Empirical and Molecular Formulas经验式与分子式

Big Idea · Scale, Proportion, and Quantity (SPQ)AP Science Practice 5 · Mathematical RoutinesAP link · Unit 1~40 min

Learning goals

  • Tell empirical formulas from molecular formulas and explain that a molecular formula is a whole-number multiple of the empirical formula
  • Find an empirical formula from percent composition or lab masses, including ratios like 1.5 and 1.33
  • Find a molecular formula from the empirical formula and the molar mass

Before you start

  • Percent composition (2.6.3)
  • Converting between mass, moles, and particles (2.6.2)

Three substances, one recipe

Formaldehyde (CH₂O), acetic acid (C₂H₄O₂, the acid in vinegar), and glucose (C₆H₁₂O₆) are completely different substances, yet their percent compositions are identical: 40.0% C, 6.7% H, and 53.3% O. So percent composition alone only tells you the simplest ratio of atoms. To know how many atoms are really in one molecule, you need one more piece of data: the molar mass.

Empirical and molecular formulas

An empirical formula经验式（实验式） gives the simplest whole-number ratio of atoms in a compound. It is called “empirical” because it comes straight from experimental data. A molecular formula分子式 gives the actual number of each kind of atom in one molecule.

  • A molecular formula is a whole-number multiple of the empirical formula: molecular formula = (empirical formula)ₙ, where n = molar mass ÷ empirical formula mass.
  • n can be 1: for H₂O, CO₂, and CH₄, the molecular formula is the empirical formula.
  • Ionic compounds have no molecules, and their formulas are already the simplest ratio, so NaCl, Fe₂O₃, and K₂Cr₂O₇ are empirical formulas.

Chinese textbooks call the empirical formula 实验式 (“experimental formula”) or 最简式 (“simplest formula”) and introduce it in the organic chemistry course, but the method is the same.

Steps for finding an empirical formula

  1. Assume a 100-g sample. Each percent becomes grams. Skip this step if the problem gives masses.
  2. Convert grams to moles. Divide each mass by the element’s molar mass.
  3. Divide by the smallest number of moles. This gives the mole ratio.
  4. If a ratio is not close to a whole number, multiply. Multiply all ratios by 2 if one is about 1.5, by 3 for about 1.33 or 1.67, by 4 for about 1.25 or 1.75, and by 5 for about 1.2.
  5. Round and write the formula. Then find n = molar mass ÷ empirical formula mass and write the molecular formula.

Step 4 is where most mistakes happen. Never round 1.5 to 2. Round only ratios within about 0.1 of a whole number (like 2.98 or 1.03); that small difference comes from measurement and rounding.

This lesson uses the atomic masses on the AP periodic table: H 1.008, C 12.01, N 14.01, O 16.00, P 30.97, and Fe 55.85.

Pick a substance and work through the steps to its empirical and molecular formulas. You can also change the percents and the molar mass:

Try it: empirical and molecular formulas from percent composition

Example 1: from percent composition to molecular formula

Hydrazine, a rocket fuel, is 87.4% N and 12.6% H by mass, and its molar mass is 32.05 g/mol. Find its empirical and molecular formulas.

Step-by-step solution

  1. Assume 100 g. The percents become masses: 87.4 g N and 12.6 g H.

  2. Convert to moles. Atom ratios are mole ratios:

    n(N)=87.4/14.01=6.238 moln(H)=12.6/1.008=12.50 mol\begin{aligned} n(\text{N}) &= 87.4 / 14.01 \\ &= 6.238\ \text{mol} \\ n(\text{H}) &= 12.6 / 1.008 \\ &= 12.50\ \text{mol} \end{aligned}

  3. Divide by the smallest.

    N:H=6.2386.238:12.506.238=1:2.00\begin{aligned} \text{N} : \text{H} &= \frac{6.238}{6.238} : \frac{12.50}{6.238} \\ &= 1 : 2.00 \end{aligned}

  4. Empirical formula. NH₂, with a formula mass of 14.01 + 2(1.008) = 16.03.

  5. Molecular formula. The molecular formula is a whole-number multiple of the empirical one:

    n=32.0516.03=2.00NX2HX4\begin{gathered} n = \frac{32.05}{16.03} = 2.00 \\ \boxed{\ce{N2H4}} \end{gathered}

  6. Check. N₂H₄ has a molar mass of 2(14.01) + 4(1.008) = 32.05 g/mol, as given.

Example 2: from lab masses to molecular formula

A 2.50-g sample of a phosphorus oxide contains 1.09 g of phosphorus. Its molar mass is 283.9 g/mol. Find its molecular formula.

