The null space's insight into chemical balance

In this paper on arxiv.org, Lawrence Thorne introduces the nullspace method for balancing chemical equations. It is an improvement upon the method of solving by inspection (which results in trial-and-error and becomes unwieldy for large equations) and the method of matrix inverses (which is only applicable for equations that can be balanced in exactly one way.) Not only is the nullspace method more systematic and more widely applicable, it also provides detailed information about the all the possible ways of balancing the equation.

The problem of balancing equations

Consider the problem of balancing each of the following three chemical equations: $$\begin{eqnarray*} \rm C_2H_4 + O_2 &\leftrightarrow& \rm CO_2 + H_2O\\ \rm CO + CO_2 + H_2 & \leftrightarrow & \rm CH_4 + H_2O\rm\\ \rm FeS_2 + HNO_3 &\leftrightarrow& \rm Fe_2 (SO_4)_3 + NO + H_2 SO_4 \\ \end{eqnarray*}$$ Which of these equations can be balanced, and what are the coefficients that balance them? One of the usual problem-solving strategies is to use guessing-and-checking, often accompanied by intelligent heuristic rules about which coefficients to balance first. But searching for the answer with guessing-and-checking can lead to a lot of dead ends. Moreover, guessing-and-checking does not make it easy to see that the three equations are fundamentally different:
• The first equation can be balanced in essentially one way, as ${\rm C_2H_4} + 3{\rm O}_2 \leftrightarrow 2\, {\rm CO}_2 + 2 \rm{H_2O}$. (The only allowed variation is you can multiply the coefficients by any amount.)
• The second equation can actually be viewed as a composite of two reactions—$\rm CO_2 + CH_4 \leftrightarrow \rm 2 CO + 2H_2$ and $\rm CO + H_2O \leftrightarrow \rm CO_2 + H_2$. You can multiply each sub-reaction by any amount, then add them together, and the result will be a balanced version of the original equation. Thus, there are infinitely many different ways to balance the original equation.
• The third equation is overdetermined; it can't be balanced at all.
In this article, I will describe the null space method for balancing chemical equations. The null space method is easy, because you can use a calculator to perform the calculations instantly, and it is powerful, because it will not only give you coefficients that balance your equation, but also provide deep insight into all possible ways of balacing the equation.

The chemical composition matrix

The null space method uses the language of matrices to describe the problem of balancing a chemical equation. Then, it uses linear algebra to solve the problem efficiently while providing you with insight into the nature of the solutions.

To start, you construct the chemical composition matrix for the substances you are combining. This matrix tabulates what elements each of the substances in your equation is made of. It has one column for each substance (such as H2SO4) and one row for every atomic element involved (such as H, S, O)— each entry records how many atoms of a particular element are in a particular substance. For example, when you construct the chemical composition matrix for the combustion of ethylene (C2H4) , this is the result: Having defined the chemical composition matrix, let's examine how we can use it. Considered as a linear transformation, the chemical composition matrix converts substance amounts into element amounts: if you specify how much of each substance you have, the chemical composition matrix converts it into a list of how much of each chemical element you have. So, referring to the combustion reaction again, if we imagine you have (for example)

• 1 mole of ethylene (C2H4)
• 4 moles of molecular oxygen (O2)
• 7 moles of carbon dioxide (CO2),
• 0 moles of water (H2O), and
then a little matrix multiplication will tell you how many moles of each element (C, H, O) you have: In other words, in order to make 1 mole of ethylene, 4 moles of molecular oxygen, 7 moles of carbon dioxide, and 0 moles of water, we require 9 moles of carbon, 4 moles of hydrogen, and 22 moles of oxygen. (Which you can confirm yourself.)

The ability to take a combination of substances and tally the amounts of each element is a key part of the balancing process. For example, suppose you have a chemical equation involving some of these substances (this equation is not properly balanced): $$10\, {\rm C_2H_4} + 3\, {\rm O}_2 \leftrightarrow 15\, {\rm CO}_2 + 4\, \rm{H_2O}$$ To find out whether the equation is balanced, you count how many of each element you have on one side and on the other side. If you have the same amount of each element on each side, the equation is balanced— otherwise, the equation is unbalanced. In this case, we have:

• C: 20 on the left side, 15 on the right side. (5 extra on the left side.)
• H: 40 on the left side, 8 on the right side. (32 extra on the left side.)
• O: 6 on the left side, 34 on the right side. (28 extra on the right side.)
In other words, this equation is very much unbalanced.

