Determine molecular and empirical formulas from percent composition by mass and molar mass. Get step-by-step solutions with detailed calculations for chemistry students and professionals.
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The experimentally determined molar mass of the unknown compound
Element SymbolPercent Composition (%)
The experimentally determined molar mass of the unknown compound
Enter the subscript counts from your known empirical formula:
Molecular Formula
—
The complete molecular formula
Empirical Formula
—
The simplest whole-number ratio
Empirical Formula Mass
—
g/mol
Multiplier (n)
—
Molar mass ÷ Empirical mass
📝 Step-by-Step Solution
Molecular Formula Examples
🧪 Glucose (C₆H₁₂O₆)
Problem: A compound has the following percent composition: C = 40.00%, H = 6.71%, O = 53.29%. Its molar mass is 180.16 g/mol. Find the molecular formula.
Step 1: Assume 100 g sample → C = 40.00 g, H = 6.71 g, O = 53.29 g
Step 2: Moles: C = 40.00/12.01 = 3.331, H = 6.71/1.008 = 6.657, O = 53.29/16.00 = 3.331
Step 3: Divide by smallest: C = 3.331/3.331 = 1, H = 6.657/3.331 = 2, O = 3.331/3.331 = 1
Problem: A hydrocarbon compound contains 82.66% carbon and 17.34% hydrogen. Its molar mass is 58.12 g/mol. Determine the molecular formula.
Given: C = 82.66%, H = 17.34%, Molar Mass = 58.12 g/mol
Moles: C = 82.66/12.01 = 6.883, H = 17.34/1.008 = 17.202
Ratio: C = 6.883/6.883 = 1, H = 17.202/6.883 = 2.5 → Multiply by 2 → C₂H₅
Empirical mass: 2(12.01) + 5(1.008) = 29.06 g/mol
n = 58.12/29.06 = 2
Molecular formula:C₄H₁₀ (Butane)
💧 Hydrogen Peroxide (H₂O₂)
Problem: A compound contains 5.90% hydrogen and 94.10% oxygen. Its molar mass is 34.02 g/mol. Find the molecular formula.
Moles: H = 5.90/1.008 = 5.853, O = 94.10/16.00 = 5.881
Ratio: H = 5.853/5.853 = 1, O = 5.881/5.853 = 1.005 ≈ 1
Empirical formula:HO
Empirical mass: 1.008 + 16.00 = 17.01 g/mol
n = 34.02/17.01 = 2
Molecular formula:H₂O₂ (Hydrogen Peroxide)
Molecular Formula Guide
Empirical Formula × n = Molecular Formula
n = Molar Mass ÷ Empirical Formula Mass
How to Find the Molecular Formula
1
Assume a 100 g sample. Convert each percent composition value directly to grams. For example, 40.00% C becomes 40.00 g of carbon.
2
Convert grams to moles. Divide the mass of each element by its atomic mass (g/mol) from the periodic table.
3
Find the mole ratio. Divide each mole value by the smallest mole value among all elements present.
4
Get whole numbers. If the ratios are close to integers (within ±0.1), round them. If not (e.g., 1.5, 0.33), multiply all ratios by a common factor (2, 3, etc.) to obtain whole numbers. This gives the empirical formula.
5
Calculate empirical formula mass. Sum the atomic masses of all atoms in the empirical formula.
6
Find the multiplier n. Divide the given molar mass by the empirical formula mass: n = Molar Mass / Empirical Mass
7
Determine the molecular formula. Multiply each subscript in the empirical formula by n. The result is the molecular formula.
Common Atomic Masses (g/mol)
Element
Symbol
Atomic Mass (g/mol)
Hydrogen
H
1.008
Helium
He
4.003
Lithium
Li
6.941
Beryllium
Be
9.012
Boron
B
10.811
Carbon
C
12.011
Nitrogen
N
14.007
Oxygen
O
15.999
Fluorine
F
18.998
Neon
Ne
20.180
Sodium
Na
22.990
Magnesium
Mg
24.305
Aluminum
Al
26.982
Silicon
Si
28.086
Phosphorus
P
30.974
Sulfur
S
32.065
Chlorine
Cl
35.453
Argon
Ar
39.948
Potassium
K
39.098
Calcium
Ca
40.078
Iron
Fe
55.845
Copper
Cu
63.546
Zinc
Zn
65.409
Bromine
Br
79.904
Silver
Ag
107.868
Iodine
I
126.904
Barium
Ba
137.327
Gold
Au
196.967
Key Concepts
📌 Empirical Formula
The empirical formula shows the simplest whole-number ratio of atoms in a compound. For example, CH₂O is the empirical formula for glucose (C₆H₁₂O₆). Different compounds can share the same empirical formula.
