In order to calculate the empirical formula of vanillin from the given percent composition data, we can pretend we have a 100 g sample vanillin and from there estimate how many grams of each element make up the sample. For example, Vanillin is  carbon; therefore, the estimated sample would contain 66.3 g of carbon. After doing this for each of the elements vanillin contains, we can use the atomic mass of carbon to find how many moles of carbon make up the sample. We then do the same for each element that composes vanillin. See calculations below:

After finding the number of moles of each element in the hypothetical 100 g sample, we look at the ratio between the moles of each element by dividing all the samples by smallest number of moles. After finding the ratios of the samples, we are able to use the empirical formula of the vanillin by rounding up to the nearest whole number for each sample. If the ratios are not close enough to round up (like in this example), we multiply the ratios all by the same number until each number is equal to a whole number nicely. See calculations below:

We need whole numbers, so multiply each of the resulting numbers by 3 to get rid of the fractions.

This gives us  as the final answer.

To find an empirical formula, take a molecular formula and divide the subscript of each element by the greatest common factor of all the subscripts. In this case, the only pair that works is , which can be verified by dividing the coefficients of the molecular formula by 6. Note that  is not correct because neither of the formulas is an empirical formula; they are both possible molecular formulas for a compound with the empirical formula .

First, calculate the molar mass of heptane, which is

We can calculate the percentage of carbon by mass in heptane by creating a fraction of the mass of carbon over the mass of the entire molecule and converting it into a percentage:

For every 100 g of heptane, 84 g must come from carbon. This means that the percentage of carbon by mass is 

Note that there are two nitrate () ions for every formula unit of calcium nitrate. This means that in one mole of calcium nitrate, there are two moles of nitrate ions. To go from moles to ions, multiply 2 by Avogadro's number. 

In one molecule of , there are 2 ammonium ions, each of which has 4 hydrogen atoms for a total of 8 hydrogen atoms. Thus, to find the number of hydrogen atoms in one mole of , one must multiply Avogadro's number by 8.

In order calculate the number of aluminum ions, we must first find the number of aluminum ions in the entire compound. In this case, there are two molecules of   for every molecule of . After understanding this, we can use Avogadro's constant to determine the number of atoms (more specifically ions) of aluminum. See equation below for specific calculations.

The formula for aluminum chloride is . The molar mass of  is 26.98 g and the molar mass of  is 35.45 g. There are three  atoms in . To calculate the molecular mass of , we need to find the sum of the mass of one aluminum atom and three chlorine atoms:

The total molecular weight is 130.33 g. Starting with the grams of , we convert to the amount of moles  using the molecular weight value we just calculated. Then we calculate the number of moles of chloride ion in . using the 3:1 ratio. Finally, we calculate the number of atoms by using Avogadro's number. See calculations below:

The coefficients of the equation, when balanced, should be 2, 13, 10, and 8. This problem wants to know the sum of the reactants, so only the coefficients on the left side of the reactants should be added. The sum of 2 and 13 is 15, so 15 is the correct answer.

First, convert from grams to moles using the molar masses of the compounds. There are 4 mol of hydrogen gas and 0.5 mol of oxygen gas. Oxygen gas is clearly the limiting reagent (since there is more than twice as much hydrogen gas as oxygen gas), which means that the theoretical yield of water is

Water contains two hydrogen atoms and an oxygen atom, so it has a molar mass of

Hence, 1 mol of water would have a mass of 18 g.

First, the equation should be balanced. This can be done with the coefficients 2, 25, 16, and 18.

Next, calculating the molar mass of octane shows that it has a molar mass of 114 g/mol. This means that 1 mole of octane is used in the reaction, and 9 times as many moles of water should be produced, meaning one of the products will be 9 mol of water. Water has a molar mass of 18 g/mol, so the total mass of water produced is .