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Fundamental Chemical Laws

Fundamental Chemical Laws. Law of Conservation of Mass. In every chemical operation an equal amount of matter exists before and after the operation. Mass is conserved, the total mass after the chemical operation must be the same as that before. 1775 - Lavoisier “Father of Modern Chemistry”.

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Fundamental Chemical Laws

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  1. Fundamental Chemical Laws

  2. Law of Conservation of Mass In every chemical operation an equal amount of matter exists before and after the operation. Mass is conserved, the total mass after the chemical operation must be the same as that before. 1775 - Lavoisier “Father of Modern Chemistry”

  3. Law of Definite Proportions In a given chemical compound, the proportions by mass of the elements that compose it are fixed, regardless of the source of the compound. The ratio of elements in a compound is fixed regardless of the source of the compound. Water is made up of 11.1% by mass of hydrogen and 88.9% oxygen. Joseph Proust

  4. Avogadro’s Hypothesis Equal volumes of different gases (at the same temperature and pressure) contain equal numbers of particles 2 volumes of hydrogen + 1 volume of oxygen  2 volumes of water vapor can be expressed as 2H2 + O2 2H2O While at this time there was no direct evidence to show that hydrogen and oxygen gas were H2 and O2, 50 years later this was proven to be the case.

  5. Dalton’s Atomic Theory • Elements are made of tiny particles called atoms. • All atoms of a given element are identical. The atoms of a given element are different from those of any other element. • Atoms of one element can combine with atoms of other elements to form compounds. A given compound always has the same relative number of types of atoms. • Atoms cannot be created, nor divided into smaller particles, nor destroyed in the chemical process. A chemical reaction simply changes the way atoms are grouped together.

  6. Evidence for sub-atomic particles 1897: J.J. Thomsen: Cathode Ray Tube Evidence for electrons: Bent a stream of rays originating from the negative electrode (cathode). Stream of particles with mass & negative charge. 1909: Ernest Rutherford: Gold Foil Evidence for protons & nucleus: Alpha particles deflected passing through gold foil 1932: James Chadwick: Beryllium Evidence for neutrons: Alpha particles caused beryllium to emit rays that could pass through lead but not be deflected

  7. Millikan’s Oil Drop Experiment Robert Millikan’s oil drop experiment calculated the charge/mass ratio of the electron, and combining Thompson’s results the mass of the electron was calculated to be 9.10 x 10-28 g. (actual mass of the electron 9.10939 x 10 -28 g) There must be a positive species which counters the electron charge.

  8. Radioactivity Henri Becquerel in 1896 discovered high-energy radiation was spontaneously emitted from uranium. Later Marie Curie and her husband Pierre further investigated this spontaneous emission of radiation which was termed radioactivity.

  9. The Structure of the Atom J.J. Thompson, realized that electrons were sub-atomic particles, and presented his theory of the model of the atom. The “PLUM-PUDDING” model

  10. Model of the Atom Since the times of Rutherford, many more subatomic particles have been discovered. However, for chemists three sub-atomic particles are all that we need to focus on – ELECTRON, PROTON, NEUTRON. Electrons are –1, protons +1 and neutrons are neutral. Atoms have an equal number of electrons and protons they are electrically neutral.

  11. Protons and neutrons make up the heavy, positive core, the NUCLEUS which occupies a small volume of the atom.

  12. 11B 10B Isotopes • Atoms of the same element but different mass number. • Boron-10 (10B) has 5 p and 5 n • Boron-11 (11B) has 5 p and 6 n

  13. Two Isotopes of Sodium.

  14. Masses of Particles Relative Isotopic Mass • Chemists as early as John Dalton, two centuries ago, used experimental data to determine the weight of different atoms relative to one another. • Dalton estimated relative atomic weights based on a value of one unit for the hydrogen atom. • In 1961, it was decided that the most common isotope of 12C would be used as the reference standard. • On this scale, the 12C isotope is given a relative mass of exactly 12 units.

