Working Backward From Percent Composition to Formula
The PGRS method for determining an empirical formula
An empirical formula expresses the simplest, smallest whole-number ratio of atoms of each element present in a compound — it may or may not be the compound's actual, true molecular formula (the exact number of each atom actually present in a single real molecule of the substance), but it always represents that same ratio reduced to its lowest possible whole-number terms.
The most common way empirical formula problems are presented is starting from percent composition data — the percentage, by mass, that each element contributes to the compound's total mass — and the PGRS method converts that percentage data systematically into the actual empirical formula: Percent (start with the given percent composition) → Grams (convert percentages directly into an equivalent mass in grams, by assuming a convenient 100 gram sample) → Ratio (convert those masses into moles, then find the simplest ratio by dividing every mole value by the smallest one) → Simplify (if the resulting ratio isn't yet made of whole numbers, multiply every value by a small whole number to clear any remaining decimals).
This method works specifically because percent composition, by its very definition, doesn't depend on the specific total sample size you happen to be considering — assuming exactly 100 grams of sample is simply a convenient, arbitrary choice that turns each given percentage directly into an equivalent number of grams, without changing the underlying ratio between elements at all.
💡 Why Assuming Exactly 100 Grams Works Every Single Time
The first step of PGRS — assuming a 100-gram sample — can seem like an arbitrary or even slightly suspicious shortcut at first, but it works precisely and reliably because percent composition is a ratio, not an absolute quantity, and ratios don't change no matter what specific total sample size you imagine.
If a compound is 40.0% carbon by mass, that statement is true regardless of whether you're holding 10 grams, 100 grams, or 10,000 grams of the compound — in every case, exactly 40.0% of that total mass is carbon. Choosing to imagine exactly 100 grams simply makes the arithmetic maximally convenient, since a percentage of 100 converts directly into that same number of grams with no additional calculation needed at all (40.0% of 100 g is simply 40.0 g, with no separate percentage-to-decimal conversion or multiplication step required). Any other sample size would work in principle and produce the exact same final mole ratio, but would require an extra multiplication step to convert the percentage into an equivalent mass first — the 100-gram assumption is chosen purely for arithmetic convenience, not because it reflects any special, physically meaningful property of the actual compound or sample being analyzed.
PG
Percent to Grams — the first conversion
Given a compound's percent composition (for example, 40.0% carbon, 6.7% hydrogen, 53.3% oxygen by mass), assume you have exactly 100 grams of the compound. Under this assumption, each given percentage converts directly into an equivalent number of grams of that specific element, with no further calculation needed for this step — 40.0% becomes 40.0 g, 6.7% becomes 6.7 g, and 53.3% becomes 53.3 g.
A compound reported as 40.0% C, 6.7% H, 53.3% O by mass is treated, for calculation purposes, as containing 40.0 g C, 6.7 g H, and 53.3 g O within an assumed 100 g total sample.
GR
Grams to moles, then finding the ratio
Convert each element's assumed mass into moles by dividing by that element's molar mass (read from the periodic table), exactly as covered in The Mole lesson. Once every element's mole value has been calculated, find the simplest ratio between them by dividing every element's mole value by whichever element had the smallest mole value — this produces a ratio where the smallest element is normalized to exactly 1.0, with every other element's value expressed relative to it.
Continuing the example: 40.0 g C ÷ 12.011 g/mol = 3.33 mol C; 6.7 g H ÷ 1.008 g/mol = 6.65 mol H; 53.3 g O ÷ 16.00 g/mol = 3.33 mol O. Dividing all three by the smallest value (3.33): C = 3.33/3.33 = 1.00, H = 6.65/3.33 = 2.00, O = 3.33/3.33 = 1.00 — giving a ratio of C:H:O = 1:2:1.
S
Simplifying to whole numbers, and finding the molecular formula
If the ratio from the previous step already comes out to whole numbers (or numbers extremely close to whole numbers, allowing for rounding), that ratio directly gives the empirical formula's subscripts. If the ratio instead includes a clear, recognizable fraction (commonly .5, .33/.67, or .25/.75), every value in the ratio must be multiplied by the same small whole number (2, 3, or 4, respectively) to clear that fraction and produce a full whole-number ratio. Once the true empirical formula is known, the actual molecular formula can be found if the compound's real molar mass is also known: calculate n = (given molar mass) ÷ (empirical formula's own calculated mass), then multiply every subscript in the empirical formula by that value of n to get the true molecular formula.
The ratio C:H:O = 1:2:1 from the previous step is already whole numbers, giving the empirical formula CH₂O (formula mass ≈ 30.03 g/mol). If the compound's actual molar mass is separately given as 60.06 g/mol, then n = 60.06 ÷ 30.03 = 2, so the true molecular formula is (CH₂O) × 2 = C₂H₄O₂ (acetic acid).
