Convert mass percent composition or mass data into empirical formulas. Use measured molar mass to determine molecular formulas. See step-by-step mole calculations, ratios, and subscript determination.
Enter the percent composition or mass data for your compound to determine its empirical and molecular formulas.
Last Updated: November 20, 2025. This content is regularly reviewed to ensure accuracy and alignment with current IUPAC atomic mass standards.
Chemical formulas are fundamental representations that tell us the composition of compounds. The empirical formula shows the simplest whole-number ratio of atoms of each element in a compound, while the molecular formula shows the actual number of atoms of each element in one molecule. For example, glucose has the molecular formula C₆H₁₂O₆, but its empirical formula is CH₂O because the ratio 6:12:6 simplifies to 1:2:1. Understanding empirical and molecular formulas is crucial for students studying general chemistry, organic chemistry, biochemistry, and analytical chemistry, as they explain compound composition, help identify unknown substances, and connect percent composition to molecular structure. Empirical and molecular formula concepts appear on virtually every chemistry exam and are foundational to understanding stoichiometry, percent composition, and chemical analysis.
Empirical formulas are determined from experimental data—either mass data or percent composition. The process involves: (1) Converting masses or percents to moles, (2) Finding the smallest number of moles, (3) Dividing all mole values by the smallest to get mole ratios, (4) Rounding ratios to whole numbers (or multiplying to eliminate fractions) to get subscripts. For example, if a compound contains 40.0% C, 6.71% H, and 53.29% O, assuming 100 g gives 40.0 g C, 6.71 g H, and 53.29 g O. Converting to moles and dividing by the smallest gives ratios of 1:2:1, so the empirical formula is CH₂O. Understanding this process helps you determine compound composition from experimental data and connect percent composition to molecular structure.
Molecular formulas are whole-number multiples of empirical formulas: Molecular formula = n × Empirical formula, where n is a positive integer. To determine the molecular formula, you need the measured molar mass (from mass spectrometry, freezing point depression, vapor density, etc.). Calculate n by dividing the measured molar mass by the empirical formula's molar mass: n = Measured Molar Mass ÷ Empirical Molar Mass. Then multiply each subscript in the empirical formula by n. For example, if the empirical formula is CH₂O (molar mass ≈ 30 g/mol) and the measured molar mass is 180 g/mol, then n = 180 ÷ 30 = 6, so the molecular formula is C₆H₁₂O₆. Understanding this relationship helps you see that many compounds can share the same empirical formula but have different molecular formulas (e.g., CH₂O is the empirical formula for glucose, formaldehyde, and acetic acid, but they have different molecular formulas).
Percent composition is the percentage by mass of each element in a compound. It can be calculated from a formula or used to determine a formula. When percents are given, assume a 100 g sample so percentages become grams directly. This simplifies calculations because 40% C means 40 g C in a 100 g sample. Understanding percent composition helps you see how experimental data (mass or percent) connects to molecular formulas, and why assuming 100 g makes calculations easier. Percent composition is also used to verify formulas, check experimental results, and understand how elements are distributed in compounds.
This calculator is designed for educational exploration and practice. It helps students master empirical and molecular formula calculations by working with mass data or percent composition, determining empirical formulas, calculating molecular formulas from molar masses, and understanding the relationship between formulas and composition. The tool provides step-by-step calculations showing how to convert masses to moles, find mole ratios, round to whole numbers, and determine molecular formulas. For students preparing for chemistry exams, analytical chemistry courses, or organic chemistry labs, mastering empirical and molecular formulas is essential—these calculations appear on virtually every chemistry assessment and are fundamental to understanding compound composition and chemical analysis. The calculator supports both mass and percent composition modes, helping students understand different approaches to formula determination.
Critical disclaimer: This calculator is for educational, homework, and conceptual learning purposes only. It helps you understand empirical and molecular formula theory, practice formula calculations, and explore compound composition. It does NOT provide instructions for actual chemical identification, forensic analysis, or material characterization, which require proper training, calibrated equipment, safety protocols, and adherence to validated analytical procedures. Never use this tool to identify unknown compounds, determine chemical safety, or make forensic or legal determinations. Real-world compound identification involves considerations beyond this calculator's scope: structural isomers (different compounds with the same molecular formula), stereoisomers, spectroscopic data, and empirical verification. Use this tool to learn the theory—consult trained professionals and proper analytical methods for practical applications.
