Chapter1 1 Dian

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Stoichiometry (1)



-- the study of thethe study of the

quantitativequantitative

aspects ofaspects ofchemicalchemical

reactions.reactions.



Counting AtomsCounting Atoms

Chemistry is a quantitativeChemistry is a quantitative

sciencescience— —we need awe need a

“counting unit.”“counting unit.”

MOLEMOLE



Lavoisier: The Law of

Conservation of MassEarly 1700’s Lavoisier: Law of

Conservation of Mass

During a chemical change,

matter is neither created nor

destroyed.



LAW OF CONSERVATION OF MASSLAW OF CONSERVATION OF MASS

In every chemical operation an equalIn every chemical operation an equalquantity of matter exists before and afterquantity of matter exists before and after

the operation.the operation. ThatThat is, the amount ofis, the amount of

matter before a reaction must equal thematter before a reaction must equal the

amount of matter after a reaction. Noamount of matter after a reaction. No

matter is lost.matter is lost.

TheThe total mass of reactants = total mass of productstotal mass of reactants = total mass of products



LL AW OF AW OF CCONSERVATION OFONSERVATION OF MM ASS ASS

When 0.0976 g of magnesium was heated in air ,

0.1618 g of magnesium oxide (MgO) was

produced. What is the mass of oxygen needed to

produce 0.1618 g MgO?

Using the LCM:Total mass reactants = total mass products

mass of Mg + mass O = mass of MgO

0.0976 g Mg + mass O = 0.1618 g MgO

mass O = 0.1618 g - 0.0976 = 0.0642 g O0.0642 g O



1799, Proust: Law of Definite

Proportions

A compound always contains the

same elements in certain definite

proportions.

Proust: The Law of

Definite Proportions



LAW OF DEFINITE PROPORTIONSLAW OF DEFINITE PROPORTIONS Elements combine in specific ratios to form compoundsElements combine in specific ratios to form compounds

Use the Generic equation for percent:Use the Generic equation for percent:

% = ( port ion / total ) 100% = ( port ion / total ) 1001. What is the experimental percent of oxygen in CO2 if42.0 g of carbon reacted completely with 112.0 g ofoxygen?

% O = (mass of O / mass of CO2) 100

% O= [112.0 g O / (42.0 g + 112.0 g) CO2] 100 = 72.7% O72.7% O

2. What is the theoretical percent of aluminum inaluminum oxide?

% Al = (Atomic mass of Al / Formula mass of Al2O3) 100

% Al = (54 amu / 102 amu) 100 = 52.9%52.9%

3. What is the percent composition of sodium chloride?

% Na = 39.3% %% Na = 39.3% % ClCl = 60.7%= 60.7%



LAW OF MULTIPLE PROPORTIONSLAW OF MULTIPLE PROPORTIONS

When two elements form a series ofcompounds, the masses of the one

element that combine with a fixed

mass of the other element stand to

one another in the ratio of small

integers.

Iron oxide exists in different ratiosIron oxide exists in different ratioswith different propertieswith different properties

FeO and FeFeO and Fe22OO33



Law of conservation of mass &Law of conservation of mass & LLaw ofaw of ddefiniteefinitepproportionsroportions

When 0.0976 g of magnesium was heated in air,

0.1618 g of magnesium oxide (MgO) wasproduced.

a) what is the percent of Mg in MgO?% Mg = (mass Mg / Mass MgO) 100

= (0.0976g / 0.1618 g) 100 = 60.3 %60.3 %

b) Using only LDP, what mass of oxygen was

needed to combine with the magnesium?% O = 100% MgO - 60.3% Mg = 39.7% O

% O = (mass O / mass MgO) 100

39.7 % = (mass O / 0.1618 g) 100

mass O = 0.397 ( 0.1618 g) = 0.0642 g O0.0642 g O

Same as using the LCM!!



PRACTICE PROBLEMSPRACTICE PROBLEMSPractic ing Law of conservation of mass:

________1. Aluminum metal combines with oxygen to producealuminum oxide. If 141.0g of aluminum yields 266.7 g of

aluminum oxide, how many grams of oxygen were needed?

