Law of Conservation of MassLaw of Conservation of MassAntoine Lavoisier, ~ 1775Antoine Lavoisier, ~ 1775

Law of Definite ProportionsLaw of Definite ProportionsJ.L. Proust, 1799J.L. Proust, 1799

Law of Conservation of MassLaw of Conservation of Mass

In a chemical reaction, the Law of In a chemical reaction, the Law of Conservation of Mass states that the Mass Conservation of Mass states that the Mass of the Reactants must equal the Mass of the of the Reactants must equal the Mass of the Products.Products.

A + BA + B C + D + EC + D + E

ReactantsReactants ProductsProducts

Mass A + Mass B = Mass ( C + D + E )Mass A + Mass B = Mass ( C + D + E )

Law of Definite ProportionsLaw of Definite Proportions

Any pure compound only contains the same Any pure compound only contains the same elements in the same proportion by mass.elements in the same proportion by mass.

HH22OO

Define proportion: the ratio that relates one Define proportion: the ratio that relates one part to another part, or relates one part to part to another part, or relates one part to the whole. the whole.

Example: A large proportion of the people Example: A large proportion of the people present in this classroom are students.present in this classroom are students.

AcidsAcids

Vinegar is an AcidVinegar is an Acid Chemical name is Acetic AcidChemical name is Acetic Acid Chemical formula:Chemical formula:

CHCH33COCO22HH

BasesBases

Baking Soda is a BaseBaking Soda is a Base Chemical name is Sodium BicarbonateChemical name is Sodium Bicarbonate Chemical formula:Chemical formula:

NaHCONaHCO33

Acids React with BasesAcids React with Bases

Reactants = ProductReactants = Product

Acid + BaseAcid + BaseA SaltA Salt

WaterWater

Gas (sometimes)Gas (sometimes)

Vinegar + Baking SodaVinegar + Baking Soda

Sodium AcetateSodium Acetate

Water (HWater (H22O)O)

Carbon DioxideCarbon Dioxide

Mass of ReactantsMass of Reactants Mass of ProductsMass of Products

==

==

==

HypothesisHypothesis

If reactant is 84 grams of baking soda, then by If reactant is 84 grams of baking soda, then by

proportion, a product is 44 g of carbon dioxide.proportion, a product is 44 g of carbon dioxide.

NaHCONaHCO33 + CH+ CH33COCO22HH 84g84g 60g60g++ == 144g144g

HH22OO + CH+ CH33COCO22Na + CONa + CO22

82g82g18g18g 44g44g++++ == 144g144gSodium AcetateSodium Acetate Carbon DioxideCarbon DioxideWaterWater

Law of Definite ProportionsLaw of Definite ProportionsCalculating Mass of Molecule ACalculating Mass of Molecule A

Atom Atom Mass Mass (g)(g)

NaNa

SodiumSodium23 g23 g

HH

HydrogeHydrogenn

1 g1 g

CC

CarbonCarbon12 g12 g

OO

OxygenOxygen16 g16 g

Baking SodaBaking SodaSodium BicarbonateSodium Bicarbonate

Na x 1Na x 1 23g23g

H x 1H x 1 1g1g

C x 1C x 1 12g12g

O X 3O X 3 16(3) = 16(3) = 48g48g

NaHCONaHCO33 84g84g

Law of Definite ProportionsLaw of Definite ProportionsCalculating Mass of Molecule BCalculating Mass of Molecule B

Atom Atom Mass Mass (g)(g)

HH

HydrogeHydrogenn

1g1g

CC

CarbonCarbon12g12g

OO

OxygenOxygen16g16g

VinegarVinegarAcetic AcidAcetic Acid

H x 4H x 4 4g4g

C x 2 C x 2 24g24g

O 2 x O 2 x 1616

32g32g

CHCH3 3

COCO22HH60g60g

Law of Definite ProportionsLaw of Definite ProportionsCalculating Mass of Molecule BCalculating Mass of Molecule B

Atom Atom Mass Mass (g)(g)

HH

HydrogeHydrogenn

1 g1 g

CC

CarbonCarbon12 g12 g

OO

OxygenOxygen16 g16 g

VinegarVinegarAcetic AcidAcetic Acid

H x 4H x 4 1(4) = 4g1(4) = 4g

C x 2C x 2 12(2) = 12(2) = 24g24g

O X 2O X 2 16(2) = 16(2) = 32g32g

CHCH3 3

COCO22HH60g60g

Law of Definite ProportionsLaw of Definite ProportionsCalculating Mass of Molecule CCalculating Mass of Molecule C

Atom Atom Mass Mass (g)(g)

HH

HydrogeHydrogenn

OO

OxygenOxygen

WaterWaterDihydrogen MonoxideDihydrogen Monoxide

H H

O O

HH2 2 OO

Law of Definite ProportionsLaw of Definite ProportionsCalculating Mass of Molecule CCalculating Mass of Molecule C

Atom Atom Mass Mass (g)(g)

HH

HydrogeHydrogenn

1g1g

OO

OxygenOxygen16 g16 g

WaterWaterDihydrogen MonoxideDihydrogen Monoxide

H x 2H x 2 = 2g= 2g

O X 1O X 1 16g16g

HH2 2 OO 18g18g

Law of Definite ProportionsLaw of Definite ProportionsCalculating Mass of Molecule DCalculating Mass of Molecule D

Atom Atom Mass (g)Mass (g)