The empirical formula shows the simplest whole-number ratio of atoms (e.g., CH₂O for glucose). 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 of the empirical formula: Molecular formula = n × Empirical formula, where n is a positive integer. For example, glucose: n = 6, so C₆H₁₂O₆ = 6 × CH₂O. Many compounds can share the same empirical formula but have different molecular formulas (e.g., CH₂O is the empirical formula for glucose C₆H₁₂O₆, formaldehyde CH₂O, and acetic acid C₂H₄O₂). Understanding this distinction helps you see that empirical formulas show composition ratios, while molecular formulas show actual atom counts.
The process involves: (1) Assume a 100 g sample so percentages become grams (e.g., 40% C = 40 g C). (2) Convert grams to moles: moles = mass ÷ atomic mass (e.g., 40 g C ÷ 12.01 g/mol = 3.33 mol C). (3) Find the smallest number of moles among all elements. (4) Divide each element's moles by the smallest to get mole ratios (e.g., if smallest is 3.33, then C: 3.33 ÷ 3.33 = 1.00, H: 6.66 ÷ 3.33 = 2.00, O: 3.33 ÷ 3.33 = 1.00). (5) Round ratios to whole numbers (or multiply to eliminate fractions) to get subscripts (e.g., 1:2:1 gives CH₂O). Understanding this process helps you determine empirical formulas from experimental data and connect percent composition to molecular structure.
You need the measured molar mass (from mass spectrometry, freezing point depression, vapor density, etc.). Calculate the multiplier n: n = Measured Molar Mass ÷ Empirical Molar Mass. Then multiply each subscript in the empirical formula by n. For example, if empirical formula is CH₂O (molar mass ≈ 30 g/mol) and measured molar mass is 180 g/mol, then n = 180 ÷ 30 = 6, so molecular formula is C₆H₁₂O₆. The multiplier n must be a positive integer (1, 2, 3, ...). If n is not close to a whole number, check for errors in the measured molar mass, percent composition, or empirical formula calculation. Understanding this relationship helps you see how empirical formulas connect to molecular formulas through molar mass.
If ratios like 1.5, 1.33, or 1.25 appear, multiply all ratios by a common factor to get whole numbers: (1) Ratios like 1.5 or 2.5 → multiply all by 2 (e.g., 1.5:2.5 becomes 3:5). (2) Ratios like 1.33 or 1.67 → multiply all by 3 (e.g., 1.33:2.67 becomes 4:8). (3) Ratios like 1.25 or 1.75 → multiply all by 4 (e.g., 1.25:3.75 becomes 5:15). The goal is to find the smallest multiplier that gives whole numbers within rounding tolerance (typically within 0.1 of an integer). Understanding this helps you handle non-integer ratios and see that empirical formulas must have whole-number subscripts.
Hill notation is the standard way to write chemical formulas: (1) For carbon compounds: C first, then H, then other elements alphabetically (e.g., C₆H₁₂O₆ for glucose, C₂H₆O for ethanol). (2) For compounds without carbon: all elements in alphabetical order (e.g., H₂O for water, NaCl for salt). Hill notation ensures consistency in formula representation and makes formulas easier to compare and search. Understanding Hill notation helps you write formulas correctly and understand why formulas are written in specific orders.
Experimental measurements always have some uncertainty. Percents that total 98–102% are typically acceptable due to rounding and measurement error. If percents are significantly off (> 5%), check for: (1) Transcription errors (wrong numbers entered), (2) Missing elements (often oxygen is calculated by difference: 100% - sum of other elements), (3) Calculation errors. Sometimes oxygen isn't measured directly but is calculated as the difference from 100%. Understanding this helps you know when percent deviations are acceptable and when to check for errors.
Yes! Many compounds can share the same empirical formula but have different molecular formulas and structures. For example, CH₂O is the empirical formula for: (1) Formaldehyde (CH₂O), (2) Acetic acid (C₂H₄O₂), (3) Glucose (C₆H₁₂O₆), (4) Fructose (C₆H₁₂O₆, same molecular formula as glucose but different structure). This is why empirical formulas alone cannot identify compounds—structural isomers (different compounds with the same molecular formula) and compounds with different molecular formulas can share the same empirical formula. Understanding this limitation helps you see why additional data (molecular formula, structure, spectroscopy) is needed for compound identification.
