A vanadium oxide compound is found to be 68 percent vanadium and 32 percent oxygen by mass. What is the empirical formula of this compound? Take the atomic masses of vanadium and oxygen to be 51 and 16, respectively.
The empirical formula is the formula of a compound with the simplest whole number ratio of atoms. For example, C4H10 has the empirical formula C2H5, and C3H6 has the empirical formula CH2. Multiple compounds can have the same empirical formula. For example, C4H8 would also have the empirical formula CH2. We have been told in the question that we want to find the empirical formula for a vanadium oxide compound. So our final answer will be V, because it’s the chemical symbol for vanadium, and a subscript value, which we’ll call 𝑥 for now — this value will tell us how many vanadium atoms there are in the empirical formula — O, because it’s the chemical symbol for oxygen and we’re finding the empirical formula for an oxide compound. This will also have a subscript value, which we’ll denote 𝑦.
The best way for us to work out the empirical formula is by drawing a table where each element in the empirical formula has its own column. First of all, we need to know the mass of vanadium and the mass of oxygen. The question tells us that the compound is 68 percent vanadium and 32 percent oxygen by mass. This means that in 100 grams of vanadium oxide, there will be 68 grams of vanadium and 32 grams of oxygen.
Since we’re finding the empirical formula and we only need to find the simplest whole number ratio of atoms, we don’t need a sample of vanadium oxide to be weighed to fill in the mass section of the table. We can just use the percentage by mass. This is because the percentages show the ratio between vanadium and oxygen atoms. It is, however, easier for the sake of calculation to have the mass in grams. So let’s just assume that we have 100 grams of vanadium oxide. Therefore, we’ll have 68 grams of vanadium and 32 grams of oxygen.
Filling in the next line of the table involves us finding the molar mass for vanadium and oxygen. The molar mass, which is often given the symbol of a capital 𝑀, is the average mass in grams per mole of species. It has the unit grams per mole, but we haven’t been given the molar mass in the question. We’ve been given the atomic mass. The atomic mass is the weighted mean mass of an atom of an element based on the natural isotopic abundance of that element. Its units are atomic mass units, denoted 𝑢, or it is not given units at all.
It’s clear that the definitions of molar mass and atomic mass are not the same, and they have different units. But the atomic mass of a species is numerically equivalent to the mass in grams per mole of that species. So the value of the atomic mass with atomic mass units or no units is the same as the molar mass in units of grams per mole. The question tells us that the atomic mass of vanadium is 51, so the molar mass of vanadium is 51 grams per mole. The question also tells us that the atomic mass of oxygen is 16, so the molar mass of oxygen would be 16 grams per mole.
To fill in the next row on the table, we need to work out the number of moles of vanadium and oxygen. We can do this by looking at units. If we look at the units that we have so far, we have grams and grams per mole. And we want to work out the number of moles. Dividing grams by grams per mole is the same as multiplying grams by the reciprocal of grams per mole, which is mole per gram. If we multiply grams by moles per grams, then the gram units cancel, leaving us with moles. So if we need to divide grams by grams per mole to get the number of moles, then we need to divide the mass, which is in grams, by the molar mass, which is in grams per mole.
We can also work this out by rearranging a well-known equation. This states that the molar mass in grams per mole is equivalent to the mass of a substance in grams divided by the number of moles. If we rearrange this equation to put in the number of moles as the subject, then we find that we have to divide the mass by the molar mass. This is the same conclusion that we came to when we looked at the units. So to find the number of moles of vanadium, we have to do 68 divided by 51. This gives us 1.3 recurring. And to find the number of moles of oxygen, we have to do 32 divided by 16. This gives us two moles.
We now know what the ratio of atoms is. For every 1.3 recurring moles of vanadium, there are two moles of oxygen. But we need a whole number ratio. So we need to divide 1.3 recurring and two by the smallest number between them. 1.3 recurring is smaller than two. So both 1.3 recurring and two need to be divided by 1.3 recurring. This gives values of one and 1.5, but this is still not a whole number ratio. Therefore, we need to turn the decimal into a fraction. So 1.5 is three over two.
We then need to multiply one and three over two by the largest denominator between them. There is only one denominator, which is two. So both one and three over two must be multiplied by two. This gives us values of two for vanadium and three for oxygen. These values are the subscript values 𝑥 and 𝑦. Therefore, the empirical formula for this compound is V2O3.