Chemical Equation Balancer with Steps | Free Online Tool

A chemical equation balancer finds the smallest whole-number coefficients that make every element appear the same number of times on both sides. For propane burning, C₃H₈ + O₂ → CO₂ + H₂O balances as:

Rule of thumb: change only the coefficients, never the subscripts. Balance the element found in one compound on each side first, hydrogen next, oxygen last. Here 204.087 g of reactants give 204.087 g of products.

A chemical equation is a short way to write a reaction with formulas: reactants on the left, products on the right, and an arrow between them. A balanced chemical equation is one where each element has the same number of atoms on both sides, and, for ions, the same total charge. A chemical equation balancer is a tool that finds the numbers in front of the formulas, called coefficients, that make this true.

Balancing follows from the law of conservation of mass, stated by Antoine Lavoisier in 1789: atoms are not created or destroyed in a chemical reaction, they are only rearranged. If 3 carbon atoms go in, 3 carbon atoms must come out. Writing that rule once for each element turns any equation into a small set of simultaneous equations, which is why chemical equations are “actually matrix equations” (Alberty, 1991) and why a chemical equation balancer can solve any of them exactly.

A worked one-liner: hydrogen burns in oxygen to make water. The skeleton H₂ + O₂ → H₂O has 2 oxygen atoms on the left and 1 on the right. Put 2 in front of H₂O, then 2 in front of H₂: 2 H₂ + O₂ → 2 H₂O. Now there are 4 H and 2 O on each side.

Who uses it: CBSE and ICSE students in Classes 9 to 12 (balancing starts in the NCERT Class 10 chapter on chemical reactions), JEE and NEET aspirants with long redox equations, IGCSE, GCSE and A level students, AP Chemistry and IB students, first-year university students, and teachers who need an answer key for a balancing chemical equations worksheet. Lab staff, pharmacy and water-testing technicians use balanced equations every day to turn masses into moles.

This chemical equation balancer uses the exact algebraic method to find the answer, then writes out the inspection method in teacher order so you can follow every step.

How accurate is this chemical equation balancer?

The chemical equation balancer solves the element matrix with exact whole-number (BigInt) arithmetic, so there is no rounding at any stage. It recounts every element and the total charge after solving, and the answer only gets the Verified badge if every row matches. Every equation on this page, 51 in all, was also balanced independently with SymPy in Python and gave identical coefficients.

Two cases trip up most online balancers, and this one reports them instead of guessing. An equation with a missing species has no solution; the tool says Cannot be balanced and shows which element will not match. An equation that hides two reactions has many solutions; the tool says Two independent reactions, because chemistry, not arithmetic, decides the ratio (Herndon, 1997; Olson, 1997). Students who balance by rule without picturing the particles make exactly these mistakes (Yarroch, 1985; Sanger, 2005), which is why the results include a particle diagram.

4.1 What a balanced chemical equation means

A balanced chemical equation is a counting statement. It says that the atoms present before the reaction are exactly the atoms present after it, only joined in new ways. In C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O there are 3 carbon, 8 hydrogen and 10 oxygen atoms on each side. Nothing appears and nothing vanishes.

That is why a chemical equation is balanced when the atom-count table has a tick in every row. If the species carry charges, the total charge must also match. Those two checks are exactly what the green Verified badge means.

4.2 Coefficients are mole ratios, not masses

The coefficients tell you how many particles, or how many moles, react. 1 mol of propane reacts with 5 mol of oxygen. They do not say 1 g reacts with 5 g. To go from moles to grams you multiply by the molar mass: 1 mol C₃H₈ is 44.097 g and 5 mol O₂ is 159.990 g. This is the significance of balancing chemical equations in stoichiometry: every calculation of reacting masses, gas volumes, limiting reagents and yields uses these ratios.

4.3 Atoms and mass are conserved, molecules are not

Students often expect the number of molecules to match too. It does not have to. Propane combustion turns 6 molecules into 7 (1 + 5 on the left, 3 + 4 on the right). Mass, however, is conserved exactly, because mass belongs to the atoms: 204.087 g of reactants give 204.087 g of products for one mole of reaction. The mass table in Section 3 shows this for your equation.

4.4 Why the smallest whole numbers

Any multiple of a balanced equation is also balanced: 2 C₃H₈ + 10 O₂ → 6 CO₂ + 8 H₂O is still correct. By convention the answer uses the smallest whole numbers with no common factor. Fractions such as ½ O₂ are accepted in thermochemistry (they make "per mole of product" equations), but exam markers expect whole numbers in an ordinary balanced equation.

4.5 State symbols do not change the coefficients

The symbols (s), (l), (g) and (aq) tell you the physical state at the reaction conditions, usually 25 °C and 1 atm for school work. They matter for enthalpy changes and for net ionic equations, but they never change how many atoms there are, so they never change the coefficients. This balancer keeps any state you type and can hide them.

4.6 When an equation cannot be balanced

Some equations have no balanced form. CO₂ + H₂O → C₆H₁₂O₆ fails because a product, O₂, is missing: the carbon and hydrogen equations force 6 CO₂ and 6 H₂O, which carry 18 oxygen atoms, but glucose holds only 6. The tool reports Cannot be balanced instead of forcing an answer, and shows the unbalanced atom count so you can find the missing or wrong formula.

Other equations have more than one independent answer. H₂ + O₂ → H₂O + H₂O₂ is really two reactions written as one (making water and making hydrogen peroxide), and any mix of them balances. The tool reports Two independent reactions. Chemistry, not arithmetic, decides how much of each happens, so split it into two equations.

4.7 Ionic and redox equations

In an ionic equation the total charge must balance as well as the atoms. 2 Ag⁺ + Cu → 2 Ag + Cu²⁺ has a charge of +2 on each side. Ionic redox equations in water usually need H⁺ and H₂O (acidic solution) or OH⁻ and H₂O (basic solution) added before they balance. Choose the medium in Section 2 and the balancer adds them, which reproduces the result of the half-reaction method: MnO₄⁻ + 5 Fe²⁺ + 8 H⁺ → Mn²⁺ + 5 Fe³⁺ + 4 H₂O.

4.8 A teacher's shortcut that the matrix explains

The classroom rule "oxygen last" is not superstition. Oxygen usually appears in the most species (O₂, CO₂ and H₂O in a combustion), so it is the element whose equation links all the others. Fixing the elements that sit in only one species on each side first, then hydrogen, leaves oxygen with a single free coefficient: O₂. That is the same order the chemical equation balancer uses to write its steps, and it is why a fraction like 13/2 O₂ for butane appears at the very end and is cleared by doubling.

4.9 How this links to the next topic

Balancing is step one of every quantitative chemistry problem. Next you convert grams to moles with the molar mass, use the coefficients to find the limiting reagent, and work out the theoretical and percent yield. Skipping the balancing step is the most common reason a stoichiometry answer is wrong by a whole-number factor.

▶ Balance an equation above to fill all five cards with your own answer.

This flowchart shows the steps for balancing chemical equations by inspection, on the same example used in the Excel, R, Python and by-hand sections: propane burning in oxygen, C₃H₈ + O₂ → CO₂ + H₂O.

Check any box yourself: the Excel sheet in Section 9, the R script in Section 10 and the Python script in Section 11 all return 1, 5, 3, 4 and 204.087 g on each side for this equation.

1. Type the unbalanced equation. Write reactants, an arrow, then products, with spaces around each plus sign, for example C3H8 + O2 -> CO2 + H2O.
2. Check the preview. The formulas appear under the box with proper subscripts. If a formula is wrong, a message in purple tells you what to fix.
3. Use capitals carefully. Element symbols start with a capital letter: Co is cobalt, CO is carbon monoxide, and NaCl is not NACL.
4. Add charges and states if you need them. Write ions as SO4^2- or Fe^3+, electrons as e-, and states as (s), (l), (g) or (aq). The symbol buttons insert them for you.
5. Choose the medium for ionic redox. Pick Acidic or Basic solution and the balancer adds H⁺ or OH⁻ and H₂O, as the half-reaction method does.
6. Pick the atomic weights. CIAAW 2024 values are the default; choose rounded exam values to match a textbook. This changes the mass check only.
7. Press Balance Equation. Nothing is calculated before you press it, and any change to the inputs clears the old result.
8. Read the answer box and the atom count. The Verified badge means every element and the charge match on both sides.
9. Follow the steps. Read the inspection method for the written answer, and the matrix method to see why the answer is unique.
10. Copy, export or batch. Copy the equation, download a report or PDF, or use the Batch tab to balance a whole worksheet at once.

Excel cannot read a formula like Ca(OH)2, but it can solve the element equations once you type the atom counts into cells. The trick is to fix one coefficient at 1 (the most complicated formula, here C₃H₈), move it to the right-hand side, and solve the square system that is left with MINVERSE and MMULT. The same sheet then checks atoms and mass with SUMPRODUCT. Everything below uses the propane example, so the answer must be 1, 5, 3, 4, the same as the chemical equation balancer gives in Section 3.

9.1 The functions you need

9.2 Step by step

9.3 Awkward cases

• More elements than unknowns. Combustion of propane has 3 unknowns after fixing a, and 3 elements, so it is square. KMnO₄ + HCl has 6 species, so 5 unknowns after fixing one, and 5 elements (K, Mn, O, H, Cl): square again. If you ever have more element rows than unknowns, leave one row out of the solve and use it afterwards as a check.
• Fractions in the answer. Butane gives b = 6.5. Multiply the whole column by 2.
• Ionic equations. Add a "charge" row with the ion charges (reactants positive, products negative), exactly like an element row.
• Brackets and hydrates. Count atoms by hand before typing: Ca₃(PO₄)₂ has 8 oxygen atoms, CuSO₄·5H₂O has 9.

9.4 Why your Excel answer might differ from R, Python or this page

9.5 Charting the atom count in Excel

1. Put the elements in one column and the left and right atom totals (F9:G11) next to them.
2. Insert → Charts → Clustered Column.
3. Equal pairs of bars mean the equation is balanced; this is Chart 2 in Section 3.
4. For the mass check, use a Stacked Column with one series per species.

9.6 Excel errors you will probably hit

R solves the element matrix exactly with the null space. MASS::Null() returns a basis for the null space, the script scales it to the smallest whole numbers, prints the balanced equation, verifies every element, checks the mass and saves one 300 dpi figure. It prints 1, 5, 3, 4 and 204.087 g for propane, matching the chemical equation balancer on this page.

10.1 The complete script

R with MASS, saves balanced_atoms.png

# ============================================================
# Chemical equation balancer in R: exact null space method
# StatsUnlock - https://statsunlock.com/chemical-equation-balancer/
# Needs only MASS (ships with every R install).
# ============================================================
library(MASS)

# ---- 1. Your equation ---------------------------------------
# One column per species, one row per element (atoms in ONE formula unit).
# Reactants positive, products negative.
#               C3H8  O2  CO2  H2O
A <- rbind(C = c(3,   0,  -1,   0),
           H = c(8,   0,   0,  -2),
           O = c(0,   2,  -2,  -1))
species <- c("C3H8", "O2", "CO2", "H2O")
n_left  <- 2                      # how many species are reactants
colnames(A) <- species

# Molar masses in g/mol from CIAAW 2024 atomic weights (C 12.011, H 1.008, O 15.999)
M <- c(C3H8 = 3*12.011 + 8*1.008, O2 = 2*15.999,
       CO2 = 12.011 + 2*15.999,   H2O = 2*1.008 + 15.999)

# ---- 2. Solve A %*% x = 0 ----------------------------------
N <- Null(t(A))                   # basis of the null space of A
if (ncol(N) != 1) stop(if (ncol(N) == 0) "Cannot be balanced: check the formulas."
                        else "More than one solution: this is two reactions added together.")
v <- N[, 1] / min(abs(N[, 1]))    # scale so the smallest entry is 1

# smallest whole numbers: try multipliers 1, 2, 3, ... until every entry is an integer
for (k in 1:1000) if (all(abs(k * v - round(k * v)) < 1e-6)) break
coef <- round(k * v)
if (coef[1] < 0) coef <- -coef
if (any(coef <= 0)) stop("No balance with every species on the side given.")
names(coef) <- species

# ---- 3. Print the answer ------------------------------------
term <- function(i) paste0(ifelse(coef[i] == 1, "", paste0(coef[i], " ")), species[i])
eq <- paste(paste(sapply(1:n_left, term), collapse = " + "), "->",
            paste(sapply((n_left + 1):length(species), term), collapse = " + "))
cat("Balanced equation:", eq, "\n")
cat("Coefficients     :", paste(coef, collapse = " : "), "\n")

# ---- 4. Verify ----------------------------------------------
resid <- A %*% coef
cat("Atom check (all zero means balanced):", resid, "\n")
cat("Verdict:", ifelse(all(resid == 0), "Verified, every element balances", "NOT balanced"), "\n")
mass_left  <- sum(coef[1:n_left] * M[1:n_left])
mass_right <- sum(coef[-(1:n_left)] * M[-(1:n_left)])
cat(sprintf("Mass check: reactants %.3f g, products %.3f g per mole of reaction\n",
            mass_left, mass_right))

# ---- 5. One 300 dpi figure: atoms per element, left vs right ---------
left_atoms  <- A[, 1:n_left, drop = FALSE] %*% coef[1:n_left]
right_atoms <- -A[, -(1:n_left), drop = FALSE] %*% coef[-(1:n_left)]
png("balanced_atoms.png", width = 2400, height = 1500, res = 300)
par(mar = c(4.5, 4.5, 3, 1))
barplot(rbind(left_atoms[, 1], right_atoms[, 1]), beside = TRUE,
        col = c("#2563eb", "#16a34a"), border = NA, las = 1,
        names.arg = rownames(A), xlab = "Element", ylab = "Number of atoms",
        main = eq, cex.main = 0.95, legend.text = c("Reactants", "Products"),
        args.legend = list(x = "topright", bty = "n"))
invisible(dev.off())
cat("Saved balanced_atoms.png\n")

Output: Balanced equation: C3H8 + 5 O2 -> 3 CO2 + 4 H2O, Atom check: 0 0 0, Mass check: reactants 204.087 g, products 204.087 g.

10.2 R traps

• Null(A) gives the null space of t(A), so pass t(A) to solve A x = 0.
• The basis vector is in floating point (for example 0.18, 0.90, 0.54, 0.72). Scale it before rounding, never round first.
• round() rounds half to even; that never matters here because the scaled values are whole numbers to within 10⁻⁶.
• Vectors recycle silently: the molar mass vector must be the same length as the species vector.

10.3 What each part does

10.4 Using a package for the formulas

Building the matrix by hand is fine for school equations. For many equations, the CHNOSZ package can parse formulas (makeup("C3H8")) and fill the matrix for you.

10.5 The figure

The script saves balanced_atoms.png at 2400 × 1500 pixels and 300 dpi: one blue and one green bar per element. Equal pairs of bars are the visual proof of a balanced chemical equation.

This script reads an equation as text, counts the atoms in each formula (brackets and hydrates included), builds the element matrix, and finds the exact null space with SymPy, so there is no rounding at all. It prints the balanced equation, verifies the atoms and the mass, and saves a 300 dpi chart. It is the same method this chemical equation balancer runs in your browser.

Install once: pip install sympy matplotlib

11.1 The complete script

Python with SymPy, saves balanced_atoms.png

# ============================================================
# Chemical equation balancer in Python: exact null space with SymPy
# StatsUnlock - https://statsunlock.com/chemical-equation-balancer/
# pip install sympy matplotlib
# ============================================================
import re
from math import lcm
from sympy import Matrix
import matplotlib
matplotlib.use("Agg")
import matplotlib.pyplot as plt

# ---- 1. Your equation (spaces around + and an arrow) ----------
equation = "C3H8 + O2 -> CO2 + H2O"

# CIAAW 2024 standard atomic weights (g/mol) for the elements you need
WEIGHTS = {"H": 1.008, "Li": 6.94, "C": 12.011, "N": 14.007, "O": 15.999,
           "F": 18.998, "Na": 22.990, "Mg": 24.305, "Al": 26.982, "Si": 28.085,
           "P": 30.974, "S": 32.06, "Cl": 35.45, "K": 39.098, "Ca": 40.078,
           "Cr": 51.996, "Mn": 54.938, "Fe": 55.845, "Co": 58.933, "Ni": 58.693,
           "Cu": 63.546, "Zn": 65.38, "Br": 79.904, "Ag": 107.87, "I": 126.90,
           "Ba": 137.33, "Pb": 207.2}   # add any element you need

def parse_formula(f):
    """Count atoms in a formula such as Ca(OH)2, Al2(SO4)3 or CuSO4.5H2O."""
    total = {}
    for i, part in enumerate(re.split(r"[.*·]", f)):
        mult = 1
        m = re.match(r"^(\d+)(.*)$", part)
        if m and i > 0:                        # the 5 in CuSO4.5H2O
            mult, part = int(m.group(1)), m.group(2)
        tokens = re.findall(r"[A-Z][a-z]?|\d+|[()\[\]]", part)
        stack, j = [{}], 0
        while j < len(tokens):
            t = tokens[j]; j += 1
            if t in "([":
                stack.append({}); continue
            n = 1
            if j < len(tokens) and tokens[j].isdigit():
                n = int(tokens[j]); j += 1
            if t in ")]":                      # close a bracket, multiply what is inside
                for k, v in stack.pop().items():
                    stack[-1][k] = stack[-1].get(k, 0) + v * n
            else:                              # an element symbol
                stack[-1][t] = stack[-1].get(t, 0) + n
        for k, v in stack[0].items():
            total[k] = total.get(k, 0) + v * mult
    return total

left_txt, right_txt = re.split(r"\s*(?:->|=|→)\s*", equation)
left = [s.strip() for s in left_txt.split(" + ")]
right = [s.strip() for s in right_txt.split(" + ")]
species = left + right
counts = [parse_formula(s) for s in species]
elements = sorted({e for c in counts for e in c})

# ---- 2. Element matrix: reactants +, products - --------------
A = Matrix([[c.get(e, 0) * (1 if i < len(left) else -1) for i, c in enumerate(counts)]
            for e in elements])
basis = A.nullspace()
if len(basis) != 1:
    raise SystemExit("Cannot be balanced" if not basis else "Two reactions added together: split them")
v = basis[0]
k = lcm(*[int(x.q) for x in v])          # clear every fraction
coef = [int(x * k) for x in v]
if coef[0] < 0:
    coef = [-c for c in coef]
assert all(c > 0 for c in coef), "No balance with every species on the side given"

# ---- 3. Print the answer -------------------------------------
term = lambda c, s: (f"{c} " if c != 1 else "") + s
eq = " + ".join(term(c, s) for c, s in zip(coef[:len(left)], left)) + " -> " + \
     " + ".join(term(c, s) for c, s in zip(coef[len(left):], right))
print("Balanced equation:", eq)
print("Coefficients     :", " : ".join(map(str, coef)))

# ---- 4. Verify: atoms and mass --------------------------------
resid = A * Matrix(coef)
print("Atom check (all zero means balanced):", list(resid))
print("Verdict:", "Verified, every element balances" if all(r == 0 for r in resid) else "NOT balanced")
missing = sorted({e for c in counts for e in c} - WEIGHTS.keys())
if missing:
    raise SystemExit(f"Add the atomic weight of {missing} to WEIGHTS for the mass check")
M = [sum(WEIGHTS[e] * n for e, n in c.items()) for c in counts]
m_left = sum(c * m for c, m in zip(coef[:len(left)], M[:len(left)]))
m_right = sum(c * m for c, m in zip(coef[len(left):], M[len(left):]))
print(f"Mass check: reactants {m_left:.3f} g, products {m_right:.3f} g per mole of reaction")

# ---- 5. One 300 dpi figure -------------------------------------
L = [sum(coef[i] * counts[i].get(e, 0) for i in range(len(left))) for e in elements]
R = [sum(coef[i] * counts[i].get(e, 0) for i in range(len(left), len(species))) for e in elements]
x = range(len(elements))
fig, ax = plt.subplots(figsize=(8, 5))
ax.bar([i - 0.2 for i in x], L, width=0.4, color="#2563eb", label="Reactants")
ax.bar([i + 0.2 for i in x], R, width=0.4, color="#16a34a", label="Products")
ax.set_xticks(list(x), elements)
ax.set_xlabel("Element"); ax.set_ylabel("Number of atoms"); ax.set_title(eq)
ax.legend(frameon=False)
fig.tight_layout(); fig.savefig("balanced_atoms.png", dpi=300)
print("Saved balanced_atoms.png")

Output: Balanced equation: C3H8 + 5 O2 -> 3 CO2 + 4 H2O, Atom check: [0, 0, 0], Mass check: reactants 204.087 g, products 204.087 g. Change the equation line to balance anything else; KMnO4 + HCl -> KCl + MnCl2 + H2O + Cl2 gives 2 : 16 : 2 : 2 : 8 : 5.

11.2 Python traps

• NumPy's linalg.svd gives a floating-point null space; SymPy's nullspace() gives exact fractions such as 13/2, which is why this script uses SymPy.
• ^ is bitwise XOR in Python; write powers with **.
• math.lcm needs Python 3.9 or newer.
• The formula parser here does not read charges; for ions add a "charge" row to A yourself, or use ChemPy.

11.3 The one-line alternative with ChemPy

ChemPy balances directly from formulas: from chempy import balance_stoichiometry, then balance_stoichiometry({'C3H8', 'O2'}, {'CO2', 'H2O'}) returns O₂: 5, CO₂: 3, H₂O: 4. Install it with pip install chempy.

11.4 What each part does

11.5 The figure

The script saves balanced_atoms.png at 300 dpi with reactant atoms in blue and product atoms in green for each element.

12.1 Atomic weights used by this balancer (common elements)

Source: CIAAW (IUPAC Commission on Isotopic Abundances and Atomic Weights), abridged standard atomic weights, 2024 edition, dimensionless; multiply by g/mol for molar mass. Rounded values are the ones used in most CBSE, GCSE and IGCSE papers. Atomic weights affect only the mass check, never the coefficients.

12.2 Common polyatomic ions

A polyatomic ion that appears unchanged on both sides (SO₄²⁻ in Al + CuSO₄) can be balanced as one unit by hand. The balancer counts its atoms one by one and reaches the same answer.

12.3 The seven diatomic elements

Writing O instead of O₂ for oxygen gas is the most common skeleton-equation error. Sulfur (S₈) and phosphorus (P₄) are usually written S and P at school level.

12.4 State symbols

12.5 Types of reaction and their general form

The reaction type shown in Section 3 comes from these simple rules and is a guide. Many reactions belong to more than one type (combustion is also redox).

12.6 Thirty balanced chemical equations with answers

Every equation in this answer key was balanced and checked by the same engine as the chemical equation balancer above. Use it as a balancing chemical equations worksheet answer key: cover the right-hand column and try each one first.

Eight examples run through the chemical equation balancer, from a textbook combustion to an equation that cannot be balanced. Each figure is drawn from the computed atom counts: blue bars are reactant atoms, green bars product atoms (orange where they do not match).

This is the written method examiners expect, on the same example as Sections 7, 9, 10 and 11: balance C₃H₈ + O₂ → CO₂ + H₂O.

14.1 The written method

1. Write the skeleton equation with correct formulas. C₃H₈ + O₂ → CO₂ + H₂O. Oxygen gas is O₂.
2. Draw an atom-count table. Left: C 3, H 8, O 2. Right: C 1, H 2, O 3.
3. Start with the most complicated formula (C₃H₈) and keep its coefficient at 1.
4. Balance carbon. 3 C on the left, so write 3 CO₂. Carbon: 3 = 3.
5. Balance hydrogen. 8 H on the left, and each H₂O holds 2, so write 4 H₂O. Hydrogen: 8 = 8.
6. Balance oxygen last. Right side: 3 × 2 + 4 × 1 = 10 O atoms. Each O₂ holds 2, so write 10 ÷ 2 = 5 O₂.
7. Clear any fractions. None here. (Butane gives 13/2 O₂: multiply every coefficient by 2.)
8. Check. C 3 = 3, H 8 = 8, O 10 = 10. The coefficients 1, 5, 3, 4 share no common factor. Answer: C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O.

For ionic redox equations use the half-reaction method instead: split into two half-equations, balance atoms other than O and H, add H₂O for oxygen, H⁺ for hydrogen and electrons for charge, multiply so the electrons cancel, then add (and in basic solution add OH⁻ to both sides).

14.2 On a scientific calculator

A calculator cannot read formulas, but models with an equation mode solve the element equations. On the Casio fx-991EX ClassWiz: press MENU, choose A Equation/Func, then 1 Simul Equation, then 3 unknowns. Enter one element per row with a = 1 already moved to the right: carbon 0 1 0 3, hydrogen 0 0 2 8, oxygen 2 (−)2 (−)1 0, pressing = after each number. Press = again to read x = 5 (b), y = 3 (c), z = 4 (d).

Menu names differ between models (the fx-991ES uses MODE then EQN). Check your model's manual. Basic calculators such as the fx-82 have no equation mode: use the inspection method.

14.3 Do it in the lab: proving the law of conservation of mass

Example data. Every change is within the 0.01 g resolution of the balance, so mass is conserved, just as the balanced equation says. Moles check: 0.0125 mol CuSO₄ and 0.010 mol NaOH; the 1 : 2 ratio means NaOH runs out first (it is the limiting reagent).

14.4 Balancing chemical equations practice problems with answers

Ten practice problems, easy to hard. Open each one to see the answer.

14.5 Mistakes that lose marks

Research on how students balance equations finds the same errors again and again: changing subscripts, and treating a coefficient as if it only applied to the first atom of a formula (Yarroch, 1985; Nakhleh, 1992). Check your own answer against this list, or paste it into the chemical equation balancer to see the atom count.

• Changing a subscript (writing H₂O₂ instead of 2 H₂O). That makes a different substance.
• Writing oxygen, hydrogen or chlorine gas as single atoms (O, H, Cl) instead of O₂, H₂, Cl₂.
• Putting a coefficient in the middle of a formula (Ca2(OH)2).
• Forgetting the atoms inside brackets: Ca₃(PO₄)₂ has 2 P and 8 O.
• Leaving a fraction or a common factor in the final answer (4 H₂ + 2 O₂ → 4 H₂O should be 2, 1, 2).
• Balancing atoms but not charge in an ionic equation.
• Not rechecking every element after the last change.

Four real examples

Decision tree

1. Are there charges, and does the reaction happen in water? Yes: is it redox? Use the half-reaction method (choose Acidic or Basic here). No: go on.
2. Six or more species? Yes: algebraic method. No: go on.
3. Does a polyatomic ion appear unchanged on both sides? Yes: balance it as one unit by inspection.
4. Otherwise: inspection, O and H last.

Limitations. A chemical equation balancer checks atoms and charge, not chemistry. A balanced equation says nothing about whether the reaction actually happens, how fast it goes, or how far it goes before equilibrium; for that you need thermodynamics, kinetics and equilibrium constants. The reaction-type label is rule based and can be wrong for unusual reactions; use an oxidation number calculation to confirm redox. Formulas for organic structures are read as molecular formulas only (C₂H₅OH is counted as C₂H₆O). The particle diagram is drawn only for small reactions.

What the answer is

A chemical equation balancer turns a skeleton equation into a balanced chemical equation with the smallest whole-number coefficients. For the headline example, C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O, every element balances (C 3, H 8, O 10) and 204.087 g of reactants become 204.087 g of products.

What it means chemically

A balanced equation is the law of conservation of mass written in symbols. The coefficients are the mole ratio of the reaction, which is why chemical equations must be balanced before any stoichiometry: an unbalanced equation gives wrong reacting masses, wrong gas volumes and wrong yields.

Watch out for

Never change a subscript. Write diatomic gases as O₂, H₂, N₂ and Cl₂. Balance charge as well as atoms in ionic equations. If an equation will not balance, check the formulas before you check the arithmetic.

Next steps

Use the mole ratio to find the limiting reagent and the theoretical yield, or practise with the thirty worked equations in Section 12 and the ten problems in Section 14. Whatever the question, start every quantitative problem the same way: with the chemical equation balancer.

Balanced equations tell you what should happen; these tools help you design experiments and judge whether your lab results show a real difference.

Peer-reviewed sources for the methods used in this chemical equation balancer: the algebraic and matrix methods, inspection methods, how students learn balancing, and the atomic-weight and constant data.

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15. Sanger, M. J. (2005). Evaluating students’ conceptual understanding of balanced equations and stoichiometric ratios using a particulate drawing. Journal of Chemical Education, 82(1), 131–134. https://doi.org/10.1021/ed082p131
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17. Smith, W. R., & Missen, R. W. (1997). Using Mathematica and Maple to obtain chemical equations. Journal of Chemical Education, 74(11), 1369–1371. https://doi.org/10.1021/ed074p1369
18. Tiesinga, E., Mohr, P. J., Newell, D. B., & Taylor, B. N. (2021). CODATA recommended values of the fundamental physical constants: 2018. Reviews of Modern Physics, 93(2), 025010. https://doi.org/10.1103/RevModPhys.93.025010
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