📌 Molecular Formula
The molecular formula shows the actual number of atoms of each element in a molecule. It is always a whole-number multiple of the empirical formula. Water is H₂O, hydrogen peroxide is H₂O₂.
📌 Percent Composition
Percent composition is the mass percentage of each element in a compound. It can be determined experimentally through combustion analysis or other analytical techniques.
📌 The Multiplier n
The value of n must be a positive integer (1, 2, 3...). If n is close to 1, the empirical and molecular formulas are the same. If n is not close to an integer, recheck your empirical formula.
Important Rules
The percentages should sum to approximately 100% (±1%). If they don't, oxygen is often the balancing element.
When dividing by the smallest mole value, watch for ratios like 0.5 (×2), 0.33 (×3), 0.25 (×4), 0.67 (×3), and 0.75 (×4) — these indicate multipliers are needed.
The multiplier n should be a whole number. If it's not close to an integer, the empirical formula or molar mass may be incorrect.
Molecular formulas for ionic compounds are the same as their empirical formulas (n = 1).
📊
Percent Composition Mode
Input up to 4+ elements with their percent compositions by mass. Automatically calculates moles, ratios, and the empirical formula from your data.
🧬
Empirical Formula Mode
Already know the empirical formula? Enter the element counts and molar mass to directly compute the molecular formula and multiplier.
📝
Step-by-Step Solutions
Every calculation includes a detailed 7-step breakdown showing mass-to-mole conversions, ratio analysis, rounding, and formula determination.
📋
Atomic Mass Database
Built-in atomic masses for all common elements. Automatically looks up atomic weights so you don't need a periodic table.
⚠️ Important Note: Percent compositions must sum to 100% (±1% tolerance). The calculator assumes accurate experimental data. Always verify molecular formulas using additional analytical techniques such as mass spectrometry or NMR spectroscopy for confirmation.
Frequently Asked Questions
What is the difference between empirical and molecular formula?
The empirical formula represents the simplest whole-number ratio of atoms in a compound (e.g., CH₂O for glucose), while the molecular formula shows the actual number of atoms per molecule (e.g., C₆H₁₂O₆ for glucose). The molecular formula is always a whole-number multiple (n) of the empirical formula: Molecular Formula = Empirical Formula × n.
How do you find the molecular formula from percent composition?
First, determine the empirical formula by assuming a 100 g sample, converting the mass of each element to moles, and finding the simplest whole-number mole ratio. Then calculate the empirical formula mass. Finally, divide the given molar mass by the empirical formula mass to get the multiplier n. Multiply all subscripts in the empirical formula by n to obtain the molecular formula.
What does it mean if the multiplier n is not a whole number?
If n is not close to a whole number (within ±0.1), it indicates that the empirical formula may be incorrect, the molar mass data may be inaccurate, or there may be experimental errors in the percent composition. The multiplier n must be a whole number because molecules contain discrete, integer numbers of atoms. Recheck your empirical formula calculations, especially the step where you convert mole ratios to whole numbers.
Can two different compounds have the same empirical formula?
Yes, many compounds share the same empirical formula but have different molecular formulas. For example, CH₂O is the empirical formula for glucose (C₆H₁₂O₆), formaldehyde (CH₂O), and several other sugars. This is why the molar mass is essential — it allows you to determine which molecular formula corresponds to your compound by finding the correct multiplier n.
What is the 100 g assumption in empirical formula calculations?
The 100 g assumption is a mathematical convenience used when working with percent composition data. Since percentages are based on 100, assuming a 100 g sample allows you to directly treat each percentage value as the mass in grams of that element. For example, 40.00% carbon becomes 40.00 g of carbon. This simplifies the calculation without affecting the mole ratios, which are dimensionless.
Why do I need the molar mass to find the molecular formula?
The empirical formula only gives the ratio of atoms, not the actual count. Multiple molecular formulas (e.g., CH₂O, C₂H₄O₂, C₃H₆O₃) can share the same empirical formula (CH₂O). The molar mass is needed to determine which multiple (n) is correct. Without the molar mass, you can only determine the empirical formula. This is why both percent composition data AND molar mass are required for molecular formula determination.