An empirical formula identifies the simplest whole-number ratio of atoms of each element in a compound. To determine an empirical formula from experimental data, convert the measured mass of each element into moles, divide each mole amount by the smallest value, and adjust the resulting ratios to whole numbers. These steps reveal the relative numbers of atoms that make up the compound, even when the actual number of atoms in a molecule is unknown.
This method is fundamental to chemistry because experimental measurements typically provide information about mass, while chemical formulas describe atoms and their proportions. Understanding how to move from one to the other makes it possible to determine the composition of an unknown compound using laboratory data.
What an empirical formula tells you
An empirical formula represents the simplest whole-number ratio of the elements in a compound. For example, hydrogen peroxide has the molecular formula H₂O₂, but its empirical formula is HO because hydrogen and oxygen occur in a 1:1 ratio.
Similarly, glucose has the molecular formula C₆H₁₂O₆ and the empirical formula CH₂O. The empirical formula shows that the ratio of carbon to hydrogen to oxygen is 1:2:1, without specifying the total number of atoms in a glucose molecule.
An empirical formula does not necessarily describe the actual structure of a substance. It provides information about elemental proportions, not the arrangement of atoms or the number of atoms in an individual molecule.
This distinction matters when interpreting experimental results. A compound with an empirical formula of CH₂O could have a molecular formula of CH₂O, C₂H₄O₂, C₃H₆O₃, or another whole-number multiple, provided the molecular composition is chemically consistent. Additional information, such as the compound’s molar mass, is needed to determine its molecular formula.
What experimental data you need
The most direct way to determine an empirical formula is to measure the mass of each element in a known sample of a compound. Experimental data may report these masses in grams or provide the percentage by mass of each element.
For example, an analysis might show that a compound contains 2.40 grams of carbon and 0.40 grams of hydrogen. These measurements describe the amounts of the elements by mass, but they do not directly reveal their atomic ratio.
Atoms of different elements have different masses. A gram of hydrogen contains far more atoms than a gram of carbon because hydrogen atoms are much lighter. Consequently, comparing the measured masses alone cannot establish the ratio of atoms in the compound.
Chemists solve this problem by converting mass measurements into moles. A mole is a unit representing a fixed number of particles, approximately 6.022 × 10²³. Because each element has a known molar mass, its measured mass can be converted into an amount of substance.
Once all the elements are expressed in moles, their relative amounts can be compared directly.
Step 1: Convert each element’s mass into moles
The first step is to calculate the number of moles of each element in the sample. Use the molar mass of each element from the periodic table.
The relationship is:
Moles = Mass ÷ Molar mass
Mass is usually measured in grams, and molar mass is expressed in grams per mole. Dividing grams by grams per mole gives moles.
Consider a compound containing 2.40 grams of carbon and 0.40 grams of hydrogen. Using molar masses of approximately 12.01 g/mol for carbon and 1.008 g/mol for hydrogen, the calculations are:
Carbon:
2.40 g ÷ 12.01 g/mol ≈ 0.200 mol
Hydrogen:
0.40 g ÷ 1.008 g/mol ≈ 0.397 mol
The sample therefore contains approximately 0.200 moles of carbon atoms and 0.397 moles of hydrogen atoms.
These values are not yet the empirical formula. They represent the amounts of each element in the sample. The next step is to determine their simplest relative proportions.
Step 2: Divide each mole amount by the smallest value
To find the simplest ratio, identify the smallest calculated mole amount and divide every mole amount by it.
In the example, carbon has approximately 0.200 moles, while hydrogen has approximately 0.397 moles. Carbon is the smaller value.
Divide both amounts by 0.200:
Carbon:
0.200 ÷ 0.200 = 1
Hydrogen:
0.397 ÷ 0.200 ≈ 1.99
The resulting ratio is approximately 1:1.99, which is effectively 1:2 given the precision of the measurements.
This means that the compound contains approximately two hydrogen atoms for every carbon atom. The simplest whole-number ratio is therefore carbon to hydrogen = 1:2.
The empirical formula is CH₂.
The key principle is that dividing every mole amount by the same smallest value preserves the relative proportions while expressing them in a form that is easier to interpret.
Step 3: Convert the ratios into whole numbers
The ratios obtained after division are often close to whole numbers, but they are not always exact. Experimental measurements, rounding, and the precision of atomic masses can produce values such as 1.98, 2.02, or 2.99 instead of precisely 2 or 3.
When a ratio is sufficiently close to a whole number, the appropriate integer is usually the best interpretation. For example, a ratio of 1:2.01 generally supports a 1:2 ratio rather than suggesting a distinct composition.
However, some calculations produce fractional ratios that cannot reasonably be rounded to the nearest integer. A result such as 1:1.5 requires further analysis because it indicates that the simplest ratio may involve more than one atom of the element represented by 1.
To remove a fractional component, multiply every ratio by the same integer until all values become whole numbers.
For example, suppose the mole amounts produce a ratio of:
A:B = 1:1.5
Multiplying both values by 2 gives:
A:B = 2:3
The empirical formula would contain two atoms of A for every three atoms of B, written A₂B₃.
