1 Answers
Call
There are 11 total parts in this ratio. So the total mass needs to be divided by 11.
6 parts are of isotope 80 so the mass of these parts is
5 parts are of isotope 100 so the mass of these parts is
The total mass units are the sum of the 11 parts.
480 + 500 = 980 mass units.
To find the average divide the total by 11
0
like
0 dislike
Talk Doctor Online in Bissoy App
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...
1 Answers 1 views
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"$
1 Answers 1 views
A sample of element X contains 100 atoms with a mass of 12.00 and 10 atoms with a mass of 14.00. What is the average atomic mass (in amu) of element X?
The average of an element is calculated by taking the weighted average of the atomic masses of its stable . In other words, each stable isotope will contribute to...
1 Answers 1 views
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 $g$...
1 Answers 1 views
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...
1 Answers 1 views
What is the average atomic mass of an element if, out of 100 atoms, 5 have a mass of 176 amu, 19 have a mass of 177 amu, 27 have a mass of 178 amu, 14 have a mass of 179 amu, and 35 have a mass of 180 amu?
The $"Average Atomic Mass"$ of an element is defined as "the weighted average mass of all naturally-occurring (occasionally radioactive) of the element." (and hence the name "average") [1] Dividing the...
1 Answers 1 views
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...
1 Answers 1 views
Naturally occurring boron is 80.20% boron-11 (atomic mass of 11.01 amu) and 19.80% of some other isotopic form of boron. What must the atomic mass of this second isotope be in order to account for the 10.81 a average atomic mass of boron?
Background Info Most elements have different isotopes, or atoms with the same number of protons, but different numbers of neutrons. Hence, it is possible to have two atoms that are...
1 Answers 1 views
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 $""^63A$...
1 Answers 1 views
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 $chi$.... $chi=(117.3-0.6543xx119.8)/(0.3457)=112.6*"amu"$
1 Answers 1 views