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The Law of Combining Volumes of Gases: When two gases react, the volumes that combine are in a ratio of small whole numbers. The ratio of the volume of.

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Presentation on theme: "The Law of Combining Volumes of Gases: When two gases react, the volumes that combine are in a ratio of small whole numbers. The ratio of the volume of."— Presentation transcript:

1 The Law of Combining Volumes of Gases: When two gases react, the volumes that combine are in a ratio of small whole numbers. The ratio of the volume of each product, if a gas, is also in the ratio of small whole numbers. Example: 1 Liter of hydrogen + 1 Liter of chlorine = 2 Liters of Hydrogen Chloride 2 Liters of hydrogen + 1 Liter of oxygen = 2 Liters of Water 3 Liters of hydrogen + 1 Liter of nitrogen = 2 Liters of Ammonia Mass is always conserved; but the volume of a gas is not. Avogadro’s Hypothesis: Equal volumes of different gases contain the same number of particles. The particles of a gas may be atoms or molecules. One liter of hydrogen = one liter of chlorine = one liter of hydrogen chloride in terms of particles (read atoms or molecules)

2 Avogadro’s Law and the reaction of hydrogen and oxygen to produce water 2 liters of hydrogen + 1 Liter of oxygen -> 2 Liters of water Dalton: H + O -> HOSimplicity of atoms Avogadro: 2 H 2 + O 2 -> 2 H 2 OEqual volumes equal particles Both equations are balanced. Which is correct (if either?) Dalton’s hypothesis provided no link between the relationships between the volumes of the reactants and products. Avogadro’s hypothesis provided a direct link between the relationships between the volumes of the reactants and products.

3 (1)Avogadro’s Number (N 0 ) is defined a the number of atoms in exactly 12 g (0.012 kg) of 12 C (2)Experimentally, N 0 is found to be equal to 6.02 x (3)The mass (expressed in grams) of Avogadro’s number of atoms of an element is equal to the relative atomic mass of that element (4)A mole of a substance equals the amount of that substance that contains Avogadro’s number of atoms (element) or molecules (compound)

4 *An Avogadro's number of standard soft drink cans would cover the surface of the earth to a depth of over 200 miles. If you had Avogadro's number of unpopped popcorn kernels, and spread them across the United States of America, the country would be covered in popcorn to a depth of over 9 miles. *If we were able to count atoms at the rate of 10 million per second, it would take about 2 billion years to count the atoms in one mole. Lorenzo Romano Amedeo Carlo Avogadro, conte di Quaregna e di Cerreto ( ) Avogadro’s number is 6.02 x (unlisted)

5 Avogadro’s interpretation of gas reactions: H 2 +Cl 2 ->2 HCl 1 L1 L2 L 2 H 2 +O 2 ->2 H 2 O 2 L1 L2 L 3 H 2 +N 2 ->2 NH 3 3 L1 L2 L Dalton’s interpretation of reactions: H+Cl->HCl H+O->HO H+N->HN

6 Conceptual basis of Avogadro’s Hypothesis The particles of any gas all occupy the same space at a given temperature and pressure, independent of the composition of the gas. Thus, a given number or “particles” (atoms or molecules) of a gas will occupy the same space of same volume. One liter of hydrogen gas, contains the same number of hydrogen “particles” as the number of oxygen “particles” in one liter of oxygen gas or the numbe of nitrogen “particles” in one liter of nitrogen gas. Consider the experiment: 1 L of hydrogen + 1 L of chloring -> 2 L of hydrogen chloride Consider the interpretations: H +Cl->2 HCl Impossible!Dalton H 2 +Cl 2 ->2 HClPossibleAvogadro

7 Amazing consequence of Avogadro’s hypothesis The molar volume of a gas is the volume of one mole of the gas (at standard temperature and pressure). This molar volume near room temperature and atmospheric pressure is 22.4 L The weight of 22.4 L of any gas is the molecular weight the gas!