NaNa

SodiumSodium23 g23 g

HH

HydrogenHydrogen1 g1 g

OO

OxygenOxygen16 g16 g

CC

CarbonCarbon12 g12 g

A SaltA SaltSodium AcetateSodium Acetate

Na x 1Na x 1 23g23g

H x 3H x 3 1(3) = 3g1(3) = 3g

O X 2O X 2 16(2) = 16(2) = 32g32g

C x 2C x 2 12(2) = 12(2) = 24g24g

CHCH3 3

COCO22NaNa 82g82g

Law of Definite ProportionsLaw of Definite ProportionsCalculating Mass of Molecule ECalculating Mass of Molecule E

Atom Atom Mass Mass (g)(g)

CC

CarbonCarbon

OO

OxygenOxygen

GasGas

Carbon DioxideCarbon Dioxide

C C

OO

COCO22

Law of Definite ProportionsLaw of Definite ProportionsCalculating Mass of Molecule ECalculating Mass of Molecule E

Atom Atom Mass Mass (g)(g)

CC

CarbonCarbon12 g12 g

OO

OxygenOxygen16 g16 g

GasGas

Carbon DioxideCarbon Dioxide

C x 1C x 1 12g12g

O X 2O X 2 16(2) = 16(2) = 32g32g

COCO22 44g44g

Mass Reactants = Mass ProductsMass Reactants = Mass Products

NaHCONaHCO33 + CH+ CH33COCO22HH Mass of 6 atomsMass of 6 atoms

84g84g 60g60g++ == 144g144g

HH22OO + CH+ CH33COCO22Na + CONa + CO22

82g82g18g18g 44g44g++++ ==Sodium AcetateSodium Acetate Carbon DioxideCarbon DioxideWaterWater

Mass of 8 atomsMass of 8 atoms

Mass of 3 atomsMass of 3 atoms Mass of 8 atomsMass of 8 atoms Mass of 3 atomsMass of 3 atoms

144g144g

ReactantsReactants14 atoms14 atoms

ProductsProducts14 atoms14 atoms

Test HypothesisTest Hypothesis

To shorten the reaction time, we want to use only To shorten the reaction time, we want to use only a small amount of baking soda.a small amount of baking soda.

If reactant is 84 grams of baking soda, then we would If reactant is 84 grams of baking soda, then we would get 44 grams of carbon dioxide.get 44 grams of carbon dioxide.

But if we use only 5 grams of baking soda, then by But if we use only 5 grams of baking soda, then by proportion, the product is 2.6 grams of carbon dioxide.proportion, the product is 2.6 grams of carbon dioxide.

5g5g Sodium Bicarbonate Sodium Bicarbonate ?? gg CO CO22

5g x 44g5g x 44g = 2.6g CO = 2.6g CO22

84g84g

How can we measureHow can we measurethe mass of gas produced?the mass of gas produced?

Subtract the mass of the bottle + cap Subtract the mass of the bottle + cap afterafter the gas is released from the mass of the the gas is released from the mass of the bottle + cap bottle + cap beforebefore the CO the CO22 is released.is released.

The value should less than 2.6 g The value should less than 2.6 g because about 10% of the CObecause about 10% of the CO22

remains dissolved in theremains dissolved in thewater solution.water solution.

How do we MeasureHow do we Measurethe Volume of a Gas?the Volume of a Gas?

If we can measure the circumference of a If we can measure the circumference of a sphere that traps the gas, such as a balloon, sphere that traps the gas, such as a balloon, then we can calculate the volume of the gas.then we can calculate the volume of the gas.

Volume CalculationVolume Calculation

What is the volume of 2.6 grams of COWhat is the volume of 2.6 grams of CO22?? The density of COThe density of CO22 is 0.001975 g/cm is 0.001975 g/cm33

V = V = mm dd V = V = 2.6g 2.6g 0.001975g/cm0.001975g/cm33

V = 1,316 cmV = 1,316 cm33

Circumference CalculationCircumference Calculation What should be the circumference of the balloon, if it holds What should be the circumference of the balloon, if it holds

1,316 cm1,316 cm33 of CO of CO22??

V = V = CC3 3 where C = Circumference where C = Circumference

66ππ22

V6V6ππ22 = C = C33

1,316 cm1,316 cm3 3 x 6 (3.1415 x 3.1415) = x 6 (3.1415 x 3.1415) = CC33

42.7 cm = C42.7 cm = C

How do I CalculateHow do I Calculatethe Mass of a Gas?the Mass of a Gas?

If we can measure the volume of the gas and If we can measure the volume of the gas and we know its density, then we use D = m/V:we know its density, then we use D = m/V:

Density (D) = Density (D) = Mass (m)Mass (m) Volume (V)Volume (V)

Volume (V) x Density (D) = Mass (m) Volume (V) x Density (D) = Mass (m)

oror

Comparing Our Measurements Comparing Our Measurements with Our Calculationswith Our Calculations

Calculated Circumference:Calculated Circumference: 42.7 cm42.7 cm

Measured Circumference:Measured Circumference:

Explain Any DifferenceExplain Any Difference

ConclusionConclusion

My hypothesis……. was supportedMy hypothesis……. was supported

by my data because the mass of all by my data because the mass of all the products of this chemical the products of this chemical reaction was equal to mass of all reaction was equal to mass of all the reactantsthe reactants

Conclusion ContinuedConclusion Continued

I I know that this reaction obeys the Law know that this reaction obeys the Law of Conservation of Mass because I used of Conservation of Mass because I used the Law of Definite Proportions to the Law of Definite Proportions to predict the mass of carbon dioxide, and predict the mass of carbon dioxide, and my results matched my prediction my results matched my prediction within the +/- margin of uncertainty within the +/- margin of uncertainty caused by the carbon dioxide that caused by the carbon dioxide that remains dissolved in the water.remains dissolved in the water.