Friday, April 3, 2020

Atomic Mass From Atomic Abundance Chemistry Problem

Atomic Mass From Atomic Abundance Chemistry Problem You may have noticed the atomic mass of an element isnt the same as the sum of the protons and neutrons of a single atom. This is because elements exist as multiple isotopes. While each atom of an element has the same number of protons, it can have a variable number of neutrons. The atomic mass on the periodic table is a weighted average of the atomic masses of atoms observed in all samples of that element. You can use the atomic abundance to calculate the atomic mass of any element sample if you know the percentage of each isotope. Atomic Abundance Example Chemistry Problem The element boron consists of two isotopes, 105B and 115B. Their masses, based on the carbon scale, are 10.01 and 11.01, respectively. The abundance of 105B is 20.0% and the abundance of 115B is 80.0%. What is the atomic mass of boron? Solution: The percentages of multiple isotopes must add up to 100%. Apply the following equation to the problem: atomic mass (atomic mass X1) Â · (% of X1)/100 (atomic mass X2) Â · (% of X2)/100 ...where X is an isotope of the element and % of X is the abundance of the isotope X. Substitute the values for boron in this equation: atomic mass of B (atomic mass of 105B Â · % of 105B/100) (atomic mass of 115B Â · % of 115B/100)atomic mass of B (10.01Â · 20.0/100) (11.01Â · 80.0/100)atomic mass of B 2.00 8.81atomic mass of B 10.81 Answer: The atomic mass of boron is 10.81. Note that this is the value listed in the periodic table for the atomic mass of boron. Although the atomic number of boron is 10, its atomic mass is nearer to 11 than to 10, reflecting the fact that the heavier isotope is more abundant than the lighter isotope. Why Arent Electrons Included? The number and mass of electrons is not included in an atomic mass calculation because the mass of the electron is infinitesimal compared to that of a proton or neutron. Basically, electrons dont significantly affect the mass of an atom.