Molecular Formula Calculator with Steps | StatsUnlock

To calculate a molecular formula, find the empirical formula, divide the molar mass by the empirical formula mass, and multiply every subscript by that whole number. A compound that is 40.0% C, 6.71% H and 53.3% O with a molar mass of 180.16 g/mol:

Rule of thumb: the empirical formula gives the ratio of atoms; only the molar mass gives the actual number. Formaldehyde (CH₂O), acetic acid (C₂H₄O₂) and glucose (C₆H₁₂O₆) all have the empirical formula CH₂O.

A molecular formula gives the actual number of atoms of each element in one molecule: glucose is C₆H₁₂O₆. An empirical formula gives only the simplest whole-number ratio of those atoms: CH₂O. The molecular formula is always the empirical formula multiplied by a whole number n, and n is the molar mass divided by the empirical formula mass.

A molecular formula calculator does this work from whatever data you have. Elemental analysis gives percent composition or the masses of CO₂ and H₂O from burning a sample; that gives the empirical formula. A molar mass from a gas density, a freezing-point measurement or a mass spectrum gives n. A high-resolution mass spectrometer gives a mass so exact that it can pick out the formula directly. This calculator handles all of these, and the reverse: molecular formula to empirical formula and molecular weight.

A worked one-liner: benzene has the empirical formula CH (13.019 g/mol) and a molar mass of 78.11 g/mol. 78.11 ÷ 13.019 = 6.00, so its molecular formula is C₆H₆.

Who uses it: CBSE and ISC students meet empirical and molecular formulas in NCERT Class 11 (“Some Basic Concepts of Chemistry”) and JEE and NEET set them every year; GCSE, A level, AP Chemistry and IB courses include combustion analysis. In research labs, accurate-mass formula assignment is the first step in identifying an unknown from an LC-MS or GC-MS run.

What this free molecular formula calculator solves

One tool covers every empirical and molecular formula calculation in a typical course, plus accurate-mass formula finding. It works as an empirical to molecular formula calculator, a percent composition to molecular formula calculator, a molar mass to molecular formula calculator, a molecular formula calculator from exact mass for mass spectrometry, and a molecular to empirical formula calculator that also gives the molecular weight. Press Try it to load an example into the calculator, then press the button 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.

En español: ¿cómo se calcula la fórmula molecular? Calcula la fórmula empírica, divide la masa molar entre la masa de la fórmula empírica y multiplica cada subíndice por ese número entero: (CH₂O) × 6 = C₆H₁₂O₆.

How accurate is this molecular formula calculator?

Moles are calculated with the CIAAW 2024 standard atomic weights (Prohaska et al., 2022), or rounded exam values if you choose them. Exact-mass searches use monoisotopic masses (C = 12 exactly, H = 1.007825, N = 14.003074, O = 15.994915). Every answer is checked before it gets the Verified badge: the formula must reproduce your percentages within one percentage point, its molar mass must match the value you gave, or it must lie inside your mass window with a whole-number ring-plus-double-bond count. The R and Python scripts in Sections 10 and 11 return the same CH₂O and C₆H₁₂O₆ for the headline problem.

Studies of how students solve formula problems find the hard part is rounding the mole ratio correctly, not the arithmetic (Schmidt & Jignéus, 2003; Pekdağ & Azizoğlu, 2013). That is why the results show the ratio before and after rounding, and chart 2 draws them side by side.

4.1 Empirical formula vs molecular formula

The empirical formula tells you the ratio of atoms; the molecular formula tells you how many atoms are in one molecule. They share the same percent composition, so elemental analysis alone can never tell them apart. CH₂O is 40.00% carbon whether it is formaldehyde, acetic acid or glucose. Only a molar mass (or an exact mass) fixes n.