The chemical composition matrix offers an easier way of counting the amounts of each element on both sides of an equation. To do so, simply list the coefficients as a vector. Then, put a minus sign on all of the products' coefficients. (As far as the math is concerned, you can put minus signs either on all the reactants or on all the products. I'm just picking the products as a convention.) We can call the resulting vector the stoichiometric vector. As an example, the coefficients of our equation $10\, {\rm C_2H_4} + 3\, {\rm O}_2 \leftrightarrow 15\, {\rm CO}_2 + 4\, \rm{H_2O}$ become the vector $$\begin{bmatrix}10 \\ 3 \\ -15 \\ -4\end{bmatrix}.$$ When we multiply our chemical composition matrix by this stoichiometric vector, we obtain the excess amount of each chemical element in our equation: $$\begin{bmatrix} 2 & 0 & 1 & 0 \\ 4 & 0 & 0 & 2 \\ 0 & 2 & 2 & 1 \\ \end{bmatrix} \begin{bmatrix}10 \\ 3 \\ -15 \\ -4\end{bmatrix} = \begin{bmatrix}5 \\ 32 \\ -28\end{bmatrix} .$$ The vector we get as an answer says that our equation has an excess of 5 moles of carbon on the left, 32 moles of hydrogen on the left, and (as indicated by the minus sign), 28 moles of oxygen on the right.

What do you expect will happen if we instead take the coefficients from a balanced equation such as ${\rm C_2H_4} + 3{\rm O}_2 \leftrightarrow 2\, {\rm CO}_2 + 2 \rm{H_2O}$? If we multiply the chemical composition matrix by this stoichiometric vector, we get $$\begin{bmatrix} 2 & 0 & 1 & 0 \\ 4 & 0 & 0 & 2 \\ 0 & 2 & 2 & 1 \\ \end{bmatrix} \begin{bmatrix}1 \\ 3 \\ -2 \\ -2\end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix} .$$ This is the signature of a balanced chemical equation— the coefficients are picked in such a way that the amount of each element on the left and right side is perfectly equal. Thus, there are no excesses on either side; when we perform the matrix multiplication, we find that each entry in the resulting vector is zero.

Here are the principles we can extract from this result.

1. When the equation is balanced, you can feed its coefficients into the chemical composition matrix and get all zeroes as output.
2. If there are no vectors you can feed into the chemical composition matrix to get all zeroes, then you know that the equation can't be balanced.
3. On the other hand, if there are several different vectors you can feed into the chemical composition matrix and get all zeroes as output, then the equation can be balanced in many different ways.
Our next question is “ how do you find all of the various inputs that produce zeros as output?”.

The null space method

Here, briefly, are the steps of the method. (I'll work through an example next to show what they mean.)
1. Form the chemical composition matrix for your reaction. $A$
2. Transpose the matrix. $A^\top$
3. Augment it with the identity matrix.$\left[A^\top | I\right]$
4. Using a calculator, put the matrix into row echelon form $\left[ E | F \right] = \text{ref} \left[A^\top | I\right]$.
5. Now check if any rows of $E$ are zero.
• If none of the rows are zero, the equation can't be balanced.
• If one of the rows is zero, the equation can be balanced in exactly one way. The corresponding row of $F$ gives you the coefficients. If the coefficients aren't in lowest terms, divide all coefficients by the smallest coefficient.
• If multiple rows are zero, the equation can be broken apart into multiple equations. The corresponding rows of $F$ tell you the coefficients of those equations. Any combination of those equations will balance the original equation.
Now I'll work through the three chemical reactions shown in the introduction.

