From percentage to formula Empirical Formula From percentage to formula

Types of Formulas Empirical Molecular (true) Name CH C2H2 acetylene The formulas for compounds can be expressed as an empirical formula and as a molecular(true) formula. Empirical Molecular (true) Name CH C2H2 acetylene CH C6H6 benzene CO2 CO2 carbon dioxide CH2O C5H10O5 ribose

An empirical formula represents the simplest whole number ratio of the atoms in a compound.

It is not just the ratio of atoms, it is also the ratio of moles of atoms In 1 mole of CO2 there is 1 mole of carbon and 2 moles of oxygen In one molecule of CO2 there is 1 atom of C and 2 atoms of O

The molecular formula is the true or actual ratio of the atoms in a compound.

Learning Check EF-1 1) CH2O 2) C2H4O2 3) C3H6O3 A. What is the empirical formula for C4H8? 1) C2H4 2) CH2 3) CH B. What is the empirical formula for C8H14? 1) C4H7 2) C6H12 3) C8H14 C. What is a molecular formula for CH2O? 1) CH2O 2) C2H4O2 3) C3H6O3

Learning Check EF-1 1) CH2O 2) C2H4O2 3) C3H6O3 A. What is the empirical formula for C4H8? 1) C2H4 2) CH2 3) CH B. What is the empirical formula for C8H14? 1) C4H7 2) C6H12 3) C8H14 C. What is a molecular formula for CH2O? 1) CH2O 2) C2H4O2 3) C3H6O3

Learning Check EF-2 If the molecular formula has 4 atoms of N, what is the molecular formula if SN is the empirical formula? Explain. 1) SN 2) SN4 3) S4N4

Learning Check EF-2 If the molecular formula has 4 atoms of N, what is the molecular formula if SN is the empirical formula? Explain. 3) S4N4

Empirical and Molecular Formulas molar mass = a whole number = n simplest mass n = 1 molar mass = empirical mass molecular formula = empirical formula n = 2 molar mass = 2 x empirical mass molecular formula = 2 x empirical formula molecular formula = or > empirical formula

Learning Check EF-3 2) C6H8O6 3) C9H12O9 A compound has a formula mass of 176.0 and an empirical formula of C3H4O3. What is the molecular formula? 1) C3H4O3 2) C6H8O6 3) C9H12O9

Learning Check EF-3 2) C6H8O6 3) C9H12O9 A compound has a formula mass of 176.0 and an empirical formula of C3H4O3. What is the molecular formula? 1) C3H4O3 2) C6H8O6 3) C9H12O9

Learning Check EF-4 1) C7H6O4 2) C14H12O8 3) C21H18O12 If there are 192.0 g of O in the molecular formula, what is the true formula if the EF is C7H6O4? 1) C7H6O4 2) C14H12O8 3) C21H18O12

Learning Check EF-4 1) C7H6O4 2) C14H12O8 3) C21H18O12 If there are 192.0 g of O in the molecular formula, what is the true formula if the EF is C7H6O4? 1) C7H6O4 2) C14H12O8 3) C21H18O12

Calculating Empirical Formula Pretend that you have a 100 gram sample of the compound. That is, change the % to grams. Convert the grams to mols for each element. Write the number of mols as a subscript in a chemical formula. Divide each number by the least number. Multiply the result to get rid of any fractions.

Example Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N. Assume 100 g so 38.67 g C x 1mol C = 3.220 mole C 12.01 gC 16.22 g H x 1mol H = 16.09 mole H 1.01 gH 45.11 g N x 1mol N = 3.219 mole N 14.01 gN

If we divide all of these by the smallest one it will give us the subscripts for the empirical formula 3.220 mol C = 1 3.219 mol N 16.09 mol H = 5 3.219 mole N = 1 Empirical formula: CH5N

Learning Check EF-5 Aspirin is 60.0% C, 4.5 % H and 35.5 O. Calculate its simplest formula. In 100 g of aspirin, there are 60.0 g C, 4.5 g H, and 35.5 g O.

Solution EF-5 60.0 g C x ___________= ______ mol C 4.5 g H x ___________ = _______mol H 35.5 g O x ___________ = _______mol O

5.00 mol C = ________________ ______ mol O Divide by the smallest # of moles. 5.00 mol C = ________________ ______ mol O 4.5 mol H = ________________ 2.22 mol O = ________________ Are are the results whole numbers?_____

Finding Subscripts A fraction between 0.1 and 0.9 must not be rounded. Multiply all results by an integer to give whole numbers for subscripts. (1/2) 0.5 x 2 = 1 (1/3) 0.333 x 3 = 1 (1/4) 0.25 x 4 = 1 (3/4) 0.75 x 4 = 3 (1/5) 0.20 x 5 = 5

Homework Worksheet C: #1-7