4.2 Reading the multiplier n

n = M ÷ M_e should come out within a few percent of a whole number. For glucose, 180.16 ÷ 30.026 = 6.000. A ratio such as 3.33 means the molar mass or the empirical formula is wrong: CH₂O with M = 100 g/mol gives 3.33, so the calculator stops rather than guess. When n = 1, the empirical and molecular formulas are the same, as for water and methane.

4.3 Rounding the mole ratio

Experimental percentages are rounded, so the mole ratios come out close to, not exactly on, whole numbers. Treat 0.98 to 1.02 as 1. A ratio near 1.5 means multiply everything by 2; near 1.33 or 1.67 means × 3; near 1.25 or 1.75 means × 4. Never round 1.5 to 2. Vitamin C’s ratio C 1 : H 1.33 : O 1 becomes C₃H₄O₃ after multiplying by 3.

4.4 Exact mass, ppm error and why it matters

A high-resolution mass spectrometer measures masses to a few parts per million (ppm). Because every isotope has a slightly different mass, formulas with the same nominal mass separate: CO (27.99491) and N₂ (28.00615) differ by 0.011 u, which is 400 ppm. The error column shows how far each candidate is from your measurement; the closer to zero, the better, but at higher masses many formulas fit even a 5 ppm window.

4.5 Rings plus double bonds (RDBE) and the nitrogen rule

RDBE = C − H/2 + N/2 + 1 (with halogens counted like H) counts rings and π bonds. A real neutral molecule must give a whole number of zero or more. Caffeine, C₈H₁₀N₄O₂, has RDBE = 8 − 5 + 2 + 1 = 6. A half-number result means an odd-electron species, which is how the nitrogen rule removes impossible formulas: a neutral molecule with an odd number of nitrogen atoms has an odd nominal mass.

4.6 The isotope pattern

Natural carbon is 1.07% ¹³C, so a molecule with 6 carbons shows an M+1 peak about 6.5% of the main peak from carbon alone (glucose, counting its H and O too, 6.9%). Nitrogen adds 0.37% per atom: caffeine, with 8 C and 4 N, shows M+1 at 10.3%. Chlorine (24% ³⁷Cl) gives an M+2 peak about one third of M; bromine gives M and M+2 almost equal. The calculator predicts this pattern for the top candidate so you can compare it with your spectrum.

4.7 Why a molar mass alone is not enough

Searching only an average molar mass of 180.16 g/mol, within ± 0.05 g/mol and with C, H, N and O allowed, returns 19 chemically valid formulas, including glucose (C₆H₁₂O₆, 180.156) and aspirin (C₉H₈O₄, 180.159). That is why textbooks combine the empirical formula from elemental analysis with the molar mass, and why research labs use exact masses and isotope patterns.

4.8 A field note on where these problems go wrong

In marking empirical and molecular formula work, the same errors appear again and again: rounding 1.5 to 2, dividing by the largest number of moles instead of the smallest, forgetting oxygen when the percentages do not add to 100, and using 16 g/mol for hydrogen in water calculations. Research on problem solving in chemistry finds students often follow the algorithm without checking that the answer makes sense (Nurrenbern & Pickering, 1987; Nakhleh, 1992). The calculator therefore checks that the formula reproduces your data.

4.9 What comes next

A molecular formula lists atoms, not how they are joined. C₂H₆O can be ethanol or dimethyl ether. The next step is structure: infrared and NMR spectroscopy, and the RDBE as a first clue. The formula’s molar mass then feeds every quantitative calculation: grams to moles, stoichiometry, concentrations and yields.

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

The formula for calculating a molecular formula is n = M ÷ M_e, then molecular formula = (empirical formula) × n. The formula to calculate molecular weight is M = Σ (atoms × atomic weight). Every formula the calculator uses is below, each with a worked number.

This flowchart shows how to calculate empirical and molecular formulas, on the same problem used in the Excel, R, Python and by-hand sections: a compound is 40.0% C, 6.71% H and 53.3% O, and its molar mass is 180.16 g/mol. What is its molecular formula?

