lavoisier: law of conservation of mass! chemical reaction mass before = mass after

Lavoisier: Law of conservation of mass!

Chemical Reaction

Mass before = Mass after.

Chapter 3: Stoichiometry

Scientific contributions cut short – Antoine Lavoisier was guillotined in 1794at the age of 60. Today, considered to be father of modern chemistry.

1

Chemical equations are descriptions of chemical reactions.

There are 2 parts to an equation: reactants and products:

H2 + O2 H2O

To balance:

change only the stoichiometric coefficients

cannot change the chemical species

Stoichiometric coefficients: numbers in front of the chemical formulas; give ratio of

reactants and products.

Chemical Equations

2

Consider the unbalanced chemical equation:

H2 + O2 H2O

The equation H2 + O2 H2O2 is a balanced equation but for a different reaction.

DO NOT change subscripts!

Balancing options

H2 + 1/2 O2 H2O, or

2H2 + O2 2H2O, or

4H2 + 2O2 4H2O, etc.

Preferably, no fractions and smallest possible common coefficients used.

3

Example 1Write a balanced equation for the combustion of octane, C8H18.

We know that our reactants include C8H18 and O2(g), and the products will be CO2 and H2O.

But in what ratios will they react and produce? Let’s balance!

Let’s consider combustion reactions which involve the burning/oxidation of hydrocarbons to produce CO2 and

H2O always.

Hydrocarbons are organic compounds made up only of carbon and hydrogen.

For burning/oxidation to occur O2 has to be present! All combustion equations will include O2(g) as a reactant!

4

C8H18 + O2(g) CO2 + H2O

We need to increase C on the product side to 8, using a stoichiometric coefficient.

C8H18 + O2(g) 8CO2 + H2O

We need to increase H on the product side to 18, using a stoichiometric coefficient.

Element Reactants Products

C 8 1

H 18 2

O 2 3


C 8 8

H 18 2

O 2 17

Start with C & H. Do O

last because it appears in 3 species.

5

C8H18 + O2(g) 8CO2 + 9H2O

We now need to increase O2 on the reactant side. It needs to change from 2 to 25. This can only be done using a fraction!

C8H18 + 12.5 O2(g) 8CO2 + 9H2O

Now to remove the fraction, and keep coefficients as small as possible we multiply ALL coefficients by 2!

2C8H18 + 25O2(g) 16CO2 + 18H2O


C 8 8

H 18 9x2 = 18

O 2 16+9 =25


C 8 8

H 18 18

O 12.5x2=25 25

===

6

C8H18 + O2 CO2 + H2OHINT: Balance O last (occurs in 3 species)

but C and H only appear in two!

Balance C

C8H18 + O2 8CO2 + H2O

Balance H

C8H18 + O2 8CO2 + 9H2O

Balance O

C8H18 + 25/2O2 8CO2 + 9H2O

2C8H18 + 25O2 16CO2 + 18H2O

We usually add the physical states of the reactants and products

2C8H18(l) + 25O2(g) 16CO2(g) + 18H2O(l)

Here is another set up of the same example:

7

In combustion reactions:

Rapid reactions that produce a flame.

Usually very “clean” and products are

predictable, but incomplete combustion

(leftover ash) can be very harmful.

8

Example 2

Write a balanced equation for the combustion of purine, C5H4N4.

C CO2 H H2O N N2

9

Patterns of chemical reactivity

Alkali metal + water

2Na + 2H2O 2NaOH + H2

2K + 2H2O 2KOH + H2

Elements in the same group of the periodic table react in a similar manner.

Lithium Sodium Potassium

10

Alkali earth metal + water

Mg + 2H2O Mg(OH)2 + H2

Ca + 2H2O Ca(OH)2 + H2

Note: These are the EXACT same products as for alkali metals, except for the stoichiometry!!!

Always look for patterns in chemistry!

11

2Mg(s) + O2(g) 2MgO(s)

Mg has combined with O2 to form MgO.Two reactants combine to form a single product.

They are characterized by having fewer products than reactants.

Metal + oxygen

Combination Reactions

12

C(s) + 2H2(g) CH4(g)

Non-metal + hydrogen

Ca(s) + Cl2(g) CaCl2(s)

Metal + halogen

2Li(s) + F2(g) 2LiF(s)

13

They are characterized by having fewer reactants than products.

