Atoms, Molecules, and Ions Chapter 2 Atoms, Molecules, and Ions

Atoms, Molecules, and Ions Dalton’s Postulates 1)Each element is composed of extremely small particles called atoms. Subatomic particles had not been discovered yet at the time Dalton proposed his theory. 2)All atoms of a given element are identical to one another in mass and other properties, but the atoms of one element are different from the atoms of all other elements. Obviously the first part of the postulate was later revised when neutrons were discovered and isotopes were found. 3)Atoms of an element are not changed into atoms of a different element by chemical reactions; atoms are neither created nor destroyed in chemical reactions. Law of conservation of matter. 4)Compounds are formed when atoms of more than one element combine; a given compound always has the same relative number and kind of atoms. Law of constant composition.

Atoms, Molecules, and Ions Law of Constant Composition Joseph Proust (1754–1826) Also known as the law of definite proportions. The elemental composition of a pure substance never varies. MgO is always 60.3% Mg and 39.7% O by mass, no matter the size of the sample, as long as the compound is pure.

Atoms, Molecules, and Ions Law of Multiple Proportions Two compounds containing nitrogen and oxygen are: NO 2 and N 2 O To find if the law of multiple proportions apply to these examples we must work with the ratio of masses: For NO 2 : 14 g N / 32 g O For N 2 O: 28 g N / 16 g O Making the mass of O = 1 NO 2 : g N / g O N 2 O: 1.75 g N / g O

Atoms, Molecules, and Ions Law of Multiple Proportions NO 2 : g N / g O N 2 O: 1.75 g N / g O Once the mass of one of the elements is set to 1, the other element can be compared: 1.75 / = 4/1 Small number ratios!!!!

Atoms, Molecules, and Ions Law of Conservation of Mass The total mass of substances present at the end of a chemical process is the same as the mass of substances present before the process took place.

Atoms, Molecules, and Ions The Electron Streams of negatively charged particles were found to emanate from cathode tubes. J. J. Thompson is credited with their discovery (1897).

Atoms, Molecules, and Ions The Electron Thompson measured the charge/mass ratio of the electron to be 1.76  10 8 coulombs/g.

Atoms, Molecules, and Ions Millikan Oil Drop Experiment Once the charge/mass ratio of the electron was known, determination of either the charge or the mass of an electron would yield the other.

Atoms, Molecules, and Ions Millikan Oil Drop Experiment Robert Millikan (University of Chicago) determined the charge on the electron in See video

Atoms, Molecules, and Ions Radioactivity: The spontaneous emission of radiation by an atom. First observed by Henri Becquerel. Also studied by Marie and Pierre Curie.

Atoms, Molecules, and Ions Radioactivity Three types of radiation were discovered by Ernest Rutherford:   particles.   particles.   rays See video

Atoms, Molecules, and Ions The Atom, circa 1900: “Plum pudding” model, put forward by Thompson. Positive sphere of matter with negative electrons imbedded in it.

Atoms, Molecules, and Ions Discovery of the Nucleus Ernest Rutherford shot  particles at a thin sheet of gold foil and observed the pattern of scatter of the particles. See video

Atoms, Molecules, and Ions The Nuclear Atom Since some particles were deflected at large angles, Thompson’s model could not be correct.

Atoms, Molecules, and Ions The Nuclear Atom Rutherford postulated a very small, dense nucleus with the electrons around the outside of the atom. Most of the volume of the atom is empty space.

Atoms, Molecules, and Ions Other Subatomic Particles Protons were discovered by Rutherford in Neutrons were discovered by James Chadwick in 1932.

Atoms, Molecules, and Ions Subatomic Particles Protons and electrons are the only particles that have a charge. Protons and neutrons have essentially the same mass. The mass of an electron is so small we ignore it.

Atoms, Molecules, and Ions Symbols of Elements Elements are symbolized by one or two letters.

Atoms, Molecules, and Ions Atomic Number All atoms of the same element have the same number of protons: The atomic number (Z)

Atoms, Molecules, and Ions Atomic Mass The mass of an atom in atomic mass units (amu) is the total number of protons and neutrons in the atom.

Atoms, Molecules, and Ions Isotopes: Atoms of the same element with different masses. Isotopes have different numbers of neutrons C 12 6 C 13 6 C 14 6 C

Atoms, Molecules, and Ions protonsneutronselectronsMass number Atomic number 32 S Co Cl - Problem 1. Complete the following table.

Atoms, Molecules, and Ions How can isotopes be detected? Mass Spectrometry Molecules are converted to ions so that they can be moved about and manipulated by external electric and magnetic fields. The three essential functions of a mass spectrometer, and the associated components, are: The Ion Source: A small sample is ionized, usually to cations by loss of an electron. M + The Mass Analyzer: The ions are sorted and separated according to their mass-to-charge ratio. The Detector: The separated ions are then measured their abundance. The results are displayed on a chart.

Atoms, Molecules, and Ions Mass Spec.

