The school chemistry course is an introductory guide to a complex science. From the very beginning, students try to understand how to solve calculation problems. Let at the first stages they have little practical application, but if a student has learned, for example, how to find the mass of substances that have reacted, then he may well claim serious achievements.
Let's consider a simple example of a problem, on the basis of which we can learn to solve more complex ones. Suppose that it took you 11.2 liters to completely burn carbon monoxide (II). How many grams of CO2 did you get?
1. Write the reaction equation.
CO + O2=CO2
2. Equalize for oxygen. There is a rule that in most cases can help you. Start setting the coefficients from that substance, the number of atoms of which is odd. In this case, it is the oxygen in the CO molecule. We put a coefficient 2 to it. Since two carbon atoms were formed on the left, and one on the right, we put 2 in front of CO2. Thus, we get:
2CO + O2=2CO2
As you can see, there are four oxygen atoms in the left and right sides. Carbon is also in balance. Therefore, equalizedright.
3. Next, you need to find the amount of O2. Determining the molecular weight for schoolchildren is too cumbersome and hard to remember, so we will use another method. Recall that there is a molar volume, which is equal to 22.4 l / mol. You need to find how many moles (n) reacted: n=V/V m. In our case, n=0.5 mol.
4. Now let's make a proportion. The amount of oxygen that has entered into the reaction is two times less than n (CO2). This follows from the fact that 0.5 mol/1=x mol/2. A simple ratio of two quantities helped to make the right equation. When we have found x=1, we can get the answer to the question of how to find the mass.
5. True, to begin with, you will have to remember one more formula: m \u003d Mn. The last variable was found, but what to do with M? Molar mass is an experimentally determined value relative to hydrogen. It is she who is denoted by the letter M. Now we know that m (CO2) u003d 12 g / mol1 mol \u003d 12 g. So we got the answer. As you can see, there is nothing complicated.
This task is quite easy compared to many others. However, the main thing is to understand how to find the mass. Imagine a molecule of some substance. It has long been known that a mole consists of 610^23 molecules. At the same time, in the Periodic system there is an established mass of an element per 1 mole. Sometimes you need to calculate the molar mass of a substance. Suppose M(H20)=18 grams/mol. That is, one hydrogen molecule has M=1 gram/mol. But water contains two H atoms. Also, do not forgetabout the presence of oxygen, which gives us another 16 grams. Summing up, we get 18 grams/mol.
Theoretical mass calculation will later have practical applications. Especially for students who are expecting a chemistry workshop. Do not be afraid of this word if you study at a non-core school. But if chemistry is your core subject, it's better not to run the basic concepts. So, now you know how to find the mass. Remember that in chemistry it is very important to be a consistent and attentive person who not only knows some algorithms, but also knows how to apply them.
