Copper has two naturally occurring isotopes. Cu−63 has a mass of 62.939 amu and relative abundance of 69.17%. Use the atomic weight of copper to determine the mass of the other copper isotope
The atomic mass of an element (given in the periodic table) is the average of the atomic masses of its isotopes. It's not just a simple average, it's a weighted average.
From the periodic table, the atomic mass of copper is 63.546 amu. The problem states that there are only two isotopes of copper. One isotope accounts for 69.17% of the copper and has an atomic mass of 62.939 amu. The other isotope has an unknown atomic mass (let's cal it x) and accounts for the remainder of the copper. The remainder is 100% - 69.17% = 30.83% This means that by calculating a weighted average of the two isotopes, you have to get the atiomic mass that is listed on the periodic table. 69.17% (62.939 amu) + 30.83% (x) = 63.546 amu 0.6917 * 62.939 amu + 0.3083x = 63.546 amu
No. You need to find x in the equation. 63.546 amu is the average atomic mass of all copper. Solve this equation for x: 0.6917 * 62.939 + 0.3083x = 63.546
is it 4320.77463
Not even close. Where do you get these guesses from?
Since you are looking of the atomic mass of an isotope of copper, it must be close the the atomic mass of the given isotope and of the atomic mass given in the periodic table. The number has to be close to 63.546 and 62.939.
No. Instead of guessing, try solving the equation I gave you.
Solve this equation for x: 0.6917 * 62.939 + 0.3083x = 63.546 Do it one step at a time. What is 0.6917 * 62.939?
43.5349063
Good. 43.534906 + 0.3083x = 63.546 Now subtract 63.546 - 43.534906
20.0110937
Good. Now you have 0.3083x = 20.0110937 Now divide 20.0110937 by 0.3083 to find x. That is your answer.
64.9078615
\( \LARGE \color{red}{B} \color{blue}{I} \color{green}{N} \color{brown}{G} \color{purple}{O} !!!!!!!!!\)
That is the answer.
can you help with oone more
Enter the appropriate symbol for an isotope of phosphorus-32 corresponding to the isotope notation
ISNT IT \[\frac{ 30 }{ 15 }P\]
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