If the ratio is 1:1.33, it may be close to 1:1⅓, or 1:4/3. Multiplying every value by 3 gives 3:4, producing the formula A₃B₄.
Common multipliers include 2 for ratios ending in approximately 0.5, 3 for ratios ending in approximately 0.33 or 0.67, and 4 for ratios ending in approximately 0.25 or 0.75. These are useful starting points, but the correct multiplier depends on the entire set of ratios.
Do not automatically round every decimal to the nearest integer. A value of 1.5 is not adequately represented by 2 when it reflects a meaningful fractional relationship. Instead, determine whether the ratios can be converted into a consistent set of small whole numbers.
Step 4: Write the empirical formula
Once the simplest whole-number ratio has been established, use the numbers as subscripts in the chemical formula.
Each element is represented by its chemical symbol, and the subscript indicates the relative number of atoms in the formula.
For example, a ratio of carbon:hydrogen:oxygen = 1:2:1 produces CH₂O.
A ratio of sodium:sulfur:oxygen = 2:1:4 produces Na₂SO₄.
A ratio of aluminum:oxygen = 2:3 produces Al₂O₃.
When an element has a ratio of 1, its subscript is omitted. For example, CH₂O is correct, rather than C₁H₂O₁.
The subscripts must express the simplest whole-number ratio. If every subscript can be divided by the same integer, the formula is not yet in empirical form.
For example, C₂H₄O₂ simplifies to CH₂O by dividing all subscripts by 2. The latter is the empirical formula.
Determining an empirical formula from percentage composition
Experimental results are not always reported as masses. Sometimes they provide the percentage by mass of each element in a compound.
The same method applies because percentages can be converted into masses by assuming a convenient sample size.
A particularly useful choice is a 100-gram sample. In a 100-gram sample, the numerical value of each percentage becomes the mass of that element in grams.
For example, suppose a compound contains 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass.
Assume that the sample has a mass of 100 grams. Its elemental masses are then:
- Carbon: 40.0 g
- Hydrogen: 6.7 g
- Oxygen: 53.3 g
Convert each mass into moles using the corresponding molar mass.
Carbon:
40.0 g ÷ 12.01 g/mol ≈ 3.33 mol
Hydrogen:
6.7 g ÷ 1.008 g/mol ≈ 6.65 mol
Oxygen:
53.3 g ÷ 16.00 g/mol ≈ 3.33 mol
Next, divide each mole amount by the smallest value, approximately 3.33.
Carbon:
3.33 ÷ 3.33 = 1
Hydrogen:
6.65 ÷ 3.33 ≈ 2
Oxygen:
3.33 ÷ 3.33 = 1
The simplest ratio is approximately 1:2:1, so the empirical formula is CH₂O.
The assumed 100-gram sample does not need to represent the actual amount of compound used in an experiment. It is simply a convenient mathematical choice that converts percentages into corresponding masses without changing the elemental ratios.
If the percentages do not add up to approximately 100%, investigate whether the data include measurement error, rounding, omitted elements, or other components. Do not automatically force incomplete or inconsistent data into a formula.
Determining an empirical formula from combustion analysis
Combustion analysis is a common method for determining the elemental composition of compounds containing carbon and hydrogen, often with oxygen as well.
In a typical experiment, a measured sample is burned completely in excess oxygen. The carbon in the sample is converted into carbon dioxide, while the hydrogen is converted into water. The quantities of carbon dioxide and water produced can then be measured.
These product masses provide indirect measurements of the amounts of carbon and hydrogen originally present in the sample.
Carbon dioxide contains one carbon atom per molecule. Therefore, the moles of carbon in the original sample equal the moles of carbon dioxide produced, assuming complete combustion and that the carbon dioxide is measured accurately.
Water contains two hydrogen atoms per molecule. Therefore, the moles of hydrogen atoms equal twice the moles of water produced.
The calculations begin by converting the measured product masses into moles:
Moles of CO₂ = Mass of CO₂ ÷ 44.01 g/mol
Moles of H₂O = Mass of H₂O ÷ 18.02 g/mol
The elemental amounts are then:
Moles of carbon = Moles of CO₂
Moles of hydrogen = 2 × Moles of H₂O
Suppose combustion produces 4.40 grams of carbon dioxide and 1.80 grams of water.
The carbon dioxide amount is:
4.40 g ÷ 44.01 g/mol ≈ 0.100 mol CO₂
This corresponds to approximately 0.100 moles of carbon.
The water amount is:
1.80 g ÷ 18.02 g/mol ≈ 0.100 mol H₂O
Because each water molecule contains two hydrogen atoms, the sample contains approximately 0.200 moles of hydrogen atoms.
The resulting carbon-to-hydrogen ratio is 0.100:0.200, or 1:2.
If the compound contains only carbon and hydrogen, its empirical formula is CH₂.
If oxygen is also present, its amount may be determined by measuring it directly or by calculating its mass by difference, provided the sample contains only the known elements and the measurements account for all of its mass. The oxygen mass is then converted into moles, and all elemental mole amounts are reduced to their simplest whole-number ratio.