2NaN3(s) 2Na(s) + 3N2(g)

The NaN3 is ignited and rapidly decomposes into Na and N2 gas

Metal azide

Metal carbonate

CaCO3(s) CaO(s) + CO2(g)

limestone,seashells

lime

Decomposition Reactions

14

Quantitative vs. Qualitative Chemical formulae and equations both have qualitative and

quantitative significance.

Chemical formula says what compounds you are working with and the equation gives a qualitative idea of how the reactants react, and what

products they produce.

N2 + 3H2 2NH3 (balanced)

Quality: We have nitrogen and hydrogen as reactants. The equation tells us that they undergo a combination reaction to form ammonia.

Subscripts in formulae and coefficients in balanced equations represent precise quantities.

Quantity: The product, NH3, is composed of 3H for every 1N atom.

The balanced equation shows that 2 molecules of NH3 are produced for every N2 and 3H2 that are used up. 16

Average Atomic MassREVISION FROM ISOTOPES!

Most elements occur as mixtures of isotopes.

To determine the average atomic mass of an element we must consider each isotopes mass as well as their respective

abundances.

Before we consider real isotopes, lets create an analogy using popcorn

I have a bowl of popcorn where 93% of the pieces weigh 1.14 mg, 2% weigh 1.09 mg and 5% weigh 1.03 mg.

Which popcorn piece mass is in the greatest and lowest abundance?

Which mass will dominate in calculating the average mass of the popcorn pieces?

So what do we expect the average mass to be? 17

If the bowl contains 1000 popcorn pieces in it. Then 930 of them would weigh 1.14 mg, 20 would weigh 1.09 mg and 50

would weigh 1.03 mg.

Let’s add those abundances together = 930 + 20 + 50

= 1000.

This shows that each isotope’s (popcorn-piece mass’) abundance must make up 100% , which is 1000 in this case.

In isotopes be sure to check that your abundances ALWAYS add up to 100% 18

......%100

%

%100

%2

2

1

1

mass

abundancemass

abundanceweightAverage

mgmgmgweightAverage 03.1

%100

%509.1

%100

%214.1

%100

%93

321

mgmgmgweightAverage 0515.00218.006.1

Now let us calculate the average weight of the popcorn pieces. (We are expecting it to

be just a little less than 1.14 mg.)

mgweightAverage 13.1

19

Example 3

Three isotopes of silicon occur in nature, calculate the average atomic weight of silicon.

28Si (92.21%) mass = 27.97693 amu

29Si (4.70%) mass = 28.97659 amu

30Si (3.09%) mass = 29.97376 amu

First make a logical prediction!

......%100

%

%100

%2

2

1

1

mass

abundancemass

abundanceweightAverage

20

Molecular and Formula Weights

Formula weight (Fr or FW) is the mass of a collection of atoms represented by a chemical formula.

So it is the sum of atomic masses (Ar) of the atoms in the chemical formula.

Fr (K2Cr2O7) = 2Ar(K) + 2Ar(Cr) + 7Ar(O)

= 2(39.10) + 2(52.00) + 7(16.00)

= 294.20 amu

For example, K2Cr2O7

Significant figures!!

21

If the chemical formula of a substance is the same as its molecular formula, then the formula weight is also called the

molecular weight.

Molecular weight (Mr or MW) is the mass of a collection of atoms represented by a chemical formula for a molecule.

MW(C6H12O6) =

What is the molecular formula for glucose?

Formula weight and molecular weight are virtually identical and are used interchangeably.

When dealing with an ionic compound (3D lattice) we do not use molecular formulas to name them, and so we could not calculate a molecular weight.

In this case we use the formula weight.22

Some compounds have water of crystallization. These are water molecules associated with the solid as it crystallizes

from solution.

CuSO4•5H2O5 water molecules per CuSO4 unit(Add them into molecular weight!)

Can often be driven off:

CuSO4•5H2O → CuSO4 + 5H2Oblue grey

anhydrous copper sulphate

Water of Crystallization

23

Mole: a convenient measure of enormous numbers.

NEW PUBLIC HOLIDAY!Mole Day: October 23 from 6:02am to 6:02pm, invented May 15, 1991.

