The elements X and Y form a compound that is 40% X and 60%Y by mass. The atomic mass of X is twice that of Y. What is the empirical formula of the compound?

Oh, it sounds like X and Y are quite the dynamic duo! With 40% X and 60% Y, it seems like Y is definitely pulling its weight. And with X having an atomic mass twice that of Y, we can say that X is the heavyweight in this compound.

Now, since we're looking for the empirical formula, we'll need to find the simplest whole number ratio between X and Y. To do that, let's pretend we have 100 grams of this compound.

We know it's 40% X, which means we have 40 grams of X. And since X has an atomic mass twice that of Y, we can infer that Y has an atomic mass half that of X. So, let's say X weighs 2 grams and Y weighs 1 gram.

In our pretend 100-gram compound, we have 40 grams (or 20 "X units") of X and 60 grams (or 60 "Y units") of Y. Simplifying these ratios gives us 2 X units to 3 Y units.

So, the empirical formula of this compound is X2Y3. X and Y are like the comedy duo Laurel and Hardy, always supporting each other!

To find the empirical formula of the compound, we need to determine the ratio of the elements X and Y in the compound. Given that the compound is 40% X and 60% Y by mass, we can assume a 100g sample.

Let's assume the atomic mass of Y is "y" and the atomic mass of X is "2y". Since the compound is 40% X and 60% Y by mass, the mass of X is 40g (40% of 100g) and the mass of Y is 60g (60% of 100g).

To determine the moles of X and Y, we need to divide the mass of each element by their respective atomic masses:

Moles of X = Mass of X / Atomic mass of X = 40g / (2y g/mol) = 20/y mol
Moles of Y = Mass of Y / Atomic mass of Y = 60g / (y g/mol) = 60/y mol

Since we need to find the empirical formula, we want to find the simplest whole-number ratio between the moles of X and Y.

Dividing both moles by the smaller value:
Moles of X: Moles of X / (20/y) = (20/y)/(20/y) = 1
Moles of Y: Moles of Y / (20/y) = (60/y)/(20/y) = 3

From the above calculations, the mole ratio between X and Y is 1:3. Therefore, the empirical formula of the compound is X1Y3.

To determine the empirical formula of the compound, you need to find the ratio of elements present in the compound. Here's how you can solve it step-by-step:

1. Assume we have a 100 gram sample of the compound. This assumption simplifies the calculations as percentages will then represent grams.

2. Let's say the mass of X in the compound is 40 grams, and the mass of Y is 60 grams.

3. Since the atomic mass of X is twice that of Y, you can assign the atomic mass of Y as 'x.' Therefore, the atomic mass of X is '2x.'

4. To find the number of moles for each element, divide their respective masses by their atomic masses:
- Moles of X = Mass of X / Atomic mass of X = 40g / (2x g/mol) = 20 / x moles of X
- Moles of Y = Mass of Y / Atomic mass of Y = 60g / (x g/mol) = 60 / x moles of Y

5. To simplify the ratio, divide both moles by the smallest number of moles. In this case, the smallest number of moles is 20 / x:
- Moles of X = (20 / x) / (20 / x) = 1
- Moles of Y = (60 / x) / (20 / x) = 3

6. Based on the moles, we can conclude that the ratio of X to Y in the empirical formula is 1:3.

7. Since we want the empirical formula, we use the ratio in its simplest form:
- Empirical formula = X1 Y3
- Combining both elements, the empirical formula of the compound is XY3.

Therefore, the empirical formula of the compound is XY3.

The mass of Y is 1.5 times the mass of X in the compound, as per the percentages.

If both molecules had the same atomic mass, they would also be present in this ratio in the compound.

However, the atomic mass of X is twice that of Y. So, for the same mass, X would have half the number of atoms as Y.

Ratio of atoms of Y to that of X = Ratio of mass given * Ratio of atomic masses
= 1.5 * 2
= 3

Hence, the compound is X(1)Y(3)