This interactive tool helps you determine empirical and molecular formulas from mass data or percent composition. Here's a comprehensive guide to using each feature:
Select how you want to enter composition data:
Percent Composition Mode
Enter the percentage by mass of each element (e.g., 40.0% C, 6.71% H, 53.29% O). The calculator assumes a 100 g sample, so percentages become grams directly.
Mass Data Mode
Enter the actual mass of each element in your sample (e.g., 40.0 g C, 6.71 g H, 53.29 g O). Use this when you have mass data instead of percentages.
For each element in your compound:
Element Symbol
Enter the element symbol (e.g., C, H, O, N). The calculator looks up atomic masses from a built-in table.
Mass or Percent
Enter the mass (in grams) or percent (by mass) for that element, depending on your selected mode.
Custom Atomic Mass (Optional)
If your problem specifies a different atomic mass, enter it here. This overrides the built-in value.
Add/Remove Elements
Use the "+ Add Element" button to add more elements, or remove elements you don't need.
If you want to determine the molecular formula:
Measured Molar Mass
Enter the measured molar mass (in g/mol) from experimental data (mass spectrometry, freezing point depression, etc.). The calculator will calculate the multiplier n and determine the molecular formula.
Click "Calculate" to determine formulas:
View Calculation Steps
The calculator shows: (a) Conversion of masses/percents to moles, (b) Mole ratios after dividing by the smallest, (c) Final subscripts (rounded or multiplied), (d) Empirical formula in Hill notation, (e) Empirical molar mass, (f) If molar mass provided: multiplier n and molecular formula.
Check for Issues
The calculator notes if percents don't add to 100%, if ratios needed multiplication, or if the multiplier n is not a whole number. Review these notes to verify your results.
Example: Determine empirical formula from 40.0% C, 6.71% H, 53.29% O
Input: Percent mode, C = 40.0%, H = 6.71%, O = 53.29%
Output: Empirical formula = CH₂O
If measured molar mass = 180 g/mol: n = 6, Molecular formula = C₆H₁₂O₆
Understanding the mathematics empowers you to calculate empirical and molecular formulas on exams, verify calculator results, and build intuition about compound composition.
Molecular Formula = n × Empirical Formula
Where:
n = positive integer (1, 2, 3, ...)
n = Measured Molar Mass ÷ Empirical Molar Mass
Key insight: The molecular formula is always a whole-number multiple of the empirical formula. For example, glucose: C₆H₁₂O₆ = 6 × CH₂O. Understanding this relationship helps you see how empirical formulas connect to molecular formulas through molar mass.
The process involves several steps:
Step 1: Assume 100 g sample
Percentages become grams: 40% C = 40 g C
Step 2: Convert to moles
moles = mass ÷ atomic mass
Example: 40 g C ÷ 12.01 g/mol = 3.33 mol C
Step 3: Find smallest number of moles
Identify the element with the smallest mole value
Step 4: Divide by smallest to get ratios
mole ratio = moles ÷ smallest moles
Example: If smallest is 3.33, then C: 3.33 ÷ 3.33 = 1.00
Step 5: Round to whole numbers
If ratios aren't whole numbers, multiply all by 2, 3, 4, etc.