6.022 x 1023 is known as Avogadro’s Number, NA

Defined as the amount of matter that contains as many objects (atoms/molecules etc.) as the number of atoms in 12g of 12C

12 g = 6.02214 x 1023 atoms

The Mole

24

1 mole of anything = 6.022 1023 of that thing.

1 mole 12C atoms = 6.022 1023 12C atoms. 1 mole H2O molecules = 6.022 1023 H2O molecules.

1 mole NO3- ions = 6.022 1023 NO3

- ions.

The Mole

Example 4

Calculate the number of carbon atoms in 0.350 mol of C6H12O6.

25

Example 4

Calculate the number of carbon atoms in 0.350 mol of C6H12O6.

Let’s look at this more mathematically now!

Avo’s number provides a conversion factor between the number of moles and molecules of a given compound.

1 mole sugar = 6.022 x 1023 molecules of sugar.

So we can use the “GIVEN and DESIRED” formula! (Given: mol, Desired molecules!)

unitDesiredunitGivenunitGiven

unitDesired

26

The mass in grams of 1 mol of substance (units g/mol).

Molar Mass

27

The molar mass (in g) of any substance is always numerically equal to its formula weight (in amu).

One HCl molecule weighs 36.46 amu One mol of HCl weighs 36.46 g.

Can you see where this comes from?

Remember we said that amu and g.mol-1 are equivalent and can be used interchangeably?

So….. HCl weighs 36.46 amu = 36.46 g.mol-1.36.46 g.mol-1 x 1 mol = 36.46 g.

1 mol of N2

1 mol of H2O

1 mol of NaCl

18 g

28 g

58.45 g

28

MAKE SURE YOU UNDERSTAND THESE RELATIONSHIPS!!!

29

Inter-converting masses, moles and number of particles:

Molar Mass

30

Converting from mass to moles and moles to mass is straight forward:

1.Calculate the molar mass of the substance2.Use the mole concept

number of moles = mass / molar mass n = m / Mr

Once the number of moles of the substance is calculated, use Avogadro’s number to attain number of molecules:

number of molecules = n x AVO

Now we use the number of molecules to solve for number of atoms, or ions within the substance.

TAKE NOTE OF THESE PARTICLES AND HOW THEY RELATE TO EACH OTHER!

Molar Mass

31

Sugar C6H12O6

This is 1 MOLECULE of sugar.

This molecule contains 24 ATOMS; 6 carbon, 12 hydrogen and 6 oxygen.

Sulphuric Acid H2SO4

Example 5

Calculate the number of H atoms in 20.00 g of C6H12O6.

Mass Mols Molecules

Strategy

n = molesm = mass (grams)MW = molecular weight or molar mass (g/mol)

n = molesNA = Avogadro’s Number

32

MUST know two things to figure this out…(1)mass (g) and (2)molecular weight (molar mass) (g.mol-1)

…of the compound in question.

33

AMOUNT number of moles (must always compare moles!)

n = mass / molar mass, or n = concentration/volume.

PERCENT COMPOSITION

r

r

No. of atoms of Element A% Element 100

F of Compound

Definitions

Calculate by dividing the atomic weight (Ar) for each element by the formula weight (Fr) of the compound, and

express as a percentage:

34

What is the % by mass (% composition) of O in K2Cr2O7?

r

r

No. of atoms of Element A% Element 100

F of Compound

First determine atomic and formula weights (or MWs).

Then apply equation (works the same in amu or g/mol).

Example 6

35

36

Percent Composition

If you are battling to understand what is meant by percentage (by mass) contributed by each element in the

compound then just think about a box of colourful smarties!

What percentage of the box of smarties is yellow?

100#

#%

smartiesTotal

smartiesyellowYellow

So it’s the same as ELEMENTAL composition, except that we need to consider mass when we use elements!!!

Mass %of elements

Grams ofeach element

in sample

Empirical formula

Strategy

Assumehave 100 g

sample Moles of each element

n = mMr

37

Empirical Formulas from Analyses

Mole ratio

Mr of that specific element!

