We began a discussion about chemical formulas with a chemical formula...C6H12O6. When I wrote this formula on the board in class I asked what you could tell me about the formula. Some said it is sugar, which was correct. I did not say it was a formula so someone said it was a formula which contained 6 carbon atoms, 12 hydrogen atoms and 6 oxygen atoms. Then someone said the molar mass of the compound was 180 grams per mol. Then we recognized it was a covalent compound. So lots of interesting details from just a formula.

But what is a formula? A formula indicates the ratio of moles of atoms in the compound. In C6H12O6 there are 6 carbon atoms, 12 hydrogen atoms and 6 oxygen atoms. We can also say in 1 mol of C6H12O6 there are 6 mol of carbon atoms, 12 mol of hydrogen atoms and 6 mol of oxygen atoms. The next thing about this compound I would like to know is the percent (by mass) composition of C6H12O6. That is I want to know the percentage of each of the elements in the compound, the percentage by mass.

To determine this we will assume some amount of C6H12O6. How much should we use to solve this problem? The easiest amount to use is 180 grams of C6H12O6. 180 grams of C6H12O6 is a nice number to assume because it is the molar mass of the compound. By assuming the molar mass, which we obtained by adding the atomic masses of the elements in the compound it is an easy task to determinet he percent (by mass) composition of the elements in the compound. The molar mass was determined by summing...

6 x atomic mass of C = 6 x 12 u

+12 x atomic mass of H = 12 x 1.01 u

+6 x atomic mass of O = 6 x 16 u

or 180 u

If we use the units grams instead of u's 180 grams is the mass of 1 mol of C6H12O6. And 72 grams of the 180 grams are grams of carbon, 12.1 grams are grams of hydrogen and 96 grams are grams of oxygen. Now we have a very easy way of determining the percentage composition of the elements in C6H12O6.

The easiest way to solve this type of problem is to assume a 1 mol of the compound and a mass equal to the molar mass of the compound, then you can use the masses of the atoms in the numerator and the molar mass in the denominator.

Determine the percentage composition of each element in Ni3(PO4)2.

The molar mass of Ni3(PO4)2 is obtained by summing,

3 x atomic mass of Ni = 3 x 58.7 g = 176 g

2 x atomic mass of P = 2 x 31 g = 62 g

8 x atomic mass of O = 8 x 16 g = 128 g

Total = 366 g

Do the calculation and click here when you have your answer.

Now lets consider a slightly different problem.

Determine the formula of a compound which is 40.0% carbon, 6.72% hydrogen and 53.3% oxygen.

To solve this problem the most important thing is to remember what a formula is...? It is the ratio of moles of the elements in the compound! Knowing this we can answer this question. We need to determine the mol of each of the elements in this compound, i.e, mol of carbon, mol of hydrogen and mol of oxygen. But how do we do that just given the percentage composition? We do it by assuming an amount of grams of the compound, and since I gave you the pecent composition of each elemetn the easiest amount to assume is 100 g!

Because by assuming 100 g of the compound you would have 40.0 g of carbon, 6.72 grams of hydrogen and 53.3 grams of oxygen. And now that we have grams of each of the elements it is easy to calculate the number of moles of each element.

These are the mol of each element in the sample of the compound. So now what do we do with these mol? The next step is to take the mol and find the simplest ratio of the mol of the elements. The formula is the ratio of the mol of the elements and so we need to determine the simplest ratio.

To do this we must divide each of the mol by the smallest number of mol, that will give us our simplest ratio.

Which is, 1 mol C : 2 mol H : 1 mol O. So the simplest ratio is CH2O. Why didn't we get C6H12O6? Because this type of calculation will only reveal the simplest, or empirical formula of the compound. To get the actual formula we must know molar mass of the compound. In this case the molar mass is 180 grams. The empirical mass is 30 grams. By determining the factor which relates the empirical mass to the molar mass we can determine the molecular formula of the compound from the empirical formula. This is done the following way,

empirical mass x n = molecular mass

30 g x n = 180 g

n = 6

This tells us that the molar mass is six times the empirical mass, for that to happen the actual formula of the compound is 6 times the empirical formula, (CH2O)6 or C6H12O6.

 

Determine the formula of a compound which is 26.6% potassium, 35.4% chromium and 38.1% oxygen.

To solve this problem the most important thing is to remember what a formula is...? It is the ratio of moles of the elements in the compound! Knowing this we can answer this question. We need to determine the mol of each of the elements in this compound, i.e, mol of potassium, mol of chromium and mol of oxygen. But how do we do that just given the percentage composition? We do it by assuming an amount of grams of the compound, and since I gave you the pecent composition of each element the easiest amount to assume is 100 g!

Because by assuming 100 g of the compound you would have 26.6 g of potassium, 35.4 grams of chromium and 38.1 grams of oxygen. And now that we have grams of each of the elements it is easy to calculate the number of moles of each element.

These are the mol of each element in the sample of the compound. So now what do we do with these mol? The next step is to take the mol and find the simplest ratio of the mol of the elements. The formula is the ratio of the mol of the elements and so we need to determine the simplest ratio.

To do this we must divide each of the mol by the smallest number of mol, that will give us our simplest ratio.

Which is, 1 mol K : 1 mol Cr : 3.5 mol O. But we can not express the formula of a compound with anything but whole numbers! The formula gives us the ratio of atoms, so we can not have a fraction of an atom. So what do we do with the ratio 1 mol K : 1 mol Cr : 3.5 mol O? Well, if we multiply each number by 2 we get a ratio of simple whole numbers...2 mol K : 2 mol Cr : 7 mol O. The formula for this compound is K2Cr2O7.

 

Determine the formula of a compound which is 81.8% carbon, and 18.2% hydrogen.

Begin by assuming 100 g of the compound you would have 81.8 g of carbon, and 18.2 grams of hydrogen. Having grams of each of the elements it is easy to calculate the number of moles of each element.

These are the mol of each element in the sample of the compound. So now what do we do with these mol? The next step is to take the mol and find the simplest ratio of the mol of the elements. The formula is the ratio of the mol of the elements and so we need to determine the simplest ratio.

To do this we must divide each of the mol by the smallest number of mol, that will give us our simplest ratio.

Which is, 1 mol C : 2.67 mol H. But we can not express the formula of a compound with anything but whole numbers and we can not round 2.67 to 3.00! Well, if we multiply each number by 3 we get a ratio of simple whole numbers...3 mol C : 8 mol H. The formula for this compound is C3H8.

 

A sample of iron oxide weighs 0.0589 grams is found to contain 0.0410 grams of iron. Determine the empirical formula of this compound.

In this case we do not have to assume 100 grams of the compound because we already have the information in grams. The mass of iron is given to us. We are told it is iron oxide so we must determine the mass of oxygen in the compound. we can obtain the mass of oxygen by suybtracting the mass of iron from the mass of iron oxide. When we do this the mass of oxygen is 0.0589 g - 0.410 g = 0.0179 g of O. Having the mass of all of the elements we can determine mol.

These are the mol of each element in the sample of the compound. So now what do we do with these mol? The next step is to take the mol and find the simplest ratio of the mol of the elements. The formula is the ratio of the mol of the elements and so we need to determine the simplest ratio.

To do this we must divide each of the mol by the smallest number of mol, that will give us our simplest ratio.

Which is, 1 mol Fe : 1.52 mol O. But we can not express the formula of a compound with anything but whole numbers and we can not round 1.52 to 2.00 or to 1! Well, if we multiply each number by 2 we get a ratio of simple whole numbers...2 mol Fe : 3 mol O. The formula for this compound is Fe2O3.