Bromine has two isotopes, Br- 79 and Br- 81. The relative atomic mass of bromine is 79.9.

CALCULATE THE PERCENTAGE of Br-79 atoms in a sample of bromine.

PLease help i'm really stuck

The total for 79 and 81 is 100% and the total for fraction is 1, so

let X = fraction 79 and 1-X = fraction of 81.
Then 79(X) + 81(1-X) = 79.9
Solve for X.

I dont get it the total is 160 and there are2 numbers so 2 fractions? and I dont get the x thing with 1-x AND 79(X)+81(1-X) iM VERY CONFUSED PLEASE HELP

To calculate the percentage of Br-79 atoms in a sample of bromine, we need to consider the relative abundance of each isotope.

Given that the relative atomic mass of bromine is 79.9, we can assign variables for the abundance of each isotope:
Let's assume x represents the percentage abundance of Br-79 and (100 - x) represents the percentage abundance of Br-81.

We can use these variables to set up an equation based on the average atomic mass:
(79.9) = (x/100) * 79 + ((100 - x)/100) * 81.

Now, we can solve this equation to find the value of x, which will give us the percentage of Br-79 atoms in the sample:

79.9 = (x/100) * 79 + ((100 - x)/100) * 81
7990 = 79x + 8100 - 81x
7990 - 8100 = -2x
-110 = -2x
x = -110 / -2
x = 55

Therefore, the percentage of Br-79 atoms in a sample of bromine is 55%.

To calculate the percentage of Br-79 atoms in a sample of bromine, you need to use the relative abundance of each isotope.

The relative abundance of an isotope is the percentage of that isotope present in a naturally occurring sample. In this case, we have two isotopes of bromine: Br-79 and Br-81.

Let's assume the relative abundance of Br-79 is x (in percentage). Since there are only two isotopes, the relative abundance of Br-81 would be 100 - x.

Given that the relative atomic mass of bromine is 79.9, we can set up the following equation:

(x/100) * 79 + ((100 - x)/100) * 81 = 79.9

Simplifying the equation, we get:

(79x + 8100 - 81x)/100 = 79.9

Now, let's solve for x:

79x + 8100 - 81x = 79.9 * 100
-2x = 7990 - 8100
-2x = -110
x = -110 / -2
x = 55

Therefore, the relative abundance of Br-79 is 55%. Hence, in a sample of bromine, 55% of the atoms would be Br-79.