Chemical Equation Balancer Calculator


This online Chemical Equation Balancer Calculator finds the stoichiometric coefficients to balance a given chemical equation, including equations with free electrons and electrically charged molecules (ions) as well as hydrated compounds. In case the original equation was unbalanced, the field with this equation is highlighted in light pink.

You can enter a chemical equation manually or paste the equation copied from a web page or text document (including DOC or PDF file). Indices denoted using <sub> and </sub> html tags (e.g. H2O) as well as denoted using the ‘tiny’ numbers, like ₂ or ₅ , (e.g. H₂O) are supported and automatically converted to normal form. In a hydrated compound the middle dot (·) or the asterisk (*) precedes the water formula (e.g. CuSO4·5H2O). In what follows is a more detailed syntax guide to our chemical equation balancer.


Equation to Balance
Output Format:
Balanced Equation


Balancing Chemical Equations

A chemical equation is a symbolic representation of a chemical reaction, with reactants on the left and products on the right hand side of the equation.

Balancing chemical equations means finding stoichiometric coefficients for reactants and products. This is important because in a chemical reaction, the amount of each element does not change (the law of conservation of mass). Thus, each side of the equation must represent the same number of atoms of any chemical element. In the case of ionic reactions, the same electric charge must be present on both sides of the equation.

There are a number of methods for balancing chemical equations. Let’s consider two of the simplest and most common methods using examples of specific chemical reactions.

Inspection Method

As the first example, we will take the combustion of ethane (C2H6) in the presence of oxygen to form water and carbon dioxide:

C2H6+O2 → H2O+CO2↑.

As we can see, atoms of only three chemical elements are involved in this reaction. Relatively simple equations like this one can be balanced by inspection and directly fitting the stoichiometric coefficients.

Let us first write down the number of atoms of each element for both sides of the equation. Consider the subscripts next to each element to determine the total number of atoms.

• On the left side we have 2 carbon atoms, 6 hydrogen atoms and 2 oxygen atoms: C=2, H=6, O=2.
• On the right side we have 1 carbon atom, 2 hydrogen atoms and 3 oxygen atoms: C=1, H=2, O=3.

Note that usually hydrogen and oxygen are part of several molecules at once, so it is better to balance them last. If it is necessary to balance several elements, we first choose one that is part of only one molecule of reactants and one molecule of reaction products. So the first thing to do in our case is to balance the carbon.

Put a factor in front of the single carbon atom on the right side of the equation to balance it with the 2 carbons on the left side of the equation:

C2H6+O2 → H2O+2CO2↑.

After that, on the left side of the equation we will have the following atoms: C=2, H=6, O=2, and on the right: C=2, H=2, O=5.

After we have balanced the number of carbon atoms on the left and right side, hydrogen and oxygen remained unbalanced. The left side of the equation contains 6 hydrogen atoms, the same number should be on the right side. Let’s achieve this using the coefficient 3:

C2H6+O23H2O+2CO2↑.

After that, on the left side of the equation we will have the following atoms: C=2, H=6, O=2, and on the right: C=2, H=6, O=7. And all we have to do is equalize the number of oxygen atoms on both sides.

Obviously, without changing the number of atoms of other types oxygen can be balanced simply by adding a factor of 7/2 = 3.5 in front of the oxygen molecule on the left side of the equation:

C2H6+3.5O23H2O+2CO2↑.

But this, of course, makes no sense, since the molecule must be present as a whole and not some part of it. Therefore you should simply multiply all the coefficients found so far by some minimum number so that they all become integers. In our case, this number is 2. In general, if any fractional coefficients arise during equation balancing, the presence of fractions may be eliminated by multiplying all coefficients by their lowest common denominator.

So after multiplying by 2, we end up with the following final equation:

2C2H6+7O2 = 6H2O+4CO2↑.

As we can see, both sides of the equation contain the same number of carbon, hydrogen and oxygen atoms. The equation is balanced!

