Tis the season to be thankful so lets thank Avogadro for math in Chemistry

Percent Composition Purpose: Can be used to figure out chemical formulas. the percentage by mass of each element in a compound

I. Percent Composition Two different types of problems: 1) Masses are given 2) No Masses are given

Masses are Given Steps to solve problem: 1) Add given masses to get total mass for one compound 2) Divide mass of each element by the total mass 3) Multiply by 100 to get the percent

%Fe = 28 g 36 g  100 = 78% Fe %O = 8.0 g 36 g  100 = 22% O Find the percentage composition of a sample that is 28 g Fe and 8.0 g O. A. Percentage Composition

No Masses Given Steps to solve problem: 1) Assume you have 1 mole of the compound 2) Calculate the molar mass of each element in the compound by multiplying the subscript by the molar mass of the element 3) Divide the molar mass for the element by the total molar mass of the compound 4) Multiply by 100 to get the percent

No Masses Given Examples 1) Calculate the percent composition of oxygen in water. H 2 O H: 2 x = g/mol O: 1 x = g/mol Total molar mass ( ) = g/mol %O = (16.00 / 18.02) x 100 = 88.79%

No Masses Given Examples 2) Calculate the percent composition of calcium carbonate g/mol %Ca = 40.08g/100.09g×100 = 40.04% %C = 12.01g/100.09g×100 =12.00% %O = 3(16.00g)/100.09g×100 = 47.96%

B. Empirical Formula C2H6C2H6 CH 3 reduce subscripts Smallest whole number ratio of atoms in a compound

B. Empirical Formula 1. Find mass (or %) of each element. 2. Find moles of each element. (divide given mass by molar mass) 3. Divide answers by the smallest # to find subscripts. 4. When necessary, multiply subscripts by 2, 3, or 4 to get whole #’s.

Remember! Percent to Mass Mass to Mole Divide by Small Multiply til Whole

B. Empirical Formula Find the empirical formula for a sample of 25.9% N and 74.1% O g 1 mol g = 1.85 mol N 74.1 g 1 mol 16.00g = 4.63 mol O 1.85 mol = 1 N = 2.5 O

B. Empirical Formula N 1 O 2.5 Need to make the subscripts whole numbers  multiply by 2 N2O5N2O5 If.0_ something or.9_ something, just round

Empirical Formulas Examples 3) Ascorbic acid (vitamin C) contains C (40.89%), H (4.56%), and O (54.55%) by mass. What is the empirical formula of ascorbic acid? C 3 H 4 O 3

C. Molecular Formula “True Formula” - the actual number of atoms in a compound CH 3 C2H6C2H6 empirical formula molecular formula ?

IV. Molecular Formulas Usually the empirical formula is the molecular formula for a compound. When it is not, the molecular formula is defined as the elements and number of atoms that are contained in a compound.

IV. Molecular Formulas Molecular formulas are always multiples of empirical formulas. CH 3 C 2 H 6

C. Molecular Formula 1. Find the empirical formula. 2. Find the empirical formula mass. 3. Divide the molecular mass by the empirical mass. 4. Multiply each subscript by the answer from step 3.

C. Molecular Formula The empirical formula for ethylene is CH 2. Find the molecular formula if the molecular mass is g/mol? g/mol g/mol = empirical mass = g/mol (CH 2 ) 2  C 2 H 4

Molecular Formulas Examples 2) The compound methyl butanoate smells like apples. Its percent composition is 58.8% C, 9.8% H, and 31.4% O. If the molecular mass is g/mol, what is the molecular formula?

Molecular Formulas Examples 3) You find 7.36 g of a compound has decomposed to give 6.93 g of oxygen. The rest is hydrogen. If the molecular mass is g/mol, what is the molecular formula?