Example 7 (p. 85 8th Ed; p. 93 9th Ed; p. 98 10th Ed)

Information given in problem

Ethylene glycol: 38.7% C 9.7% H

51.6% O by mass

Molar mass (from mass spectrometry): 62.1 g mol-1

Calculate:(a) empirical formula(b) molecular formula

Ethylene glycol, the substance used in automobile antifreeze,is composed of 38.7% C, 9.7% H, and 51.6% O by mass. Itsmolar mass is 62.1g/mol.

38

C H O

38.7% 9.7% 51.6%

38.7g 9.7g 51.6g

3.22 mol 9.62 mol 3.23 mol

3.22/3.22 9.62/3.22 3.23/3.22

1 2.99 1.003

assume 100 g

number of moles

38.7g 9.7g 51.6g12.01 g mol-1 1.008 g mol-1 16.00 g mol-1

mole ratio (divide by smallest no. mols)

39

Experimental error is to be expected, so

C H O1 2.99 1.003

C H O1 3 1

is likely to be

The empirical formula is therefore CH3O

which has the formula weight 1 x 12.01 + 3 x 1.008 + 1 x 16.00= 31.03 g/mol

But molecular weight = 62.1 g/mol, which is twice this value

The molecular formula must be C2H6O2.40

CxHyOz Catalysis: completeoxidation of C to CO2

Hydrogen in CxHyOz isoxidized to H2O and absorbed in the H2Oabsorber

C in CxHyOz isabsorbed here

O in CxHyOz is determined by mass difference

Combustion Analysis to Determine Experimental Data

41

Example 88th ed., p. 99, 3.54(a); 9th ed., p. 107, 3.48(a); 10th ed., p. 114, 3.52(a)

Combustion of 2.78 mg of ethyl butyrate produced 6.32 mg CO2 and 2.58 mg of H2O. What is the empirical formula of this compound?

Consider the given information separately. Let’s look at CO2 first. Recallthat we can only determine C and H content from information given!

42

Ask yourself: How much of the 6.32 mg of CO2 is due to C?


Now let’s deal with H2O, noting we can only determine H content.

43

Example 8

How much of the 2.58 mg of H2O is due to H?


Finally, same strategy as before…we have all three masses and can determine the

empirical formula.

Now, determine how much O is present from the difference in mass.

44

Example 8

START HERE

THIS TIME!

45

C H O

1.73 x 10-3 g 0.289 x 10-3 g 0.76 x 10-3 g12.01 g mol-1 1.008 g mol-1 16.00 g mol-1

1.44 x 10-4 mol 2.87 x 10-4 mol 4.75 x 10-5 mol4.75 x10-5 mol 4.75 x10-5 mol 4.75 x10-5 mol

Determine moles of each element

Divide each by smallest # moles to determine ratios

46

Quantitative Information from Balanced Equations

In a balanced equation the coefficients can be interpreted as the relative numbers of molecules involved in the

reaction AND as the relative number of moles.

When given any chemical equation you always ensure that you BALANCE it 1st!

Then you use the given information (usually a mass in grams, or a number of moles in mol), together with your calculated molar masses, to solve for

the unknown.

47

Quantitative Information from Balanced Equations

Example 9

What mass of CO2 is produced when 1.00 g of propane, C3H8, is burned?

First, write the balanced equation

Then, look at what is given…grams of reactant. In orderto determine how much CO2 is produced, you need to

consider reaction STOICHIOMETRY.

48

STRATEGY

49

What mass of CO2 is produced when 1.00 g of propane, C3H8, is burned?

C3H8 + 5O2 → 3CO2 + 4H2O

1.00 g ??? g

Calculate moles of propane:

Next, consider the MOLAR relationship between propane and carbon dioxide.It is NOT 1:1, but rather 1:3 (THIS is stoichiometry!).

Stoichiometric coefficients are MOLE relationships!!50

Finally, calculate mass (g) of CO2 produced…need molar mass!

Follow-up problem: What mass of O2 was consumed in the process? Ans: 3.63 g

51

Notice how we link or compare the number of moles of each substance!! Not their masses,

or anything else!!

If the reactants are not present in stoichiometric amounts, then at the end of a reaction some reactants are still present.

EXCESS!

52

Limiting Reactants

Limiting ReactantOne reactant that is consumed

completely in the reaction.

The limiting reactant/reagent hampers us from continuing

with the reaction!

Non-Chemistry Limiting Reagent

Cheese Sandwich

2B + 1C = B2C

Ideally…

More likely…

4B + 1C = B2C + 2BYou’ve run out of cheese,

time to go to the store.

53

The cheese limits us from making another sandwich! Therefore we call it the limiting reagent!

Suppose we have 10 mol H2 and 7 mol O2

H2(g) + O2(g) → H2O(l)

54

Strategy

Example10Al reduces Fe2O3 to Fe. How much Fe is produced from the reaction of 30.0 g of Al and 100 g Fe2O3?

Mass of product

Moles of product basedon limiting reagent

Massreactants

Identify limitingreagent

molesreactants

55

Solving Limiting Reagant Problems

Mass ofreactants


Mass of product

Moles ofreactants

Moles product based on

limiting reagent

1

2

3

4

5

Example 10 Al reduces Fe2O3 to Fe. How much Fe is produced from the reaction of 30.0 g of Al and 100 g Fe2O3?

56

Massreactants


Mass of product

molsreactants

Mols of product based on

limiting reagent

1

2

3

4

5

From reaction stoichiometry

2 Al : 1 Fe2O3

__________is in excess, and __________ is the limiting reagent!

57

Massreactants


Mass of product

molsreactants

Mols of product based on

limiting reagent

1

2

3

4

5

Now compare stoichiometry of Fe to Al.

m = n x Mr

=

Finally, calculate mass of Fe produced.

58

= of Fe

100yield lTheoretica

yield ActualYield %

Theoretical yield

The percent yield relates the actual yield (amount of material recovered in the laboratory) to the theoretical

yield:

59

Percentage Yield

So, we conduct an experiment in the lab:Let us make a metal oxide, calcium oxide!

From theory we now know that we need to do a combination reaction using a metal and O2 as our reactants.

The equation: Ca + O2 CaOBalance: 2Ca + O2 2CaO

60

Percentage Yield

We are given 2.053 g of calcium, and the oxygen used is atmospheric oxygen, and therefore is in excess (unlimited

quantity).

From this information we can calculate how much product the stoichiometry of our equation predicts we will produce

theoretical yield.

61

Percentage YieldExample 11Preparation of calcium oxide is achieved by reacting 2.053 g of calcium with an excess of oxygen. Calculate the theoretical yield of calcium oxide, and solve for the percentage yield, if the real mass of the product was measured to be 2.26 g.

2Ca + O2 2CaO

m = 2.053 g m = excess m = desired quantity!

Mr = 40.078g/mol Mr = 32.00 g/mol Mr = 56.08 g/mol

n= mass/molar mass n= n(Ca) /2 n(CaO) = n(Ca)

n = 2.053 g = 0.05123 mol

40.078 g.mol-1n = 0.05123 mol

= 0.02561 mol 2

n(CaO) = n(Ca)

n = 0.05123 mol

62

2Ca + O2 2CaO

Example 11Calculate the theoretical yield of calcium oxide, and solve for the percentage yield, if the real mass of the product was measured to be 2.26 g.

m = 2.053 g m = excess m = desired quantity!

Mr = 40.078g/mol Mr = 32.00 g/mol Mr = 56.08 g/mol

n= 0.05123 mol n= 0.02561 mol n= 0.05123 mol

mass(CaO) = n x Mr

= 0.05123 mol x 56.08 g/mol= 2.873 g

100

yield lTheoreticayield Actual

Yield %

63

Expressing Concentration

There are many ways in which you can express the concentration of a solution.

Qualitatively OR Quantitatively

•Mass percentage•Mole fraction•Molarity•Molality

DILUTE or CONCENTRATED

MANY WAYS TO DESCRIBE THIS!

•Dilute = Low concentration of solute.•Concentrated = High concentration of solute.

64

Quantitative Concentration

Mass Percentage

A 24.00 % NaCl solution by mass contains 24.00 g of NaCl in a 100.0 g solution.

So if our solution has a total mass of 634.0 g, then what mass of NaCl does it contain?

100% solutionmassTotal

solutionincomponentofmasscomponentofMass

24.00 % of the total mass is due to the NaCl.

Ans: 152.2 g of NaCl

The remaining mass is due to the SOLVENT (water).