Algebraic Method

As the number of chemical elements involved in a reaction increases, the problem of balancing the equation becomes more complicated. In this case, the algebraic method of finding stoichiometric coefficients can be used.

As an example of the application of this method, consider the reaction of phosphorus(V) fluoride with water to form phosphoric acid and hydrogen fluoride:

PF5+H2O → H3PO4+HF.

Atoms of four chemical elements are already involved in this reaction. Let’s start by denoting the unknown coefficients in this equation with letters:

aPF5+bH2O → cH3PO4+dHF.

Next, we equate the number of atoms of each element in the left and right sides of the equation. For example, on the left we have 2b hydrogen atoms (2 in each H2O molecule), while on the right we have 3c+d hydrogen atoms (3 in each H3PO4 molecule and 1 in each HF molecule). Since the left and right sides must contain the same number of hydrogen atoms, we get: 2b =3c+d.

Let’s do this for all the elements:
P: a = c
F: 5a = d
H: 2b = 3c + d

Note that for oxygen we get O: b=4c. As can be easily checked by substituting in the previous equations, this equality is a consequence of them, not an independent relation.

It remains for us to solve the above system of linear equations in order to find the numerical values of the coefficients. This system has several solutions, since there are more variables than equations. It is necessary to find such a solution that all the coefficients have the form of the smallest possible integers.

To quickly solve this system of equations, we assign a numerical value to one of the variables. Suppose a=1.

Let’s solve the system by direct substitution and find the values of the remaining variables:
P: a = c, so c = 1
F: 5a = d, so d = 5
H: 2b = 3c + d, so b = 4

Thus, we have the following coefficients: a=1, b=4, c=1, d=5

As the result we have the following balanced equation:

PF5+4H2O = H3PO4+5HF.

However, in cases where we are dealing with more complex reactions involving many compounds, it is preferable to balance the equations using more advanced algebraic methods based on solving a system of linear equations.

Our chemical equation balancer calculator uses the Gauss-Jordan elimination algorithm for solving a system of linear equations. The method is modified for finding integer coefficients.

Syntax Guide

• Reactants and products in a chemical reaction are separated by an equal sign (=). The substances are separated by a plus sign (+).

• The formula of a substance should be entered using the upper case for the first character in the element’s name and the lower case for the second character (compare: Co – cobalt and CO – carbon monoxide).

• Indices should be entered as normal numbers after the appropriate elements or groups, e.g. H2O for a water molecule or (NH4)2SO4 for ammonium sulfate.

• Parentheses ( ), square brackets [ ] and braces (curly brackets) { } can be used in the formulas. Nested brackets are also allowed, e.g. [Co(NH3)6]Cl3. The degree of nesting is unlimited but all the brackets should be balanced.

• In a hydrated compound the middle dot (·) or the asterisk (*) must go before the water molecule formula (e.g. CuSO4·5H2O).

• To denote an ion specify its charge in curly brackets after the compound: {+2} or {2+}. Example: H{+}+CO3{2-}=H2O+CO2.

• To include an electron into a chemical equation use {-}, e.g. Fe{+3}+{-}=Fe.

• Do not enter the state of compounds such as solids (s), liquids (l) or gases (g).

Using Chemical Equation Balancer Calculator

When entering a chemical equation manually or pasting the copied equation it converts automatically to the ‘normal’ form according to the above rules. All the spaces are ignored and symbol → is converted to =. But symbols ↑ and ↓ remain in place.

Note, that both indices and charges can be denoted in the source document using <sub></sub> and <sup></sup> html tags, e.g. SO4-2, or denoted using the ‘tiny’ symbols, e.g. SO₄⁻². They are also converted automatically to the ‘normal’ form.

A single electron can be denoted as e (e<sup>-</sup>) or e⁻ (with ‘tiny’ minus).

