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Equation Balancer

1
H
hydrogen
1.008
2
He
helium
4.0026
3
Li
lithium
6.94
4
Be
beryllium
9.0122
5
B
boron
10.81
6
C
carbon
12.011
7
N
nidivogen
14.007
8
O
oxygen
15.999
9
F
fluorine
18.998
10
Ne
neon
20.180
11
Na
sodium
22.990
12
Mg
magnesium
24.305
13
Al
aluminum
26.982
14
Si
silicon
28.085
15
P
phosphorus
30.974
16
S
sulfur
32.06
17
Cl
chlorine
35.45
18
Ar
argon
39.948
19
K
potassium
39.098
20
Ca
calcium
40.078
21
Sc
scandium
44.956
22
Ti
titanium
47.867
23
V
vanadium
50.942
24
Cr
chromium
51.996
25
Mn
manganese
54.938
26
Fe
iron
55.845
27
Co
cobalt
58.933
28
Ni
nickel
58.693
29
Cu
copper
63.546
30
Zn
zinc
65.38
31
Ga
gallium
69.723
32
Ge
germanium
72.63
33
As
arsenic
74.922
34
Se
selenium
78.96
35
Br
bromine
79.904
36
Kr
krypton
83.798
37
Rb
rubidium
85.468
38
Sr
sdivontium
87.62
39
Y
ytdivium
88.906
40
Zr
zirconium
91.224
41
Nb
niobium
92.906
42
Mo
molybdenum
95.96
43
Tc
technetium
[97.91]
44
Ru
ruthenium
101.07
45
Rh
rhodium
102.91
46
Pd
palladium
106.42
47
Ag
silver
107.87
48
Cd
cadmium
112.41
49
In
indium
114.82
50
Sn
tin
118.71
51
Sb
antimony
121.76
52
Te
tellurium
127.60
53
I
iodine
126.90
54
Xe
xenon
131.29
55
Cs
cesium
132.91
56
Ba
barium
137.33
72
Hf
hafnium
178.49
73
Ta
tantalum
180.95
74
W
tungsten
183.84
75
Re
rhenium
186.21
76
Os
osmium
190.23
77
Ir
iridium
192.22
78
Pt
platinum
195.08
79
Au
gold
196.97
80
Hg
mercury
200.59
81
Tl
thallium
204.38
82
Pb
lead
207.2
83
Bi
bismuth
208.98
84
Po
polonium
[208.98]
85
At
astatine
[209.99]
86
Rn
radon
[222.02]
87
Fr
francium
[223.02]
88
Ra
radium
[226.03]
104
Rf
rutherfordium
[265.12]
105
Db
dubnium
[268.13]
106
Sg
seaborgium
[271.13]
107
Bh
bohrium
[270]
108
Hs
hassium
[277.15]
109
Mt
meitnerium
[276.15]
110
Ds
darmstadtium
[281.16]
111
Rg
roentgenium
[280.16]
112
Cn
copernicium
[285.17]
113
Uut
unundivium
[284.18]
114
Fl
flerovium
[289.19]
115
Uup
ununpentium
[288.19]
116
Lv
livermorium
[293]
117
Uus
ununseptium
[294]
118
Uuo
ununoctium
[294]
57
La
lanthanum
138.91
58
Ce
cerium
140.12
59
Pr
praseodymium
140.91
60
Nd
neodymium
144.24
61
Pm
promethium
[144.91]
62
Sm
samarium
150.36
63
Eu
europium
151.96
64
Gd
gadolinium
157.25
65
Tb
terbium
158.93
66
Dy
dysprosium
162.50
67
Ho
holmium
164.93
68
Er
erbium
167.26
69
Tm
thulium
168.93
70
Yb
ytterbium
173.05
71
Lu
lutetium
174.97
89
Ac
actinium
[227.03]
90
Th
thorium
232.04
91
Pa
protactinium
231.04
92
U
uranium
238.03
93
Np
neptunium
[237.05]
94
Pu
plutonium
[244.06]
95
Am
americium
[243.06]
96
Cm
curium
[247.07]
97
Bk
berkelium
[247.07]
98
Cf
californium
[251.08]
99
Es
einsteinium
[252.08]
100
Fm
fermium
[257.10]
101
Md
mendelevium
[258.10]
102
No
nobelium
[259.10]
103
Lr
lawrencium
[262.11]
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Sarah Taylor

Sarah Taylor

Last Updated September 16 - 2021

I am a professional Chemist/Blogger & Content Writer. I love to research chemistry topics and help everyone learning Organic & Inorganic Chemistry and Biochemistry. I would do anything to spend vacations on a Hill Station.