When ratios aren't whole numbers, multiply by a common factor:
If ratio ≈ 1.5 or 2.5:
Multiply all ratios by 2
Example: 1.5:2.5 → 3:5
If ratio ≈ 1.33 or 1.67:
Multiply all ratios by 3
Example: 1.33:2.67 → 4:8
If ratio ≈ 1.25 or 1.75:
Multiply all ratios by 4
Example: 1.25:3.75 → 5:15
Given empirical formula and measured molar mass:
Step 1: Calculate empirical molar mass
Sum atomic masses in empirical formula
Example: CH₂O = 12.01 + 2(1.008) + 16.00 = 30.03 g/mol
Step 2: Calculate multiplier n
n = Measured Molar Mass ÷ Empirical Molar Mass
Example: n = 180 ÷ 30.03 = 5.99 ≈ 6
Step 3: Multiply empirical subscripts by n
Molecular formula = n × Empirical formula
Example: C₆H₁₂O₆ = 6 × CH₂O
Given: 40.0% C, 6.71% H, 53.29% O
Find: Empirical formula
Step 1: Assume 100 g sample
40.0 g C, 6.71 g H, 53.29 g O
Step 2: Convert to moles
C: 40.0 ÷ 12.01 = 3.33 mol
H: 6.71 ÷ 1.008 = 6.66 mol
O: 53.29 ÷ 16.00 = 3.33 mol
Step 3: Find smallest (3.33)
Smallest = 3.33 mol
Step 4: Divide by smallest
C: 3.33 ÷ 3.33 = 1.00
H: 6.66 ÷ 3.33 = 2.00
O: 3.33 ÷ 3.33 = 1.00
Step 5: Write empirical formula
Empirical formula: CH₂O
Given: Empirical formula = CH₂O, Measured molar mass = 180 g/mol
Find: Molecular formula
Step 1: Calculate empirical molar mass
CH₂O = 12.01 + 2(1.008) + 16.00 = 30.03 g/mol
Step 2: Calculate multiplier n
n = 180 ÷ 30.03 = 5.99 ≈ 6
Step 3: Multiply empirical subscripts by n
C: 1 × 6 = 6
H: 2 × 6 = 12
O: 1 × 6 = 6
Step 4: Write molecular formula
Molecular formula: C₆H₁₂O₆
Understanding empirical and molecular formulas is essential for students across chemistry coursework. Here are detailed student-focused scenarios (all conceptual, not actual chemical procedures):
Scenario: Your general chemistry homework asks: "A compound contains 40.0% C, 6.71% H, and 53.29% O. What is the empirical formula?" Use the calculator: enter percent mode, C = 40.0%, H = 6.71%, O = 53.29%. The calculator shows: empirical formula = CH₂O. You learn: the process of converting percents to moles, finding ratios, and writing formulas. The calculator helps you check your work and understand each step of the calculation.
Scenario: An exam asks: "The empirical formula of a compound is CH₂O and its molar mass is 180 g/mol. What is the molecular formula?" Use the calculator: enter empirical formula data and measured molar mass = 180 g/mol. The calculator calculates: n = 180 ÷ 30 = 6, molecular formula = C₆H₁₂O₆. You learn: how to use molar mass to determine the multiplier n and find the molecular formula. The calculator makes this relationship concrete—you see exactly how empirical and molecular formulas are related.
Scenario: Your analytical chemistry lab report asks: "Based on combustion analysis, a compound contains 85.6% C and 14.4% H. Determine the empirical formula." Use the calculator: enter percent mode, C = 85.6%, H = 14.4%. The calculator shows: empirical formula = CH₂. Understanding this helps explain why the compound might be an alkene or alkane. The calculator helps you verify your calculations and understand how experimental data leads to formulas.
Scenario: Problem: "A compound contains 50.0% C and 50.0% H. What is the empirical formula?" Use the calculator: enter C = 50.0%, H = 50.0%. The calculator calculates: C: 50.0 ÷ 12.01 = 4.16 mol, H: 50.0 ÷ 1.008 = 49.6 mol. Ratios: C: 4.16 ÷ 4.16 = 1.00, H: 49.6 ÷ 4.16 = 11.9 ≈ 12. Empirical formula: CH₁₂ (or C₁H₁₂, but typically written as CH₁₂). This demonstrates how to handle calculations and rounding.
Scenario: Your organic chemistry homework asks: "Both ethanol and dimethyl ether have the molecular formula C₂H₆O. Explain why they are different compounds." Use the calculator: enter data to verify both have C₂H₆O. Understanding this helps explain why structural isomers (different connectivity) can have the same molecular formula. The calculator makes this relationship concrete—you see that formulas don't uniquely identify compounds.
Scenario: Problem: "A 2.50 g sample contains 1.20 g C, 0.20 g H, and 1.10 g O. What is the empirical formula?" Use the calculator: enter mass mode, C = 1.20 g, H = 0.20 g, O = 1.10 g. The calculator calculates: empirical formula = CH₂O. This demonstrates how to work with actual mass data instead of percentages, which is useful when you have experimental mass measurements.
Scenario: Your instructor recommends practicing different types of formula problems. Use the calculator to work through: (1) Percent composition to empirical formula, (2) Empirical formula to molecular formula, (3) Mass data to empirical formula, (4) Problems with non-integer ratios. The calculator helps you practice all problem types, identify common mistakes, and build confidence. Understanding how to solve different types of formula problems prepares you for exams where you might encounter various scenarios.