Worked example: The combustion of ethylene has one solution

For an example equation, let's take the combustion of ethylene ${\rm C_2H_4} + {\rm O}_2 \leftrightarrow {\rm CO}_2 + \rm{H_2O}$ and show that the nullspace method produces the answer we expect, namely $${\rm C_2H_4} + 3{\rm O}_2 \leftrightarrow 2\, {\rm CO}_2 + 2 \rm{H_2O}.$$
1. First, we form the chemical composition matrix .
2. Second, we take its transpose. This means that we simply turn it sideways, making every column a row and every row a column. $$\begin{bmatrix} 2 & 4 & 0\\ 0 & 0 & 2 \\ 1 & 0 & 2 \\ 0 & 2 & 1 \end{bmatrix}$$
3. An identity matrix is a matrix that has all ones along its diagonal, and zeroes everywhere else. They come in different sizes, such as $\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$ and $\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0& 0& 1\end{bmatrix}$, and $\begin{bmatrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0& 1 & 0 \\ 0& 0& 0& 1\end{bmatrix}.$ In this step, we attach an appropriately sized identity matrix to the right hand side of our transposed matrix. (An appropriate size means it is the same height.) $$\left[ \begin{array}{ccc|ccc} 2 & 4 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 2 & 0 & 1 & 0 & 0\\ 1 & 0 & 2 & 0 & 0 & 1 & 0\\ 0 & 2 & 1 & 0 & 0 & 0 & 1\\ \end{array}\right]$$
4. Fourth, we put the matrix into row echelon form using a calculator. (We could perform this calculation by hand, of course, but it's often tedious.) Here, I've used the GNU Octave software to convert the matrix to reduced row echelon form. (Alternatively, I could have used a scientific calculator that can perform matrix math, and it doesn't matter whether you convert the matrix into row echelon form or reduced row echelon form.) $$\text{rref} \left[ \begin{array}{ccc|ccc} 2 & 4 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 2 & 0 & 1 & 0 & 0\\ 1 & 0 & 2 & 0 & 0 & 1 & 0\\ 0 & 2 & 1 & 0 & 0 & 0 & 1\\ \end{array}\right] = \left[ \begin{array}{ccc|ccc} 1 & 0 & 0 & 0 & -1 & 1 & 0\\ 0 & 1 & 0 & 0 & -1/4 & 0 & 1/2\\ 0 & 0 & 1 & 0 & 1/2 & 0 & 0\\ 0 & 0 & 0 & 1 & 3 & -2 & -2\\ \end{array}\right]$$
5. Fifth, we examine the row echelon form's left side for rows of all zero. Here, there's exactly one— the last row: $$\left[\begin{array}{ccc|cccc}0&0&0& 1 & 3 & -2 & -2\end{array}\right].$$ This means that the right hand part contains the coefficients that balance our equation. Putting all the positive terms on the left and the negative terms on the right, we get: $$1\cdot \rm C_2H_4 + 3 \cdot O_2 \leftrightarrow 2 \cdot CO_2 + 2 \cdot H_2O.$$ This balances our equation!

Worked example: Many solutions for some carbon-containing compounds

(Details to be added later.) $$\rm CO + CO_2 + H_2 \leftrightarrow \rm CH_4 + H_2O\rm$$ When the augmented chemical composition matrix is put in row echelon form, there are two rows of zeros. Each corresponds to the coefficients of a smaller reaction. By combining various multiples of the smaller reactions, you can obtain infinitely many different balanced forms of the original equation.

Worked example: No way to balance

(Details to be added later.) $$\rm FeS_2 + HNO_3 \leftrightarrow \rm Fe_2 (SO_4)_3 + NO + H_2 SO_4$$ When the augmented chemical composition matrix is put in row echelon form, there are no rows of zeros. Hence, the equation can't be balanced.

You can balance electric charge, too

When balancing charges, e.g. in redox reactions, simply add a row to the chemical composition table to account for charge. (Here, charge behaves as if it were a kind of chemical element, except the amount of charge in a substance can be negative.) The null space method will balance the equation as usual, taking into account both charge and the amount of each chemical elements.

For example, let's balance the equation $$\rm H^+ + Cr_2O_7^{2-} + C_2H_5OH \leftrightarrow Cr^{3+} + CO_2 + H_2O.$$ We can write the chemical composition matrix as $$\begin{bmatrix} 1 & 0 & 6 & 0 & 0 & 2\\ 0 & 2 & 0 & 1 & 0 & 0\\ 0 & 7 & 1 & 0 & 2 & 1\\ 0 & 0 & 2 & 0 & 1 & 0\\ +1 & -2 & 0 & +3 & 0 & 0\\ \end{bmatrix},$$ where the last row denotes electrical charge. (There are many other ways of writing this matrix by putting the columns and rows in a different order.)

According to the nullspace method, we transpose and augment the chemical composition matrix with an identity matrix, yielding $$\left[ \begin{array}{ccccc|cccccc} 1&0&0&0&+1&1&0&0&0&0&0\\ 0&2&7&0&-2&0&1&0&0&0&0\\ 6&0&1&2&0&0&0&1&0&0&0\\ 0&1&0&0&+3&0&0&0&1&0&0\\ 0&0&2&1&0&0&0&0&0&1&0\\ 2&0&1&0&0&0&0&0&0&0&1\\ \end{array}\right].$$

A quick calculation reduces this to row echelon form (not shown here). In row echelon form, we notice exactly one row containing zeros: $$\left[ \begin{array}{ccccc|cccccc} 0&0&0&0&0&1&1/8&1/16 & -1/4 & -1/8 & -11/16 \end{array}\right].$$ This mean that the equation can be balanced, and that essentially the only way to balance it is to use the coefficients on the right. To get rid of the fractions and put the coefficients in lowest terms, we will divide all of the coefficients by the smallest coefficient (1/16), obtaining: $$\left[ \begin{array}{ccccc|cccccc} 0&0&0&0&0&16&2&1 & -4 & -2 & -11 \end{array}\right].$$ Hence the balancing act is complete; we can confirm by hand that all chemical elements and electric charge are balanced by these coefficients. $$\rm 16\, H^+ + 2\, Cr_2O_7^{2-} + C_2H_5OH \leftrightarrow 4\,Cr^{3+} + 2\,CO_2 + 11\, H_2O.$$ Moreover, we know that this is essentially the only way to balance the equation.

