Section 12.2: Using Moles (part 3)

Mass Percent Steps:1) Calculate mass of each element2) Calculate total mass3) Divide mass of element/ mass of compound

Calcualte the mass % of C and H in C2H6 2 mol C x 12 g C = 24 g C6 mol H x 1 g H = 6 g H Total mass = C2H6 = 30 g% C = 24 g C x 100 = 80% 30 g C2H6

 % H = 6 g H x 100 = 20% 30 g C2H6

* Should add up to 100

Example 2 Calculate the mass % of C, H, Br in C6H5Br6 mol C x 12.011 = 72.0665 mol H x 1.0079 = 5.03951 mol Br x 79.904 = 79.904= 157.0095 g % C = 72.066 g C x 100 = 45.9% 157.0095 g C2H5 Br % H = 5.0395 g H x 100 = 3.2% 157.0095 g C2H5 Br % Br = 79.904 g Br x 100 = 50.9% 157.0095 g C2H5 B

Chemical FormulasEmpirical Formulas Molecular Formulas

Empirical: NaClMolecular: K2C2O4 Empirical: KCO2Molecular: C20H20O4 Empirical:

C5H5O

Empirical formula Formula that shows the smallest whole-number ratio of atoms for the compound

Steps:1) Switch % into grams2) Convert grams to moles for each element3) Divide by the smallest mole

EMPIRICAL FORMULA   A compound is ~26 % N and 74 % O. What is the empirical formula?26 g N x 1 mol N = 1.86 mol N / 1.86 = 1 14.007 g N 74 g O x 1 mol O = 4.63 mol O / 1.86 = 2.5 15.999 g O Can’t have decimal #s(N1O2.5) 2 = N2O5

 

What is the empirical formula of a compound that is 80% C and 20% H of 100 g compound? 80 g C x 1 mol C = 6.67 mol C 12 g C 20 g H x 1 mol H = 20 mol H 1 g H Divide both by smallest:6.67 mol C = 1.0 mol C 6.67 20 mol H = 3.0 mol H 6.67The mole ratio for this compound is 1 mol C: 3 mol HFormula: CH3 (Empirical formula)

Molecular formulaTo determine the molecular formula you need the mass of the compound Steps:1) First find the empirical formula2) Divide the total mass by the empirical formula mass3) Multiply the empirical formula by this #

Example continued:Molar mass is 30 g/mol →molecular mass is 30u

Molecular mass of CH3 = 15 uBy dividing the molecular mass you can find the multiple:30 u = 2 15 u Molecular formula is C2H6

An oxide of Cr is 68.4% Cr by mass. The molar mass is 152 g/mol. What is the formula?100-68.4 = 31.6 g O 68.4 g Cr x 1 mol Cr = 1.32/1.32 = 1 51.996 g Cr 31.6 g O x 1 mol O = 1.98/1.32 = 1.5 15.999 g OCr1O1.5 Cr= 51.996 O= 23.9985 = 75.9945 152/ 75.99945 = 2(Cr1O1.5) 2 = Cr2O3

What is the molecular formula for C4H4O with a mass = 136 g/mol?C: 4 x 12.011 = 48.044 H: 4 x 1.0079 = 4.0316O: 1 x 15.999 = 15.999 = 68.07136/68.07 = 2(C4H4O)2 = C8H8O2

What % of O is NO?N:14O:16= 30 16/30 = 53.3 % O

A nitride of Cr is 84.8 %. It has a mass of ~ 184 g. What is the formula?100 - 84.8 = 15.2 % N84.8 g Cr x 1 mol Cr = 1.63/1.09 = 1.5 51.996 g Cr15.2 g N x 1 mol N = 1.09/1.09 = 1 14.007 g N

Cr1.5N 77.994 + 14.007 = 92.001184/92.001 = 1.99 (2)(Cr1.5N) 2 = Cr3N2