Empirical and molecular formula problems involve percents, masses, moles, ratios, and rounding that are error-prone. Here are the most frequent mistakes and how to avoid them:
Mistake: Trying to work directly with percentages instead of converting to grams first.
Why it's wrong: The calculation requires masses (in grams) to convert to moles. When given percents, you must assume a 100 g sample so percentages become grams directly (e.g., 40% C = 40 g C). Working directly with percentages gives wrong mole calculations. For example, if you try to divide 40% by 12.01, you get wrong units and wrong values.
Solution: Always assume a 100 g sample when given percents. This makes percentages become grams directly, simplifying calculations. The calculator does this automatically—observe it to reinforce the assumption.
Mistake: Forgetting to divide all mole values by the smallest to get ratios, or dividing by the wrong value.
Why it's wrong: The empirical formula requires the simplest whole-number ratio. To get ratios, you must divide all mole values by the smallest. If you forget this step or divide by the wrong value, you get wrong ratios and wrong formulas. For example, if you have 3.33 mol C, 6.66 mol H, and 3.33 mol O, dividing by 3.33 gives 1:2:1 (CH₂O), but if you divide by 6.66, you get wrong ratios.
Solution: Always identify the smallest number of moles first, then divide all mole values by that smallest value. The calculator shows this step—use it to reinforce the process.
Mistake: Rounding ratios like 1.5, 1.33, or 1.25 directly to 2, 1, or 1 instead of multiplying all ratios by a common factor.
Why it's wrong: Empirical formulas must have whole-number subscripts. If ratios aren't whole numbers, you must multiply all ratios by a common factor (2, 3, 4, etc.) to get whole numbers. Rounding directly gives wrong formulas. For example, if ratios are 1.5:2.5, rounding to 2:3 gives wrong formula. Multiplying by 2 gives 3:5 (correct).
Solution: Always check if ratios are whole numbers. If not, multiply all ratios by 2, 3, 4, etc. until they are. The calculator does this automatically—observe it to reinforce the multiplication step.
Mistake: Using incorrect atomic mass values or using atomic numbers instead of atomic masses.
Why it's wrong: Atomic masses are needed to convert grams to moles (moles = mass ÷ atomic mass). Using wrong values gives wrong mole calculations and wrong formulas. For example, using C = 12 (atomic number) instead of 12.01 (atomic mass) gives slightly wrong results. Using very wrong values (e.g., C = 6) gives completely wrong formulas.
Solution: Always use correct atomic masses from a reliable source. The calculator uses standard IUPAC values—use them to verify your values. Memorize common ones: C = 12.01, H = 1.008, O = 16.00, N = 14.01.
Mistake: Accepting a multiplier n that is not close to a whole number (e.g., n = 2.7 instead of 3).
Why it's wrong: The multiplier n must be a positive integer (1, 2, 3, ...) because molecular formulas are whole-number multiples of empirical formulas. If n is not close to a whole number, there's an error in the measured molar mass, percent composition, or empirical formula calculation. For example, if n = 2.7, something is wrong—check your inputs.
Solution: Always check that n is close to a whole number (within 0.1). If not, double-check: (1) Measured molar mass, (2) Percent composition, (3) Empirical formula calculation. The calculator shows n—use it to verify it's reasonable.
Mistake: Using the empirical formula when asked for the molecular formula, or vice versa.
Why it's wrong: Empirical formulas show ratios (simplest), while molecular formulas show actual atom counts. They're different! For example, glucose has empirical formula CH₂O but molecular formula C₆H₁₂O₆. Using the wrong one gives wrong answers. If a problem asks for molecular formula and you give empirical formula, you're missing the multiplier step.
Solution: Always identify what's being asked: empirical (ratios) or molecular (actual counts). If molecular is asked, you need the measured molar mass to calculate n. The calculator shows both—use it to reinforce the distinction.
Mistake: Writing formulas in random order instead of Hill notation (C first, H second, then alphabetical).
Why it's wrong: Hill notation is the standard way to write formulas. While the formula is technically correct in any order, Hill notation ensures consistency and makes formulas easier to compare and search. Writing formulas in wrong order (e.g., H₆C₆O₆ instead of C₆H₁₂O₆) is not wrong but not standard.
Solution: Always write formulas in Hill notation: C first (if present), H second (if present), then other elements alphabetically. The calculator uses Hill notation—observe it to reinforce the standard order.
