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

Someone said:
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 are 2 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.
I need a simple clear explanation please.

It’s 55%

Don't worry, I'm here to help! Let's break it down step by step.

First, we need to find the percentage of Br-79 atoms in the sample of bromine.

We can assign the fraction of Br-79 atoms as X, and the fraction of Br-81 atoms as 1-X. This is because the sum of the fractions should be equal to 1, or 100%.

Now, we can set up an equation based on the relative atomic mass of bromine:

79(X) + 81(1-X) = 79.9

The left side of the equation represents the weighted average of the two isotopes, where 79(X) corresponds to the contribution of Br-79 atoms, and 81(1-X) corresponds to the contribution of Br-81 atoms. The right side of the equation represents the relative atomic mass of bromine.

To solve for X, we need to simplify the equation:

79X + 81 - 81X = 79.9
79X - 81X = 79.9 - 81
-2X = -1.1

Now, we can divide both sides of the equation by -2 to solve for X:

X = -1.1 / -2
X = 0.55

So, the fraction of Br-79 atoms (X) in the sample of bromine is 0.55, or 55%.

I hope that helps clear things up! If you have any more questions, feel free to ask.

Sure! I'll try to explain it in a simpler way.

Let's break it down step by step:

1. The first thing we need to understand is that the relative atomic mass is a weighted average of the masses of different isotopes of an element. In this case, bromine has two isotopes: Br-79 and Br-81.

2. We want to calculate the percentage of Br-79 atoms in a sample of bromine. Let's assume that the fraction of Br-79 atoms is represented by X.

3. Since there are only two isotopes, the fraction of Br-81 atoms can be represented by (1 - X), which means the remaining fraction is Br-81.

4. Now, we need to set up an equation using the information we have. We know that the sum of the masses of Br-79 and Br-81 isotopes should give us the relative atomic mass of bromine, which is 79.9.

The equation is: (mass of Br-79)(X) + (mass of Br-81)(1 - X) = 79.9

5. Now, we know the masses of Br-79 and Br-81 isotopes are 79 and 81, respectively. Plugging those values into the equation, we get:

79(X) + 81(1 - X) = 79.9

6. Simplify the equation. Expand the terms:

79X + 81 - 81X = 79.9

7. Combine like terms:

-2X + 81 = 79.9

8. Move the constant term to the right side:

-2X = 79.9 - 81

9. Simplify:

-2X = -1.1

10. Divide both sides by -2 to solve for X:

X = -1.1 / -2

X = 0.55

11. Now, we need to convert the fraction X into a percentage to find the percentage of Br-79 atoms:

Percentage of Br-79 atoms = X * 100

Percentage of Br-79 atoms = 0.55 * 100

Percentage of Br-79 atoms = 55%

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

is it 40.1 percent??

assume you have x percent of 79. Then you have 100-x percent of 81, and you have 100 percent of the mixture.

x*79+(100-x)81=79.9*100
solve for x, then 100-x.
Of course there are two fractions, each isotope has a differing content. Now if you cant solve that equation above, you have very serious chemistry issues ahead, this is algebra I.