[logan3]

The problem says:

"The ratio of atoms of Y to atom of X in a certain compound is 2:3. The mass of X in a certain sample is 2.50 g, and the mass of Y in the sample is 3.40 g. What is the ratio of their atomic masses?"

The answer given is as follows:

"Let N = number of atoms of Y; then 1.5N = number of atoms of X. A_y and A_x represent their atomic masses.

So

[tеx]\frac{NA_{y}}{1.5NA_{x}} = \frac{3.40g}{2.50g}[/tеx]

The Ns cancel, and the ratio of the atomic masses is A_y/A_x = 2/1"

I think I understand the reasoning, which is that the number of atoms multiplied by their atomic mass should equal the sample's mass. For example, N * A_y = 3.40 g, and 1.5 * N * A_x = 2.50 g; and the compound would equal  (N * A_y) + (1.5 * N * A_x) =  3.40 g + 2.50 g. But I guess I'm not understanding the final ratio of 2/1. The closest I can get is to take the equation

N(A_y)/1.5N(A_x) = 3.40 g / 2.50 g

cancel the Ns, and multiple by 1.5 to give:

A_y/A_x = 3.40 g *(1.5) / 2.50 g

But that equals 5.1 g/ 2.50 g, which is not the integer ratio of 2/1.

[Hunter2]

You use n = m/M

You have X₃Y₂ => 3 X + 2 Y

You take M = m/n   M(X) = 2,5 g/3 mol = 0,833333 g/mol and M(Y) = 3,4 g/2 mol = 1,7 g/mol

Ratio is 1,7/0,833333333333 =  2,04/1 ~ 2/1

[logan3]

We haven't been introduced to moles or molar mass yet. How do I understand this problem without referencing those concepts and units?

Also, I see that you did some rounding in the end, was that for significant figures? If so, maybe in the final sentence I posed "5.1 g / 2.50 g" = 2.04 / 1 was rounded to 2 / 1. But in my example I counted the s.f. and found that it was three s.f., not two, so why would it be rounded as such?

[Hunter2]

What is the ratio of their atomic masses?"

Strange why this question then, if not introduced.

[logan3]

They'll be introduced in four chapters. Right now we're going over the law of definite and multiple proportions, atomic masses, isotopes and the basic atomic structure.