Web-based chemical reaction equation calculators use this method to calculate the stoichiometric coefficients for most reactions that a beginning chemistry student will encounter. Illustrated below the steps of this algebraic method of balancing chemical equations are some examples showing how the method works for some reactions typically encountered in high school and first year college chemistry courses, ranging in total atoms from 2-5 and total compounds from 3-6.

Step 1

Give each reactant and product an algebraic coefficient, starting from the left with \(a\).

Step 2

Set up a system of equations, one for each element in the reaction, starting from the left. The equations will each be a number balance for the atom in question – the number of atoms on the left side of the chemical equation equals the number of that same atom on the right side.

Step 3

Solve the system of equations, usually by setting \(a = 1\).

Step 4

If any coefficients are not whole numbers, then multiply all the intermediate results obtained by the smallest number that results in all coefficients being whole numbers, otherwise use the results from Step 3 directly to write down the balanced equation.

Step 5

Check result, just to make sure you solved the simultaneous equations correctly.

\[\ce{ N2 + H2 -> NH3 }\]

Step 1

\[\ce{ $a$ N2 + $b$ H2 -> $c$ NH3 }\]

Step 2

For N: \(2a = c\)

For H: \(2b = 3c\)

Step 3

\(a:=1\)

\(\therefore c=2\)

\(2b = 3(2) = 6; b=3\)

and \(2b-3c=0\)

Using matrices to solve: \[\begin{bmatrix} 0 & 1 \\ 2 & -3 \end{bmatrix} \begin{bmatrix} b \\ c \end{bmatrix} = \begin{bmatrix} 2 \\ 0 \end{bmatrix}\]

\[\begin{bmatrix} 3/2 & 1/2 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 2 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 2 \end{bmatrix}\]

\[b=3, c=2\]

Step 4

\[\ce{ N2 + 3 H2 -> 2 NH3 }\]

Step 5

For N: \(2(1) = 2(1)\), correct

For H: \(3(2) = 2(3)\), correct

\[\ce{ Cu + HNO3 -> Cu(NO3)2 + NO + H2O }\]

Step 1

\[\ce{ $a$ Cu + $b$ HNO3 -> $c$ Cu(NO3)2 + $d$ NO + $e$ H2O}\]

Step 2

For Cu: \(a = c\)

For H: \(b = 2e\)

For N: \(b = 2c + d\)

For O: \(3b = 6c + d + e\)

Step 3

\(a:=1\)

\(\therefore c=1\)

\(\therefore b = 2 + d; d = b - 2\)

\(\therefore 3b = 6 + b - 2 + \frac {b}{2} = 4 + b + \frac {b}{2} = 4 + \frac {3b}{2}\)

\(\therefore \frac {3b}{2} = 4; 3b = 8; b = \frac {8}{3}\)

\(\therefore e = \frac {8}{6} = \frac {4}{3}\)

\(\therefore d = \frac {8}{3} - 2(1) = \frac {2}{3}\)

and \(b-2e=0\)

\(b-2c-d=0\)

\(3b-6c-d-e=0\)

Using matrices to solve: \[\begin{bmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & -2 \\ 1 & -2 & -1 & 0 \\ 3 & -6 & -1 & -1 \end{bmatrix} \begin{bmatrix} b \\ c \\ d \\ e \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}\]

\[\begin{bmatrix} 8/3 & -1/3 & -2/3 & 2/3 \\ 1 & 0 & 0 & 0 \\ 2/3 & -1/3 & -5/3 & 2/3 \\ 4/3 & -2/3 & -1/3 & 1/3 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 8/3 \\ 1 \\ 2/3 \\ 4/3 \end{bmatrix}\]

\[b= \frac {8}{3}, c=1, d= \frac {2}{3}, e= \frac {4}{3}\]

Step 4

Multiply all coefficients obtained by 3.

