Chem 116 - Fall 1997

Answer Key
August 28, 1998

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2. To determine the number of carbons in each cluster, simply divide the cluster mass by the atomic mass of carbon which is 12.011 amu. The mass spectrum shows clusters of mass 673, 697, 721, 769, and 841.

n_Carbon(cluster #1) = 673/12.011 = 56; Formula = C₅₆

n_Carbon(cluster #2) = 697/12.011 = 58; Formula = C₅₈

n_Carbon(cluster #3) = 721/12.011 = 60; Formula = C₆₀

n_Carbon(cluster #4) = 769/12.011 = 64; Formula = C₆₄

n_Carbon(cluster #5) = 841/12.011 = 70; Formula = C₇₀

The most stable cluster is the one with the highest intensity. This would be C₆₀.

3. When ¹³¹I emits an alpha particle, which is a helium nucleus (), it will lose two neutrons and two protons (because the helium nucleus contains two neutrons and two protons). Since the iodine nucleus loses two protons, its atomic number changes from 53 to 51. The resulting nucleus with Z=51 is antimony, Sb. The appropriate reaction is

4. a. We have four isotopes: ²⁰⁸Pb, ²⁰⁷Pb, ²⁰⁶Pb, and ²⁰⁴Pb. For ²⁰⁸Pb, number of protons = 82, number of neutrons = 126, number of electrons = 82. For ²⁰⁷Pb, number of protons = 82, number of neutrons = 125, number of electrons = 82. For ²⁰⁶Pb, number of protons = 82, number of neutrons = 124, number of electrons = 82. For ²⁰⁴Pb, number of protons = 82, number of neutrons = 122, number of electrons = 82.

b. Atomic Mass = 0.014(203.9731 amu) + 0.241(205.9745 amu) + 0.221(206.9759 amu) + 0.524(207.9766 amu)
Atomic Mass = 207.2 amu