POINT > Define empirical formula POINT > Relate empirical formula to formula unit POINT > Relate empirical formula to molecular formula POINT > Determine empirical and molecular formulas from percent composition

POINT > Define empirical formula The empirical formula is the lowest whole-number ratio of the atoms in a compound

POINT > Relate empirical formula to formula unit For ionic compounds the empirical formula is the same as the formula unit CaCl 2 Na 3 PO 4 KBr Fe 2 O 3 SrO

WB CHECK: Determine the formula unit of a compound formed between copper(II) and chlorine. What is the empirical formula of this compound? What is the empirical formula of aluminum sulfide?

POINT > Relate empirical formula to molecular formula For molecular compounds the empirical formula might be the same as the molecular formula: Ex. H 2 O CH 4 CO 2 CO NH 3

POINT > Relate empirical formula to molecular formula For molecular compounds the empirical formula might be different than the molecular formula: Mol. formula: H 2 O 2 Emp. Formula: HO Mol. formula: C 2 H 4 Emp. formula CH 2 Mol. formula: C 2 H 6 Emp. formula CH 3

POINT > Relate empirical formula to molecular formula The molecular formula is either the same as the empirical formula or a whole number multiple of it Mol. formula: C 2 H 4 Emp. formula CH 2 Mol. formula: C 2 H 6 Emp. formula CH 3

WB CHECK: Identify whether the following are empirical formulas: (yes or no) Ca(OH) 2 H2O2H2O2 C 10 H 14 N 2 C3H6O3C3H6O3 C3H8C3H8 C 12 H 22 O 11

WB CHECK: What is the empirical formula for propanoic acid, C 3 H 6 O 3 ? What is the molecular formula of glucose? What is the empirical formula of glucose?

1. Given percent composition of a compound, assume the mass to be 100g (since it’s a percent) Ex. A compound is 30.4 % N and 69.6% O. Find the empirical formula Assume 30.4g N and 69.6g O 2. How many moles of each?

Assume 30.4g N and 69.6g O 2. How many moles of each? (divide by molar mass) 2.17 mol N and 4.35 mol O 3. To get whole numbers, divide each by the smallest number (2.17) 1.00 mol N 2.00 mol O NO 2

Ex. A sample is 38.8% Cl and 61.2% O. Find the empirical formula 38.8g Cl = 1.09 mol Cl 61.2g O = 3.83 mol O Divide both by 1.09 = 1.00 mol Cl 3.51 mol O Need whole numbers, so x2 = 2 mol Cl 7 mol O Cl 2 O 7

Find the empirical formula of a compound that is 25.45% sulfur and 74.55% flourine SF 5

If you know the empirical formula and the molar mass of a compound you can determine the molecular formula. The empirical mass is the mass of the empirical formula (CH 2 O = 30g/mol) The molar mass must be a whole number multiple of the empirical mass

Ex. If a compound has the empirical formula CH 2 O and a molar mass of 120g/mol, what is the molecular formula? CH 2 O empirical mass = 30g/mol Must multiply by four to get 120g/mol Molecular formula is C 4 H 8 O 4

What is the molecular formula of a compound with the empirical formula CClN and a molar mass of 184.5? C 3 Cl 3 N 3

Homework: Read Practice #1-3 page 235 Practice #1-2 page 237 F.A. #1-5 page 237 (challenge yourself with #5)