The course of school chemistry is an introductory manual in complex science. From the very beginning, students are trying to understand how to solve design problems. Suppose that they have little practical use at the first stages, but if a schoolchild learns about, for example, how to find a mass of reacted substances, then he may well claim serious achievements.
Consider a simple example of a problem, on the basis of which you can learn to solve more complex ones. Suppose you needed 11.2 liters to completely burn carbon monoxide (II). How many grams of CO2 did you manage to get?
1. We make up the reaction equation.
CO + O2 = CO2
2. Equalize by oxygen. There is some rule that in most cases can help you. Begin to arrange the coefficients with the substance, the number of atoms of which is odd. In this case, it is oxygen in the CO molecule. We put a coefficient of 2 to it. Since two carbon atoms formed on the left and one carbon atom on the right, then 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 parts. Carbon is also in balance. Consequently, equalized correctly.
3. Next, you need to find the amount of O2. The determination of molecular weight for schoolchildren is too cumbersome and hard to remember, so we will use another method. Recall that there is a molar volume that is 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 make a proportion. The amount of reacted oxygen is half that of n (CO2). This follows from the fact that 0.5 mol / 1 = x mol / 2. The simple ratio of the two quantities helped make the right equation. When we found x = 1, we can get an answer to the question of how to find the mass.
5. True, for starters you will have to remember one more formula: m = M * n. Found the last variable, but what to do with M? Molar mass is an experimentally established value relative to hydrogen. It is it that is designated by the letter M. Now we know that m (CO2) = 12 g / mol * 1 mol = 12 g. So we got the answer. As you can see, there is nothing complicated.
This task is very easy relative 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 6 * 10 ^ 23 molecules. At the same time, in the Periodic system there is a fixed element mass per 1 mol. Sometimes it is necessary 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 the water contains two H atoms. We also do not forget about the presence of oxygen, which gives us another 16 grams. Summing up, we get 18 grams / mol.
Theoretical calculation of mass will subsequently be of practical use. Especially for students who expect a chemical workshop. Do not be afraid of the word if you are in a non-core school. But if chemistry is your core subject, it’s best not to run 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.