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The relative of carbon is 12.011, which is extremely close to 12.0. This means that the masses C-13, and C-14 are practically negligible when contributing to the relative atomic mass of carbon.
In fact, the C-12 isotope makes up 98.9% of carbon atoms, C-13 makes up 1.1% of carbon atoms, and C-14 makes up just a trace of carbon atoms as they are found in nature.
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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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Calcium has three different isotopes. One has a mass of 35.00 amu; another has a mass of 41.00 amu; and another has a mass of 40.00 amu. Which isotope is the most abundant of the three?
It follows that whatever atomic mass is quoted on the Periodic Table, the most abundant isoptope of that element must make the greatest statistical contribution. Calcium has an atomic mass...
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Gallium has two naturally occurring isotopes. The mass of gallium-69 is 68.9256 amu and it is 60.108% abundant. The mass of gallium-71 is 70.9247 amu and it is 39.892% abundant. What is the atomic mass of gallium?
To find the average ($"A"_r$) of an element, given the percentage rarities of its , simply multiply each isotope's $"A"_r$ by its percentage that is expressed as a decimal...
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One isotope of bromine has an atomic mass of 78.92 amu and a relative abundance of 50.69%. The other major isotope of has an atomic mass of 80.92 amu and a relative abundance of 49.31%? What is the average atomic mass of bromine?
Multiply the atomic mass of each isotope times its percent abundance in decimal form. Add them together. $"Average atomic mass of Br"$$=$$(78.92xx0.5069)+(80.92xx0.49331)="79.92 u"$
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Argon has three naturally occurring isotopes: argon-36, argon-38, and argon-40. Based on argon's reported atomic mass, which isotope do you think is the most abundant in nature?
The mass of Argon on is 39.948. The periodic table's is the AVERAGE weight of ALL its . If one isotope is MORE abundant than the others, the average will...
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The atomic weight of a newly discovered e ement is 98.225 amu. It has two naturally occuring isotopes. One isotope has a mass of 96.780 amu. The second isotope has a percent abundance of 41.7%. What is the mass of the second isotope?
$M_r=(sum(M_ia))/a$, where: $M_r$ = relative attomic mass ($g$ $mol^-1$) $M_i$ = mass of each isotope ($g$ $mol^-1$) $a$ = abundance, either given as a percent or amount of $g$...
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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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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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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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