Step-by-step solution

  1. Mass of each element. These are already masses, so there is no need to assume 100 g:

    m(P)=1.09 gm(O)=2.50−1.09=1.41 g\begin{aligned} m(\text{P}) &= 1.09\ \text{g} \\ m(\text{O}) &= 2.50 - 1.09 \\ &= 1.41\ \text{g} \end{aligned}

  2. Convert to moles (mol).

    n(P)=1.09/30.97=0.03520n(O)=1.41/16.00=0.08813\begin{aligned} n(\text{P}) &= 1.09 / 30.97 \\ &= 0.03520 \\ n(\text{O}) &= 1.41 / 16.00 \\ &= 0.08813 \end{aligned}

  3. Divide by the smallest. P : O = 1 : 2.504. That is not close to a whole number, so don’t round.

  4. Multiply by 2. P : O = 2 : 5, so the empirical formula is P₂O₅, with a formula mass of 2(30.97) + 5(16.00) = 141.94.

  5. Molecular formula.

    n=283.9141.94=2.000PX4OX10\begin{gathered} n = \frac{283.9}{141.94} = 2.000 \\ \boxed{\ce{P4O10}} \end{gathered}

    This is tetraphosphorus decoxide; its common name, “phosphorus pentoxide,” comes from the empirical formula.

  6. A common mistake. Forcing 1 : 2.50 to 1 : 2 or 1 : 3. When a ratio ends in .5, .33, or .25, multiply by 2, 3, or 4.

Practice: after dividing by the smallest number of moles, you get this ratio. What next?
  1. 1 : 1.50
  2. 1 : 1.33
  3. 1 : 2.98
  4. 1 : 1.25
  5. 1 : 1.67
  6. 1 : 2.50
  7. 1 : 2.01 : 0.99
  8. 1 : 1.20
  9. 1 : 1.75

Common mistakes

Mistake 1: “1 : 1.5 rounds to 1 : 2.” 1.5 is exactly halfway between 1 and 2, and rounding it gives the wrong formula. Only ratios within about 0.1 of a whole number can be rounded. Ratios like 1.5, 1.33, and 1.25 need a multiplier.

Mistake 2: “The percents are the mole ratio.” Glucose’s mass percents are 40.0 : 6.7 : 53.3, but its mole ratio is 1 : 2 : 1. Elements have different molar masses, so you must convert grams to moles first.

Mistake 3: “Same empirical formula, same substance.” Formaldehyde (CH₂O), acetic acid (C₂H₄O₂), and glucose (C₆H₁₂O₆) all have the empirical formula CH₂O, but they are very different substances. You need the molar mass to pin down the molecular formula.

Quick check

What is the empirical formula of benzene, C₆H₆?

A compound has the empirical formula CH₂O and a molar mass of 60.05 g/mol. What is its molecular formula?

Summary

  • An empirical formula is the simplest whole-number ratio of atoms. A molecular formula is the actual count in one molecule, n times the empirical formula.
  • To find the empirical formula: assume 100 g → convert to moles → divide by the smallest → multiply if needed → round. Use × 2 for 1.5, × 3 for 1.33, and × 4 for 1.25.
  • To find the molecular formula: n = molar mass ÷ empirical formula mass, and the molecular formula is (empirical formula)ₙ.
On to AP (for later)

AP Chemistry Unit 1 asks you to find empirical formulas from mass data. The data may be a percent composition, or the masses of CO₂ and H₂O produced when a sample burns, from which you first find the masses of carbon and hydrogen. The steps for finding and rounding the ratio are exactly the ones in this lesson.

Practice this chapter: 2.6 · The Mole and Stoichiometry · 10 questions

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