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?

Md Ariful Islam

0 like 0 dislike
1 views
Share with your friends

1 Answers

Md Ariful Islam

Call

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 the average mass of the element proportionally to its abundance.

$color(blue)("avg. atomic mass" = sum_i ("isotope"_i xx "abundance"_i)$

To calculate an isotope's decimal abundance, you need to divide the number of atoms of that isotope by the total number of atoms present in the sample.

In your case, you know that the sample contains $100$ atoms with an atomic mass of $"12.00 u"$ and $10$ atoms with an atomic mass of $"14.00 u"$.

The total number of atoms in the sample will thus be

$100 + 10 = "110 atoms"$

The decimal abundances for the two will be

$(100 color(red)(cancel(color(black)("atoms"))))/(110color(red)(cancel(color(black)("atoms")))) = 0.90909 ->$ for the first isotope

$(10color(red)(cancel(color(black)("atoms"))))/(110color(red)(cancel(color(black)("atoms")))) = 0.090909 ->$ for the second isotope

This means that the average atomic mass of the element will be equal to

$"avg atomic mas" = "12.00 u" xx 0.90909 + "14.00 u" xx 0.090909$

$"avg. atomic mass " = " 12.18181 u"$

I think I'll leave this rounded off to four , the number of sig figs you have for the atomic masses of the two isotopes

$"avg. atomic mass " = color(green)(" 12.18 u")$

0 like 0 dislike
1 views
Talk Doctor Online in Bissoy App
The atomic weight of a newly discovered element is 110.352 amu. It has two naturally occurring isotopes. One has a mass of 111.624 amu. The other has an isotopic mass of 109.75 amu. What is the percent abundance of the last isotope (109.75 amu)?

The idea here is that each isotope will contribute to the average of the element proportionally to their respective abundance. Now, the key to this problem lies in how you...

1 Answers 1 views
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
What is the average atomic mass of titanium if we have an isotopic distribution of $78.5%$ of $""^46Ti$, $45.95263*"amu"$, $12.3%$ of $""^48Ti$, $47.94795*"amu"$, and $""^50Ti$, $49.94479*"amu"$?

$"Atomic mass of titanium"={0.785xx45.95263+0.123xx47.94795+0.092xx49.94479}*"amu"=??$ Is this the same as on this planet?

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
The atomic mass of Cl is 35.45 amu and the atomic mass of Al is 26.98 amu. What are the masses in grams of 4.000 mol of Al atoms and of 4.000 mol of Cl atoms?

The molar mass of an element is its atomic weight (relative atomic mass) on in g/mol. Molar masses $"Cl":$$"35.45 g/mol"$ $"Al":$$"26.98 g/mol"$ The molar mass is the mass of one...

1 Answers 1 views
An element has two naturally occurring isompes widx atomic masses of 10.01 amu and 11.01 amu. The relative abundances of these elements are 19.8% and 80.2%, respectively. What is the average atomic mass of the element?

Now I hope you have been exposed to the concept of weighted average; If not here is a video Now lets get our data; Lets call our element...

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
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: $(62.93xx69.2%+64.93xx30.8%)*"amu"$ $=$ $63.55$ $"amu"$

1 Answers 1 views
What is the average atomic mass of an element composed of a series of isotopes, $X, Y, Z$ with respective masses, $18.473*"amu", 19.962*"amu", 21.469*"amu"$, and respective abundances, $X=28.812%$, $Y=28.757%$...?

$"Average atomic mass"={18.473_"X"xx28.812%+19.962_"Y"xx 28.757% +21.469_"Z" xx(100-28.812-28.757)%}*"amu"$ $-=20.15*"amu"$...if I have done my 'rithmetic right... How did I get the percentage abundance of $Z$? And it is likely that we got the...

1 Answers 1 views
X