Output format

Using the appropriate drop-down menu one can choose an output format for the balanced chemical equation:

Html – The balanced equation is represented using html tags for indices and charges. A single electron is denoted as e. Clicking the ‘Copy to clipboard’ button ( ) you can copy the result ‘as is’, including all the tags, and then you can paste it to any html-page. However, clicking Ctrl-A and Ctrl-C you can copy the result without the tags and paste it to a DOC document keeping duly formatted indices and charges.

Small indices – The balanced equation is represented using ‘tiny’ symbols for indices and charges. For example, CO₃²⁻ where unicode characters are used: ₃ = (\u2083), ² = (\u00B2), ⁻ = (\u207B). A single electron is denoted as e⁻.

Normal – The balanced equation is represented according to the above syntax guide.

Error notifications

There are a number of obvious notifications in case of error detected in the course of initial inspection of the entered equation, like Unexpected character or Brackets not balanced. The following notifications deserve special attention:

Improper equation – The entered equation has chemical elements on the left hand side that are missing on its right hand side, or vice versa.

Impossible reaction – The entered equation represents an impossible reaction. For example, the equation (NH4)2SO4=NH4OH+SO2 has only trivial solution when all the coefficients set to zero.

Multiple independent solutions – The entered equation can be balanced in an infinite number of ways. Usually it is a combination of a few different independent reactions. For example, the equation H+O=H2O+HO has no unique solution because, for instance, two solutions are 3H+2O=H2O+HO and 4H+3O=H2O+2HO which are not multiples of each other. The equation can be separated as H + O = H2O and H + O = HO, each of which does have a unique solution.

Sometimes an equation can not be uniquely balanced because it contains immutable groups in chemical compounds. For example, equation C6H5C2H5 + O2 = C6H5OH + CO2 + H2O can not be balanced, but if we replace the immutable group C6H5 with X then XC2H5 + O2 = XOH + CO2 + H2O can be easily balanced.
Note: An element’s name entered in accordance with the above syntax guide does not have to indicate a real chemical element.

Examples of Balancing Chemical Equations

Our Chemical Equation Balancer Calculator is a sophisticated online tool that makes it easy to solve even the most complex chemical equation balancing problems.

Example 1.

For example, in an article in the Journal of Chemical Education, were given three naturally occurring reactions that were described as either “difficult or incredibly challenging to balance”. One of those reactions was considered as impossible to balance. Here it is:

[Cr(N2H4CO)6]4[Cr(CN)6]3+KMnO4+H2SO4=K2Cr2O7+MnSO4+CO2+KNO3+K2SO4+H2O.

It took the author of the article several hours to balance this equation manually, and another hour to check the answer. With our equation balancer, you can get the result in a fraction of a second:

10[Cr(N2H4CO)6]4[Cr(CN)6]3+1176KMnO4+1399H2SO4=35K2Cr2O7+1176MnSO4+420CO2+660KNO3+223K2SO4+1879H2O.

Example 2.

Our chemical equation balancer easily copes with the tasks of balancing equations that also include ions and free electrons. As an example, let’s try to balance the half-equation for the reduction of dichromate(VI) ions:

Cr2O72-+H++e → Cr3++H2O.

We enter this equation into our calculator according to our syntax guide in the form:

Cr2O7{2-}+H{+}+{-}=Cr{3+}+H2O.

And, as a result, we immediately get the following balanced equation:

Cr2O72-+14H++6e=2Cr3++7H2O.

Example 3.

As an example of a reaction involving a hydrated compound, consider the reaction of pure natron, which is sodium carbonate decahydrate (Na2CO3·10H2O), with sulfuric acid (H2SO4):

Na2CO3·10H2O+H2SO4 → Na2SO4+CO2↑+H2O.

We can simply copy this equation as is and paste it into our chemical equation balancer. It is only important to make sure that there is the midpoint (·) or the asterisk (*) in front of the formula for the water molecule. As a result, we easily get the following balanced equation:

Na2CO3*10H2O+H2SO4=Na2SO4+CO2↑+11H2O.

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