________2. Sodium metal reacts with chlor ine gas to produce thesalt, sodium chloride. If 15.0 g of chlorine yields 26.5 g of salt,how much sodium metal is needed?

Practicing the law of definite proportions:

________3. What is the experimental percent of oxygen in a copperoxide if 10.0 g of copper reacted completely wi th 2.52 g ofoxygen?

_______ 4. Based on question #1, what is the experimental percentcomposition of aluminum oxide?

_______ 5. Calculate the theoretical percent composi tion foraluminum chloride and sodium oxide.



PRACTICE PROBLEMSPRACTICE PROBLEMSPractic ing percents:

________1. Pure gold is too sof t a metal for many uses, so it is

alloyed to give it more mechanical strength. One particular alloy is

made by mixing 29.17 g of gold, 3.81 grams of si lver, and 5.91 g of

copper. What is the percent of gold in this mixture?

________2. If 255 g of a meat sample contains 21.9 g of fat, what

percentage of fat is present?

Using the LAWS:

________3. How many grams of CuO can be obtained from 1.80 g of

copper (use the theoretical percent composi tion)?

4. When aluminum combines with bromine gas, they produce the

substance aluminum bromide, AlBr 3. Write a chemical equationdescribing this reaction.

_______ If 56.88 g of aluminum bromide is formed f rom 5.75 g of

aluminum, how many grams of bromine was needed?

75.0%

8.6%

2.25 g

2Al + 3Br 2 2AlBr 351.13 g



Connecting Mass to Moles

One of the greatest challenges early chemists

faced was trying to find a way to connect the

mass of a substance to the number of particles

in the sample. It was determined that “elementary particles”

combined in fixed ratios by weight.



Connecting Mass to Moles

This led Dalton to the “atomic model” of matter

Example: The mass ratio of oxygen to

hydrogen in water is 8:1

This does not tell us how many atoms of eachelement are involved

It could tell us this if we knew the relative mass of

each kind of atom



Relative Mass

At the time, chemists did not know which was

true and tended to think the latter was more

likely

Example: A bucket of baseballs has fewer ballsthat an identical bucket of golf balls

If this is true in the macroscopic world, why

wouldn’t it be true in the sub-microscopic one?



Relative Mass

Consider earlier this year when we studied

density: was iron more dense than aluminum

because iron had more particles per given

volume than aluminum or because iron’sindividual particles were more massive than

aluminum’s? Could it be some combination of

both?



Relative Mass

The truth is, based on the experiments we

conducted earlier in the year, we couldn’t say

which was true.

Dalton did not know what was true during histime either.

Since the mass of individual atoms could not be

determined, a system of atomic masses had to

be determined by comparison.



Relative Mass

To determine a system of masses by

comparison, one element would have to be

chosen as the basis of comparison for all

others Dalton chose hydrogen and assigned it a mass

of 1.



Relative Mass

To find the mass of another element like

oxygen:

Compare the masses of equal number of

oxygen and hydrogen atoms ORFind the combining masses of oxygen and

hydrogen in water



Relative Mass

Dalton thought that the former approach was

invalid because he thought identical volumes of

hydrogen and oxygen gases would have

different numbers of particles He thought the latter was valid but did not take

into account that it is valid ONLY if the ratio of

atomic combination is known



Reactions of Gases

Research conducted by Gay-Lussac

suggested that equal volumes of gases, at the

same temperature and pressure, contain equal

numbers of particles



Reactions of Gases

Gay-Lussac noted that gases

appear to react in simple integer

ratios

Example: Two volumes of

hydrogen reacted with one volumeof oxygen to produce two volumes

of water

These findings appeared to

contradict the idea that equalvolumes of gases have equal

numbers of particles



Reactions of Gases

Why? Well, if water is was H2O,

then two volumes of hydrogen and

one volume of oxygen should

make one volume of water

+ + =



Avogadro’s Hypothesis

Avogadro assumed

Equal volumes of gases have

equal numbers of molecules

These molecules can be split into

half-molecules during chemical

reactions

That molecules of elemental gases

could contain more than a singleatom




Two volumes of hydrogen react

with one volume of oxygen to

produce two volumes of water

when hydrogen and oxygen can besplit into half-molecules!

