Final Review Day 2 Ch 11, 12 & 13

Percent Composition refers to the percentage by mass. does not tell us the relative rate by which the atoms combine, or what portion of the volume is made up by an individual element.

Calculating Percent Composition percent composition = part x 100 whole

What is the percentage composition of glucose (C6H12O6) ? find the mass of the entire molecule 2. Use the formula to calculate the percent composition of each element

Formula Mass C = 12.0 u x 6 atoms   = 72.0 u H = 1.01 u x 12 atoms = 12.1 u O = 16.0 u x 6 atoms   = 96.0 u                                    ----------                                       180 u

Answer 72.0 u % for Carbon = ---------------- x 100 = 40.0% 180 u                                12.1 u % for Hydrogen = ------------ x 100 = 6.7 %                                180 u                               96.0 u % for Oxygen = ----------------  x 100  = 53.3%                               180 u

Empirical Formulas Empirical formulas: smallest whole number ratio of atoms present in a substance Molecular formula: actual number of each type of atom present in a molecule Many molecular compounds have different empirical and molecular formulas N2O4 (molecular) NO2 (empirical)

Calculating Empirical Formulas Percent composition data is commonly used to determine the empirical formula of a compound: Four steps: percent to mass mass to mole divide by smallest multiple ‘til whole

Calculating Empirical Formulas Example: An iron compound contains 69.943 % Fe and 30.057 % O. Calculate its empirical formula. Step 1: % to mass 69.943 % Fe  69.943 g Fe 30.057 % O  30.057 g O

Calculating Empirical Formulas Step 2: Mass to Moles mol Fe = 69.943 g Fe x mol Fe = 1.2524 mol Fe 55.845 g Fe mol O = 30.057 g O x mol O = 1.8786 mol O 15.9994 g O

Calculating Empirical Formulas Step 3: Divide by smallest O: 1.8786 mol O = 1.5 1.2524 mol Fe Fe: 1.2524 mol Fe = 1.0

Calculating Empirical Formulas Step 4: Multiply ‘til whole FeO1.5 doesn’t make sense. 1.5000 mol O 1.0000 mol Fe x 2 = 3.000 mol O Fe2O3 x 2 = 2.000 mol Fe

Using Empirical Formulas to Find Molecular Formulas Steps: Find the empirical formula Calculate the formula weight for the empirical formula. MW = the whole number FW (empirical formula) ratio between MW and FW Multiply the subscripts in the empirical formula by the whole number ratio.

Using Empirical Formulas to Find Molecular Formulas Example: A certain substances has an empirical formula of CH2O. If its molecular weight is 180.0 amu, what is its molecular formula? Step 1: Empirical Formula = given = CH2O

Using Empirical Formulas to Find Molecular Formulas Step 2: FW = 1 (12.0 amu) + 2 (1.0 amu) + 1 (16.0 amu) = 30.0 amu Step 3: Whole number ratio = 180.0 amu = 6 30.0 amu Step 3: Molecular formula: C(1x6)H(2x6)O(1x6) = C6H12O6

The Mole–Volume Relationship 10.2 The Mole–Volume Relationship At STP, 1 mol or, 6.02  1023 representative particles, of any gas occupies a volume of 22.4 L. The quantity 22.4 L is called the molar volume of a gas. This box, with a volume of 22.4 L, holds one mole of gas at STP.

10.2 The Mole Road Map The map shows the conversion factors needed to convert among volume, mass, and number of particles. Interpreting Diagrams How many conversion factors are needed to convert from the mass of a gas to the volume of a gas at STP?