In chemistry, we deal with the chemical equations because chemical equations help us to determine the identity of substances that are being reacted (reactants), also the substances which are being formed by their reactions (products).

It is surprising that we can predict the product of the following reaction only by seeing the L.H.S of the equation. For this we have to remember some key points. Chemical equation balancer helps you complete the process digitally.

Chemical Equation Balancer

Related: How To Write Net Ionic Equations Step by Step?

How to predict the product of the following reaction?

For predicting the nature of substance during a chemical reaction, we need to look at to the nature of reacting species and the type of chemical reaction occurring.

Related: Complete demonstration of metal displacement reaction in an aqueous medium.

  • The chemical reaction between Metal and halogens results in the formation of salt. The example is the formation of NaCl such as

    2Na(metal) + Cl2(halogen) → 2NaCl(salt)

  • The chemical reaction between acid and base results in the formation of salt and water. It happens due to neutralization reaction such as

    HCl + NaOH → NaCL + H2O

  • In the synthesis reactions, the final product is the combination of reactants involved in the chemical reaction such as follows

    2Al + 3Cl2 → 2AlCl3

  • In simple displacement reactions, one part of a compounds gets replace by the other part such as follows

    PbCl4 + 2F2 → PbF4 + 2Cl2

  • For predicting the product of a reaction, the equation balancer is available on the internet. While you enter the reactants, complete equation would be displayed in a few seconds. The chemical equation product calculator works faster and is the best alternative of manual calculations.

Also learn how to determine heat of combustion and how to calculate percent yield of a reaction in chemistry manually or you can use percent yield calculator to calculate yield percentage of a chemical reaction by using online tool.

Chemical equation identity substances and determine the quantity (the number of atoms of each element) of each substance involved in the reaction. This makes balancing chemical equations calculator very significant to use.

For determining of quantity correctly, chemical equations must be completely balanced. Balancing equations calculator allows this without errors. Before we move towards the methods of balancing chemical equations, let's recall the definitions of balanced and unbalanced chemical equations.

Related: Learn what are gas laws and how many gas laws are there.

What is an unbalanced equation?

Unbalanced chemical equation has unequal number of atoms at both sides of the equations. Let's consider a general reaction for illustrating equation balancing

2A + 2B → AB

The above equation in the reaction mixture has two atoms of A and two atoms of B. Both atoms react and form a product AB, which contains only one atom of A and one atom of B.

As per the law of mass conservation, it is not possible so the given equation is unbalanced. Balance chemical equations calculator highlights the user if the equation is unbalanced. Lets consider a chemical equation this time.

Na2+Cl2 → NaCl(unbalanced equation)

The example has two atoms of sodium and two atoms of chlorine at L.H.S. Atoms of sodium and one atom of chlorine are on R.H.S making the equation unbalanced. You can find the atomic weight of substances using atomic weight calculator. Also find molecular weight calculator on this website to calculate molar mass, atomic mass and molecular weight.

While we are studying chemical reactions in chemistry, there are unbalanced chemical equations needed to be balanced. Not balancing such equations may result in ruining the entire research work due to wrong observations. Therefore, it is necessary to learn how equation balancer works & how to balance chemical equations with chemical equation calculator.

What is chemical equation balancing?