\(\therefore a = 3, b = 8, c = 3, d = 2, e = 4\)

\[\ce {3 Cu + 8 HNO3 -> 3 Cu(NO3)2 + 2 NO + 4 H2O}\]

Step 5

For Cu: \(3 = 3\), correct

For H: \(8 = 4(2)\), correct

For N: \(8 = 6 + 2\), correct

For O: \(24 = 18 + 2 + 4\), correct

\[\ce {FeCr2O4 + Na2CO3 + O2 -> Na2CrO4 + Fe2O3 + CO2}\]

Step 1

\[\ce { $a$ FeCr2O4 + $b$ Na2CO3 + $c$ O2 -> $d$ Na2CrO4 + $e$ Fe2O3 + $f$ CO2 }\]

Step 2

For Fe: \(a = 2e\)

For Cr: \(2a = d\)

For O: \(4a + 3b + 2c = 4d + 3e + 2f\)

For Na: \(2b = 2d\)

For C: \(b = f\)

Step 3

\(a:=1\)

\(\therefore 2e=1\)

\(d=2\)

\(3b+2c-4d-3e-2f= -4\)

\(2b-2d=0\)

\(b-f=0\)

Using matrices to solve: \[\begin{bmatrix} 0 & 0 & 0 & 2 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 3 & 2 & -4 & -3 & -2 \\ 2 & 0 & -2 & 0 & 0 \\ 1 & 0 & 0 & 0 & -1 \end{bmatrix} \begin{bmatrix} b \\ c \\ d \\ e \\ f \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ -4 \\ 0 \\ 0 \end{bmatrix}\]

\[\begin{bmatrix} 0 & 1 & 0 & 1/2 & 0 \\ 3/4 & 3/2 & 1/2 & -1/4 & -1 \\ 0 & 0 & 0 & 0 & 0 \\ 1/2 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1/2 & -1 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ -4 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 2 \\ 7/4 \\ 2 \\ 1/2 \\ 2 \end{bmatrix}\]

\[b= 2, c= \frac {7}{4}, d= 2, e= \frac {1}{2}, f= 2\]

Step 4

Multiply all coefficients by 4.

\(a = 4, b = 8, c = 7, d = 8, e = 2, f = 8\)

\[\ce {4 FeCr2O4 + 8 Na2CO3 + 7 O2 -> 8 Na2CrO4 + 2 Fe2O3 + 8 CO2}\]

Step 5

For Fe: \(4 = 2(2)\), correct

For Cr: \(4(2) = 8\), correct

For O: \(16 + 24 + 14 = 32 + 6 + 16\), correct

For Na: \(8(2) = 8(2)\), correct

For C: \(8 = 8\), correct

\[\ce{ C8H18 + O2 -> CO2 + H2O }\]

Step 1

\[\ce{ $a$ C8H18 + $b$ O2 -> $c$ CO2 + $d$ H2O }\]

Step 2

For C: \(8a = c\)

For H: \(18a = 2d\)

For O: \(2b = 2c + d\)

Step 3

\(a:=1\)

\(\therefore c=8\)

\(2d=18\)

\(2b-2c-d=0\)

Using matrices to solve: \[\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 2 \\ 2 & -2 & -1 \end{bmatrix} \begin{bmatrix} b \\ c \\ d \end{bmatrix} = \begin{bmatrix} 8 \\ 18 \\ 0 \end{bmatrix}\]

\[\begin{bmatrix} 1 & 1/4 & 1/2 \\ 1 & 0 & 0 \\ 0 & 1/2 & 0 \end{bmatrix} \begin{bmatrix} 8 \\ 18 \\ 0 \end{bmatrix} = \begin{bmatrix} 25/2 \\ 8 \\ 9 \end{bmatrix}\]

\[b= \frac {25}{2}, c= 8, d= 9\]

Step 4

Multiply all coefficients by 2.

\(a = 2, b = 25, c = 16, d = 18\)

\[\ce {2 C8H18 + 25 O2 -> 16 CO2 + 18 H2O}\]

Step 5

For C: \(2(8) = 16\), correct

For H: \(2(18) = 18(2)\), correct

For O: \(25(2) = 16(2) + 18\), correct

There are some very complicated chemical reactions that the algebraic method doesn't work to balance, but most reactions that are covered in high school and college chemistry courses can be balanced this way. Typically, matrices haven't been involved when using this method, but their use permits a standardized approach, and many tools online (i.e. here) and elsewhere are available for learning about and using matrices to solve a system of algebraic equations in any amount of variables.

KaTeX, a Khan Academy LaTeX environment, was used to render the math symbols in this paper, and mhchem (which is included in the KaTeX distribution) was used to produce the chemical symbols.