6.7 Empirical Formula

Formulas Percent composition allows you to calculate the simple ratio of the atoms in a compound. Empirical Formula: formula of a compound that expresses lowest whole number ratios of atoms Molecular Formula: Actual formula of a compound showing the number of atoms present

Formulas Examples: MolecularEmpirical C 8 H 18  C 6 H 12 O 6 

Formulas Examples: MolecularEmpirical C 8 H 18  C 4 H 9 C 6 H 12 O 6 

Formulas Examples: MolecularEmpirical C 8 H 18  C 4 H 9 C 6 H 12 O 6  CH 2 O

Formulas H 2 O 2. Empirical or molecular? MOLECULAR!!!! It can be reduced to HO HO = Empirical formula

Example An oxide of aluminum is formed by the reaction of g of aluminum with g of oxygen. Calculate the empirical formula. 1.Determine the number of grams of each element Al: 4.151gand O: 3.692g

Example 2. Convert mass to moles n Al = m/MM =4.151/26.98 = mol Al n O = m/MM =3.692/16.00 = mol O

Example 3. Find ratio by dividing each element by the smallest amount 4. Multiply by a factor to get whole numbers moles Al = mol Al moles O = mol O O = x 2 = 3 Al = x 2 = 2 therefore, Al 2 O 3

Example g of Co reacts with g Cl to form a binary compound. Determine the empirical formula g of iron metal is heated in air. It reacts with oxygen to achieve a final mass of g. Determine the empirical formula.

Example 3.The most common form of nylon (Nylon-6) is 63.38% carbon, 12.38% nitrogen, 9.80% hydrogen and 14.14% oxygen. Calculate the empirical formula for Nylon-6.

Exit Ticket A sample of lead arsenate, an insecticide used against the potato beetle, contains g lead, g of hydrogen, g of arsenic, and g of oxygen. Calculate the empirical formula for lead arsenate.

Homework Read section 6.7 Questions p. 292 #1 p. 293 #2-9