When methane is burned with oxygen the products are carbon dioxide and water. if you produce 36 grams of water and 44 grams of carbon dioxide from 16 grams of methane, how many grams of oxygen were needed for the reaction?

First of all we need to convert the masses to moles so that we can see the ratios of what we have been given.

36 grams of water divided by its molecular weight (18g/mol) gives us 2 moles of water.

44 grams of carbon dioxide divided by its molecular weight (44g/mol) is 1 mole of carbon dioxide.

16 grams of methane divided by its molecular weight (16 grams) is 1 mole of methane.

We can look at a balanced chemical equation of methane burning such as the one below to find how much oxygen we need:

CH₄ + 2O₂ ⇒ CO₂ + 2H₂O

Therefore to burn one mole of methane we need two moles of oxygen molecules to produce one mole of carbon dioxide and two moles of water molecules. Because the amounts of water, methane and carbon dioxide in this equation match the ones we were given, we can say that 2 moles of oxygen molecules were required.

To convert this to grams we need to know the molecular weight of oxygen molecules, not atoms. By looking at a table of values you can see it is 32g/mol. Therefore two have two moles of oxygen molecules we need 64 grams of oxygen, which is the answer.