__Finding Empirical formula__

__Finding Empirical formula__

General information about various types of formulas follows:

- In a
*chemical*formula, the elements are represented by symbols, and a subscript number represents the number of each element.

*Example: *The chemical formula for ethane is C_{2}H_{6}.

- An
*empirical (simplest)*formula represents the relative number of atoms of each element in the compound.*Example:*The empirical formula for ethane is CH_{3}. Or the lowest common multiple of the subscripts.

- A
*molecular*formula represents the actual number of atoms of each element rather than a ratio of atoms.

*Example: *The molecular formula for ethane is C_{2}H_{6}.

Another way of representing this is: Molecular formula = (empirical formula)_{n}, where “n” is a whole number such as n = 1, etc. In this example, n = 2 C_{2}H_{6} = (CH_{3})_{2}

__Steps in calculating empirical formula:__

Q- a compound is found to contain the following % by mass: 69.58% Ba, 6.090% C, 24.32% O. What is the simplest (i.e. empirical) formula?

**Step** 1: Change your percentage so it is out of 100g of the substance. Thus, % will become mass in grams. E.g. 69.58 % Ba becomes 69.58 g Ba. (Some questions will give grams right off, instead of %)

**Step** 2: Calculate the # of moles for each.

**Step** 3: Divide the moles of each compound by the lowest number of moles calculated in your different types of compounds.

**Step** 4: Round the answer to the nearest whole number

**Step** 5: Write the simplest formula based on the whole number from the mol ratios.

__Sample Problem:__

If a compound has a composition of 40.0% carbon (C), 6.714% hydrogen (H), and 53.29% oxygen (O), determine the *empirical *formula of the compound.

Assume that there is a 100.0 g sample of the compound, in which case:

**Step** 1: 40.0% = 40.0g of C, 6.714g of H, 52.29g of O

**Step** 2: Calculate moles

No. of Moles of C = 40.0g/12.00g/mol = 3.33mol

No. of Moles of H = 6.714/1.00g/mol = 6.7mol

No. of Moles of O = 53.29/16.00g/mol = 3.33mol

**Step** 3: Divide by lowest mol number

C = 3.33mol/3.33mol = 1

H = 6.7mol/3.33mol = 2

O = 3.33 mol /3.33mol = 2

**Step** 4:

C_{1}H_{2}O_{2}

Variation #2:

If a compound contains 71.65% chlorine (Cl), 24.27% carbon (C), and 4.07% hydrogen (H), determine the *molecular *formula if the molar mass is 98.96 g/mol.

**Step** 1: Assume that there is a 100.0 g sample of the compound, in which case:

**Step** 2:By dividing by the smallest number of moles, the ratio between the elements is C1H2Cl1, and the formula mass would then be 49.5 g/mol.

**Step** 3: If the molar mass is 98.96 g/mol, the molecular formula would be a multiple of the simplest formula, or The molecular formula is therefore C2H4Cl2.

Topics Covered

- Calculating average atomic mass and what that means.
- Importance and applications of isotopes
- Know how to name covalent/ionic compounds with polyatomic ions and multivalent elements.
- Know how to balance chemical reactions
- Be able to describe the concept of the mole and its importance to measurement in chemistry.
- Mole conversions (Molar volume, mass, and particles)
- Know how to identify the type of reaction and be able to predict products and reactants.
- Know how to use mol-mol ratio to calculate moles of another or mass of another
- Calculate your limiting reagent and excess
- Know how to determine your empirical formulas from percent composition