A balanced chemical equation possesses an equal number of atoms at both sides of the equation. Such as

2A + 2B → 2AB

It can be seen that there are two atoms of A and two atoms of B which on reacting with each other, changes into two molecules of AB (2AB:2A1=2•1=2 atoms, 2B1=2•1=2) Now, let us consider a chemical equation to clear the concerns

Na2 + Cl2 → 2NaCl

Just as general equation, there are two atoms of sodium (1 Na2 = 1•2 = 2) and two atoms of chlorine (1 Cl2 = 1•2 = 2) which reacts with each other to form 2 molecules of NaCl (2 NaCl:2Na = 2•1 = 2, 2 C1 = 2•1=2)

It is proved with the examples that the process of balancing chemical equations requires to balance the equal number of atoms of reactants and products. This will help you understand what happens when sodium and chlorine react with each other. This also signify the use of balanced equation calculator.

Related: Percent composition calculator helps you to calculate the percentage of each element in a compound.

How to balance chemical equations?

Balancing complex equations has many ways but each method is specific to a specific type of reaction. Ispection method is commonly used to balance chemical equations. In this method, you have to count the number of atoms of each element at both sides of equations.

This method is used to balance the number of atoms at each side of equations and another method includes the use of chemistry equation balancer.

We can determine the number of atoms by multiplying the subscript value of each element present in the compound with the coefficient value such as

AB4 + 2C2 → AC + BC(unbalanced)

We need to determine the number of atoms taking part in the reaction to balance this equation.

1A1 = 1•1 = one atom of A

1B4 = 1•4 = four atoms of B

2C2 = 2•2 = four atoms of C

  1. Since there is only one atom of A at L.H.S, there is no need to add any coefficient or subscript value with A at R.H.S.
  2. Since there are four atoms of B and four atoms of C, the product BC would be converted to 2B2C. We cannot write the product as 2B2C2 because it would show the four atoms of B and five atoms of C since there is another atom of C present in product AC (2B2C2+AC=4+1=5 unbalanced). Find this blog useful for learning more about conversion factors and chemical factors.
  3. To balance the value of C, change the subscript value of C in AC as AC2

Thus, the final equation would be such as

AB4 + 2C2 → AC2 + 2B2C(Balanced)

To balance the number of atoms at both sides of atoms, the chemical equation balancer is also available on the internet.

You just have to write the unbalanced equations, and in a few seconds, it will balance equations for you. The balancing equations calculator usually make your work error free.

We must not depend on equation balancer completely because we can't use this tool in the examination hall.

Chemical equation product calculator and the limiting reactant calculator widely used by many people. Method of these balance equation calculator are almost same however, the final equation is shown with the exact value of coefficient and subscript.

So you can predict the limiting reactant easily by seeing the equation. Due to this reason, such a balance chemical equation calculator is more preferred by the researchers.

Test your skills with balancing chemical equations practice free online.

How to balance equation?

HNO3 + Ca(OH)2 → Ca(NO3)2 +H2O

First of all, compare the total number of atoms of each element taking part in the chemical reaction.

  • Since there are two N atoms at the product side, multiply HNO3 with 2. The equation would become

    2HNO3 + Ca(OH)2 → Ca(NO3)2 + H2O(unbalanced)

  • Since the reactant side contains four atoms of hydrogen, balance the number of atoms at the product side by multiplying H2O with 2. The equation would become

    2HNO3 + Ca(OH)2 → Ca(NO3)2 + 2H2O(unbalanced)

Since both sides are containing equal numbers of atoms now, thus the given equation is balanced.

Equation Balancer

Can you balance the equation with the fraction?

While balancing the number of atoms in an unbalanced chemical equation, you might be subjected to the cases where you have to use a fractional coefficient other than a whole number coefficient.

This is being done in cases when there is no way to balance chemical equations completely as it also work on the functionality of equation balancer. Moreover, this fractional value is being used temporarily and never mentioned in the final equation.

To clarify the point, let us consider an example

C4H10 → H2O + CO2(unbalanced)

First, we have to balance the number of atoms on both sides - starting with the hydrogen atom.

