The **molecular formula** indicates the actual number of atoms in a compound; The **empirical formula **indicates the smallest whole-number ratio of atoms in a compound.

The molecular formula for a compound such as phosphorous pentoxide is P_{4}O_{10}. We can see this as a ratio of 4 phosphorus atoms to 10 oxygen atoms, which can be written as 4:10. This ratio can be reduced to 2:5, as both numbers are divisible by 2. Therefore, the empirical formula of P_{4}O_{10} can be written as P_{2}O_{5}.

Empirical formulas can be determined using mass composition data. If we know the mass or **percentage composition** of each element in a compound, this information can be converted into moles and then used to work out the empirical formula.

Suppose you are given a compound such as methyl acetate, a solvent commonly used in paints, inks, and adhesives. When methyl acetate was chemically analyzed, it was discovered to have 48.64% carbon (C), 8.16% hydrogen (H), and 43.20% oxygen (O). For the purposes of determining empirical formulas, we assume that we have 100 g of the compound. If this is the case, the percentages will be equal to the mass of each element in grams.

Step 1: Change each percentage to an expression of the mass of each element in grams. That is, 48.64% C becomes 48.64 g C, 8.16% H becomes 8.16 g H, and 43.20% O becomes 43.20 g O because we assume we have 100 g of the overall compound.

Step 2: Convert the amount of each element in grams to its amount in moles by dividing by the **molecular weight **

Step 3: Divide each of the mole values by the smallest of the mole values.

Step 4: If necessary, multiply these numbers by integers in order to get whole numbers; if an operation is done to one of the numbers, it must be done to all of them.

Thus, the empirical formula of methyl acetate is C_{3}H_{6}O_{2}.

Given the molecular weight of the unknown, the molecular formula of the unknown compound can be determined. If for example our substance has a mass of 118 g/mol and we worked out the molecular formula to be C_{2}H_{3}O_{2} this would have a mass of 59 g/mol which goes into 118 twice, so I need to multiply through my empirical formula by a factor of 2 to get my molecular formula C_{4}H_{6}O_{4}.

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Practice Questions

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MCAT Official Prep (AAMC)

Chemistry Online Flashcards Question 15

Chemistry Question Pack Question 69

Chemistry Question Pack Passage 15 Question 82

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Key Points

• Empirical formulas are the simplest form of notation.

• The molecular formula for a compound is equal to, or a whole-number multiple of, its empirical formula.

• Like molecular formulas, empirical formulas are not unique and can describe a number of different chemical structures or isomers.

• To determine an empirical formula, the mass composition of its elements can be used to mathematically determine their ratio.

Key Terms

**Per cent composition**: elemental composition by mass per cent

**Molecular weight**: the total mass of atoms in a molecule, represented by g/mol

**Empirical formula**: a notation indicating the ratios of the various elements present in a compound, without regard to the actual numbers

**Molecular formula**: a formula that describes the exact number and type of atoms in a single molecule of a compound