atomic structure - Atomic Structure 1 F321 THE STRUCTURE OF...

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Unformatted text preview: Atomic Structure 1 F321 THE STRUCTURE OF ATOMS ATOMS Atoms consist of a number of fundamental particles, the most important are ... Mass / kg Charge / C Relative mass Relative Charge PROTON NEUTRON ELECTRON MASS NUMBER & ATOMIC NUMBER Atomic Number (Z) Number of protons in the nucleus of an atom Mass Number (A) Q.1 23 Sum of the protons and neutrons in the nucleus Protons Neutrons Electrons A 19 21 20 + 6 E 92 F Mass No. Neutral C D Atomic No. Na 6 6 11 23 Neutral 235 6 16 Symbol 40 Neutral G 13 2- 16 27 H Relative Atomic Mass (Ar) Atomic number (protons) 19 B Charge 11 Mass number (protons + neutrons) Al 3+ The mass of an atom relative to the 12C isotope having a value of 12.000 Ar = average mass per atom of an element × 12 mass of one atom of carbon-12 Relative Isotopic Mass Similar, but uses the mass of an isotope 238U Relative Molecular Mass (Mr) Similar, but uses the mass of a molecule CO2, N2 Relative Formula Mass Used for any formula of a species or ion NaCl, OH¯ © KNOCKHARDY PUBLISHING 2008 2 Atomic Structure F321 ISOTOPES Definition Atoms with ... the same atomic number but different mass number or the same number of protons but different numbers of neutrons. Properties Chemical properties of isotopes are identical Theory Relative atomic masses measured by chemical methods rarely produce whole numbers but they should do (allowing for the low relative mass of the electron). This was explained when the mass spectrograph revealed that atoms of the same element could have different masses due to the variation in the number of neutrons in the nucleus. The observed mass was a consequence of the abundance of each type of isotope. Example P N 1 0 1 1 H 1 2 P N % 18 75 20 25 1 H 1 2 H 1 3 1 There are two common isotopes of chlorine. 35 17 Calculate the average relative atomic mass of chlorine atoms Cl 17 37 17 Cl 17 Method 1 Three out of every four atoms will be chlorine-35 Average = 35 + 35 + 35 + 37 = 35.5 4 Method 2 Out of every 100 atoms 75 are 25 are 35Cl 37Cl Average = (75 x 35) + (25 x 37) = 35.5 100 Q.2 Calculate the average relative atomic mass of sulphur from the following 33S 1% 34S 4% isotopic percentages... 32S 95% Q.3 Bromine has isotopes with mass numbers 79 and 81. If the average relative atomic mass is 79.908, calculate the percentage of each isotope present. Need help:- See example calculation on the next page © KNOCKHARDY PUBLISHING 2008 Atomic Structure Mass spectra 3 F321 An early application was the demonstration by Aston, (Nobel Prize, 1922), that naturally occurring neon consisted of three isotopes ... Ne, 21 Ne and 22 Ne. 90.92% Abundance 20 • positions of peaks gives atomic mass • peak intensity gives relative abundance 8.82% 0.26% • highest abundance is scaled up to 100% - other values are adjusted accordingly. 19 20 21 22 23 mass to charge ration m/z CALCULATIONS Example 1 Calculate the average relative atomic mass of neon using the above information. Out of every 100 atoms 90.92 are Average = 20Ne , 0.26 are 21Ne 22Ne and 8.82 are (90.92 x 20) + (0.26 x 21) + (8.82 x 22) = 20.179 100 Ans. = Example 2 20.18 Naturally occurring potassium consists of potassium-39 and potassium-41. Calculate the percentage of each isotope present if the average is 39.1. Assume there are x nuclei of so thus Q.4 39 K in every 100; there will then be (100-x) of 39x + 41 (100-x) = 39.1 100 - 2x = - 190 so x = 95 therefore 39 x + 4100 - 41x ANSWER 62 20 63 25 64 100 95% 39K and 5% 65 5 Mass spectra can also be used to find the relative molecular mass of compounds © KNOCKHARDY PUBLISHING 2008 K. = 3910 Calculate the average relative atomic mass of an element producing the following peaks in its mass spectrum... m/z Relative intensity 41 41 K 4 Atomic Structure F321 MASS SPECTROMETER A mass spectrometer consists of ... an ion source, an analyser and a detector. 20 Ne ANALYSER 21 Ne ELECTRON GUN & ION SOURCE 22 Ne DETECTOR Ion source • gaseous atoms are bombarded by electrons from a gun and are IONISED • sufficient energy is given to form ions of 1+ charge • resulting ions can be ACCELERATED out of the ion source by an electric field Analyser • • • • • charged particles will be DEFLECTED by a magnetic or electric field the radius of the path depends on the value of their mass/charge ratio (m/z) ions of heavier isotopes with larger m/z values follow a larger radius curve as most ions are singly charged (1+), the path depends on their mass if an ion acquires a 2+ charge it will be deflected more; its m/z value is halved 75% Abundance 75% 2+ ion + ion Changes when the ion is 2+ 25% 25% • M/Z VALUE IS HALVED • ABUNDANCE IS THE SAME 4 Detector 6 8 10 mass to charge ratio m/z 12 • by electric or photographic DETECTION methods • mass spectra record the mass/charge values relative abundance of each ion © KNOCKHARDY PUBLISHING 2008 and Q.1 Protons 13 H 16 G 6 F 92 E 6 D 11 C 20 B 19 A 79Br Q.3 Neutrons 21 20 12 6 143 7 16 14 Electrons 19 20 10 6 92 6 18 10 Charge 27 13 32 16 13 6 235 92 12 6 23 11 40 20 40 19 Mass No. Atomic No. Neutral Neutral + Neutral Neutral Neutral 23+ Symbol 40 40 23 K Ca Na+ 12 C 235 13 U C 32 2- S 27 Al3+ 32.49 Q.2 © KNOCKHARDY PUBLISHING 2008 Q.4 54.6% 81Br 46.4% 63.6 ANSWERS TO QUESTIONS F321 Atomic Structure 5 ...
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This note was uploaded on 03/12/2012 for the course CHEMISTRY Stubbs taught by Professor Stubbs during the Summer '09 term at Al Ahliyya Amman University.

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