a*% abundace + b*%abundance + c*%abundance
Element A has a greater atomic mass than element B if the atomic mass value of A is higher. The atomic mass represents the average mass of an element’s isotopes based on their abundance in nature.
(mass of A) x (%of A) + (mass of B) x (% of B) + (mass of C) x (% of C)
Each isotope of an element has a different Atomic Mass, so an average is taken of all the isotopes, but the average is weighted because the natural abundance (%) of each isotope is factored in. If hydrogen-1 is much more abundant than deuterium and tritium, then the weighted average will be closer to 1 than 2 or 3 but not a whole number. The following equation shows how percent abundance factors into the weighted average. (atomic mass A)(X% abundance) + (atomic mass B)(Y% abundance)...=(weighted average of all isotopes of the element)(100% abundance)
The element with isotopes of approximately 10 amu and 11 amu is boron (B) on the periodic table. Its isotopes include Boron-10 and Boron-11.
To calculate the atomic weight of an element with multiple isotopes, you multiply the mass of each isotope by its relative abundance (as a decimal), then sum the products. The result is the atomic weight of the element, which is a weighted average of the masses of its isotopes based on their abundance in nature.
(mass of A) x (%of A) + (mass of B) x (% of B) + (mass of C) x (% of C)
(mass of A) x (%of A) + (mass of B) x (% of B) + (mass of C) x (% of C)
Element A has a greater atomic mass than element B if the atomic mass value of A is higher. The atomic mass represents the average mass of an element’s isotopes based on their abundance in nature.
(mass of A) x (%of A) + (mass of B) x (% of B) + (mass of C) x (% of C)
Each isotope of an element has a different Atomic Mass, so an average is taken of all the isotopes, but the average is weighted because the natural abundance (%) of each isotope is factored in. If hydrogen-1 is much more abundant than deuterium and tritium, then the weighted average will be closer to 1 than 2 or 3 but not a whole number. The following equation shows how percent abundance factors into the weighted average. (atomic mass A)(X% abundance) + (atomic mass B)(Y% abundance)...=(weighted average of all isotopes of the element)(100% abundance)
The atomic mass of boron-11 (B-11) can be calculated by setting up an equation based on the given average atomic mass of boron and the known atomic mass and abundance of boron-10 (B-10). Since the average atomic mass is a weighted average of the isotopes, the equation would be: (mass of B-10 * % abundance of B-10) + (mass of B-11 * % abundance of B-11) = average atomic mass of boron. Solving this equation will give you the atomic mass of B-11.
The element with isotopes of approximately 10 amu and 11 amu is boron (B) on the periodic table. Its isotopes include Boron-10 and Boron-11.
To calculate the atomic weight of an element with multiple isotopes, you multiply the mass of each isotope by its relative abundance (as a decimal), then sum the products. The result is the atomic weight of the element, which is a weighted average of the masses of its isotopes based on their abundance in nature.
The atomic mass of an element is based upon the weighted average of the atomic masses of the naturally occurring isotopes of that element. We need to know what isotopes of a given element occur in nature, and what percentage of the natural abundance of an element each of those naturally occurring isotopes represents. What that means might be more easily seen if we just run through some examples. So let's run through some.We have an element with 3 naturally occurring isotopes. They occur equally. That is, each one represents a third of the natural atoms, or 33 1/3 % of them. If the atomic masses of our isotopes are 17, 18 and 19, and all occur equally, then it's 17 + 18 + 19 (which equals 54) divided by 3 which is 54 divided by 3 which is 18. Our atomic mass for the element cited is 18, and that represents all the natural isotopes and the percentage in which they occur. Pretty easy.We have an element with 4 naturally occurring isotopes. Here are their atomic masses and percent abundance: a = 46 (25%), b = 47 (50%), c = 48 (15%), and d = 49 (10%). There is a (fairly) simple process one can use to find the weighted average. Multiply the percent natural abundance times the atomic mass of each one and add them all together. Not to difficult, is it? Roll up your sleeves and let's have at it.46 x 25% = 11.547 x 50% = 23.548 x 15% = 7.249 x 10% = 4.911.5 + 23.5 + 7.2 + 4.9 = 47.1 atomic mass unitsSo now we have our atomic mass for our mystery element in example two. We calculated it by determining a weighted average of the naturally occurring isotopes. And that's very important information to take to the lab where we'll be measuring (weighing) some of this stuff.
b. the atomic number is constant. The mass number is not constant because there are isotopes of an element that have the same number of protons but different number of neutrons. Likewise, there are isotopes of the same element with different atomic weights.
yes it does b/c the isotopes have different number of neutrons so therefore isotopes cause a change in mass.
Isotopes of the same element have the same number of protons but different numbers of neutrons in the nucleus. This results in variations in atomic mass. Isotopes can exhibit different physical properties and may have different levels of stability due to their differing numbers of neutrons.