🔬 Applied Scenario — Empirical Formula Determination in Practice
Determining an unknown compound's empirical (and eventually molecular) formula from experimental data is a routine, essential task across analytical chemistry.
A
Identifying a newly synthesized or newly isolated compound. When chemists synthesize a genuinely new compound, or isolate an unknown natural substance, determining its percent composition experimentally (often via combustion analysis, covered in the next lesson) and then applying PGRS is typically the very first step toward identifying exactly what that compound actually is.
B
Verifying a compound's purity and identity against a known reference. Measuring a sample's percent composition and calculating its empirical formula provides a direct way to confirm that a manufactured or purchased chemical actually matches its expected, labeled identity, or to detect that it's been contaminated with an impurity that would shift the measured percent composition away from the expected values.
C
Empirical formula alone is often not enough — molar mass data is essential for determining the true molecular formula. Since multiple different molecular formulas can share the exact same empirical formula (for example, CH₂O, C₂H₄O₂, and C₆H₁₂O₆ all reduce to the same 1:2:1 ratio), a separate molar mass measurement (commonly obtained through mass spectrometry in modern analytical chemistry) is required to determine which specific multiple of the empirical formula matches the compound's true molecular identity.
D
Empirical formula determination underlies combustion analysis specifically. As covered in the next lesson, combustion analysis of an unknown organic compound produces exactly the kind of mass data (converted first into percent composition, or directly into moles of each element) that the PGRS method is specifically designed to process into a final empirical formula.
⚠️ Most Common Empirical Formula Mistakes
The empirical formula is NOT automatically the same as the true molecular formula — they can differ, sometimes considerably. Students sometimes assume that once PGRS produces a formula, that formula must be the compound's actual, complete molecular formula. The empirical formula only gives the simplest whole-number ratio; the true molecular formula could be that same ratio multiplied by any whole number n, which can only be determined using additional molar mass data.
A ratio like 1.5 must be cleared by multiplying by 2 (not rounded to the nearest whole number) — rounding a clearly fractional ratio produces an incorrect empirical formula. Students sometimes round a ratio value like 1.5 up to 2 or down to 1, rather than recognizing it as a genuine fraction requiring the entire ratio to be multiplied through by 2 to clear it properly. A ratio value of 1.5 specifically signals multiplying every element's ratio value by 2; a value near 1.33 or 1.67 signals multiplying by 3; a value near 1.25 or 1.75 signals multiplying by 4.
Dividing by the smallest mole value is a required step, not an optional shortcut — skipping it produces a ratio in the wrong units entirely. Students sometimes try to write an empirical formula directly from raw mole values without first normalizing by dividing by the smallest one. This normalization step is what actually produces the simplest whole-number ratio; raw mole values alone don't represent the correct empirical formula ratio.
✓ Quick Self-Test
1. What is an empirical formula, and how does it differ from a molecular formula?
2. What does each letter in the PGRS method stand for?
3. Why does assuming a 100 gram sample work correctly, regardless of how much of the compound you actually have?
4. If a mole ratio calculation produces values like C = 1.00, H = 1.33, O = 2.00, what should you do to find the correct empirical formula?
5. How do you determine a compound's true molecular formula once you know its empirical formula?
Answers:
1. An empirical formula expresses the simplest, smallest whole-number ratio of atoms of each element in a compound. It differs from the molecular formula, which states the exact, true number of each atom actually present in one real molecule of the substance — the molecular formula is always a whole-number multiple of the empirical formula (possibly the same, or possibly larger).
2. PGRS stands for Percent (start with given percent composition data), Grams (convert percentages directly to grams by assuming a 100 g sample), Ratio (convert grams to moles, then divide all values by the smallest to find the simplest ratio), and Simplify (multiply through by a small whole number if needed to clear any remaining fractional values).
3. Assuming a 100 gram sample works because percent composition is a ratio (a percentage of the whole), and that ratio remains exactly the same no matter what total sample size is actually being considered — 100 grams is simply the most arithmetically convenient choice, since it converts each percentage directly into an equal number of grams with no extra calculation.
4. A ratio value of 1.33 (close to 1⅓) signals that the entire ratio should be multiplied by 3 to clear the fraction, giving C = 3, H = 4, O = 6 — producing the empirical formula C₃H₄O₆.
5. Divide the compound's true, separately known molar mass by the mass calculated from the empirical formula alone, giving a whole number n. Multiply every subscript in the empirical formula by that value of n to obtain the true molecular formula.