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Law of Conservation of Mass Antoine Lavoisier, ~ 1775 Law of Definite Proportions J.L. Proust, 1799.

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Presentation on theme: "Law of Conservation of Mass Antoine Lavoisier, ~ 1775 Law of Definite Proportions J.L. Proust, 1799."— Presentation transcript:

1 Law of Conservation of Mass Antoine Lavoisier, ~ 1775 Law of Definite Proportions J.L. Proust, 1799

2 Law of Conservation of Mass In a chemical reaction, the Law of Conservation of Mass states that the Mass of the Reactants must equal the Mass of the Products. A + B C + D + E Reactants Products Mass A + Mass B = Mass ( C + D + E )

3 Law of Definite Proportions Any pure compound only contains the same elements in the same proportion by mass. H2OH2OH2OH2O Define proportion: the ratio that relates one part to another part, or relates one part to the whole. Example: A large proportion of the people present in this classroom are students.

4 Acids Vinegar is an Acid Vinegar is an Acid Chemical name is Acetic Acid Chemical name is Acetic Acid Chemical formula: Chemical formula: CH 3 CO 2 H

5 Bases Baking Soda is a Base Baking Soda is a Base Chemical name is Sodium Bicarbonate Chemical name is Sodium Bicarbonate Chemical formula: Chemical formula: NaHCO 3

6 Acids React with Bases Reactants = Product Reactants = Product Acid + Base A Salt Water Gas (sometimes) Vinegar + Baking Soda Sodium Acetate Water (H 2 O) Carbon Dioxide Mass of Reactants Mass of Products = = =

7 Hypothesis If reactant is 84 grams of baking soda, then by proportion, a product is 44 g of carbon dioxide. If reactant is 84 grams of baking soda, then by proportion, a product is 44 g of carbon dioxide. NaHCO 3 + CH 3 CO 2 H 84g60g + =144g H 2 O+ CH 3 CO 2 Na + CO 2 H 2 O + CH 3 CO 2 Na + CO 2 82g18g44g++=144g Sodium Acetate Carbon Dioxide Water

8 Law of Definite Proportions Calculating Mass of Molecule A Atom Mass (g) NaSodium 23 g HHydrogen 1 g CCarbon 12 g OOxygen 16 g Baking Soda Sodium Bicarbonate Na x 1 23g H x 1 1g C x 1 12g O X 3 16(3) = 48g NaHCO 3 84g

9 Law of Definite Proportions Calculating Mass of Molecule B Atom Mass (g) HHydrogen 1g CCarbon 12g OOxygen 16g Vinegar Acetic Acid H x 4 4g C x 2 24g O 2 x 16 32g 32g CH 3 CO 2 H 60g

10 Law of Definite Proportions Calculating Mass of Molecule B Atom Mass (g) HHydrogen 1 g CCarbon 12 g OOxygen 16 g Vinegar Acetic Acid H x 4 1(4) = 4g C x 2 12(2) = 24g O X 2 16(2) = 32g CH 3 CO 2 H 60g

11 Law of Definite Proportions Calculating Mass of Molecule C Atom Mass (g) HHydrogen OOxygen Water Dihydrogen Monoxide H O H 2 O

12 Law of Definite Proportions Calculating Mass of Molecule C Atom Mass (g) HHydrogen 1g OOxygen 16 g Water Dihydrogen Monoxide H x 2 = 2g = 2g O X 1 16g H 2 O 18g

13 Law of Definite Proportions Calculating Mass of Molecule D Atom Mass (g) NaSodium 23 g HHydrogen 1 g OOxygen 16 g CCarbon 12 g A Salt Sodium Acetate Na x 1 23g H x 3 1(3) = 3g O X 2 16(2) = 32g C x 2 12(2) = 24g CH 3 CO 2 Na 82g

14 Law of Definite Proportions Calculating Mass of Molecule E Atom Mass (g) CCarbon OOxygen Gas Carbon Dioxide C O CO 2

15 Law of Definite Proportions Calculating Mass of Molecule E Atom Mass (g) CCarbon 12 g OOxygen 16 g Gas Carbon Dioxide C x 1 12g O X 2 16(2) = 32g CO 2 44g

16 Mass Reactants = Mass Products NaHCO 3 + CH 3 CO 2 H Mass of 6 atoms 84g60g + =144g H 2 O+ CH 3 CO 2 Na + CO 2 H 2 O + CH 3 CO 2 Na + CO 2 82g18g44g++ = Sodium Acetate Carbon Dioxide Water Mass of 8 atoms Mass of 3 atoms Mass of 8 atoms Mass of 3 atoms 144g Reactants 14 atoms Products

17 Test Hypothesis To shorten the reaction time, we want to use only a small amount of baking soda. If reactant is 84 grams of baking soda, then we would get 44 grams of carbon dioxide. If reactant is 84 grams of baking soda, then we would get 44 grams of carbon dioxide. But if we use only 5 grams of baking soda, then by proportion, the product is 2.6 grams of carbon dioxide. But if we use only 5 grams of baking soda, then by proportion, the product is 2.6 grams of carbon dioxide. 5g Sodium Bicarbonate ? g CO 2 5g Sodium Bicarbonate ? g CO 2 5g x 44g = 2.6g CO 2 5g x 44g = 2.6g CO 2 84g 84g

18 How can we measure the mass of gas produced? Subtract the mass of the bottle + cap after the gas is released from the mass of the bottle + cap before the CO 2 is released. Subtract the mass of the bottle + cap after the gas is released from the mass of the bottle + cap before the CO 2 is released. The value should less than 2.6 g because about 10% of the CO 2 remains dissolved in the water solution.

19 How do we Measure the Volume of a Gas? If we can measure the circumference of a sphere that traps the gas, such as a balloon, then we can calculate the volume of the gas. If we can measure the circumference of a sphere that traps the gas, such as a balloon, then we can calculate the volume of the gas.

20 Volume Calculation What is the volume of 2.6 grams of CO 2 ? What is the volume of 2.6 grams of CO 2 ? The density of CO 2 is 0.001975 g/cm 3 The density of CO 2 is 0.001975 g/cm 3 V = m V = m d V = 2.6g V = 2.6g 0.001975g/cm 3 0.001975g/cm 3 V = 1,316 cm 3 V = 1,316 cm 3

21 Circumference Calculation What should be the circumference of the balloon, if it holds 1,316 cm 3 of CO 2 ? What should be the circumference of the balloon, if it holds 1,316 cm 3 of CO 2 ? V = C 3 where C = Circumference V = C 3 where C = Circumference 6π 2 6π 2 V6π 2 = C 3 V6π 2 = C 3 1,316 cm 3 x 6 (3.1415 x 3.1415) = C 3 1,316 cm 3 x 6 (3.1415 x 3.1415) = C 3 42.7 cm = C 42.7 cm = C

22 How do I Calculate the Mass of a Gas? If we can measure the volume of the gas and we know its density, then we use D = m/V: If we can measure the volume of the gas and we know its density, then we use D = m/V: Density (D) = Mass (m) Volume (V) Volume (V) Volume (V) x Density (D) = Mass (m) or

23 Comparing Our Measurements with Our Calculations Calculated Circumference: Calculated Circumference: 42.7 cm 42.7 cm Measured Circumference: Measured Circumference: Explain Any Difference Explain Any Difference

24 Conclusion My hypothesis……. was supported My hypothesis……. was supported by my data because the mass of all the products of this chemical reaction was equal to mass of all the reactants

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


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