Appendix: The mathematics behind the method

This section is intended for mathematically-minded readers who want to see an explanation of the methods we used to find a basis for the null space of a matrix.

Linear algebra as a language for chemistry

The nullspace of a matrix is the set of all input vectors that produce a zero vector as output. In particular, when the matrix is a chemical composition matrix, the nullspace is the set of all coefficients of properly balanced equations.

In chemical terms, a basis for the nullspace is a set of independently balanced sub-equations. Every balanced equation you can make using the substances in the chemical composition matrix can be written as a combination of those basis-equations.

The size (dimension) of the null space is called the nullity of the matrix. The nullity of a chemical composition matrix is the number of independent balanced equations it can be split into. If zero, the equation can't be balanced. If one, the equation can be balanced in exactly one way. If more than one, the equation can be balanced in infinitely many ways by combining different multiples of the basis vectors.

A matrix is invertible just when its nullity is zero. Roughly speaking, an invertible chemical composition matrix can't be balanced because "all of the atoms can be traced back to the particular substance they came from". This means that no reaction of any kind is possible.

The rank-nullity theorem says that the rank of a matrix plus the nullity of the matrix is equal to the dimension of the matrix. In chemical terms, the rank is roughly "the number of building block components." As mentioned already, the nullity is the number of independent reactions into which the reaction can be divided. The dimension of the matrix is the number of substances involved.

The augmentation method

(Borrowed from math.stackexchange.com.) One efficient way to produce a basis for the null space of a matrix $M_{m\times n}$ is as follows: Form the augmented matrix $$\begin{bmatrix}I_{n\times n} \\ \hline \\ M_{m\times n}\end{bmatrix},$$ then perform operations on the augmented matrix so as to put $M_{m\times n}$ into column echelon form. In echelon form, some of the rightmost columns of $M_{m\times n}$ may be filled with zeros. Schematically, we can represent this as $$\left[ \begin{array}{cccc|ccc} \vec{v}_1 & \vec{v}_2 & \cdots & \vec{v}_p & \vec{v}_{p+1} & \cdots & \vec{v}_{n}\\ \hline \vec{w}_1 & \vec{w}_2 & \cdots & \vec{w}_p & \vec{0} & \cdots & \vec{0} \end{array}, \right]$$ where all of the $\vec{w}_i$ are nonzero, and where it's possible that $p=n$ so that in echelon form $M_{m\times n}$ doesn't actually have any zero columns.

Expressed in terms of this representation, the vectors $\vec{v}_1 ,\cdots, \vec{v}_p$ form a basis of the column space of $M$, and the vectors $\vec{v}_{p+1},\cdots, \vec{v}_{n}$ form a basis of the null space of $M$.

To see that the $\vec{v}_1, \cdots, \vec{v}_{p}$ form a basis for the column space of $M$, recall that the columns of any matrix form a basis for the column space of that matrix. Next, note that column operations (used to put matrices into column echelon form) never alter the column space of a matrix. Hence, $M$ has the same column space as its column echelon form. The columns of the echelon form are a basis for its own column space, which is the same as $M$'s column space, which completes the explanation.

To see that the $\vec{v}_{p+1}, \cdots, \vec{v}_{n}$ form a basis for the null space of $M$, note that in the augmented matrix, every column operation we perform on $M$ will be “recorded” in the identity matrix. (This is exactly analogous to the way you can compute the inverse of a matrix by augmenting it with the identity matrix and using Gauss-Jordan elimination.) Each $\vec{v}_i$ is a linear combination of columns of $M$ which yields $\vec{w}_i$. When $\vec{w}_i$ is zero, therefore, $\vec{v}_i$ is a member of the null space of $M$. These vectors are independent because they started out as the identity matrix (whose columns are all independent), and column operations do not alter the column space.The fact that these vectors span the entire null space is a consequence of the rank-nullity theorem.

Thorne's method

Thorne offers, without proof, an alternative method for computing a basis for the null space. I believe it is based on QR decomposition, although I don't entirely understand how it works.

Footnotes

 To see how to balance this equation in the more typical way, you can watch this YouTube tutorial. The video is quite thorough, but the process takes nearly fifteen minutes by hand and involves balancing half-reactions first.