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 CH₂O and C₆H₁₂O₆ for this problem.

1. Choose the data you have. Pick percent composition, an empirical formula, combustion analysis, an exact or average mass, or a molecular formula to reduce.
2. Type the percentages or grams. Write each element and its value, for example C 40.0, H 6.71, O 53.3. Tick the box to add oxygen by difference.
3. Or type the empirical formula. Write it as CH2O, or tap a common empirical formula.
4. Or enter the combustion data. Give the sample mass and the masses of CO2 and H2O, and nitrogen percent if you have it.
5. Add the molar mass. Type it in g/mol to get the molecular formula; leave it empty to stop at the empirical formula.
6. For mass spectrometry, set the search. Enter the m/z, the ion type, the ppm tolerance and the elements allowed, and optionally the M+1 peak height.
7. Choose the atomic weights. Keep CIAAW 2024 values or switch to rounded exam values.
8. Press the button. Nothing is calculated before you press it, and changing any input clears the old answer.
9. Read the steps, the ratio chart and the check. The Verified badge means the formula reproduces your data.
10. Copy, share or batch. Copy the answer, share a link, download a report, or solve a whole worksheet on the Batch tab.

Excel handles the whole method in one small table: one row per element with its percentage and atomic weight, then columns for moles, the ratio and the whole number. Two more cells give the empirical formula mass and n. The sheet below uses the headline problem, so it must return CH₂O and n = 6, the same as the molecular formula calculator above.

9.1 The functions you need

9.2 Step by step

9.3 Awkward cases

• Oxygen by difference. Put =100-SUM(B2:B3) in the O row when only C and H are given.
• Combustion data. Make B2 =CO2_mass*12.011/44.009 and B3 =H2O_mass*2.016/18.015, and O by difference from the sample mass.
• Ratios like 1.33. Set H1 to 3 and check every row in column E × 3 is near a whole number.

9.4 Why your Excel answer might differ

9.5 Charting in Excel

1. Select the elements, the whole numbers and the molecular subscripts (A2:A4, F2:G4).
2. Insert → Charts → Clustered Column to compare the empirical and molecular formulas, as in Chart 2 above.

9.6 Excel errors you will probably hit

This base R script turns percent composition into the empirical formula, tries multipliers 1 to 8 until every ratio is within 0.1 of a whole number, finds n from the molar mass, checks that the molecular formula reproduces the percentages, and saves one 300 dpi chart. For the headline data it prints CH2O and C6H12O6, matching the molecular formula calculator on this page.

10.1 The complete script

R, base only, saves molecular_formula.png

# ============================================================
# Molecular formula calculator in R:
# percent composition -> empirical formula -> molecular formula
# StatsUnlock - https://statsunlock.com/molecular-formula-calculator/
# Base R only. No packages to install.
# ============================================================

# ---- 1. Your data --------------------------------------------------
percent    <- c(C = 40.0, H = 6.71, O = 53.3)   # % by mass (or grams)
molar_mass <- 180.16                           # g/mol, or NA for the empirical formula only

# CIAAW 2024 standard atomic weights (g/mol); add any element you need
W <- c(H = 1.008, 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, Fe = 55.845, Cu = 63.546, Br = 79.904)

# Hill order: C, then H, then the rest alphabetically
hill <- function(counts) {
  k <- sort(names(counts))
  if ("C" %in% k) k <- c("C", intersect("H", k), setdiff(k, c("C", "H")))
  paste0(k, ifelse(counts[k] == 1, "", counts[k]), collapse = "")
}

# ---- 2. Grams -> moles -> ratio --------------------------------------
moles <- percent / W[names(percent)]
ratio <- moles / min(moles)
print(data.frame(element = names(percent), grams = as.vector(percent),
                 moles = round(as.vector(moles), 4), ratio = round(as.vector(ratio), 3)))