Once you've mastered basics, these advanced strategies deepen understanding and prepare you for complex formula determination problems:
Conceptual insight: When given percents, assuming a 100 g sample makes percentages become grams directly (40% = 40 g). This works because percent composition is independent of sample size—the ratio of elements stays the same whether you have 1 g or 100 g. The mole ratios (which determine the empirical formula) are the same regardless of sample size. Understanding this helps you see why the 100 g assumption is valid and why it simplifies calculations. This provides deep insight beyond memorization: percent composition is scale-invariant.
Quantitative insight: Common multipliers for non-integer ratios: 1.5 or 2.5 → multiply by 2, 1.33 or 1.67 → multiply by 3, 1.25 or 1.75 → multiply by 4. These patterns come from common fractions: 1.5 = 3/2, 1.33 ≈ 4/3, 1.25 = 5/4. Understanding these patterns helps you quickly identify the multiplier needed without trial and error. This provides quantitative insight into why certain multipliers work for certain ratio patterns.
Practical framework: Remember: C first (if present), H second (if present), then alphabetical. For compounds without C, all elements alphabetical. Examples: C₆H₁₂O₆ (glucose), C₂H₆O (ethanol), H₂O (water), NaCl (salt). These mental shortcuts help you write formulas correctly and consistently. Understanding Hill notation builds intuition about standard formula representation.
Unifying concept: Many compounds can share the same molecular formula but have different structures (structural isomers). For example, C₂H₆O represents both ethanol (CH₃CH₂OH) and dimethyl ether (CH₃OCH₃). Understanding this helps you see that formulas don't uniquely identify compounds—structural information is needed. This connects formula determination to structural chemistry and explains why additional data (spectroscopy, properties) is needed for compound identification.
Exam technique: For quick checks: common atomic masses: C ≈ 12, H ≈ 1, O ≈ 16, N ≈ 14. Common empirical formulas: CH₂O (carbohydrates), CH₂ (alkenes), CH (alkynes). These mental shortcuts help you quickly estimate formulas and check calculator results. Understanding approximate relationships builds intuition about formula determination.
Advanced consideration: This calculator works only for discrete molecular compounds. It cannot: (a) Distinguish structural isomers (same molecular formula, different structure), (b) Handle polymers (repeating units), (c) Work for ionic compounds (formula units, not molecules), (d) Account for isotopic variations (uses averaged atomic masses). Understanding these limitations shows why empirical/molecular formulas are just one piece of compound identification, and why additional analytical techniques (spectroscopy, chromatography) are needed for complete identification, especially for complex or unknown compounds.
Advanced consideration: Percent composition and formulas are two sides of the same coin: given a formula, you can calculate percent composition; given percent composition, you can determine a formula. This bidirectional relationship helps you verify formulas (calculate percent composition and compare to experimental data) and understand how experimental data leads to formulas. Understanding this connection provides insight into how analytical chemistry connects experimental measurements to molecular structure.
• Whole Number Subscripts: Empirical formula derivation rounds mole ratios to whole numbers. Experimental error in percent composition data can lead to ambiguous ratios (e.g., 1.5 could be 3:2 or measurement error in 1:1). Judgment is required for borderline cases.
• No Structural Information: Empirical and molecular formulas show only elemental composition, not arrangement. Isomers (same formula, different structure) cannot be distinguished—additional spectroscopic data is needed.
• Molecular Formula Requires Molar Mass: Going from empirical to molecular formula requires independently determined molar mass (from mass spectrometry, freezing point depression, etc.). Without this, only the simplest ratio is determinable.
• Discrete Molecules Only: Formula concepts apply to molecular compounds. Ionic compounds, network solids, and polymers are better described by formula units or repeat units, not molecular formulas in the traditional sense.
Important Note: This calculator is strictly for educational and informational purposes only. It demonstrates empirical/molecular formula determination for learning. For unknown compound identification, combine with spectroscopic methods (MS, NMR, IR) and professional analytical chemistry techniques.
The empirical and molecular formula principles referenced in this content are based on authoritative chemistry sources:
Empirical formulas represent simplest whole-number ratios. Molecular formula determination requires additional molar mass data from techniques like mass spectrometry.
Calculate molecular weight from chemical formulas
Calculate percent composition of compounds
Work with mole-to-mole conversions in reactions
Balance chemical equations automatically
PV = nRT calculations for gases
Analyze bond polarity and electronegativity