Since there are ten atoms of hydrogen at L.H.S, there would be five molecules of H2O. Moreover, since there are four atoms of Carbon at L.H.S, there would be four molecules of CO2

The final equation would be written such as

C4H10 + O2 → 5H2O + 4CO2(unbalanced)

Since there are a total of 13 atoms of oxygen at R.H.S, there is not any whole number coefficient that could be multiplied by the subscript 2 to form 13, so we wrote the coefficient in fraction 13/2 such as follows

C4H10 + 13/2 O2 → 5H2O + 4CO2(balanced)

It is a rule that the final equation cannot be written in fractional coefficient, thus, to solve this fractional quantity, we multiply the entire equation with 2 to convert the fraction into the whole number which results in the final equation such as

2C4H10 + 13O2 → 10H2O + 8CO2(unbalanced)

Since putting the fractional quantity in the equation is an intermediate step, whenever you would use an online equation solver, this intermediate step would not be shown there. The balancing chemical equations calculator highlights you whether your equation is balanced or not.

How to balance the equation by oxidation number method?

This method is mainly being used to balance redox equations on the basis of oxidation numbers. The method is as follows

  1. Write the complete equation and oxidation states of each element of reactants and product

    +1*2K2 +6*2Cr2-2*7O7 + -1H-1CL

    +3Cr-1*3Cl3 + -1K-1Cl + +1*2H2-2O

  2. Now consider the atoms which do not undergo any kind of change in their oxidation numbers. From the above equation, it can be seen that the oxidation state of Cl in +1H-1Cl at reactant side is -1 which changes into zero in the product side 0Cl2.

    However in +1K-1Cl, the oxidation state of Cl is again being mentioned as -1. It predicts that there are two types of Cl molecules present in the reactant mixture - one which undergo change in oxidation states and other which do not undergo change in their oxidation state.

    Thus, we can write equation such as

    +1*2K2 +6*2Cr2-2*7O7 + -1H-1CL + HCL

    +3Cr-1*3Cl3 + 0Cl2 + -1K-1Cl + +1*2H2-2O

    • Since there are two atoms of Cr at reactant side which possess oxidation state of +6 (+1×2K2+6×2Cr2-2×7O7). After the reaction, the oxidation state of Cr changes to +3 in +3Cr-1×3Cl3.

      Thus to balance equation, multiply CrCl3 with 2 which result in equation as follows

      +1*2K2 +6*2Cr2-2*7O7 + -1H-1CL + HCL

      2+3Cr-1*3Cl3 + 0Cl2 + -1K-1Cl + +1*2H2-2O

  3. The above equation predicts that there should be 6 HCl atoms at L.H.S which undergo change in oxidation state by losing six electrons and obtain an oxidation state of zero in 0Cl2. Try using chemistry equation balancer for online solution. To balance this, multiply HCl by six such as

    +1*2K2 +6*2Cr2-2*7O7 + 6-1H-1CL + HCL

    2+3Cr-1*3Cl3 + 0Cl2 + -1K-1Cl + +1*2H2-2O

  4. Since there are two atoms of potassium at reactant side +1×2K2 +6×2Cr2-2×7O7 , multiple KCl with 2 to balance number of K at both sides of equation such as

    +1*2K2 +6*2Cr2-2*7O7 + 6-1H-1CL + HCL

    2+3Cr-1*3Cl3 + 0Cl2 + 2-1K-1Cl + +1*2H2-2O

  5. To count the total number of oxidized Cl atoms at the reactant side, sum up the total number of Cl in 2KCl and 2CrCl3 which are equal to eight. Thus, put this figure in equation such as

    +1*2K2 +6*2Cr2-2*7O7 + 6-1H-1CL + 6HCL

    2+3Cr-1*3Cl3 + 0Cl2 + 2-1K-1Cl + +1*2H2-2O

  6. Since the reactant side contains seven oxygen atoms, multiply H2O with seven.

    +1*2K2 +6*2Cr2-2*7O7 + 6-1H-1CL + 6HCL

    2+3Cr-1*3Cl3 + 0Cl2 + 2-1K-1Cl + 7+1*2H2-2O

  7. Finally, simplify the equation by removing the oxidation states and summing up the total unoxidized and oxidized Cl molecules such as

    K2Cr2O7 + 12HCl

    2CrCl3 + Cl2 +2KCl + 7H2O

The balancing chemical equations calculator becomes the best option which ensures the correctness of the final equation. For specifically calculating the oxidation number using online tools, we provide you oxidation number calculator to get the job done easily with few clicks.

