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Then, you do the multiplied by its percentage, and add those two together and you'll get the of that element. If there are more isotopes, just use the same method for all of the isotopes.
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An element has three naturally occurring isotopes with the following masses and abundances. How do you calculate the atomic mass of this element? What is the identity of the element?
The atomic mass is the weighted average of the atomic masses of each isotope. In a weighted average, we multiply each value by a number representing its relative importance. In...
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Oxygen is composed of three isotopes 16/8 O (15.995 u), 17/8 O (16.999 u) and 18/8 O (17.999 u). One of these isotopes, 17/8 O, comprises of 0.037% of oxygen. What is the percentage abundance of the other two isotopes, using the average atomic mass of 15.9994 u.
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 $""_8^17"O"$...
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Vanadium has two naturally occurring isotopes, 50-V with an atomic mass of 49.9472 amu and 51-V with an atomic mass of 50.9440. The atomic weight of vanadium is 50.9415. What is the percent abundance of the vanadium isotopes?
The idea with that have two naturally occurring is that the percent abundances of those two must add up to give $100%$. In calculations, it is often easier to...
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The element copper has naturally occurring isotopes with mass numbers of 63 and 65. The relative abundance and atomic masses are 69.2% for a mass of 62.93 amu and 30.8% for a mass of 64.93 amu. What is the average atomic mass of copper?
$""^63Cu$ has $69.2%$ abundance. $""^65Cu$ has $30.8%$ abundance. So, the weighted average is $62.93xx69.2%$ $+$ $64.93xx30.8%$ $=$ $63.55$ $"amu"$. If we look at , copper metal (a mixture of isotopes...
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All isotopes of a particular element have the same atomic number. How then do the isotopes of a particular element differ?
: The number of protons in an atom, NOT the number of neutrons Isotope: An element with a varying number of neutrons Remember, we CANNOT change the number of protons...
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An average sample of chlorine contains 75.77% chlorine-35, with an atomic mass of 34.969 amu (atomic mass units), and 24.23% chlorine-37, with an atomic mass of 36.966 amu. What is the average atomic mass of chlorine?
The weighted average is the sum of each individual nuclide, multiplied by its isotopic abundance. So, we simply do the arithmetic: $0.7577xx34.969+0.2423xx36.966 = ???$ amu Then look at to...
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2. The element copper has naturally occurring isotopes with mass numbers of 63 and 65. The relative abundance and atomic masses are 69.2% for a mass of 62.93 amu and 30.8% for a mass of 64.93 amu. How do you calculate the average atomic mass of copper?
And thus the $"weighted average"$ is: $(62.93xx69.2%+64.93xx30.8%)*"amu"$ $=$ $63.55$ $"amu"$
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Why is it necessary to use the average atomic mass of all isotopes, rather than the mass of the most commonly occurring isotope, when referring to the atomic mass of an element?
Why? Because accounting, and even more so chemistry, is a highly quantitative exercise. Mass is always conserved in a chemical reaction, and accurate masses, which are the weighted averages of...
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The atomic mass of element A is 63.6 atomic mass units. The only naturally occurring isotopes of element A are A-63 and A-65. What is the percent abundance in a naturally occurring sample of element A closest to?
The average isotopic mass is the weighted average of the mass of the individual . Because the quoted average, $63.6$ $"amu"$, is closer to $""^63A$ than $""^65A$, the $""^63A$...
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A fictitious element that has two isotopes has an average atomic mass of 117.3. If 65.43% of the atoms of the element have an atomic mass of 119.8, what is the atomic mass of the other isotope?
The weighted average of the is equal to $117.3*"amu"$. And thus.........with units of $"amu"$ $[65.43%xx119.8+34.57%xxchi] =117.3$ Where $chi-="mass of the other isotope"$.... And so we solve for $chi$.... $chi=(117.3-0.6543xx119.8)/(0.3457)=112.6*"amu"$
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