculate the volume of carbon dioxide produced at

STP when 1.00g of calcium carbonate reacts with 25.0cm3 of 2.00 mol/dm3 hydrochloric acid.

CaCO3 + 2HCl CaCl2 + H2O + CO2

Wow! Another limiting reactant problem with mixed units given. Both givens have to be converted to moles:

1.00gCaCO3 x mol/100.09g = 0.0100mol CaCO3 given

25.0cm3 x 1dm3/1000cm3 x 2.00mol/dm3 = 0.0500mol HCl0.0100mol CaCO3 given x 2 mol HCl = 0.0200 mol HCl needed

1molCaCO3

0.0500mol HCl given x 1 mol CaCO3 = 0.025mol CaCO3 needed

2 mol HCl

You need 0.0250 mol CaCO3 and are only given 0.0100 mol so this is your limiting reactant.

0.0100 mol CaCO3 x 1 mol CO2 = 0.0100 mol CO2

1mol CaCO3

0.0100 mol CO2 x 22.4 dm3/mol = .224 dm3

.224 dm3 x 1000 cm3/dm3 = 224 cm3

CitationsCitationsBrown, Catrin, and Mike Ford. Brown, Catrin, and Mike Ford. Higher Level Higher Level ChemistryChemistry. 2nd ed. N.p.: Pearson Baccalaureate, . 2nd ed. N.p.: Pearson Baccalaureate, 2014. Print.2014. Print.

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