The molecular weight of K₂Cr₂O₇ is 2(39 g/mol) + 2(52 g/mol) + 7(16 g/mol) = 294 g/mol. 

Out of this total weight, the percent composition of K is: (2(39 g/mol)/294 g) = 26.5%

An empirical formula represents a molecule with the simplest ratio. C₆H₆, ethyene (C₂H₂) and C₁₀₀H₁₀₀ can all be reduced to CH, however, C₃H₄ cannot because it does not have C and H atoms in a 1 : 1 ratio.

If we assume that we have a sample of 100g of the compound, then that would mean we have 37.47 grams (3.12 moles) of carbon, 12.61 grams (12.48 moles) of hydrogen, and 49.92 grams (3.12 moles) of oxygen.

The molar ratio of C:H:O is 3:12:3, reducing to 1:4:1, giving the empirical formula of CH₄O.

To solve this problem, we will consider a sample that is 100g. We can assume that 75g of that sample is carbon and 25g of that sample is hydrogen.

First, we will determine the molar ratio of carbon to hydrogen.

Now we will determine the whole number ratio.

The correct answer is . We found that element Y is aluminum due to its discrepancy between IE₃ and IE₄. Aluminum contains three valence electrons, and it needs to shed all three to reach a noble gas configuration. This makes it Al(III), or Al³⁺. Oxygen must gain two electrons to achieve a noble gas configuration. This makes it O^2-. The only way to balance the opposing charges of the aluminum oxide is to have two aluminum ionically bonded to three oxygen.

Aluminum: charge +3; Oxygen charge: -2

2(+3) + 3(-2) = 0

When finding empirical formulas, take the percentages of each atom and convert them to grams by imagining a 100-gram sample of the compound.

Next, divide each mass by the molar mass of the atom.

Finally, divide each number by the smallest number out of the three. Remember that ratios must be a whole number.

Since dividing the number for oxygen by the smallest number gives a value of 2.5, we must multiply each atom by 2 in the empirical formula.

This results in an empirical formula of .

In order to find the molecular formula, we must first find the empirical formula. We start by imagining a sample of the compound weighing 100 grams, so the percentages can be seen as grams. 25g of the sample is carbon, 8.3g of the sample is hydrogen, and 66.7g of the sample is oxygen.

Then, each of the atoms' masses is divided by the atom's molar mass, the numbers are then compared to each other in order to find the ratio of atoms.

There is a ratio of two carbon to eight hydrogen to four oxygen, which reduces to a 1:4:2 ratio between carbon, hydrogen, and oxygen respectively.

As a result, the empirical formula is . Since the molecular formula is three times as heavy, we simply multiply each portion of the empirical formula by three. The molecular formula is .

To calculate the percent mass of a particular element within a substance I like to use a formulaic approach.  

First multiply the number of moles of the atom of interest (C) by its molar mass.

3moles* 12 g/mol= 36g

Then divide by the overall molar mass of the molecule, obtain this by adding the atomic weights of each atom present in the molecule:

36g/58g= 62%

The formula for aluminum hydroxide is:

To find the mass percentage, divide the total molar mass of oxygen by the formula mass of aluminum hydroxide.

Convert the ratio to a percentage to get our final answer.

To find the empirical formula, we have to determine a mole ratio of the elements in the substance, which we can get by dividing the mass percents of the elements by their respective atomic masses. Approximating:

To get an empirical formula we want to have whole number ratios, so we have to find a common whole number multiple or dividend that will make all of these pieces whole. A common strategy is to divide everything by the smallest number (in this case, 2.22).

We can now write our empirical formula using the above ratio for our subscripts:

To find the molecular formula, we need to use the molar mass. We are given that 100 grams of this substance is equal to 1.1 moles; dividing grams by moles gives a molar mass of 91 grams per mole.

The molar mass of is 45 grams per mole.

We divide the molar mass of the actual substance by the molar mass of the empirical formula:

Hence, 2 is the constant multiplier we apply to every subscript in our empirical formula, obtaining the final molecular formula: