Stoichiometry Calculator with Steps | StatsUnlock

A stoichiometry calculator uses a balanced equation to turn the amount of one substance into the amount of another, in three steps: convert to moles, multiply by the mole ratio, convert to the unit you want. For 4.00 g of hydrogen burning, 2 H₂ + O₂ → 2 H₂O:

Rule of thumb: never go from grams to grams directly. Only moles follow the coefficients, so every stoichiometry calculation passes through moles. Here 4.00 g of H₂ needs 31.7 g of O₂ and makes 35.7 g of water.

Stoichiometry is the calculation of quantities in chemical reactions: how much product a given amount of reactant makes, or how much of one reactant is needed to use up another. The word comes from the Greek stoicheion (element) and metron (measure). A stoichiometry calculator does these calculations from a balanced chemical equation and one known amount. This one is a chemistry stoichiometry calculator, an equation balancer and a unit conversion calculator in one: type the reaction as you would write it, and it balances the equation, finds every coefficient and mole ratio, and shows the work for each step.

The key fact is that the coefficients in a balanced equation are mole ratios. In 2 H₂ + O₂ → 2 H₂O, 2 mol of hydrogen react with 1 mol of oxygen to make 2 mol of water. They are not mass ratios: 2 g of hydrogen does not react with 1 g of oxygen. That is why every stoichiometry problem is solved through moles.

A worked one-liner: how many moles of ammonia form from 3.00 mol of hydrogen in N₂ + 3 H₂ → 2 NH₃? n(NH₃) = 3.00 × 2/3 = 2.00 mol.

Who uses it: CBSE and ICSE students meet stoichiometry in NCERT Class 10 and Class 11 (“Some Basic Concepts of Chemistry”); JEE and NEET use it in nearly every physical chemistry problem; IGCSE, GCSE, A level, AP Chemistry and IB students use it for every reacting-mass and titration question. In chemical engineering the subject is usually taught as stoichiometry and process calculations, where the same mole ratios feed material balances; lab staff use it to scale a recipe or work out a yield.

What this free online stoichiometry calculator solves

One page covers every stoichiometry word problem in a typical course. Press Try it to load an example into the calculator, then press Calculate to see the full working.

Free, with no sign-up. Everything runs in your browser: nothing you type is sent to a server or stored.

How accurate is this stoichiometry calculator?

The calculator balances the equation with exact whole-number arithmetic, so the mole ratio is never rounded. Molar masses use the CIAAW 2024 standard atomic weights for all 118 elements (Prohaska et al., 2022), or rounded exam values if you choose them. Every step keeps full precision; only the final answer is rounded, to the significant figures of the numbers you typed. Each answer is checked two ways before it gets the Verified badge: the mass of reactants used must equal the mass of products formed, and the mole ratio is applied in reverse to give back the starting moles. The R and Python scripts in Sections 10 and 11 return the same 35.7 g for the headline problem.

Research on how students learn stoichiometry finds that the hardest part is not the arithmetic but knowing which ratio to use and when to convert to moles (Schmidt & Jignéus, 2003; Fach et al., 2007). That is why every result shows the three steps separately and as one factor-label line.

4.1 The mole ratio is the heart of every stoichiometry calculation

All stoichiometry calculations involve one important step: converting to moles so you can use the mole ratio. The ratio comes straight from the coefficients. In 2 H₂ + O₂ → 2 H₂O the ratios are 2 mol H₂ : 1 mol O₂ : 2 mol H₂O, so 1.984 mol of hydrogen needs 0.992 mol of oxygen and makes 1.984 mol of water. Write the ratio as a fraction with the substance you want on top and the substance you know underneath, and the units cancel.

4.2 Why grams do not follow the coefficients

Each substance has a different molar mass, so equal moles are unequal grams. In the headline problem 4.00 g of hydrogen reacts with 31.7 g of oxygen, eight times the mass, although only half the moles. Chart 1 (moles) and Chart 2 (grams) in your results show this side by side: the bars have the coefficient pattern in Chart 1 and a completely different pattern in Chart 2.

4.3 Reading the stoichiometry table

The table under the steps lists every substance in the equation with its coefficient, molar mass, moles and grams. Reactants show the amount used, products the amount formed. It answers several questions at once: how much oxygen is needed, how much water forms and how much of every other product appears, all from one known amount. The totals row checks the law of conservation of mass: grams used equals grams formed.

4.4 Theoretical yield, actual yield and percent yield

The answer a stoichiometry calculator gives for a product is the theoretical yield: the most that can form if every particle of the limiting reactant reacts. Real experiments lose some product to side reactions, incomplete reaction and transfer losses, so the actual yield is lower. Percent yield = actual ÷ theoretical × 100. For 100.0 g of Fe₂O₃ the theoretical yield of iron is 69.94 g; if you collect 60.0 g, the percent yield is 85.8%. A percent yield above 100% means the product is wet or impure, not that you made extra.

4.5 The limiting reagent decides the yield

With amounts of two reactants, divide each one’s moles by its coefficient. The smallest result belongs to the limiting reagent; it runs out first and sets the yield. For 10.0 g H₂ (4.960 mol ÷ 2 = 2.480) and 64.0 g O₂ (2.000 mol ÷ 1 = 2.000), oxygen is limiting even though there are fewer grams of hydrogen. The water formed is 4.000 mol, or 72.1 g, and 1.94 g of hydrogen is left over. Chart 3 shows this test as bars.

4.6 Gas volumes and the molar volume

For a gas product the calculator multiplies moles by the molar volume at the conditions you chose. The same 0.2498 mol of CO₂ fills 5.67 L at IUPAC STP (22.711 L/mol), 5.60 L at the older STP of 0 °C and 1 atm (22.414 L/mol), and 6.19 L at SATP. Always state the conditions. Most CBSE and older US textbooks still use 22.4 L/mol, while IUPAC and many university courses use 22.7 L/mol.

4.7 Significant figures in stoichiometry

Coefficients are exact counts and molar masses have four or more figures, so the measured amount you type usually limits the precision. 4.00 g has three significant figures, so the answer is 35.7 g, not 35.744 g. For solution problems both the volume and the molarity count: 25.0 mL of 0.100 M gives three figures. The calculator keeps full precision through every step and rounds once, so the factor-label line and the step-by-step working agree.

4.8 A field note on where stoichiometry goes wrong

In marking stoichiometry work, the same three errors appear again and again: using an unbalanced equation, multiplying grams by the mole ratio, and turning the ratio upside down. Studies of students solving algorithmic stoichiometry problems report the same pattern: many can carry out the steps without understanding why the ratio applies to moles (Schmidt & Jignéus, 2003; Sanger, 2005). The calculator balances first, labels every factor with its unit and substance, and shows the ratio with the wanted substance on top, so each of these errors becomes visible.

4.9 Multi-step and reaction-sequence stoichiometry

When one reaction feeds another, the product moles of the first become the given moles of the second. For the blast furnace sequence 2 C + O₂ → 2 CO, then Fe₂O₃ + 3 CO → 2 Fe + 3 CO₂, find the moles of CO from the carbon, then use the second ratio. You can chain problems in this stoichiometry calculator by choosing Answer in: moles for the first and typing that value as the given amount of the second, or multiply the two mole ratios together (Section 6, formula ⑤).

▶ Solve a problem above to fill all five cards with your own answer.

These seven cards are the stoichiometry calculation notes in one place: every formula the calculator uses, with a worked number for each, covering stoichiometry calculations with chemical formulas and equations.

This flowchart shows how to do stoichiometry calculations, on the same problem used in the Excel, R, Python and by-hand sections: what mass of water forms when 4.00 g of hydrogen burns in oxygen?

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 1.984 mol and 35.7 g for this problem.

1. Type the chemical equation. Write it as H2 + O2 -> H2O with spaces around each plus sign. It does not need to be balanced; the calculator balances it for you.
2. Check the capital letters and charges. Co is cobalt and CO is carbon monoxide. Write ions as Fe^3+ or SO4^2-, and pick acidic or basic for ionic redox equations.
3. Choose the problem type. Pick one known amount for mass to mass, mole to mole, gas and solution problems, or two or more reactants for a limiting reagent problem.
4. Pick the substance you know. The list fills from your equation. Choose its amount type: mass, moles, gas volume, solution or number of particles.
5. Enter the amount and unit. Type the number exactly as measured, including trailing zeros such as 4.00, because they set the significant figures. For a solution, add its molarity.
6. Pick the substance to find. Choose the answer unit: grams, moles, litres of gas, volume of a solution or number of particles.
7. Set the gas conditions and atomic weights. Choose IUPAC STP, the older 22.4 L/mol STP, SATP, RTP or custom temperature and pressure, and CIAAW or rounded exam weights.
8. Add the actual yield if you have one. Type the mass you collected and the calculator gives the percent yield.
9. Press Calculate Stoichiometry. Nothing is calculated before you press it, and changing any input clears the old answer.
10. Read the steps, table and charts, then copy or share. Copy the 3-step working or the factor-label line, download a report, share a link, or solve a whole worksheet on the Batch tab.

Excel cannot balance an equation, so you type the balanced coefficients yourself and let the sheet do the three steps. Lay out one row per substance with its coefficient and molar mass; then each step is one formula. The sheet below uses the headline problem, 4.00 g of hydrogen burning, so it must return 1.98413 mol and 35.7 g of water, the same as the stoichiometry calculator above.

9.1 The functions you need

9.2 Step by step

9.3 Awkward cases

• Gas volumes. Multiply the moles by the molar volume for your conditions, or use =G4*0.08314462618*(T+273.15)/P_bar for any temperature and pressure.
• Solutions. Convert mL to L before multiplying by molarity: =25/1000*0.1 gives 0.0025 mol.
• Reaction sequences. Put each reaction in its own block and point the second block’s given moles at the first block’s product moles.
• Particles. Divide by 6.02214076E+23 to get moles first.

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

9.5 Charting in Excel

1. Select the substance names and their moles (A2:A4, G2:G4).
2. Insert → Charts → Column. The bar heights follow the coefficients, as in Chart 1 above.
3. Repeat with the grams column (H2:H4) to see that masses do not follow the coefficients.

9.6 Excel errors you will probably hit

This base R stoichiometry calculator program takes the balanced equation as named coefficients, checks that it really is balanced, works out each molar mass from the formula, runs the three steps, prints a stoichiometry table, checks that mass is conserved, finds the limiting reagent, and saves one 300 dpi chart. For 4.00 g of hydrogen it prints 1.9841 mol and 35.7 g of water, matching the stoichiometry calculator on this page.

10.1 The complete script

R, base only, saves stoichiometry.png

# ============================================================
# Stoichiometry in R: grams -> moles -> mole ratio -> grams
# StatsUnlock - https://statsunlock.com/stoichiometry-calculator/
# Base R only. No packages to install.
# ============================================================

# ---- 1. Your problem: the BALANCED equation as coefficients -----
reactants <- c(H2 = 2, O2 = 1)     # 2 H2 + O2 -> 2 H2O
products  <- c(H2O = 2)
given     <- "H2"; given_g <- 4.00 # the amount you know, in grams
wanted    <- "H2O"                 # the substance you want
sig_figs  <- 3                     # significant figures in your data

# CIAAW 2024 standard atomic weights (g/mol); add any element you need
W <- c(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, Mn = 54.938,
       Fe = 55.845, Cu = 63.546, Zn = 65.38, Br = 79.904, Ag = 107.87,
       I = 126.90, Ba = 137.33, Au = 196.97, Pb = 207.2)

# Count atoms, multiplying through brackets and the hydrate dot
parse_formula <- function(f) {
  total <- c()
  parts <- strsplit(f, "[.*]")[[1]]
  for (i in seq_along(parts)) {
    part <- parts[i]; mult <- 1
    if (i > 1 && grepl("^[0-9]+", part)) {
      mult <- as.numeric(sub("^([0-9]+).*", "\\1", part))
      part <- sub("^[0-9]+", "", part)
    }
    tok <- regmatches(part, gregexpr("[A-Z][a-z]?|[0-9]+|[][()]", part))[[1]]
    stack <- list(numeric(0)); j <- 1
    while (j <= length(tok)) {
      t <- tok[j]; j <- j + 1
      if (t %in% c("(", "[")) { stack[[length(stack) + 1]] <- numeric(0); next }
      n <- 1
      if (j <= length(tok) && grepl("^[0-9]+$", tok[j])) { n <- as.numeric(tok[j]); j <- j + 1 }
      top <- length(stack)
      if (t %in% c(")", "]")) {
        inner <- stack[[top]] * n; stack[[top]] <- NULL; top <- top - 1
        for (k in names(inner)) stack[[top]][k] <- sum(stack[[top]][k], inner[k], na.rm = TRUE)
      } else {
        stack[[top]][t] <- sum(stack[[top]][t], n, na.rm = TRUE)
      }
    }
    for (k in names(stack[[1]])) total[k] <- sum(total[k], stack[[1]][k] * mult, na.rm = TRUE)
  }
  total
}

molar_mass <- function(f) {
  counts <- parse_formula(f)
  if (any(!names(counts) %in% names(W))) stop("Add the atomic weight of ", paste(setdiff(names(counts), names(W)), collapse = ", "))
  sum(counts * W[names(counts)])
}

# ---- 2. Check the equation is balanced ---------------------------
side_atoms <- function(side) {
  tot <- c()
  for (f in names(side)) {
    a <- parse_formula(f) * side[[f]]
    for (k in names(a)) tot[k] <- sum(tot[k], a[k], na.rm = TRUE)
  }
  tot[order(names(tot))]
}
if (!isTRUE(all.equal(side_atoms(reactants), side_atoms(products)))) stop("The equation is not balanced")
coef <- c(reactants, products)
M <- sapply(names(coef), molar_mass)

# ---- 3. The three steps ------------------------------------------
if (given_g <= 0) stop("The amount must be greater than zero")
n_given  <- given_g / M[[given]]                 # step 1: grams -> moles
ratio    <- coef[[wanted]] / coef[[given]]       # step 2: mole ratio
n_wanted <- n_given * ratio
m_wanted <- n_wanted * M[[wanted]]               # step 3: moles -> grams
cat(sprintf("Step 1  n(%s) = %.2f g / %.3f g/mol = %.4f mol\n", given, given_g, M[[given]], n_given))
cat(sprintf("Step 2  n(%s) = %.4f mol x %g/%g = %.4f mol\n", wanted, n_given, coef[[wanted]], coef[[given]], n_wanted))
cat(sprintf("Step 3  m(%s) = %.4f mol x %.3f g/mol = %s g\n", wanted, n_wanted, M[[wanted]], signif(m_wanted, sig_figs)))

# ---- 4. Stoichiometry table and mass balance ---------------------
xi <- n_given / coef[[given]]                    # extent of reaction, mol
tab <- data.frame(substance = names(coef), coef = as.vector(coef), M = round(M, 3),
                  mol = round(coef * xi, 4), grams = round(coef * xi * M, 3), row.names = NULL)
print(tab)
m_in  <- sum(reactants * xi * M[names(reactants)])
m_out <- sum(products  * xi * M[names(products)])
stopifnot(abs(m_in - m_out) < 1e-3 * m_in)
cat(sprintf("Verdict: Verified, reactants %.3f g = products %.3f g\n", m_in, m_out))

# ---- 5. Limiting reagent (optional): grams of each reactant -------
start_g <- c(H2 = 10.0, O2 = 64.0)
test <- (start_g / M[names(start_g)]) / reactants[names(start_g)]
lim <- names(which.min(test))
cat(sprintf("Limiting reagent: %s; theoretical yield of %s = %s g\n", lim, wanted,
            signif(min(test) * coef[[wanted]] * M[[wanted]], sig_figs)))

# ---- 6. One 300 dpi figure: moles of every substance --------------
png("stoichiometry.png", width = 2100, height = 1350, res = 300)
par(mar = c(4.5, 5, 3, 1))
barplot(coef * xi, names.arg = names(coef), las = 1, border = NA,
        col = ifelse(names(coef) %in% names(reactants), "#2563eb", "#16a34a"),
        xlab = "Substance (blue used, green formed)", ylab = "Amount (mol)",
        main = sprintf("%.2f g %s makes %s g %s", given_g, given, signif(m_wanted, sig_figs), wanted))
invisible(dev.off())
cat("Saved stoichiometry.png\n")

Output: Step 1 n(H2) = 4.00 g / 2.016 g/mol = 1.9841 mol, Step 3 m(H2O) = 1.9841 mol x 18.015 g/mol = 35.7 g, Verdict: Verified, reactants 35.744 g = products 35.744 g, Limiting reagent: O2; theoretical yield of H2O = 72.1 g.

10.2 R traps

• coef[wanted] keeps the name; coef[[wanted]] gives the plain number. Use double brackets inside arithmetic.
• signif() rounds to significant figures; round() rounds to decimal places.
• R prints 4.00 as 4. Keep track of significant figures yourself with sig_figs.
• The balance check compares sorted atom counts; an unbalanced equation stops the script before any number is printed.

10.3 What each part does

10.4 Balancing in R

The script expects a balanced equation. To balance one in R, build the element-by-species matrix and find its null space with MASS::Null(), then scale to whole numbers; or balance it with the calculator in Section 2 and copy the coefficients from the balanced equation it shows.

10.5 The figure

The script saves stoichiometry.png at 2100 × 1350 pixels and 300 dpi: one bar per substance showing the moles used (blue) and formed (green), 1.98, 0.99 and 1.98 mol for the headline problem.

This Python stoichiometry calculator program does the same job: it checks the equation is balanced, computes the molar masses, runs the three stoichiometry steps, prints a table of every substance, verifies conservation of mass, finds the limiting reagent and saves a 300 dpi chart. It is the same method the stoichiometry calculator runs in your browser.

Install once: pip install matplotlib

11.1 The complete script

Python, saves stoichiometry.png

# ============================================================
# Stoichiometry in Python: grams -> moles -> mole ratio -> grams
# StatsUnlock - https://statsunlock.com/stoichiometry-calculator/
# pip install matplotlib
# ============================================================
import re
import matplotlib
matplotlib.use("Agg")
import matplotlib.pyplot as plt

# ---- 1. Your problem: the BALANCED equation as coefficients -----
reactants = {"H2": 2, "O2": 1}     # 2 H2 + O2 -> 2 H2O
products  = {"H2O": 2}
given, given_g = "H2", 4.00        # the amount you know, in grams
wanted = "H2O"                     # the substance you want
sig_figs = 3                       # significant figures in your data

# CIAAW 2024 standard atomic weights (g/mol); add any element you need
W = {"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, "Mn": 54.938,
     "Fe": 55.845, "Cu": 63.546, "Zn": 65.38, "Br": 79.904, "Ag": 107.87,
     "I": 126.90, "Ba": 137.33, "Au": 196.97, "Pb": 207.2}

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

def molar_mass(f):
    counts = parse_formula(f)
    missing = [e for e in counts if e not in W]
    if missing:
        raise SystemExit(f"Add the atomic weight of {missing} to W")
    return sum(n * W[e] for e, n in counts.items())

# ---- 2. Check the equation is balanced ---------------------------
def atoms(side):
    tot = {}
    for f, c in side.items():
        for e, n in parse_formula(f).items():
            tot[e] = tot.get(e, 0) + c * n
    return tot
if atoms(reactants) != atoms(products):
    raise SystemExit(f"Not balanced: {atoms(reactants)} vs {atoms(products)}")
coef = {**reactants, **products}
M = {f: molar_mass(f) for f in coef}

# ---- 3. The three steps ------------------------------------------
if given_g <= 0:
    raise SystemExit("The amount must be greater than zero")
n_given = given_g / M[given]                       # step 1: grams -> moles
ratio = coef[wanted] / coef[given]                 # step 2: mole ratio
n_wanted = n_given * ratio
m_wanted = n_wanted * M[wanted]                    # step 3: moles -> grams
print(f"Step 1  n({given}) = {given_g:.2f} g / {M[given]:.3f} g/mol = {n_given:.4f} mol")
print(f"Step 2  n({wanted}) = {n_given:.4f} mol x {coef[wanted]}/{coef[given]} = {n_wanted:.4f} mol")
print(f"Step 3  m({wanted}) = {n_wanted:.4f} mol x {M[wanted]:.3f} g/mol = {m_wanted:.{sig_figs}g} g")

# ---- 4. Stoichiometry table and mass balance ---------------------
xi = n_given / coef[given]                         # extent of reaction, mol
print(f"{'Substance':10}{'coef':>5}{'M g/mol':>10}{'mol':>10}{'g':>10}")
for f, c in coef.items():
    print(f"{f:10}{c:>5}{M[f]:>10.3f}{c * xi:>10.4f}{c * xi * M[f]:>10.3f}")
m_in = sum(c * xi * M[f] for f, c in reactants.items())
m_out = sum(c * xi * M[f] for f, c in products.items())
assert abs(m_in - m_out) < 1e-3 * m_in, "mass is not conserved"
print(f"Verdict: Verified, reactants {m_in:.3f} g = products {m_out:.3f} g")

# ---- 5. Limiting reagent (optional): grams of each reactant -------
start_g = {"H2": 10.0, "O2": 64.0}
test = {f: (g / M[f]) / reactants[f] for f, g in start_g.items()}
lim = min(test, key=test.get)
print(f"Limiting reagent: {lim}; theoretical yield of {wanted} = "
      f"{test[lim] * coef[wanted] * M[wanted]:.{sig_figs}g} g")

# ---- 6. One 300 dpi figure: moles of every substance --------------
names = list(coef)
fig, ax = plt.subplots(figsize=(7, 4.5))
ax.bar(names, [coef[f] * xi for f in names],
       color=["#2563eb" if f in reactants else "#16a34a" for f in names])
ax.set_xlabel("Substance (blue used, green formed)")
ax.set_ylabel("Amount (mol)")
ax.set_title(f"{given_g:.2f} g {given} makes {m_wanted:.{sig_figs}g} g {wanted}")
fig.tight_layout()
fig.savefig("stoichiometry.png", dpi=300)
print("Saved stoichiometry.png")

Output: Step 1 n(H2) = 4.00 g / 2.016 g/mol = 1.9841 mol, Step 2 n(H2O) = 1.9841 mol x 2/2 = 1.9841 mol, Step 3 m(H2O) = 1.9841 mol x 18.015 g/mol = 35.7 g, Limiting reagent: O2; theoretical yield of H2O = 72.1 g.

11.2 Python traps

• Integer coefficients divide to floats in Python 3 (2/3 is 0.667), so the ratio is safe. In Python 2 it would be 0.
• f"{m:.3g}" rounds to 3 significant figures; round(m, 3) rounds to 3 decimal places.
• 6.02214076e23 is how Python writes 6.022 × 10²³; 6.022*10^23 is wrong because ^ is bitwise XOR.
• Dictionary keys are case sensitive: "CO" and "Co" are different substances, as they should be.

11.3 Using packages

With pip install sympy, sympy.Matrix(A).nullspace() balances an equation from its element matrix. The chempy package offers balance_stoichiometry({'H2','O2'}, {'H2O'}), which returns the coefficients 2, 1 and 2 directly.

11.4 What each part does

11.5 The figure

The script saves stoichiometry.png at 300 dpi with the moles of each substance, reactants in blue and products in green.

12.1 Mole ratios in common reactions

Each balanced equation and number below was computed by the same engine as the stoichiometry calculator. The last column is a quick check for gram to gram stoichiometry: the grams of the main product made from exactly 1.00 g of the first reactant.

12.2 Molar masses of common reactants and products

Molar masses from CIAAW 2024 standard atomic weights. Exam papers often use rounded values (H 1, C 12, O 16, Cl 35.5), which change only the last digit.

12.3 Molar volume of a gas for gas stoichiometry

V_m = RT ÷ P with R = 8.314 462 618 J mol⁻¹ K⁻¹. CBSE and older textbooks use 22.4 L at 0 °C and 1 atm; IUPAC STP since 1982 is 0 °C and 1 bar.

12.4 Stoichiometry conversion factors at a glance

12.5 Stoichiometry worksheet with answers: mole-mole, gram to mole and gram to gram calculations

Twenty-five stoichiometry calculations in the three classic worksheet groups: mole-mole calculations, gram to mole calculations and mass-mass (gram to gram) calculations. Every answer was checked by the calculator engine with CIAAW 2024 atomic weights. Cover the last two columns and try them first; answers with rounded exam weights may differ in the last digit.

Eight stoichiometry calculation examples run through the calculator, from a textbook gram to gram problem to an Avogadro’s number problem. Each figure is drawn from the computed numbers: one bar per substance, length proportional to its moles, blue for reactants used and green for products formed.

This is the written method examiners expect, on the same problem as Sections 7, 9, 10 and 11: what mass of water forms when 4.00 g of hydrogen burns in oxygen?

14.1 The written method

1. Write and balance the equation. 2 H₂ + O₂ → 2 H₂O.
2. Write what you know and what you want under the equation. 4.00 g under H₂, ? g under H₂O.
3. Work out the molar masses you need. M(H₂) = 2 × 1.008 = 2.016 g/mol; M(H₂O) = 2 × 1.008 + 15.999 = 18.015 g/mol.
4. Step 1: convert to moles. n(H₂) = 4.00 g ÷ 2.016 g/mol = 1.984 mol.
5. Step 2: use the mole ratio. n(H₂O) = 1.984 mol × 2/2 = 1.984 mol.
6. Step 3: convert to the unit asked for. m(H₂O) = 1.984 mol × 18.015 g/mol = 35.74 g.
7. Round once, at the end. 4.00 has 3 significant figures, so m = 35.7 g.
8. Check. Oxygen used = 0.992 mol × 31.998 = 31.7 g, and 4.00 + 31.7 = 35.7 g. Mass is conserved.

For a gas, replace step 3 with V = n × V_m. For a solution, use n = c × V in step 1 or V = n ÷ c in step 3. For a limiting reagent problem, do step 1 for every reactant and divide each by its coefficient before step 2.

14.2 How to calculate moles in stoichiometry

Turn whatever you are given into moles first: n = m ÷ M for a mass, n = V ÷ V_m for a gas, n = c × V for a solution and n = N ÷ N_A for particles. Then multiply by the mole ratio. For 4.00 g of H₂: n = 4.00 ÷ 2.016 = 1.984 mol, and n(H₂O) = 1.984 × 2/2 = 1.984 mol.

14.3 How to calculate mass in stoichiometry

Multiply the moles of the wanted substance by its molar mass: m = n × M. Water from the example: 1.984 mol × 18.015 g/mol = 35.7 g. Oxygen used: 0.992 mol × 31.998 g/mol = 31.7 g.

14.4 How to calculate volume in stoichiometry

For a gas, multiply moles by the molar volume: 0.2498 mol of CO₂ × 22.711 L/mol = 5.67 L at 0 °C and 1 bar. For a solution, divide moles by molarity: 1.25 mmol of H₂SO₄ ÷ 0.0500 mol/L = 25.0 mL.

14.5 How to calculate theoretical yield using stoichiometry

Find the limiting reagent, take its moles through the mole ratio to the product, and convert to grams. With 10.0 g of H₂ and 64.0 g of O₂, oxygen limits and the theoretical yield is 4.000 mol × 18.015 = 72.1 g of water. Percent yield is then actual ÷ theoretical × 100; collecting 60.0 g of iron when 69.94 g was possible is an 85.8% yield.

14.6 On a scientific calculator

On a Casio fx-991EX or fx-82MS the whole calculation is one line: 4 ÷ 2.016 × 2 ÷ 2 × 18.015 = gives 35.744. Working left to right keeps full precision; do not round 1.984 before multiplying.

For particles, use the ×10^x (or EXP) key for 6.022 × 10²³, never × 10 ^ 23 typed by hand. Key names differ between models; check your manual.

14.7 Do it in the lab: testing stoichiometry with baking soda and acid

Example data. A yield of 92.7% is typical for this experiment: some CO₂ stays dissolved in the acid. The mass loss (1.02 g) has three significant figures, so the yield is good to about ±1%.

14.8 Stoichiometry practice problems with answers (word problems)

Ten stoichiometry word problems, easy to hard, covering every problem type the calculator solves, including a multi-step reaction sequence (problem 9). Open each one to see the worked answer.

14.9 Mistakes that lose marks

Research on stoichiometry finds students often apply the mole ratio to masses, or confuse the subscripts in a formula with the coefficients in front of it (Yarroch, 1985; Sanger, 2005; Dori & Hameiri, 2003).

• Using an unbalanced equation, so the mole ratio is wrong.
• Multiplying grams by the mole ratio instead of moles.
• Turning the mole ratio upside down: the substance you want goes on top.
• Using 16 g/mol for oxygen gas instead of 32 g/mol for O₂, or 1 g/mol for H₂ instead of 2.
• Forgetting to change mL to L before using molarity.
• Using the reactant in excess instead of the limiting reagent to find the yield.
• Rounding the moles in the middle of the calculation.

Four real examples

Decision tree

1. Is there a balanced (or balanceable) reaction? No: this is a single-substance conversion. Yes: go on.
2. Do you know the amount of only one substance? Yes: one known amount mode. No: go on.
3. Do you know amounts of two or more reactants? Yes: limiting reagent mode.
4. Do you have an actual yield? Yes: type it in to get the percent yield.

Limitations. A stoichiometry calculator gives the amounts a balanced equation allows; it cannot tell you whether a reaction actually happens, how fast it goes or where its equilibrium lies. When an equation has more than one independent set of coefficients (two reactions written as one), the calculator reports it rather than guessing. Molar masses of ions ignore the electron mass.

What the answer is

A stoichiometry calculator converts a known amount to moles, applies the mole ratio from the balanced equation, and converts back to the unit you need. For the headline problem, 4.00 g of hydrogen makes 35.7 g of water and uses 31.7 g of oxygen.

What it means chemically

The coefficients of a balanced equation count particles, so moles, not grams, follow them. Every reacting-mass, gas-volume, titration and yield calculation rests on that one idea.

Watch out for

Balance the equation first, convert to moles before using the ratio, put the substance you want on top of the ratio, find the limiting reagent when two amounts are given, and round only once at the end.

Next steps

Practise with the 25 worked stoichiometry calculations in Section 12.5 and the ten word problems in Section 14.8, or paste a whole worksheet into the Batch tab to check every answer at once.

Rearranging a stoichiometry formula is algebra, and judging whether two measured yields really differ is statistics; these tools cover both.

Peer-reviewed sources for the methods and data used in this stoichiometry calculator: the mole and the Avogadro constant, standard atomic weights, and research on how students learn stoichiometry, mole ratios and limiting reagents.

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