in this lesson we're going to talk about the law of definite proportions and the basic concepts behind this law is that compounds they have a constant chemical composition by mass an example of this is carbon dioxide if you go to the periodic table you'll find that the atomic mass of carbon is 12 and the atomic mass of oxygen is 16. in a single molecule of carbon dioxide we have two atoms of oxygen one atom of carbon so in a sample of carbon dioxide if there's 12 grams of carbon in that sample then there's going to be 32 grams of oxygen 16 times 2 is 32. now let's say if we double the amount of carbon so let's say if there's 24 grams of carbon in a sample of carbon dioxide how many grams of oxygen will there be well 12 times 2 is 24 so 32 times 2 is going to be 64. now what if there was 36 grams of carbon in a sample of carbon dioxide how many grams of oxygen should be present in that same sample 12 times 3 is 36 and 32 times 3 is going to be 96 so it's going to be 96 grams of oxygen the ratio between the mass of oxygen and the mass of carbon will remain the same it's going to have a constant composition by mass so 32 grams of oxygen divided by 12 grams of carbon if you divide those two numbers you should get the same ratio which is about 2.67 if you take 96 grams of oxygen and divided by 36 grams of carbon that too will give you the same ratio of 2.67 now let's talk about how to apply this information what if there were 30 grams of carbon in the sample of co2 how many grams of oxygen are present now before i give you some nice numbers like 24 36 grams of carbon multiples of 12. so it would be easy to find the number of grams of oxygen you can multiply by two or three but what if i gave you a number that's not so nice like 30 how can you find the grams of oxygen in this case well we can do so by a conversion process let's start with the number of grams of carbon that we have now the conversion is based on these numbers it's based on the atomic weights that you could find in a periodic table so in carbon dioxide for every 12 grams of carbon that's present there's going to be 32 grams of oxygen so notice that the unit grams of carbon cancels so it's going to be 30 times 32 divided by 12. so 30 grams of carbon would be present with 80 grams of oxygen in a sample of carbon dioxide now let's work on some practice problems a pure sample of sodium fluoride contains 35 grams of sodium how many grams of fluorine are present in this sample now the first thing we should do is we need to use the periodic table let's find the atomic masses sodium has an atomic mass of 23. fluorine has an atomic mass of 19. so in a sample of sodium fluoride this is the conversion factor that you need 23 grams of sodium will be present with 19 grams of fluorine so if there are 23 grams of sodium there's going to be 19 grams of fluorine that's the proportion that you want to use in order to answer this question so if we have 35 grams of sodium how many grams of fluorine should be present now you can set it up as a proportion or you could solve it as a conversion personally i like the conversion method but for this example we could do it both ways so first start with what you're given that's 35 grams of sodium in the second fraction use your conversion factor which is this now you want the units grams of sodium to cancel so you're going to place that in the bottom of this fraction and on the top we're going to put 19 grams of fluorine so these units will cancel and this will give us the answer so it's going to be 35 times 19 divided by 23 which is about 28.9 grams of fluorine so that's the answer now if you want to set up as a proportion here's what you need to do so let's write two fractions separated by an equal sign 23 grams of sodium corresponds to 19 grams of fluorine how many grams of fluorine correspond to 35 grams of na so that's how you could set it up and just cross multiply so that we're going to have 23 times x and that's going to be equal to 35 times 19. 35 times 19 is 665. so to find the value of x we need to divide both sides by 23 and this is going to give us the same answer 28.9 so that's how many grams of fluorine are present in this sample number two if there are 42 grams of hydrogen in the sample of pure methane how many grams of carbon are present so feel free to pause the video and try this problem so let's focus on methane looking at the periodic table the atomic mass of carbon is 12 and the atomic mass of hydrogen is about one now in methane we have one carbon atom so 1 times 12 is 12. in methane we have four hydrogen atoms you got to do four times one which is four so in methane for every 12 grams of carbon there's going to be four grams of hydrogen this is our conversion factor so once you have it now you can answer the question so if we have 42 grams of hydrogen how many grams of carbon are there so let's convert so i'm going to put the 4 grams of hydrogen on the bottom so that the units grams of hydrogen will cancel and then the other stuff the 12 grams of carbon i'm going to put that on top so it's 42 times 12 divided by 4 which is 126. so in this sample there's going to be 126 grams of carbon number three if there are 19 grams of oxygen in a sample of aluminum oxide how many grams of aluminum are present so using the periodic table aluminum has an atomic mass of approximately 27 and oxygen is about 16 that's the mass number now in the sample there are two aluminum atoms in this compound so 2 times 27 that's 54. and we have three oxygen atoms per formula unit of aluminum oxide so three times 16 is 48 so this is the ratio that we need to be concerned with in the sample of aluminum oxide there are 48 grams of oxygen for every 54 grams of aluminum so now that we have this conversion factor we can now answer the problem so if we have 19 grams of oxygen we can convert it to the grams of aluminum so let's put the grams of oxygen on the bottom so that those units will cancel and then let's put 54 grams of aluminum on the top so it's going to be 19 times 54 divided by 48. so this is going to be 21.5 grams of aluminum you