An empirical formula is the simplest form of a molecular formula that still retains the ratio of the elements. If a formula can be divided by a whole number, it is a molecular formula and not an empirical formula. A molecular formula is the exact identity of a compound, showing the total number of atoms used to create the compound.

The only given answer that is fully divisible by a whole number is . This means it is not an empirical formula. Any given formula, however, will be a molecular formula, provided that the atoms are capable of forming the appropriate bonds. This is the molecular formula for glucose, which reduces to the empirical formula of . is the empirical formula for a generalized carbohydrate molecule.

To find the empirical formula, use the mass percentage of each element to find mole ratios based on a hypothetical sample of .

We see that there is a 1:1 mole ratio for carbon to hydrogen, making the empirical formula .

The next step will be to find molecular formula by dividing the molar mass of the compound by the molar mass of the empirical formula.

Multiply the subscripts of the empirical formula for each element by six to get the molecular formula: .

To find the empirical formula, we must first find the mole ratios by using molar masses for each element:

Rounding these numbers, we get five moles of carbon, ten moles of hydrogen, and two moles of oxygen, making the formula . The formula cannot be simplified by any common denominator, and thus represents the final empirical formula.

We know the molecular mass, given in the question. To find the molecular formula, we need to find the ratio of the mass of the empirical formula to the molecular mass.

The mass of the empirical formula is equal to the molecular mass, meaning that the empirical and molecular formulas are the same.

The molecular formula is the same as the empirical formula if it cannot be reduced by any whole number. Any formula containing a single atom of any given element must be an empirical formula as well. The formula for DEET is . Since this contains a single atom each of oxygen and nitrogen, it cannot be further reduced and must be an empirical formula as well.

The molecular formula for ribose is , which can be reduced by a factor of five. The empirical formula for ribose (and most other carbohydrates) is .

The other two options require us to calculate pass percentages based on the given molecular formulas.

Although we can look at the formulas for chlorophyll and ethyl butyrate to deduce that oxygen makes up a larger percentage of the latter, we can double check mathematically. In order to find which compound contains oxygen in a larger percentage, divide the molar mass of the oxygen in the compound by the molar mass of the entire compound.

For ethyl butyrate:

For chlorophyll:

We see that it is in fact true that there is a larger percentage of oxygen in ethyl buyrate.

Next, find out if hydrogen and carbon make up the largest mass percentage of ribose by using the same method:

This amounts to 46.7% of the molecular mass, meaning that oxygen must account for the remaining 53.3%. Oxygen thus makes up a greater mass percentage of ribose than hydrogen and carbon combined.

Given the percentages by mass of the compound, we can convert the percentages to grams by considering a 100-gram sample of the compound. The next step involves dividing each mass by the element's molar mass.

By dividing each value by the smallest molar quantity, we can determine the ratio of elements in the compound.

As a result, the empirical formula for the compound is .

Given that the molar mass of oxygen is about 16g, and molar mass of manganese is about 55g,  contains 165g of manganese and 64g of oxygen, for a total of 229g.

165g/229g = 0.72

So, the ratio of manganese to oxygen in this compound is 72% manganese by mass.

In order to find the mass percentage of an atom in a molecule, start by finding the total mass of one mole of the molecule. Glucose has 180 grams per mol.

Next, we determine the mass of the carbon atoms in one mole of the molecule. One carbon mole has a mass of 12 grams. Multiplied by the six carbon atoms in glucose gives a mass of 72 grams.

Finally, we divide the mass of carbon by the mass of the molecule.

So, 40% of glucose's mass is made up of carbon.

Aluminum oxide has the formula .

Aluminum has a molecular weight of , and oxygen has a weight of . Using these values, we can calculate the molecular weight of aluminum oxide.

The mass percentage is given by the mass of aluminum divided by the total molecular weight.

Sodium sulfate is given by the formula:

To find the percentage by weight, we will need to divide the mass of sodium in the molecule by the total molecular mass.

Convert the ratio to a percentage.

Start by calculating the percentages of hydrogen in each compound by using molar mass ratios.

Next, take the percentages of the given sample masses to determine the total mass of hydrogen in each.

We see that there is a larger mass of hydrogen in the hydrogen cyanide sample.