+ + = +




If we accept Avogadro’s

Hypothesis, we can compare the

mass of various gases and deduce

the relative mass of the molecules

To do this, we pick a weighable

amount of the lightest element

(how about 1.0 g?) then use mass

ratios to assign atomic masses tothe other elements



Implications

If two volumes of hydrogencombine with one volume ofoxygen gas, it is reasonable to

assume that two molecules ofhydrogen are reacting with eachmolecule of oxygen

The word chosen to represent thestandard weighable amount of

stuff, the mole, comes from theLatin “mole cula” or little lump



Counting AtomsCounting Atoms

Chemistry is a quantitativeChemistry is a quantitative

sciencescience— —we need awe need a

“counting unit.”“counting unit.”

1 mole is the amount of substance that1 mole is the amount of substance that

contains as many particles (atoms,contains as many particles (atoms,

molecules) as C atoms in 12.0 g ofmolecules) as C atoms in 12.0 g of1212C.C.

MOLEMOLE



Molar MassMolar Mass

1 mol of1 mol of 1212CC= 12.00 g of C= 12.00 g of C= 6.022 x 10= 6.022 x 102323 atomsatoms

of Cof C

12.00 g of12.00 g of 1212C is itsC is its

MOLAR MASSMOLAR MASS

Taking into account all ofTaking into account all of

the isotopes of C, thethe isotopes of C, the

molar mass of C ismolar mass of C is

12.011 g/mol12.011 g/mol



OneOne--mole Amountsmole Amounts



PROBLEM: What amount of Mg isPROBLEM: What amount of Mg isrepresented by 0.200 g? How manyrepresented by 0.200 g? How many

atoms?atoms?

PROBLEM: What amount of Mg isPROBLEM: What amount of Mg isrepresented by 0.200 g? How manyrepresented by 0.200 g? How many

atoms?atoms?Mg has a molar mass of 24.3050 g/mol.

0.200 g •1 mol

24.31 g = 8.23 x 10

-3 mol

8.23 x 10-3 mol •6.022 x 1023 atoms

1 mol

= 4.95 x 10= 4.95 x 102121 atoms Mgatoms Mg

How many atoms in this piece of Mg?



MOLECULARMOLECULARWEIGHT ANDWEIGHT AND

MOLAR MASSMOLAR MASSMolecular weightMolecular weight = sum of the= sum of the

atomic weights of all atoms inatomic weights of all atoms in

the molecule.the molecule.

Molar massMolar mass = molecular weight= molecular weight

in grams per mol.in grams per mol.



What is theWhat is the

molar mass ofmolar mass ofethanol, Cethanol, C22HH66O?O?

1 mol contains1 mol contains

2 moles of C (12.01 g C/1 mol) = 24.02 g C2 moles of C (12.01 g C/1 mol) = 24.02 g C

6 moles of H (1.01 g H/1 mol) = 6.06 g H6 moles of H (1.01 g H/1 mol) = 6.06 g H

1 mol of O (16.00 g O/1 mol) = 16.00 g O1 mol of O (16.00 g O/1 mol) = 16.00 g O

TOTAL =TOTAL = molar mass = 46.08 g/molmolar mass = 46.08 g/mol



How manyHow many molesmoles of alcohol (Cof alcohol (C22HH66O)O)are there in a “standard” can of beerare there in a “standard” can of beer

if there are 21.3 g of Cif there are 21.3 g of C22HH66O?O?

(a) Molar mass of C2H6O = 46.08 g/mol

(b) Calc. moles of alcohol

21.3 g •1 mol

46.08 g = 0.462 mol



How manyHow many atoms of Catoms of C are there inare there ina “ standard” can of beer if there are 21.3 ga “ standard” can of beer if there are 21.3 g

of Cof C22HH66O?O?

= 5.57 x 1023 C atoms

There are 2.78 x 1023 molecules.

Each molecule contains 2 C atoms.

Therefore, the number of C atoms is

2.78 x 1023 molecules •2 C atoms

1 molecule

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