Combustion analysis depends on complete combustion, accurate product collection, and correct interpretation of the measured quantities. Incomplete combustion or loss of products can produce misleading results.
How to check whether the result is reasonable
A calculated empirical formula should satisfy both the mathematics of the mole ratios and the known composition of the sample.
First, check the mole conversions. Each mass should have been divided by the correct molar mass, and each product in a combustion experiment should have been interpreted according to its chemical composition.
Second, verify that every mole amount was divided by the same smallest value. Using a different divisor for different elements would distort the relative proportions.
Third, confirm that fractional ratios were handled consistently. If a ratio of 1:1.5 was multiplied by 2, every component of the ratio must be multiplied by 2, not just the fractional value.
Fourth, check whether the subscripts share a common factor. A formula such as C₂H₆O₂ reduces to CH₃O and therefore is not written in its simplest empirical form. Whether that simplified ratio corresponds to a chemically plausible compound depends on the substance and its bonding, but the mathematical reduction is straightforward.
Finally, consider the precision of the measurements. A calculated ratio of 1.99 may reasonably indicate 2, but an unexplained ratio of 1.47 should not automatically be treated as 1.5. The appropriate interpretation depends on the uncertainty in the experimental data and whether a simple integer ratio is supported.
Chemical knowledge can help identify an implausible result, but it should not be used to force data into a preferred formula. If the ratios remain inconsistent, recheck the calculations and consider whether the measurements are sufficiently accurate.
Why experimental error matters
An empirical formula is inferred from measured quantities, so its reliability depends on the quality of the experimental data.
Errors in mass measurements can affect the calculated mole amounts and the resulting ratios. Contamination may introduce additional elements or alter the measured composition. Incomplete drying can cause a sample to appear heavier than it is, while loss of material during handling can change the measured mass.
In combustion analysis, incomplete combustion can prevent all carbon from being converted into carbon dioxide. Water may also be lost or collected incompletely. Either problem can distort the calculated elemental proportions.
Rounding introduces another source of uncertainty. Molar masses and measured quantities are not always exact, and intermediate rounding can cause small differences in the final ratios. Carrying sufficient digits through the calculations and rounding the final result appropriately helps reduce this problem.
Experimental results should also be evaluated in light of the expected measurement uncertainty. A small deviation from a simple integer ratio may be consistent with ordinary experimental error. A substantial deviation may indicate a calculation mistake, an unrecognized component, a mixture of substances, or an inadequately measured sample.
When repeated measurements are available, agreement between independent determinations provides stronger evidence that the calculated formula accurately represents the compound.
Empirical formulas and molecular formulas
An empirical formula and a molecular formula describe different aspects of a compound’s composition.
The empirical formula gives the simplest whole-number ratio of atoms. The molecular formula gives the actual number of atoms of each element in a molecule.
For a molecular compound, the molecular formula must be a whole-number multiple of its empirical formula. Determining that multiple requires the compound’s molar mass.
The empirical formula mass is calculated by adding the molar masses of all atoms represented in the empirical formula.
For example, the empirical formula CH₂O has an approximate formula mass of:
12.01 + (2 × 1.008) + 16.00 ≈ 30.03 g/mol
If the compound’s experimentally determined molar mass is approximately 180.16 g/mol, divide the molecular molar mass by the empirical formula mass:
180.16 ÷ 30.03 ≈ 6
This indicates that the molecular formula contains six times as many atoms of each element as the empirical formula.
Multiplying the subscripts in CH₂O by 6 gives C₆H₁₂O₆.
Thus, the molecular formula is C₆H₁₂O₆, while the empirical formula is CH₂O.
If the ratio of molecular molar mass to empirical formula mass is close to a small whole number, that integer can identify the molecular formula. If the ratio is not close to a whole number, the measurements, assumptions, and calculations should be checked.
For ionic compounds, formulas generally express the simplest ratio of ions rather than the contents of discrete molecules. For example, NaCl represents a 1:1 ratio of sodium ions to chloride ions in the ionic solid.
A reliable method for solving empirical formula problems
The essential process is the same whether experimental data provide elemental masses, percentages, or combustion products.
Begin by identifying what the measurements represent. If the data give elemental masses, use those masses directly. If they give percentages, assume a convenient total sample mass, often 100 grams. If they give combustion products, first calculate the amounts of the elements represented by those products.
Convert each elemental mass into moles using its molar mass. Divide every mole amount by the smallest amount to establish the relative atomic ratio. If the resulting values are fractional, multiply the entire ratio by an appropriate small integer to obtain whole numbers.
Use those integers as subscripts to write the empirical formula, omitting any subscript equal to 1. Then verify that the formula is in its simplest form and that the result is consistent with the precision and reliability of the experimental data.
The underlying principle is simple: mass measurements become mole amounts, mole amounts become relative ratios, and relative ratios become the simplest whole-number chemical formula. By following these steps carefully, experimental composition data can reveal the elemental proportions of an unknown compound without requiring prior knowledge of its molecular structure.