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# How do you balance chemistry equation?

to balance an equation in chemistry, you have to understand what valency is. the valency is the charge of an atom, for example, lithium has a valency (charge) of +1.

this charge is worked by working out by using the periodic table. for reasons of simplicity, we shall ignore all transition metals for the time being.

In the 1st column (the alkali metals, from hydrogen down to francium) has a valency of +1; the 2nd column (the alkaline earth metals, from beryllium to radium) has a valency of +2;

the 3rd column (the boron family, from boron to thallium) has a valency of +3;

the 4th column (the carbon family, from carbon to lead) has a valency of 4, but depending on its use, the 4 can either be +ve or -ve.

the 5th column (the pnictogens or nitrogen family, from nitrogen to bismuth) has a valency of -3

the 6th column (the chalcogens or oxygen family, from oxygen to polonium) has a valency of -2

the 7th column (the halogens, from fluorine to astatine) has a valency of -1.

the 8th and final column (the noble gases, from helium to radon) doesn't really have a valency, as all noble gases have full outer shells, and thus, do not readily react with other elements.

now, to balance an equation, for example, Al + F --> AlF

you take the valencies, Al (being in the 3rd column) has a valency of +3; and F (being in the 7th column) has a valency of -1. once you know the valencies, you cross them over between the 2 elements, therefore Al takes on the -1 valency, and the F takes on the +3 valency. so when these 2 elements are put together, you write down the exchanged valencies as subscripts, e.g. Al-1F+3, however, the one does not really need to be written, when there is no number present, is it always assumed that the integer is 1, thus its should look like AlF3.

now, the whole equation looks like this Al + F --> AlF3

but it is not finished yet,; in chemistry, matter is never destroyed nor created, thus, you must have equal amounts of each substance on either side of the equation. this equation is not balanced, on the left hand side we have 1 amount of Al and 1 amount of F but on the right hand side we have 1 amount of Al and 3 amounts of F. this may be better expressed in a table

Left hand side Right hand side

Amount of Al 1 1

Amount of F 1 3

so as can be seen from the table, the amount of Al is balanced, but the amount of F is not.

Therefore, to balance out the amount of F, we place a 3 in front of the F on the left hand side (as the left hand side in this example has the lowest number, this is not always the case, sometimes the number is written on the right hand side; this all depends on the individual equation).

Thus, the equation now looks like

this

Al + 3F --> AlF3

this equation is balanced, and we can show how it is;

Al + 3F --> AlF3

as can be seen, there is 1 part of Al on both the left and right sides of the equation, and there is 3 parts of F on both sides of the equation, thus, the equation is now balanced.

NOTE: you can only combine elements with OPPOSITE valencies; that is you can only put a +ve valence element with a -ve valence elements. You CANNOT combine 2 +ve elements or 2 -ve elements; they must always be OPPOSITE!

another, harder, example is, AlCl + LiO

once again, work out the valencies of each element.

+3 -1 + +1 -2

Al Cl + Li O

then swap the valencies between each compound (that is between Al & Cl, and then between Li & O), and write the swapped valencies as subscripts;

Al-1Cl+3 + Li-2O+1

but the 1 does not need to be written, therefore the equation is:

AlCl3 + Li2O

now in each compound there is 2 elements, AlCl contains Aluminium and Chlorine; LiO contains Lithium and Oxygen.

Lets call the 1st element in a compound "A" and the second element "B", thus both compounds are made of an A and a B;

Al Cl3 + Li2 O

now, when working out the product, you swap the 2 "B" elements, thus the Cl and O swap, so the equation is

AlCl3 + Li2O --> AlO + LiCl

now we have to work out and swap the valencies again for the products, so we get

AlCl3 + Li2O --> Al2O3 + LiCl

now we have to see how many of each element we have either side of the equation;

Left side Right side

Amount of Al 1 2

Amount of Cl 3 1

Amount of Li 2 1

Amount of O 1 3

as can be seen the equation is not balanced.

so, it can be seen on the right side that there is 1 amount of both Li and Cl, but on the left side, there is 3 of Cl and 2 of Li, and when a number is placed in front of a compound, the number effects both elements. for the Al, it has 1 on the left side, and 2 on the right side, therefore a 2 is put in front of the AlCl compound. and for the O, it has 1 on left side and 3 on right side, so a 3 is put in front of the LiO compound.

now with the LiCl, on the left hand side, the Cl is affected by the 2 in front of the compound and the 3 subscript, so the combination of the two numbers give the Cl an amount of 6 (3x2). the same applies to the Li, 3 in front and 2 subscript combine to make 6. thus, there is 6 Cl and 6 Li on the left side, so a 6 is written in front of the LiCl compound on the right side. so the final balanced equation is