### Video Transcript

In an experiment, mercury oxide is heated in a container producing oxygen gas and mercury in its pure elemental form. The results of the experiment are shown below. We have been given the molar mass for mercury and oxygen. What is the mass of oxygen gas lost?

The question tells us that mercury oxide is heated in a container producing oxygen gas and mercury in its pure elemental form, which is a liquid. We want to find out what mass of oxygen gas was lost. To do this, we need to use the law of conservation of mass, which states that mass is neither created nor destroyed in chemical reactions. Therefore, the mass of the reactants equals the mass of products. We can use this to work out the mass of oxygen gas lost. The mass of the empty container and mercury oxide is 73.4 grams, and the mass of the container and mercury is 70.2 grams. Since the mass of reactants must equal the mass of products, 70.2 grams plus the unknown mass must equal 73.4 grams.

We want to make the unknown mass the subject, so we need to subtract 70.2 grams from both sides of the equation. 70.2 grams minus 70.2 grams is zero grams, so this part will cancel out. Therefore, the mass of oxygen gas lost is 73.4 grams minus 70.2 grams, which is 3.2 grams. So the answer to the first part of this question “What is the mass of oxygen gas lost?” is 3.2 grams.

What is the mass of mercury produced?

The question tells us that the mass of the container and mercury is 70.2 grams, but we only want to know the mass of mercury produced. So we need to subtract the mass of the empty container. The question tells us that the mass of the empty container is 30.0 grams. If we subtract 30.0 grams from 70.2 grams, we get 40.2 grams. So the answer to the question “What is the mass of mercury produced?” is 40.2 grams.

What is the empirical formula for the compound of mercury oxide?

The empirical formula is the formula of a compound with the simplest whole number ratio of atoms. For example, C3H6 would have the empirical formula CH2, and C4H10 would have the empirical formula C2H5. In order for us to find out the empirical formula for mercury oxide, we first need to know the mass of mercury that we have and the mass of oxygen that we have. We have already worked these out to be 40.2 grams and 3.2 grams. We then need to know the molar mass of mercury and oxygen. The molar mass is the average mass in grams per mole of species. It has the units grams per mole. And the values for the molar masses of mercury and oxygen have been given to us in the question. The molar mass of mercury is 201 grams per mole, and the molar mass of oxygen is 16 grams per mole.

Our next step in working out the empirical formula is to work out how many moles we have of each element. We can work out how many moles there are by looking at the units for the values that we have already. We have the mass in grams, the molar mass in grams per mole, and we want to work out how many moles there are. Dividing grams by grams per mole is equivalent to multiplying grams by the reciprocal of grams per mole, which is moles per gram. If we multiply grams by moles per gram, we can cancel the unit grams, leaving us with the number of moles.

If dividing grams by grams per mole gives us the number of moles, then dividing the mass by the molar mass will give us the number of moles. So for mercury, we need to do 40.2 divided by 201. This gives 0.2 moles. And to work out how many moles of oxygen we have, we need to do 3.2 divided by 16. This also gives 0.2 moles.

We now know that the molar ratio between mercury and oxygen is 0.2 to 0.2. But to work out the empirical formula, we need the simplest whole number ratio of atoms. Therefore, we need to divide each of the moles values, which are both 0.2, by the smallest number of moles. Since both numbers are the same, we need to divide each of them by themselves. 0.2 divided by 0.2, or in fact any number divided by itself, is equal to one. So for every one atom of mercury, we have one atom of oxygen. These values become the subscript values in the empirical formula, so the formula would be Hg1O1. But since we tend not to write the subscript value one, the empirical formula for the compound of mercury oxide is HgO.