  15. Relative Isotopic Mass cont… “The relative isotopic mass (Ir) of an isotope is the mass of an atom of the that isotope relative to the mass of an atom of 12C taken as 12 units exactly.” Know that 1.0 amu is defined as exactly 1/12 the mass of a atom. Carbon-12 has 6 protons and 6 neutrons, therefore 1 proton or 1 neutron = ~1 amu 1 amu = 1.6606 x 10 -24 grams

  16. Average relative atomic mass: is the weighted average for all of the isotopes of a given element, based on the percent abundance of each • Need masses of each isotopes • Need abundance (percentage) of each isotope • This is the value shown on the periodic table

  17. Isotopic Masses example… • Chlorine has two isotopes. • These have different masses as they have different amounts of neutrons. • Using the 12C isotope as a standard, the relative isotopic masses of these two isotopes are 34.969 (35Cl) and 36.966 (37Cl). • Naturally occurring chlorine is made of 75.80% of the lighter isotope and 24.20% of the heavier isotope.

  18. Isotopic Composition of Some Common Elements

  19. Mass Spectrometer • Relative isotopic masses of elements can be obtained using an instrument called a mass spectrometer. • This separates the individual isotopes in a sample of the element and determines the mass of each isotope. • The information is presented graphically and is known as a mass spectrum.

  20. Schematic Diagram of a Mass Spectrometer

  21. Stages • Vaporization: sample is heated to gas state • Ionization: turned into ions by blasting electrons to knock out electrons from the atoms, creating positively charged ions • Acceleration: increases the speed of particles, using an electric field • Deflection: by a magnetic field • amount of deflection depends on mass and charge of the ion • Detection: measures both mass and relative amounts (abundance) of all the ions present

  22. Results show the abundance for each isotope of an element • 90.92% is neon-20 • 0.26% is neon-21 • 8.82% is neon-22

  23. Neon Gas

  24. Mass Spectrometer cont… • In a mass spectrum showing the isotopes of an element: • The number of peaks indicates the number of isotopes • The position of each peak on the horizontal axis indicates the relative isotopic mass • The relative heights of the peaks correspond to the relative abundance of the isotopes

  25. Calculating Relative Isotopic Mass • To calculate average atomic mass you need to know 3 things: • # of stable isotopes • Mass of each isotope • % abundance of each isotope • Each isotope is a piece of fruit and the isotope’s mass is the weight of each piece of fruit.

  26. Example: Chlorine Calculation • mass of isotope X relative abundance + mass of isotope X relative abundance =_______amu • (34.969)(.7553) + (36.935)(.2447) = • That’s the same value on the periodic table! 35.4500amu

  27. Example: Copper Calculation (62.9298)(.6909)+(64.9278)(.3091)= 63.5464 amu

  28. Average Relative Mass example… • Imagine taking 100 atoms from a sample of chlorine of chlorine – there will be 75.80 atoms of 35Cl and 24.20 of 37Cl. Find the relative atomic mass… Equation to use: ((relative isotopic mass1 x % abundance1) + (relative isotopic mass2 x % abundance2)) /100 OR Ar = ∑(relative isotopic mass x %abundance) / 100

  29. Average Atomic Mass cont… • Imagine taking 100 atoms from a sample of chlorine of chlorine – there will be 75.80 atoms of 35Cl and 24.20 of 37Cl. Find the relative atomic mass… Ar = Ar = Ar = 35.452

  30. Calculating Relative Abundance • To Calculate % Abundance: • Make a Chart • Isotopic Mass X %Abundance of each isotope • Set-up equation • Solve for “x” • Plug in “x” value to solve for “y”

  31. Example x 1- x 1.00 x + y = 1.00 y = 1 – x 10.103 (x) + 11.009 (1 –x) = 10.811 10.103x + 11.009 -11.009x = 10.811 -0.996x = -0.198 x = .1987 y= 1-.1987 y= .8013 B-10 = 19.87% B-11 = 80.13%

  32. Percentage Abundance example… • Copper has two isotopes. 63Cu has a relative isotopic mass of 62.95 and 65Cu has a relative isotopic mass of 64.95. The relative atomic mass of copper is 63.54. Calculate the percentage abundance of the two isotopes. • Let x be the percentage abundance of 63Cu • So, 100-x is the percentage abundance of 65Cu

  33. Percentage Abundance example… Ar(Cu) = ∑(relative isotopic mass x %abundance) 100 So 63.54 = 6354 = 62.95x + 6495 – 64.95x 6354 = 6495 – 2x 2x = 6495 – 6354 2x = 141 x = 70.5

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