# ---- 3. Multiply until every ratio is within 0.1 of a whole number ----
k <- 1
while (k <= 8 && any(abs(ratio * k - round(ratio * k)) > 0.1)) k <- k + 1
if (k > 8) stop("No whole-number ratio up to x8: check the data")
empirical <- round(ratio * k)
M_emp <- sum(empirical * W[names(empirical)])
cat(sprintf("Empirical formula %s (x%d), M = %.3f g/mol\n", hill(empirical), k, M_emp))

# ---- 4. Molecular formula from the molar mass -----------------------
if (!is.na(molar_mass)) {
  n_raw <- molar_mass / M_emp; n <- round(n_raw)
  stopifnot(abs(n_raw - n) <= 0.05 * n)
  molecular <- empirical * n
  M_mol <- sum(molecular * W[names(molecular)])
  cat(sprintf("n = %.2f / %.3f = %.3f -> %d\n", molar_mass, M_emp, n_raw, n))
  cat(sprintf("Molecular formula %s, M = %.3f g/mol\n", hill(molecular), M_mol))

  # ---- 5. Check: the formula reproduces the percentages ----------
  pct <- 100 * molecular * W[names(molecular)] / M_mol
  stopifnot(all(abs(pct - 100 * percent / sum(percent)) < 1))
  cat("Verdict: Verified\n")

  # ---- 6. One 300 dpi figure: empirical vs molecular atoms -------
  png("molecular_formula.png", width = 2100, height = 1350, res = 300)
  par(mar = c(4.5, 5, 3, 1))
  barplot(rbind(empirical, molecular), beside = TRUE, names.arg = names(molecular), las = 1, border = NA,
          col = c("#94a3b8", "#2563eb"), ylab = "Atoms per formula",
          legend.text = c(hill(empirical), hill(molecular)),
          main = sprintf("(%s) x %d = %s", hill(empirical), n, hill(molecular)))
  invisible(dev.off())
  cat("Saved molecular_formula.png\n")
}

Output: Empirical formula CH2O (x1), M = 30.026 g/mol, n = 180.16 / 30.026 = 6.000 -> 6, Molecular formula C6H12O6, M = 180.156 g/mol, Verdict: Verified. With caffeine’s percentages and 194.2 it prints C8H10N4O2.

10.2 R traps

• round(2.5) gives 2 in R (round half to even). It does not matter here because ratios are never exactly .5 after multiplying, but use care elsewhere.
• Named vectors must match: W[names(percent)] picks the right weight for each element.
• The while loop stops at 8; if it reaches 9, the data are inconsistent.

10.3 What each part does

10.4 Packages for exact masses

For mass spectrometry, the enviPat package calculates isotope patterns and the Rdisop package decomposes an exact mass into candidate formulas, the same search the exact-mass mode of this calculator runs.

10.5 The figure

The script saves molecular_formula.png at 2100 × 1350 pixels and 300 dpi: grouped bars comparing the atoms in the empirical formula with those in the molecular formula.

This script does the same job in plain Python: moles, ratio, the multiplier search, the empirical formula, n from the molar mass, the molecular formula in Hill order, a check against your percentages and a 300 dpi chart. It is the same method the molecular formula calculator runs in your browser.

Install once: pip install matplotlib

11.1 The complete script

Python, saves molecular_formula.png

# ============================================================
# Molecular formula calculator in Python:
# percent composition -> empirical formula -> molecular formula
# StatsUnlock - https://statsunlock.com/molecular-formula-calculator/
# pip install matplotlib
# ============================================================
import matplotlib
matplotlib.use("Agg")
import matplotlib.pyplot as plt

# ---- 1. Your data --------------------------------------------------
percent = {"C": 40.0, "H": 6.71, "O": 53.3}   # % by mass (or grams)
molar_mass = 180.16                           # g/mol, or None for the empirical formula only

