Back in my September 15<\/a> article, I opened the topic of what a mole means in chemistry. Specifically I pointed out that a mole is the number of atoms that gives the mass of each element as reported on the periodic table.<\/p>\n Molecules react on a molecule-to-molecule basis – or, on a larger scale, a mole-to-mole basis. However, chemists measure the chemicals in grams. Stoichiometry explores the quantitative amounts of chemicals needed – usually in grams – and the grams of products obtained from a chemical reaction.<\/p>\n Example<\/strong>: Consider the reaction<\/p>\n C3<\/sub>H8<\/sub> + 5O2<\/sub> \u2192 3CO2<\/sub> + 4H2<\/sub>O<\/p>\n How many grams C3<\/sub>H8<\/sub> (which is propane, as you’ll know from my post here<\/a>) are required to produce 1 litre of water?<\/p>\n Solution<\/strong>:<\/p>\n (Before proceeding, you may want to take a look at my posts on significant figures. There are a number of them, but this one<\/a> might be good to start with.)<\/em><\/p>\n First of all: a litre of water is 1000. grams at room temperature. Therefore, we need to produce 1000. grams of water.<\/p>\n First, we find the molar mass of water using the masses of H and O from the periodic table:<\/p>\n mass H2<\/sub>O=2(mass H) + mass O=2(1.0)+16.0=18.0 grams<\/p>\n Similarly for propane:<\/p>\n mass C3<\/sub>H8<\/sub> =3(mass C)+8(mass H)=3(12.0)+8(1.0)=44.0 grams<\/p>\n Let’s examine the chemical equation again, in a slightly different way:<\/p>\n C3<\/sub>H8<\/sub> + 5O2<\/sub> \u2192 3CO2<\/sub> + 4H2<\/sub>O<\/p>\n If not written, the coefficient of a term in a chemical reaction is 1. Notice the relationship implied:<\/p>\n 1 mole C3<\/sub>H8<\/sub> yields 4 moles H2<\/sub>O<\/p>\n We need to figure out how many moles of water we need; we will need 1\/4 that many moles propane.<\/p>\n Well, we know from above that a mole of water is 18.0 grams, and that we need 1000. grams of it. We can find the moles of water in 1000. grams as follows:<\/p>\n 1000.grams x 1mole\/18.0grams=1000.mole\/18.0=55.6moles H2<\/sub>O<\/p>\n By the 1:4 ratio earlier observed between the moles of propane and the moles of water, we know we need (1\/4) x 55.6moles=13.9moles propane. All we need now is to determine how many grams that is. From our earlier calculation, we know that the molar mass (mass of one mole) of propane is 44.0 grams. We proceed:<\/p>\n 13.9moles x 44.0grams\/mole=611.6 grams C3<\/sub>H8<\/sub><\/p>\n Of course, in significant figures, the answer is 612 grams propane.<\/p>\n I will have much more to say about stoichiometry in future posts. Cheers:)<\/p>\n