Percent Composition  The percent by mass of an element in a compound is the number of grams of the element divided by the mass in grams of the compound multiplied by 100%.  % Mass of element = mass of element/mass of compound x 100%  % Mass = mass of element in 1 mol compound / molar mass of compound x 100%

Percent Composition  A compound is formed when 9.03 g Mg combines completely with 3.48 g N. What is the percent composition of this compound?

Percent Composition  When 14.2 g sample of mercury (II) oxide is decomposed into its elements by heating, 13.2 g Hg is obtained. What is the percent composition of the compound?

Percent Composition  Calculate the percent composition of these compounds:  Ethane ( C 2 H 6 )  Sodium hydrogen sulfate ( NaHSO 4 )

Percent Composition  Calculate the percent nitrogen in these common fertilizers:  NH 3  NH 4 NO 3

Empirical Formula  The empirical formula of a compound shows the smallest whole-number ratio of the atom in the compound.  EX: CO not C 2 O 4 (molecular formula) CH………not C 6 H 6(molecular formula) CH………not C 6 H 6(molecular formula) CH 2 O……not C 6 H 12 O 6(molecular formula) CH 2 O……not C 6 H 12 O 6(molecular formula)

Empirical Formula- STEPS  Convert each percent to grams.  Convert grams to moles of each element  Divide by the lowest number of moles  The whole numbers you get are the subscripts in the formula

Empirical Formula  Calculate the empirical formula of each compound:  94.1% O, 5.9% H  67.6% Hg, 10.8% S, 21.6% O

Empirical Formula  1,6 – diamionhexane is used to make nylon. What is the empirical formula of this compound if it is 62.1% C, 13.8% H, and 24.1% N?

Molecular Formula  The molecular formula of a compound is either the same as its experimentally determined empirical formula, or it is a simple whole-number multiple of its empirical formula.  EX: CO 2 (empirical) C 2 O 4 (molecular formula) CH (empirical) C 6 H 6 (molecular formula) CH (empirical) C 6 H 6 (molecular formula) CH 2 O (empirical) C 6 H 12 O 6 (molecular formula) CH 2 O (empirical) C 6 H 12 O 6 (molecular formula)

Molecular Formula - STEPS  Calculate the molar mass of the empirical formula given  Use the molar mass of the molecular formula that is given and divide by the empirical formula mass  This will provide you with the whole number that you use to multiply each element in the empirical formula by

Molecular Formula  Find the molecular formula of ethylene glycol, which is used as antifreeze. The molar mass is 62 g/mol and the empirical formula is CH 3 O.

Molecular Formula  Which pair of molecules has the same empirical formula? :  C 2 H 4 O 2, C 6 H 12 O 6  NaCrO 4, Na 2 Cr 2 O 7