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Stoichiometry

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AP Chemistry. Stoichiometry. In the thermite reaction, a mixture of powdered aluminum and powdered iron(III) oxide react to yield iron and aluminum oxide. The reaction burns hot enough to be useful in under- water welding. 2 Al + Fe 2 O 3  2 Fe + Al 2 O 3 + energy. - PowerPoint PPT Presentation
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Stoichiometry AP Chemistry In the thermite reaction, a ixture of powdered aluminum nd powdered iron(III) oxide to yield iron and aluminum ide. The reaction burns hot ough to be useful in under- water welding. + Fe 2 O 3 2 Fe + Al 2 O 3 + en
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Page 1: Stoichiometry

StoichiometryAP Chemistry

In the thermite reaction, amixture of powdered aluminum

and powdered iron(III) oxidereact to yield iron and aluminum

oxide. The reaction burns hotenough to be useful in under-

water welding.

2 Al + Fe2O3 2 Fe + Al2O3 + energy

Page 2: Stoichiometry

In a reaction:

atoms are rearranged

AND mass

energy

charge

are conserved

Balancing Chemical Equations

=law of

conservationof mass

same # of atomsof each type on each

side of equation

Hint: Start with most complicated substances firstand leave simplest substances for last.

Chemical Equations

Page 3: Stoichiometry

2

Solid lithum reacts w/oxygento form solid lithium oxide.

O2(g) Li(s) Li2O(s)+

Li+ O2–

+

4

Aqueous aluminum sulfate reacts w/aqueous bariumchloride to form a white precipitate of barium sulfate.The other compound remains in solution.

Al3+ SO42– Ba2+ Cl–

Al2(SO4)3(s)+ BaCl2 BaSO4 AlCl3(aq) (aq) (aq)+33 2

Page 4: Stoichiometry

Methane gas (CH4) reacts withoxygen to form carbon dioxidegas and water vapor.

CH4(g) + O2(g) CO2(g) H2O(g)+2 2

Furnaces burn primarily methane.

CaSi2 + SbI3 Si + Sb + CaI2

Al + CH3OH Al(CH3O)3 + H2

CaC2(s) + H2O(l) C2H2(g) + CaO(s)

3 2 26 3

3262 3

Page 5: Stoichiometry

2 C2H2(g) + O2(g) CO2(g) + H2O(l) 5 4 22

C3H8 + O2 CO2 + H2O 5 3 4

Complete combustion of a hydrocarbon,or of a compound containing C, H, and O

(e.g., methanol, CH3OH) yields CO2 and H2O.

C4H10 + O2 CO2 + H2O 13 4 52 8 10

Another pattern of reactivity:

e.g.,

alkalimetal

metalhydroxide

hydrogengas

water + +

2 Rb(s) + 2 H2O(l) 2 RbOH(aq) + H2(g)

Page 6: Stoichiometry

Two (of the several) Types of Reactions

combination (synthesis): simpler substances combine to form more complex substances

sodium + chlorine gas sodium chloride

A + B AB AB + C ABC A + B + C ABC

Na Cl2+ NaCl2 2

-- form:

Page 7: Stoichiometry

decomposition: complex substances are broken down into simpler ones

lithium chlorate lithium chloride + oxygen

water hydrogen gas + oxygen gas

Li+ ClO3– Li+ Cl–

LiClO3 LiCl + O22 2 3

H2O2 H2 + O22

-- form:

AB A + B ABC AB + C ABC A + B + C

Page 8: Stoichiometry

formula weight: the mass of all of the atoms in a chemical formula (unit is amu)

-- If the substance is a molecular substance (e.g., C3H8), then the term molecular weight is also used.

molar mass: the mass of one mole of a substance (unit is usually grams)

-- recall that 1 mole of any substance

6.02 x 1023 particlesof that substance

=

Page 9: Stoichiometry

(NH4)3PO4

H: 12 (1.0 g) = 12.0 g

N: 3 (14.0 g) = 42.0 g

m.m. =149.0 g P: 1 (31.0 g) = 31.0 g

NH4+ PO4

3–

O: 4 (16.0 g) = 64.0 g

Find the molar mass and formula weight ofammonium phosphate.

f.w. = 149.0 amu

Page 10: Stoichiometry

percentage composition: the mass % of each element in a compound

% of element =g element

molar mass of compoundx 100

Find the percentage of oxygen, by mass, incalcium nitrate.

Ca(NO3)2

% O = = 58.5% O6(16.0) 2(14.0) 40.1

6(16.0)

equation:

Page 11: Stoichiometry

Empirical Formula and Molecular Formula

Compound Molecular Formula

EmpiricalFormula

glucose C6H12O6

propane C3H8

butane C4H10

naphthalene C10H8

sucrose C12H22O11

octane C8H18

lowest-termsformula

shows the true number andtype of atoms in a m’cule

CH2O

C3H8

C2H5

C5H4

C12H22O11

C4H9

Page 12: Stoichiometry

“What’s your flavorof ice cream?”

Finding an Empirical Formula from Experimental Data

1. Find # of g of each element.

2. Convert each g to mol.

3. Divide each “# of mol” by the smallest “# of mol.”4. Use ratio to find formula.

Page 13: Stoichiometry

A ruthenium/sulfur compound is 67.7% Ru.Find its empirical formula.

RuS1.5 Ru2S3

Ru g 67.7

Ru g 101.1Ru mol 1

Ru mol 0.670 0.670 1

S g 32.3

S g 32.1S mol 1

S mol 1.006 0.670 1.5

Page 14: Stoichiometry

A sample of a compound contains 4.63 g lead, 1.25 gnitrogen, and 2.87 g oxygen. Name the compound.

PbN4O8

Pb(NO2)4

Pb? 4 NO2–

lead(IV) nitrite

?

Pb g 4.63

Pb g 207.2Pb mol 1

Pb mol 0.0223 0.0223 1

N g 1.25

N g 14.0N mol 1

N mol 0.0893 0.0223 4

O g 2.87

O g 16.0O mol 1

O mol 0.1794 0.0223 8

?


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