How to balance chemical equation by ion-electron method

This method is also being used to balance redox equations which possess ions and an aqueous medium. The basic principle is to eliminate those ions which do not undergo a change in oxidation states by giving or gaining valence electrons. The chemical equation balancer includes this functionality as well.

We consider only those which show the change in oxidation states either by giving or gaining electrons in the valence shell. Therefore, the method is being called the ion-electron method. To understand the method, let us discuss the redox reaction between HCl and KMnO4 in which Cl and MnO4 change oxidation state such as

Cl- + MnO4- → Cl20 + Mn2+

  1. In this reaction, Cl is oxidized by losing electrons while MnO4 is reduced by gaining these electrons. So, we can divide the equation into two parts (oxidizing part and reducing part).

    Oxidizing part :

    Cl- → Cl20

    Reducing part :

    MnO4- → Mn2+

  2. Since there are two atoms of Cl on the product side, the oxidation equation would become

    2Cl- → Cl20

  3. Since the reaction is taking place in an acidic medium, add H+ at L.H.S and balance these H+ ions by adding H2O at R.H.S such as

    8H+ + MnO4- → Mn2+ + 4H2O

  4. Add the number of electron lost and gained by each part such as

    2Cl- → Cl20 + -2e

    8H+ + MnO4- + 5e → Mn2+ + 4H2O

  5. To balance the number of electrons lost and gained in both parts, multiply both parts with a specific coefficient to get the final equation. Since the reduction part is getting 5e, multiply oxidation part with 5 and since oxidation part is losing 2 electrons, multiply reduction part with 2.

    (2Cl- → Cl20 + -2e)*5

    (8H+ + MnO4- + 5e → Mn2+ + 4H2O) * 2

  6. The final equation would be obtained by summing up both parts.

    10Cl- + 16H + 2MnO4- → 5Cl2 + 2Mn2+ + 8H2O

Related:

Theoretical yield calculator can help you finding the reaction yield of a chemical reaction.

The same method is being used for a reaction occurring in basic media. However, you have to add OH ions to balance both sides other than H+ ions. For further convenience, the online way of doing it using chemical equation product calculator is also helpful.

Related: Find the difference between endothermic and exothermic reactions.

How to balance equation in the basic medium?

Let us consider a redox reaction taking place in a basic medium such as follows

H2O + MnO4-1 + C2O4-2 → MnO2 + CO2 + OH-1

  1. Since

    MnO4-1

    and

    C2O4-2

    are showing changes in oxidation state, we would consider only them.

  2. Since

    C2O4-2

    is being oxidized by giving electrons, it would constitute the oxidation part. Moreover, since

    MnO4-1

    is getting reduced by gaining electrons, it would constitute a reduction part.

  3. We will change the oxidation part for balancing the number of C atoms on both sides

    C2O4-2 → 2CO2

  4. To balance the reduction part, add two H2O molecules on the reactant side and balance it on the product side by adding four OH ions such as

    MnO4-1 + 2H2O → MnO + 4OH-

  5. Add the number of electrons gained and lose in both parts such as

    C2 O-24 → 2CO2 + -2e

    MnO4-1 + 2H2O + 3e → MnO + 4OH-

  6. Balance the number of electrons at both sides by multiplying the reduction part with 2 and the oxidation part with 3. Then add both parts to get the final equation such as

    2MnO4-1 + 3C2O4-2 +4H2O → 2MnO + 6CO2 + 8OH-

To get accurate and fast results, you can use an advanced balancing chemical equation calculator to balance chemical equations. You can use redox reaction calculator on this website for balancing redox equations and reactions.

How to use Equation Balancer?

This balance chemical equations calculator helps to balance chemical equations quickly. If you are using balancing chemical equations calculator with steps, you'll know how easy it is to operate.

There are many sample equations in this chemical equation balance calculator so that you can practice and balance equations. Also there is a chemistry periodic table under the chemical equation balancer calculator so that you can add values from there.

Once you entered your equation in the field, the balancing equations calculator will balance your equation immediately. Share this with your class mates & others so that they could also get benefit from it.

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