Isotopes 436 Objectives: 5.2 Identify the advantages and disadvantages of using isotopes in industry, medical science, basic research, and in the environment.

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Presentation transcript:

Isotopes 436 Objectives: 5.2 Identify the advantages and disadvantages of using isotopes in industry, medical science, basic research, and in the environment. 5.3 Analyze the irregularities in the progression of the atomic masses of the elements in the periodic table.

Irregularities Of Mass In The Periodic Table Elements do not have their atomic masses as whole numbers. The atomic mass of an element is the weighted average of all the different isotopes of that element This is also known as Relative Mass.

Calculating Relative Mass To calculate relative mass do the following: STEP # 1: You need to know the relative abundance of each isotope (expressed as a percent %) and the atomic mass of each isotope. Use the formula below to solve for relative mass. STEP # 2: Relative Mass = (amu X % abundance) + (amu X % abundance) 100

EXAMPLE # 1 Using the information given below about Lithium, find its relative mass. amu % Abundance Li-6 6u 7.42% Li-7 7u 92.58% Lithium Mass = (amu X % abundance) + (amu X % abundance) 100 Relative Mass = (6u X 7.42) + (7 X 92.58) 100 Mass = 6.93u

EXAMPLE # 2 Using the information given below about oxygen, find its relative mass. amu % Abundance O-16 16u 99.762% O-17 17u 0.038% O-18 18u 0.200% Oxygen = (16u X 99.762) + (17 X 0.038) + (18 X 0.200) 100 Relative Mass = 16.008

Step # 1 EXAMPLE # 3 Step # 2 Isotope Atomic Mass % Abundance C-12 Find % abundance of second isotope…. 100% - 98.89 = ? = 1.11% EXAMPLE # 3 The atomic mass (relative mass) of carbon is 12.01113719 amu. The chart below shows data on the two isotopes of carbon. What is the atomic mass of the second isotope of carbon? Use the formula to solve for the atomic mass (working backwards) Step # 2 Relative Mass = (amu X %abun.) + (amu X %abun.) 100 12.01113719 = (12u X 98.89) + (? X 1.11%) 100 Atomic mass = 13.00335 Isotope Atomic Mass % Abundance C-12 12.00000 98.89% C-? ? 13.00335 1.11%

EXAMPLE # 4 Step # 1 Si-28 99.762% Si-29 0.038% Si-30 0.200% TRICKY! EXAMPLE # 4 Silicon (Si) has 3 naturally occurring isotopes. One isotope has 14 neutrons and a % abundance of 92.2, another isotope has 15 neutrons and a % abundance of 4.70, and finally the last isotope has 16 neutrons and a % abundance of 3.09. What is the approximate average atomic mass of Silicon? Step # 1 Get the atomic mass for each isotope…. % Abundance Si-28 99.762% Si-29 0.038% Si-30 0.200% NOTE: The question only gave the neutrons….so you add them to the protons!

Step # 2 Relative Mass =28.106u Si-28 92.2% Si-29 4.70% Si-30 3.09% Plug information into formula and solve % Abundance Si-28 92.2% Si-29 4.70% Si-30 3.09% = (28u X 92.2) + (29u X 4.70) + (30u X 3.09) 100 Relative Mass =28.106u

Key Points to Remember Calculating relative mass: Relative Mass = amu x % abundance + amu x % abundance 100 Atomic mass