Atoms, Molecules, and Ions The mass spectrum A mass spectrum will usually be presented as a vertical bar graph, in which each bar represents an ion having a specific mass-to-charge ratio (m/z) and the length of the bar indicates the relative abundance of the ion. The most intense ion is assigned an abundance of 100, and it is referred to as the base peak. Most of the ions formed in a mass spectrometer have a single charge, so the m/z value is equivalent to mass itself. Modern mass spectrometers easily distinguish (resolve) ions differing by only a single atomic mass unit (amu), and thus provide completely accurate values for the molecular mass of a compound. The highest-mass ion in a spectrum is normally considered to be the molecular ion (and therefore the molecular mass), and lower-mass ions are fragments from the molecular ion, assuming the sample is a single pure compound.

Atoms, Molecules, and Ions The mass spectrum of bromine The most abundant peak has a mass of 160 indicating 79 Br- 81 Br. The less common 158 and 162 will be formed by the same isotopes. ( 79 Br- 79 Br and 81 Br- 81 Br respectively) The fragmentation of 160 generates almost equal amounts of 79 Br and 81 Br. This suggests that both isotopes are almost equally abundant.

Atoms, Molecules, and Ions The mass spectrum of chloro ethene or vinyl chloride MW: 62.5 amu The peak at 62 represents the compound with the isotope 35 Cl, while the peak at 64 represents the compound with 37 Cl. Notice that the % abundance (peak’s height) is about 75% for 35 Cl and about 25% for 37 Cl. Infer the fragment responsible for peak at 27.

Atoms, Molecules, and Ions Chlorine: 75.77% 35 Cl and 24.23% 37 Cl Bromine: 50.50% 79 Br and 49.50% 81 Br Actual % abundances.

Atoms, Molecules, and Ions Average Mass Because in the real world we use large amounts of atoms and molecules, we use average masses in calculations. Average mass is calculated from the isotopes of an element weighted by their relative abundances. Average atomic mass = ∑(atomic mass x % abundance) for each isotope 100

Atoms, Molecules, and Ions Problem 2. Determine the atomic mass and the identity of an element with the following mass spectrum data. Isotope Atomic mass M amu M amu M amu M amu

Atoms, Molecules, and Ions Mass percent Mass % element = (# of atoms of element) ( element’s atomic mass) x 100 molar mass of compound Problem 3. Calculate the mass % of elements in (NH 4 ) 2 SO 4.

Atoms, Molecules, and Ions Law of multiple proportion H2OH2OH2O2H2O2 Mass O16.00 amu2(16.00 amu) = amu Mass H2(1.01 amu) = 2.02 amu Mass O/mass H16.00 amu/2.02 amu32.00 amu/2.02 amu 32.00/ /2.02 = 2 This tells us that H 2 O 2 has twice as many oxygen atoms than H 2 O. This is too easy!!

Atoms, Molecules, and Ions Problem 4: Use the Law of multiple proportions to prove the formulas for NO 2 and N 2 O

Atoms, Molecules, and Ions Mass ratio of elements in a pure sample of a compound The ratio of isotopes of a certain element in any pure sample of a given compound is constant. The ratio 12 C, 13 C, 14 C in pure CO 2 is constant. As the ratio of the different atoms of each element that form a given compound is constant, the ratio of the masses of the constituent elements in a pure sample of that compound will also be constant.

Atoms, Molecules, and Ions Example: Two separate pure samples of carbon dioxide were analyzed. Both samples were found to contain 17.29% carbon by mass. Justify these findings on the basis of atomic molecular theory. All carbon dioxide molecules contain exactly one carbon and two oxygen atoms. Although different isotopes of carbon and different isotopes of oxygen have different masses, the ratio of isotopes of carbon and the ratio of isotopes of oxygen in any pure sample of carbon dioxide is constant. If we could calculate the average mass of all the atoms of an element in a pure sample, we would obtain the average atomic mass of that element, which is given in the periodic table. Since the ratio of carbon atoms to oxygen atoms in both pure samples is the same, the ratio of the masses of carbon atoms to oxygen atoms in that pure sample will also be the same.

Atoms, Molecules, and Ions Nuclear Chemistry Nuclear decay: An unstable atom (because of too many subatomic particles in the tiny nucleus) tends to lose particles from the nucleus.  Alpha decay (α): An atom loses 2 protons and 2 neutrons and its nuclear mass decreases by 4 (similar to the mass of He). It usually happens to heavy isotopes. 238 U  234 Th + 4 He  Beta decay (β): A neutron in the nucleus changes into a proton, thus increasing the atomic number but not the mass and turning into a more stable isotope of another element. It looks like an electron is lost. It usually happens to light isotopes. 14 C  14 N + -1 e  Gamma decay (γ): The unstable atom loses energy but retains its mass and charge. 97* Tc  97 Tc + γ

Atoms, Molecules, and Ions Nuclear equations Problem 5. Complete the following nuclear reactions 3 H  ________ + 0 e 232 U  228 Th + _______ 144 Ce  144 Pr + _________ 7 Be*  7 Be + ____________