RUMUS KIMIA : RUMUS EMPIRIS DAN RUMUS MOLEKUL

The chemical formula is defined as a formula of a substance that uses the symbol and the number of atoms of the material order element. In the chemical formula, the number of elements stated is written in the form of the lower index after the symbol of the element.

For example, carbon dioxide is made up of one carbon atom which is labeled as C and two oxygen atoms which are labeled as O, so the chemical formula is CO2. Well, this number 2 is the index of oxygen which shows that the number of oxygen atoms in the accumulator is 2, while for carbon, the index is 1 and does not need to be written.

Next, we can differentiate the chemical formula into two, namely the empirical formula and the molecular formula. So what are the differences between the two? Let's discuss one by one. The empirical formula is the simplest formula of a molecule that only shows the type and the smallest comparison of the elements that form the substance.

Well, we can determine this empiric formula by looking for the simplest mol comparison of the components of the compiler, where the number of the comparison is made as the index of the components. In order to be understood well, let's take a look at the following examples. The first example, a number of substance samples contains 11.2 grams of iron and 4.8 grams of oxygen.

If it is known that the mass of iron atoms or Fe is 56 grams per mole and oxygen is 16 grams per mole, determine the formula of the material. Well, to determine the formula of the material, we first determine the mole of the accumulator, namely the iron mole and the oxygen mole. Remember that the mole of an element is equal to the mass divided by the relative atom or its air.

For iron or Fe, the question is known that its mass is 11.2 grams and the relative atom or its water is 56 grams per mole, so the mole is equal to 11.2 grams divided by 56 grams per mole, the result is 0.2 moles.

Then for oxygen, the question is known that its mass is 4.8 grams and its relative atom mass or AR is 16 grams per mole, so the mole is equal to 4.8 grams divided by 16 grams per mole, the result is 0.3 moles.

Next, remember that the formula for the empirical value of a substance can be determined from the simplest mol comparison of the components of the accumulator. And if we compare, the mol comparison of iron and oxygen is 0.2: 0.3.

Well, this number of comparisons, if we multiply 10, it becomes 2/3, which means that the sample consists of 2 iron atoms and 3 oxygen atoms, so the iron index is 2 and the oxygen index is 3, then the empiric formula is Fe2O3.

The second question, determine the formula for the substance of the empery that contains 26.53% potassium, 35.37% chromium and the rest is oxygen. It is known that the relative atom or water from potassium is 39 grams per mole, chromium 52 grams per mole and oxygen 16 grams per mole.

For a question like this, we assume that the mass of the substance is 100 grams. Well for the potassium, it is known that the percentage of the mass or the amount is 26.53%, so the mass of potassium per 100 grams of substance is 26.53 grams.

so the molkalium is 26.53 grams divided by the relative atom mass which is 39 grams per mol the result is 0.68 then for the chromium, it is known that the percentage or the rate is 35.37% then the chromium mass in 100 grams of substance is 35.37 grams

so the chromium mole is 35.37 grams divided by the relative atomic mass, which is 52 grams per mole, the result is 0.68.

Next for the oxygen, the question is that the percentage or the amount is the rest, which means the percentage or the percentage of oxygen is equal to 100%, minus the percentage of potassium plus the percentage of chromium, equal to 100%, minus 26.53%, plus 35.37%.

equal to 100% minus 61.9% the result is 38.1% Well, it means the mass of oxygen in 100 grams of the substance is 38.1 grams so the mole of oxygen is equal to 38.1 grams divided by the relative atomic mass, which is 16 grams per mole the result is 2.38

Next, based on the mol of the element that we have obtained, then the comparison of mol of potassium, chromium, and oxygen is equal to 0.68 compared to 0.68 compared to 2.38. Well, we simplify it by dividing 0.68 equally, so that the comparison becomes 1:1:3.5.

Then to get the number of comparisons in the form of the number of months, then the number of comparisons is multiplied by 2, so it becomes 2/2/7, which means that the sample consists of 2 potassium atoms, 2 chromium atoms, and 7 oxygen atoms, so the empiric formula is K2Cr2O7.

