unit 7 n.
Download
Skip this Video
Loading SlideShow in 5 Seconds..
Unit 7 PowerPoint Presentation
Download Presentation
Unit 7

Loading in 2 Seconds...

play fullscreen
1 / 7

Unit 7 - PowerPoint PPT Presentation


  • 133 Views
  • Uploaded on

Unit 7. John Dalton’s atomic theory (2.4) Mass relationships from Dalton’s atomic theory (2.4). John Dalton. Quaker was a school teacher at the turn of the 18 th century

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

Unit 7


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
    Presentation Transcript
    1. Unit 7 John Dalton’s atomic theory (2.4) Mass relationships from Dalton’s atomic theory (2.4)

    2. John Dalton Quaker was a school teacher at the turn of the 18th century Provided an explanation (theory) for the laws of conservation of mass, definite proportions, and multiple proportions (his law) The framework for the explanation is known as Dalton’s atomic theory

    3. Dalton’s Atomic Theory (2.4) Four premises in Dalton’s atomic theory All matter is composed of extremely small indivisible particles called atoms All atoms of a given element are alike, and atoms of different elements are different from each other Compounds are formed when atoms of different elements combine in fixed proportions Chemical changes are the result of the rearrangement of atoms – no atoms are created or destroyed

    4. Today’s Status of Dalton’s Atomic Theory Today we are aware that Dalton’s atomic theory as originally stated in not entirely correct. However, it still serves as a solid basis for interpreting chemical changes. Some of the differences, of which we will see more soon include: Atoms are not indivisible – they are made of subatomic particles Not all elements of the same element are identical – we will encounter species called isotopes

    5. How Does Atomic Theory Help? Consider the Law of Definite Proportions: If each atom of a specific element has the same mass, but different elements are different from each other, the requirement for definite proportions becomes clear. Consider our old standby water. The ratio of oxygen by mass to hydrogen by mass is 8:1. If the mass of an oxygen atom is sixteen times that of hydrogen, then the observed ratio in water is explained by having two hydrogen atoms for every oxygen atom. ( oxygen is red, hydrogen gray) It is not possible to have fractions of atoms, so the mass ratios in a compound will always depend on the relative masses and the numbers of atoms of each kind in the compound.

    6. How Does Atomic Theory Help? Consider the Law of Multiple Proportions: If only integral numbers of atoms can combine, the number of possible compounds formed may be greater than one. For example, consider hydrogen and oxygen: There is an infinite number of atom combinations for even two atoms, but not all are chemically viable. For oxygen and hydrogen only the combinations in the second and third rows above exist as actual compounds.

    7. Mass Relationships from Dalton’s Atomic Theory (2.4) An example may be used to illustrate how to use mass relationships to find useful information Octane, a compound often used to simulate gasoline, can be decomposed to give 5.3 parts by mass of carbon to 1.0 parts by mass of hydrogen. How many grams of carbon are produced when 150 g of gasoline is decomposed? Solution: Consider that if the compound is composed of 5.3 parts by mass carbon and 1.0 parts by mass hydrogen, that every 6.3 parts of the compound has 5.3 parts of carbon (the 6.3 is the sum of the parts = 5.3 +1.0). Thus, the fraction of the compound that is carbon is 5.3/6.3 = 0.84. If one starts with 150 g, the grams of carbon in it is 0.84 x 150 g = 126 g carbon.