# CIAAW 2024 standard atomic weights (g/mol); add any element you need
W = {"H": 1.008, "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, "Fe": 55.845, "Cu": 63.546, "Br": 79.904}

def hill(counts):
    """Write a formula in Hill order: C, then H, then the rest alphabetically."""
    keys = sorted(counts)
    if "C" in counts:
        keys = ["C"] + (["H"] if "H" in counts else []) + [k for k in keys if k not in ("C", "H")]
    return "".join(k + ("" if counts[k] == 1 else str(counts[k])) for k in keys)

# ---- 2. Grams -> moles -> ratio --------------------------------------
moles = {e: g / W[e] for e, g in percent.items()}
smallest = min(moles.values())
ratio = {e: n / smallest for e, n in moles.items()}
for e in percent:
    print(f"{e}: {percent[e]} g / {W[e]} = {moles[e]:.4f} mol  ratio {ratio[e]:.3f}")

# ---- 3. Multiply until every ratio is within 0.1 of a whole number ----
for k in range(1, 9):
    if all(abs(r * k - round(r * k)) <= 0.1 for r in ratio.values()):
        break
else:
    raise SystemExit("No whole-number ratio up to x8: check the data")
empirical = {e: round(r * k) for e, r in ratio.items()}
M_emp = sum(n * W[e] for e, n in empirical.items())
print(f"Empirical formula {hill(empirical)} (x{k}), M = {M_emp:.3f} g/mol")

# ---- 4. Molecular formula from the molar mass -----------------------
if molar_mass:
    n = molar_mass / M_emp
    assert abs(n - round(n)) <= 0.05 * round(n), f"M / Me = {n:.3f} is not a whole number"
    n = round(n)
    molecular = {e: c * n for e, c in empirical.items()}
    M_mol = sum(c * W[e] for e, c in molecular.items())
    print(f"n = {molar_mass} / {M_emp:.3f} = {molar_mass / M_emp:.3f} -> {n}")
    print(f"Molecular formula {hill(molecular)}, M = {M_mol:.3f} g/mol")

    # ---- 5. Check: the formula reproduces the percentages ----------
    for e, c in molecular.items():
        pct = 100 * c * W[e] / M_mol
        print(f"  {e}: {pct:.2f}% (data {percent[e]}%)")
        assert abs(pct - 100 * percent[e] / sum(percent.values())) < 1, "formula does not fit the data"
    print("Verdict: Verified")

    # ---- 6. One 300 dpi figure: empirical vs molecular atoms -------
    els = list(molecular)
    fig, ax = plt.subplots(figsize=(7, 4.5))
    x = range(len(els))
    ax.bar([i - 0.2 for i in x], [empirical[e] for e in els], 0.4, label=hill(empirical), color="#94a3b8")
    ax.bar([i + 0.2 for i in x], [molecular[e] for e in els], 0.4, label=hill(molecular), color="#2563eb")
    ax.set_xticks(list(x), els)
    ax.set_ylabel("Atoms per formula")
    ax.set_title(f"({hill(empirical)}) x {n} = {hill(molecular)}")
    ax.legend()
    fig.tight_layout()
    fig.savefig("molecular_formula.png", dpi=300)
    print("Saved molecular_formula.png")

Output: C: 40.0 g / 12.011 = 3.3303 mol ratio 1.000, Empirical formula CH2O (x1), M = 30.026 g/mol, Molecular formula C6H12O6, M = 180.156 g/mol, Verdict: Verified.

11.2 Python traps

• round() rounds half to even; ratios are compared with a 0.1 tolerance instead.
• Dictionary keys are case sensitive: "Co" is cobalt, "CO" is not an element.
• The for ... else block runs only when no multiplier from 1 to 8 works.

11.3 Packages

pip install pyteomics gives mass.calculate_mass(formula="C8H10N4O2") for monoisotopic masses, and molmass gives average masses, compositions and isotope spectra for any formula.

11.4 What each part does

11.5 The figure

The script saves molecular_formula.png at 300 dpi with the empirical and molecular atom counts side by side.

