Here it states that Iron possesses 2 naturally occurring isotopic forms. So we have Iron 54 with an isotopic mass of 53.939615 atomic mass units and Iron 56 with an isotopic mass of 55.934942 atomic mass units. What is the percent abundance of Iron 56? Alright. So here, the way we approach this question is:

Step 1: We make the first isotope's fractional abundance \( x \). And remember, together, they add up to 100% of all the iron that exists in the universe. That's its natural percent abundance. If we were to divide this by 100, we get 1. So we know that the first isotope is \( x \), so the second isotope must be \( 1 - x \), which is equivalent to 100% - \( x \) here. And we're just changing our percent abundance into our fractional abundance. That's where the \( 1 - x \) comes from.

Step 2: Now we're going to plug in the information that we know into the atomic mass formula. We know the atomic mass of iron based on the periodic table, and that's 55.845 atomic mass units. This is going to equal the isotopic mass of our first isotope, iron 54, which is 53.939615 AMU. We don't know its fractional, so it's \( x \). And plus the second isotope, 55.934942 atomic mass units, which would be \( 1 - x \). So what we're doing here is plugging your given variables into the atomic mass formula in order to isolate our missing variable.

We need to find the percent abundance of iron 56. So what we're gonna do here is we're gonna solve this mathematically. We have the first isotopic mass times \( x \), and then we're gonna distribute this isotopic mass to the one and to the minus \( x \). So our new equation becomes:

55.845 = 53.939615 x + 55.934942 - 55.934942 x

This number has an \( x \) variable. Combine them together:

55.845 = 53.939615 x - 1.995327 x + 55.934942

We need to isolate our \( x \) variable. Subtract \( 55.934942 \) from both sides. We’ll get:

- 0.089097 = - 1.995327 x

Divide both sides to isolate the \( x \) by \( -1.995327 \). We'll get \( x = 0.04508 \). But what exactly is this \( x \)? Well, this \( x \), if we go back up to the formula, represents the fractional abundance of our first isotope, giving us a fractional abundance of iron 54. But here, they're not asking us for information on iron 54. They want iron 56. So we need to find the fractional abundance of iron 56:

1 - 0.04508

This is \( 0.95492 \). To convert this fractional abundance to a percent abundance, multiply by 100:

This gives me 95.492%, which I'll just round to 95.5%. So this is a rough approximation of what the percent abundance will be of iron based on these isotopic masses that I've given to you. Alright, so working it out, we get this final answer.