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9 “In 1757, being at sea in a fleet of 6 sail bound against Louisburg, I observed the wakes of two of the ships to be remarkably smooth, while all the others were ruffled by the wind, which blew fresh. Being puzzled with the differing appearance, I at last pointed it out to our captain, and asked him the meaning of it? "The cooks," says he, "have, I suppose, been just emptying greasy water through the scuppers, which has greased the sides of those ships a little," and this answer he gave me with an air of some little contempt as to a person ignorant of what everybody else knew. It occurred to me the learned are apt to slight too much the knowledge of the ancients and the vulgar. This art of smoothing the waves with oil is an instance of both! -Benjamin Franklin, Philosophical Transactions of the Royal Society of London, 1776

10 "...at Clapham I observed a large pond very rough with the wind. I fetched a cruet of oil and dropt a little of it on the water. The oil, though not more than a teaspoonful, produced an instant calm over a space of several yards square, and then spread amazingly till it filled a quarter of the pond, perhaps half an acre, as smooth as a looking glass..... If a drop of oil is put on a polished marble table, or on a looking-glass that lies horizontally, the drop remains in place, spreading very little. But when put on water it spreads instantly many feet around, becoming so thin as to produce the prismatic colors, for a considerable space, and beyond them so much thinner as to be invisible, except in its effect of smoothing the waves." ---Benjamin Franklin, letter to William Brownrigg, November 7, Clapham Pond

11 This incredibly simple experiment provides a means of understanding fundamental facts about molecules and the forces between them. It even leads to a means of determining molecular size and shape! For example, if the teaspoon were 2 cc of oil and the area of a half an acre is approximately 2000 m 2, the film thickness (volume/area) would be ca cm (1 nanometer, 10 Å), which is right on the molecular dimensions of an "olive oil" molecule!!! This is clearly within the experimental uncertainty of measuring acreage!

12 Chapter 2 Stoichiometry Learning Goals: (1)How to translate the coefficients of the atoms involved in balanced chemical equations into moles of elements and compounds and how to translate moles into mass of elements and compounds. (2)For gases, how to translate the coefficients of the atoms involved in balanced chemical equations into moles of elements and compounds and how to translate moles into volumes of elements and compounds which are gases. (3)How to determine the limiting reagent in a reaction from the balanced chemical equation and the available masses of the reagents.

13 Empirical Formulas and Molecular Formulas (1)Empirical formula: a formula which displays the simplest ratios of the number of atoms of different elements that make up a substance (2)Molecular formula: a formula which displays the exact number and kinds of atoms that are present in one molecule. The molecular formula is the compositional structure of a molecule (3)Lewis structure: a structure which displays how the atoms of a molecule are connected. The Lewis structure is the constitutional structure of a molecule (4)Molecular structure: a three dimensional representation of how the atoms of a molecular are arranged in space. The molecular structure displays the configurational structure of atoms about each other in space.

14 Percentage atomic composition from empirical or molecular formulas Empirical or molecular formula -> % composition of elements Example: H 2 O Need: relative masses of H and O From periodic table: H = 1.00, O = 16.0 One mole of H 2 O consists of 2 moles of H atoms and one mole of O atoms. 2 moles of H atoms have a mass of 2 g and 1 mole of O atoms have a mass of 16. The mass of one mole of water is 18 g. The percent of H in water is 2/18 x 100 = 11%; the percent of O in water is 16/18 = 89 % O.

15 Section 2.2:Using balanced chemical equations Hydrocarbon + Oxygen->Carbon dioxide + Water What is the balanced chemical equation for complete combustion of C 4 H 10 ? ? C 4 H 10 + ? O 2 ->? CO 2 +? H 2 O Answer (by inspection and molecular weights of reactants and products): One mol of C 4 H 10 = g; one mol of O 2 = g; one mol of CO 2 = g; one mol of H 2 O = g. 2 C 4 H O 2 ->8 CO H 2 O g C 4 H g O 2  g CO g H 2 O 2 mol of C 4 H mol of O 2 8 mol CO 2 10 mol H 2 O Example of the law of mass balance g g = g g = g REACTANTSPRODUCTSSAME MASS


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