12.1 Common compounds: empirical formula, molecular formula and n

Each row was computed by the same engine as the molecular formula calculator.

12.2 Monoisotopic masses for mass spectrometry

Monoisotopic masses from the 2020 Atomic Mass Evaluation as tabulated by NIST; abundances are representative natural values. The proton is 1.007276 u and the electron 0.000549 u.

12.3 Exact masses of common molecules and ions

12.4 Calculating empirical and molecular formulas: worksheet with answers

Twenty-five problems in three groups: empirical to molecular formula, percent composition to the molecular formula, and molecular to empirical formula. Every answer was checked by the calculator engine. Cover the last column and try them first.

Eight examples run through the calculator, from glucose to an accurate-mass assignment. Each figure is drawn from the computed numbers: the grey bars are the empirical formula, the blue bars the molecular formula.

This is the written method examiners expect, on the same problem as Sections 7, 9, 10 and 11: 40.0% C, 6.71% H, 53.3% O, molar mass 180.16 g/mol.

14.1 How to calculate molecular formula from percent composition, step by step

1. Assume 100 g. C 40.0 g, H 6.71 g, O 53.3 g.
2. Convert to moles. C 40.0 ÷ 12.011 = 3.330; H 6.71 ÷ 1.008 = 6.657; O 53.3 ÷ 15.999 = 3.331.
3. Divide by the smallest (3.330). C 1.000, H 1.999, O 1.000.
4. Round, or multiply first if needed. 1 : 2 : 1, so the empirical formula is CH₂O.
5. Empirical formula mass. 12.011 + 2 × 1.008 + 15.999 = 30.026 g/mol.
6. Find n. 180.16 ÷ 30.026 = 6.00.
7. Multiply. (CH₂O)₆ = C₆H₁₂O₆.
8. Check. M(C₆H₁₂O₆) = 180.156 g/mol, and its percentages, 40.00% C, 6.71% H and 53.28% O, match the data.

14.2 How to calculate the molecular formula from the empirical formula

If the empirical formula is given, start at step 5. Example: the empirical formula of a compound is C₂H₅ and its molar mass is 58.12 g/mol. M_e = 2 × 12.011 + 5 × 1.008 = 29.062 g/mol; n = 58.12 ÷ 29.062 = 2.00; the molecular formula is C₄H₁₀ (butane).

14.3 How to calculate the molecular formula from combustion analysis

Burning 1.000 g of vitamin C gives 1.500 g CO₂ and 0.405 g H₂O. Carbon: 1.500 × 12.011 ÷ 44.009 = 0.4094 g; hydrogen: 0.405 × 2.016 ÷ 18.015 = 0.04532 g; oxygen by difference: 1.000 − 0.4094 − 0.0453 = 0.5453 g. Moles: C 0.03408, H 0.04496, O 0.03408; ratio 1 : 1.319 : 1; × 3 gives C₃H₄O₃ (88.062 g/mol). With M = 176.12 g/mol, n = 2 and the molecular formula is C₆H₈O₆.

14.4 How to calculate the empirical formula from the molecular formula

Divide every subscript by their greatest common divisor. C₆H₁₂O₆: gcd(6, 12, 6) = 6, giving CH₂O. C₈H₁₀N₄O₂: gcd = 2, giving C₄H₅N₂O. If the gcd is 1, as for C₉H₈O₄, the two formulas are the same.

14.5 How to calculate the molecular mass from the empirical formula

Multiply the empirical formula mass by n: M = n × M_e. For (NO₂)₂, M = 2 × 46.005 = 92.010 g/mol. To find n you still need one experimental value: a molar mass, a gas density or an exact mass.

14.6 On a scientific calculator

On a Casio fx-991EX or fx-82MS: 40 ÷ 12.011 = gives 3.330 (store it with STO A). Repeat for H and O. For n: 180.16 ÷ ( 12.011 + 2 × 1.008 + 15.999 ) = gives 6.0001.