65

Parts per Million (ppm)

Example: A 3 ppm solution

contains 3 particles of solute, for every 1 million (106) particles of solution.

3 g of solute for every million grams of solution.

610solutionmassTotal

solutionincomponentofmasscomponentofppm

very dilute solutions!

The concentration of a solution in grams of solute per 106 (million) grams of solution.

This is equivalent to milligrams (mg) of solute, per litre of solution (for aqueous solutions).

66

Parts per Billion (ppb)

Example: 329 ppb corresponds to 329 µg of solute per litre of solution.

even more dilute solutions!


solutionincomponentofmasscomponentofppb

The concentration of a solution in grams of solute per 109 (billion) grams of solution.

This is equivalent to micrograms (µg) of solute, per litre of solution (for aqueous solutions).

67

Example 12Calculate the mass percentage of NaCl in a solution containing 1.50 g of NaCl in 50.0 g of water.

100% solutionmassTotal

solutionincomponentofmasscomponentofMass

68

Example 13A bleaching solution contains 3.62 mass percent sodium hypochlorite. Calculate the mass of sodium hypochlorite in a bottle containing 2500 g of bleaching solution.

ASIDE what is the chemical formula of sodium hypochlorite?

NaOCl

69

Example 14In an experiment 2.569 mg of CuSO4 was dissolved in water to form a 2.000 kg solution. Express the concentration of the solution in parts per million.


solutionincomponentofmasscomponentofppm

RULE: Both masses MUST be in the same unit of mass!

Convert 2 kg into mg:1st convert kg g and then g mg

70

Mole Fraction

Quantitative Concentrations in terms of moles

componentsallofmolesTotal

componentofmolesComponentoffractionMole

We often express mole fraction using an X.Example: the mole fraction of HCl in a solution of HCl XHCl. 1.00 mol of HCl, dissolved in 8.00 mol of water, gives a mole fraction of HCl of 0.111.

71

Mole Fraction


The sum of the mole fractions of all the

components in a solution must equal 1!

Let’s test this theory using our previous example!1.00 mol of HCl dissolved in 8.00 mol of water gives a

mole fraction of HCl of 0.111. So lets calculate the mole fraction of the water!

72

Molarity


Example 15: Dissolving 2.36 g of CuSO4 in enough water to fill a 250 ml volumetric flask.

solutionofLitres

soluteofMolesMolarity Units mol.L-1(or more

commonly: M).

1. Weigh out 2.36 g of these crystals.

2. Place in volumetric flask.

3. Fill to the line with water.

73

Molarity


Example 15: Dissolving 2.36 g of CuSO4 in enough water to fill a 250 ml volumetric flask.

1. Calculate the number of copper sulphate moles (n) 2. Remember to convert your volume of solution (ml L).

74

Molality


There are two major differences between Molality and Molarity!

They both appear in the denominator of the equation:

NB: Notice molaLity not molaRity!

solventofkg

soluteofMolesMolality

MOLARITYConsider the VOLUME of the entire solution

(always in L)!

MOLALITYConsider the MASS of the solvent (always in kg)!

75

Be sure that you are able to

calculate concentrations using each

of these methods.

Example 16:A solution containing equal masses of glycerol (C3H8O3) and water has a density of 1.10 g/ml. Calculate the a)Molality of glycerolb)Mole fraction of glycerolc)Molarity of glycerol

76

What information has been given?

•We have 1 solution. It contains 2 components.•Each component has the same mass.

We don’t know what mass though!•The solution’s density is 1.10 g/mol.

Because we know the masses of glycerol and water are equal….we can make an assumption!

Volume

MassDensity

Assume that each component has a mass of 10.00 g, and therefore the mass of the entire solution must be 20.00 g.

77


We can now calculate the number of moles of glycerol.

From our assumption:Mass of solvent = 10.00 g

= 20.00 x 10-3 kg

a) Molality involves the number of moles of the component in question, and the mass in kg of the solvent.

78

b) Calculating the mole fraction of glycerol involves calculating the number of moles of glycerol, and the total number of moles within the solution (nglycerol + nwater).


79

c) To calculate the molarity we need the number of moles of glycerol, and the volume of the solution in litres!


lavoisier: law of conservation of mass! chemical reaction mass before = mass after

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