Oxygen is composed of three isotopes: One has a a mass of 16.999 amu. This isotope makes up 0.037% of oxygen. Of the other two, one has a mass of 15.995 amu, and the other has a mass of 17.999 amu. Calculate the abundance of the other two isotopes, using the average atomic mass of 15.9994 amu.
Answers
Answer: The abundance in % for other two isotopes of the oxygen is 99.762% and 0.201%
Explanation :
The average atomic mass of an element is the total of isotopic mass times the percent abundance for all the isotopes.
Oxygen has 3 isotope, 160, '70 and 180.
The average atomic mass of oxygen can be calculated as follows.
Average Atomic Mass = (Isotopic mass of 160 x % abundance of 160) + (isotopic mass of "0 x % abundance of "O) + (Isotopic mass of 180 x % abundance of 180)
17 The mass of "O = 16.999 amu and its % abundance is 0.037% which is 0.00037
The mass of 160 = 15.995 amu. Let us assume its % abundance as a
The mass of 180 = 17.999 amu
The total % abundance for all the isotopes is 100% or 1.
Therefore % abundance for 180 would be 1- (a + 0.00037) = 0.99963 a =
Let us plug in the above values in the average atomic mass formula.
Average atomic mass = ( 15.995)(a) + = (16.999)(0.00037) + (17.999)(0.99963 - a)
The average atomic mass of element O is 15.9994
Therefore we have,
15.9994 = 15.995a + 0.0062896 + 17.99234 - 17.999a
15.9994 = -2.004a + 17.99863
-1.99923 = -2.004a
a = 0.99762
a = 99.762%
The percent abundance of 16 O is
99.762%
The percent abundance of 180 is 0.99963 - 0.99762 = 0.00201 or 0.201%
50.
Explanation:
The abundance of #""_8^16"O"# is 99.762 %, and the abundance of #""_8^18"O"# is 0.201 %.
Assume you have 100 000 atoms of O. Then you have 37 atoms of #""_8^17"O"# and 99 963 atoms of the other isotopes.
Let x = the number of atoms of #""_8^16"O"#. Then the number of atoms of #""_8^18"O"# = 99 963 - x
The total mass of 100 000 atoms is
x × 15.995 u + (99 963 – x) × 17.999 u + 37 × 16.999 u = 100 000 × 15.9994 u
15.995 x + 1 799 234.037 – 17.999 x + 628.963 = 1 599 940
2.004 x = 199 123
x = 199 123/2.004 = 99 762
So there are 99 762 atoms of #""_8^16"O"# or 99.762 %.
The number of #""_8^18"O"# atoms is 99 963 – 99 762 = 201 atoms or 0.201 %.
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