Keep four significant figures in the moles and ratios; rounding to two figures too early can turn 1.33 into 1.3 and hide the × 3. Key names differ between models; check your manual.

14.7 Do it in the lab: the empirical formula of magnesium oxide

Example data. Ratio O : Mg = 0.00988 ÷ 0.00987 = 1.000, so the empirical formula is MgO. A ratio below about 0.9 usually means magnesium oxide smoke escaped or some magnesium was not burned.

14.8 Molecular formula practice problems with answers

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

14.9 Mistakes that lose marks

Research on chemistry problem solving finds students often apply an algorithm correctly but cannot say why each step works, so errors go unnoticed (Nurrenbern & Pickering, 1987; Gabel & Sherwood, 1984).

• Rounding 1.5 to 2 instead of multiplying everything by 2.
• Dividing by the largest number of moles instead of the smallest.
• Forgetting oxygen when the percentages add to less than 100.
• Using the percentages directly as a ratio without dividing by atomic weights.
• Rounding the moles to two figures before taking the ratio.
• Giving the empirical formula when the question asks for the molecular formula.
• Writing C₆H₁₂O₆ as 6CH₂O: a coefficient is not a subscript.

Four real examples

Decision tree

1. Do you have percentages or element masses? Yes: percent composition mode. No: go on.
2. Do you have masses of CO₂ and H₂O from burning? Yes: combustion mode.
3. Do you have an empirical formula? Yes: add the molar mass in the empirical mode.
4. Do you have only a mass spectrum? Yes: exact-mass mode, then confirm with the isotope pattern.

Limitations. A molecular formula calculator gives the number of each atom, never the structure: C₂H₆O is both ethanol and dimethyl ether. The exact-mass search lists every formula that fits; the right one must be confirmed with the isotope pattern, fragmentation or NMR. The ratio search tries multipliers up to 8 with a tolerance of 0.1.

What the answer is

The molecular formula is the empirical formula multiplied by n = M ÷ M_e. For 40.0% C, 6.71% H, 53.3% O and M = 180.16 g/mol, that is (CH₂O) × 6 = C₆H₁₂O₆.

What it means chemically

The formula fixes how many atoms of each element are in one molecule, and so its molar mass, its unsaturation (RDBE) and its isotope pattern. It does not fix the structure.

Watch out for

Divide by the smallest number of moles, multiply instead of rounding ratios like 1.5 or 1.33, add oxygen by difference when it is missing, and never skip the molar mass when the question asks for the molecular formula.

Next steps

Practise with the worksheet in Section 12.4 and the ten problems in Section 14.8, then use the molar mass of your formula for grams-to-moles and stoichiometry calculations.

Finding n is algebra, and deciding whether a measured percentage really differs from the formula’s is statistics; these tools cover both.

Peer-reviewed sources for the methods and data used in this molecular formula calculator: standard atomic weights, the mole, accurate-mass formula assignment, and research on how students learn formulas and the mole concept.

1. Dori, Y. J., & Hameiri, M. (2003). Multidimensional analysis system for quantitative chemistry problems: Symbol, macro, micro, and process aspects. Journal of Research in Science Teaching, 40(3), 278–302. https://doi.org/10.1002/tea.10077
2. Furió, C., Azcona, R., & Guisasola, J. (2002). The learning and teaching of the concepts ‘amount of substance’ and ‘mole’: A review of the literature. Chemistry Education Research and Practice, 3(3), 277–292. https://doi.org/10.1039/B2RP90023H
3. Güttler, B., Bettin, H., Brown, R. J. C., Davis, R. S., Mester, Z., Milton, M. J. T., ... Stock, M. (2019). Amount of substance – the Avogadro constant and the SI unit “mole”. Metrologia, 56(4), 044002. https://doi.org/10.1088/1681-7575/ab1fae
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