Molecular formulae indicate the simple numbers of each type of atom in a molecule, with no information on structure. The simplest types of chemical formulae are called empirical formulae, which use letters and numbers indicating the numerical proportions of atoms of each type. Chemical formulae can fully specify the structure of only the simplest of molecules and chemical substances, and are generally more limited in power than chemical names and structural formulae.
#Chemistry calculator formula full#
Although a chemical formula may imply certain simple chemical structures, it is not the same as a full chemical structural formula.
A chemical formula is not a chemical name, and it contains no words. These are limited to a single typographic line of symbols, which may include subscripts and superscripts.
#Chemistry calculator formula plus#
Examples of other chemical formulae for butane are the empirical formula C 2H 5, the molecular formula C 4H 10 and the condensed (or semi-structural) formula CH 3CH 2CH 2CH 3.Ī chemical formula is a way of presenting information about the chemical proportions of atoms that constitute a particular chemical compound or molecule, using chemical element symbols, numbers, and sometimes also other symbols, such as parentheses, dashes, brackets, commas and plus (+) and minus (−) signs. J. 2 ( SO 4 ) 3 Structural formula for butane. I hope you never see one like this during your exams: To really test the calculator try this monster.
It does not care if the reaction is probable from the chemical point of view, for example:īut if the reaction is found to be impossible for mathematical reasons one can be sure that it cannot occur in a test tube: Furthermore, it can find the coefficients for only just hypothetically possible reactions. Amazing that this mathematical procedure is able to group substrates apart from products. The calculator can only analyze the given molecules and their elements by means of simple algebraic approach. The calculator is not supposed to predict what the products will be. I do not want to underestimate chemistry. It still works! I do hope this proves that chemical knowledge is not so important as mathematics when finding coefficients. You cannot see where the error is at first glance. One could never find the coefficients for this equation using the classical approach. What would happen if we mistakenly mixed it up a bit again? Imagine that such compounds exist and react as follows:
But I can assure you that it really does not need any chemical analysis to find the true shape of the process. And what about the one below? Is it right or wrong?: Algebra decides what will be the product and what will be the substrate! If you are an experienced chemist you will easily find the mistake in the mixed equation thanks to your chemical knowledge because the example is simple. But the calculator still works! Following the methodology described you will find that the value of some variables will be negative so they should be placed on the opposite side of the equation. If one did not notice the mistake one could not find the solution by means of the classical approach. What would happen if we mistakenly mixed it up a bit? To solve this system is to find the value of variables a, b, c which are the desired coefficients. With mathematics in mind you can write such a simple equation with variables a, b, c denoting the coefficients to be found:Īll you have to do is balance the quantity for each element on the left and the right side of the equation. Use the calculator with the explanation box checked to see the output. To see how the program finds the solution let`s start from a simple example.Īll the examples can be cut and pasted into the calculator That students will find it very useful in determining the To prove that this algebraic technique works. It was a real shock when I realized the power of It provides better solutions than chemistry itself! SuchĪn idea might seem a bit strange for a chemist. Mathematics provides a general way to find reaction coefficients. Luckily it turns out that oxidation numbers and half-reaction are unnecessary! Another method exists that uses algebra to find the In some cases however it is hard to use this method because oxidation numbers cannot be unequivocally attributed. The classical approach uses oxidation numbers and many chemists cannot imagine any way to find the solution other than balancing half-reactions for the process. Finding coefficients for chemical reactions is often complicated, especially in the case of redox processes.
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