The molecular formula of substance reflects its structure. Sometimes by the name it is possible to write a formula. In other case the formula is calculated, proceeding from the percentage of atoms in substance.
Instruction
1. Sort out what it is necessary to begin drawing up a formula of substance with. All atoms have oxidation level. At some it has constant value, at others – can change. Knowing oxidation level, make formulas. If it is necessary to define a molecular formula of substance which part potassium and chlorine is, act this way: potassium has the oxidation level +1, and chlorine-1, write down K(+1) of Cl (-1). Though chlorine also has variable oxidation level, but in chlorides, and in this case, it is obvious that it it, oxidation level is equal to-1. The sum of all oxidation levels of substance has to be equal to zero, therefore, in this example it is not necessary to put any additional indexes. The received substance – chloride of potassium (KCl).
2. Other example: write a sodium sulfate formula. The cation of sodium and sulfate-anion are its part. Sodium has oxidation level +1 (since is alkaline metal, and at them it is constant), and sulfate ion-2. Na (+1)SO4(-2), consider +1-2 =-1. And there has to be zero. Therefore one more cation of sodium is necessary for equalizing. Therefore, final type of a formula such: Na2SO4.
3. It should be noted that in such a way it is possible to make not all formulas since in certain cases only by means of formulas and oxidation level is calculated.
4. There are tasks in which condition it is given percentage ratios of the substances which are a part of a formula. For their decision adhere to the following algorithm. For calculations you will choose a sample weighing 100 g. Then transfer percent to grams by means of a formula: m (matter) = m (obshch) *w where w is a mass fraction. Further calculate amounts of substances of atoms. Make a ratio, thereby define a substance formula.
5. Example: the mass fraction of sulfur is equal in sulfur oxide to 40%, and oxygen – 60%. Define a formula of this oxide. Decision: Choose the mass of oxide of equal 100 g. Then you receive: m(S) = m (obshch) *w = 100g*0,4 = 40 g of m(O) = 100g*0,6 = 60 g. Find amounts of substances of atoms on a formula: n = m/M where m is the mass of substance, M – the molar mass of substance. The molar mass of substance is specified in D.I. Mendeleyev's table under designation of an element. n(S) = 40/32 = 1.25 mol. n(O) = 60/16 = 3.75 mol. Make a ratio 1.25: 3.75 = 1: 3. Therefore, you receive a formula: SO3.