CHEMISTRY NOTES Atoms and Elements Terms: atom proton neutron electron nucleus atomic number atomic mass unit (amu) mass number Isotopes If you examine a pure sample of the element boron, you find that, while all the boron atoms have 5 protons in their nucleus, some atoms have 5 neutrons and others have 6. That is, you have a collection of 1°B and "B atoms. These are isotopes of boron. Isotopes are atoms having the same atomic number but different mass numbers, because the atoms have different numbers of neutrons. Most elements have at least two stable isotopes. There are very few elements with only one isotope (Al, F, and P, for example), and there are others with many isotopes (Sn has 9 stable isotopes). We refer to a particular isotope by giving its mass number (for example boron-10). Hydrogen's isotopes are so important that they have their own special names. All hydrogen atoms have one proton. If that is their only nuclear particle, the element is called simply "hydrogen". If one neutron is added to Ove 2H, it is called "deuterium", and adding two neutrons gives radioactive hydrogen, H, "tritium". 1. Silicon has three isotopes with 14, 15, and 16 neutrons, respectively. What are the mass numbers and symbols of these three isotopes? Atomic Mass Elements exist as atoms of different isotopes, each having the same number of protons and electrons, but different numbers of neutrons. Boron atoms have two different isotopic masses, 10.0129 and 11.0093 amu. This means that the average mass of a collection of boron atoms will be neither 10 nor 11 but somewhere in between, the actual value depending on the proportion of each kind of isotope Percent Abundance The percentage of atoms in a natural sample of the pure element represented by a particular isotope is called the isotope's percent abundance. Percent abundance = number of atoms of a given isotope X 100 total number of atoms of all isotopes of that element If these percent abundances can be determined for each isotope, then the average mass can be determined. Such determinations can be done with a mass spectrometer, i t and 11 B have percent abundances of 19.91 and 80.09, respectively. That means that if you could count out 10,000 atoms of boron from an average natural sample, 1991 of them would have a mass of 10.0129 amu and 8009 of them would have a mass of 11.0093 amu. The average mass of a representative sample of atoms is called the atomic mass or atomic weight. For boron, the atomic mass is 10.81 amu. This can be calculated as follows. Average atomic mass = (fractional abundance of isotope 1)(mass of isotope 1) + (fractional abundance of isotope 2)(mass of isotope 2) For the boron sample, the calculation would be: Atomic mass of boron = (0.1991)(10.0129 amu) + (0.8009)(11.0093 amu) = 10.811 2. Determine the atomic mass of chlorine given the following information: 35 C1, exact mass = 34.96885, percent abundance = 75.77 37 Cl, exact mass = 36.96590, percent abundance = 24.23 The atomic mass of each element has been determined . In the periodic table, each element's box contains the atomic number, the element symbol, and the atomic mass. If the masses are known for the isotopes of a given element, it is possible to compute the percent abundances by "reversing" the procedure used above. Considering elements having two isotopes, calculating the percent abundance for the isotopes would involve two unknowns. The sum of the fractional isotope abundances must be equal to 1. Therefore, if one of them is assigned the value x, then the other is 1 — x. 3. Calculate the percent abundances of copper-63 and copper-65 if the atomic mass is 63.546 and the exact masses of the isotopes are 63Cu = 62.93 amu and 65C1.1 = 64.93. Assume no other isotopes exist. Ions Atoms are electrically neutral because the number of protons in their nucleus equals the number of electrons. Cations (positively charged ions) form when atoms lose electrons. Anions (negatively charged ions) form when atoms gain electrons. Charges are written as superscripts following the element symbol. F has one extra electron, Na* has one less. PRE—AP CHEMISTRY PRACTICE PROBLEMS Isotopes and Atomic Mass i. Name Determine the number of protons, neutrons, and electrons in each of the following species: 24 x A iv.vig a) b) :521SC ..140zr c) d) e) D 27 A 13+ 131'11 6530zn2+ 1°847Ag+ Of the following atoms, which are isotopes of the same element? 20 21 ioX 10X, b) 2 11X, c) 2210X, d) 199X, e) a) 3. Bromine has two stable isotopes. The isotopic masses are 78.9183 amu (relative abundance 50.69%) and 80.9163 amu (relative abundance, 49.31%) Calculate the average atomic mass of bromine. 4. Silicon is found in nature combined with oxygen to give sand, quartz, agate, and similar materials. The element has three stable isotopes. Percent Abundance Exact Mass 92.23 27.97693 4.67 28.97649 3.10 29.97376 Calculate the average atomic weight of silicon from this data. 5. The metallic element chromium has four stable isotopes. Percent Abundance Exact Mass 4.35 49.9461 83.79 51.9405 9.50 52.9407 2.36 53.9389 Calculate the average atomic weight of chromium. 6. Gallium has two naturally occurring isotopes, 69Ga and 71Ga, with masses of 68.9257 and 70.9249, respectively. Calculate the percent abundances of the two isotopes. 7. Antimony, one of the elements known to the ancient alchemists, has two stable isotopes: 12I Sb (mass, 120.90 amu) and I23Sb (mass, 122.90 amu). Calculate the percent abundances of the two isotopes. Magnesium is commonly extracted from seawater. Magnesium-24 is its most abundant isotope (78.70%); its exact mass is 23.985 amu. If the atomic mass of magnesium is 24.305, what are the relative abundances of magnesium-25 (mass, 24.986 amu) and magnesium-26 (mass, 25.983 amu)? 9. The atomic weight of lithium is 6.941 amu. The two naturally occurring isotopes of lithium have the following masses: 6Li, 6.01512 amu; 71,i, 7.01600 amu. Calculate the percent of 6Li in naturally occurring lithium. 10. The atomic weight of rubidium is 85.4678 amu. The two naturally occurring isotopes of rubidium have the following masses: 85Rb, 84.9118 amu; 87Rb, 86.9092 amu. Calculate the percent of 85Rb in naturally occurring rubidium.
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