The molecule formula is the addition of the empiric formula. For example, let's look at the following table. An atom with the CH2 empiric formula, if multiplied by 2, will form an atom with the C2H4 molecule formula with the name of ethene.

Then if multiplied by 3, then it will form a substance with the formula C3H6 with the name of propene and if multiplied by 4, then the formula is C4H8 with the name of butane.

Based on this example, it is clear that the molecule formula is the addition of n from the empiric formula which can be written as the empiric formula n times the molecule formula with n is a round number.

Well, we can determine the value of n if the empirical formula and the relative molecule mass or MRZ are known, where the molecular formula MR is equal to n times the empirical formula MR. Well, to be understood, let's continue to the examples of questions.

The first example of a substance with the formula CH-mperis has a relative mass of 26 grams per mole. If the relative mass of carbon is 12 grams per mole and hydrogen is 1 gram per mole, determine the formula of the substance. Remember that the molecular formula MR is equal to N times the formula MR-mperis.

So for this question, the MR of the molecule formula is equal to N times the MR of CH, where MRCH is equal to 1 times the relative atom or carbon water plus 1 times the relative atom or hydrogen water.

Well, the question is already known the mass of the relative molecule or MR, which is 26 grams per mole and also known the mass of the relative atom or AR of carbon and hydrogen. Let's just substitute it, so 26 grams per mole = N multiplied by 12 grams per mole + 1 gram per mole.

then 26 grams per mole is equal to N multiplied by 13 grams per mole. We multiply by the cross, then N is equal to 26 grams per mole divided by 13 grams per mole. The result is 2. Then the formula for the molecule of the substance is a 2-fold formula for the empiric formula, which we can write CH2.

which means the amount of carbon and hydrogen is multiplied by 2 so that the formula of the molecule is C2H2 The second question, the analysis results of an oxygen-nitrogen substance show that the substance contains 30% of nitrogen

If the mass of the relative molecule or MR of the substance is 92 grams per mole, determine the formula for the molecule of the substance. Well for this question, we first determine the empiric formula. Still remember, if we know the percentage of the mass or the amount of the accumulator element, then we just assume that the mass of the substance is 100 grams.

For nitrogen, the percentage of time or the amount is 30%, which means the nitrogen mass in 100 grams of the substance is 30 grams, so the mole of the nitrogen is 30 grams divided by the relative atomic mass, which is 14 grams per mole. The result is 2.142 mol.

Next for oxygen, the percentage or the amount is 100% minus the amount or percentage of nitrogen, equal to 100% minus 30%, the result is 70%. So the mass of oxygen in 100 grams of the substance is 70 grams.

so the mole of oxygen is 70 grams divided by the relative atom of oxygen, which is 16 grams per mole the result is 4.375

then the comparison of moles of nitrogen and oxygen is 2.142 / 4.375. Well, we simplify it by dividing it equally by 2.142 so that the comparison becomes 1 / 2.04 which we can round it into 1 / 2. So the empiric formula is NO2.

Next, the relative mass or molecular mass of the molecule is equal to N times the relative mass or the empirical mass of the molecule, so the molecular mass of the molecule is equal to N times the NO2 of the molecule, where the NO2 of the molecule is 1 times the relative mass or nitrogen, plus 2 times the relative mass or oxygen.

Well, it is known that the mass of the relative molecule or MR is 92 grams per mole and the mass of the relative atom or AR of nitrogen and oxygen is also known

then we just substitute it so that 92 g/mol = n * 1 * 14 g/mol + 2 * 16 g/mol then 92 g/mol = n * 46 g/mol

Then we multiply by the cross so that N = 92 g/mol divided by 46 g/mol, the result is 2.

then the formula for the molecule is the 2nd multiplication of the empiric formula which we can write NO22, which means the amount of nitrogen and oxygen is multiplied by 2 so that the formula for the molecule is N2O4.