# Ch. 7 – The Mole Formula Calculations.

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Ch. 7 – The Mole Formula Calculations

A. Percentage Composition
EMPIRICAL and MOLECULAR FORMULAS A. Percentage Composition the percentage by mass of each element in a compound

A. Percentage Composition
Find the % composition of Cu2S. g Cu g Cu2S %Cu =  100 = 79.852% Cu 32.07 g S g Cu2S %S =  100 = 20.15% S

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

A. Percentage Composition
How many grams of copper are in a 38.0-gram sample of Cu2S? Cu2S is % Cu (38.0 g Cu2S)( ) = 30.3 g Cu

A. Percentage Composition
Find the mass percentage of water in calcium chloride dihydrate, CaCl2•2H2O? 36.04 g H2O g H2O %H2O =  100 = 24.51% H2O

C2H6 CH3 B. Empirical Formula
Smallest whole number ratio of atoms in a compound C2H6 reduce subscripts CH3

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

What ?????????? How can I remember? There is a poem to help
Percent to Mass Mass to Mole Divide by Small Multiply ‘til Whole Sometimes the mole values are still not whole numbers so multiply by the smallest factor that will make them whole numbers

B. Empirical Formula Find the empirical formula for a sample of 25.9% N and 74.1% O. 25.9 g N 1 mol N 14.01 g N = 1.85 mol N = 1 N 1.85 mol N 1.85 mol 74.1 g 1 mol 16.00 g = 4.63 mol O = 2.5 O

N2O5 N1O2.5 B. Empirical Formula
Need to make the subscripts whole numbers  multiply by 2 N2O5

Example of Determining Empirical Formula
Ex. What is the empirical formula of a compound that is 25.9% nitrogen and 74.1% Oxygen? 25.9g N 1 mol N = 1.85 mol N 14.0g N 74.1g O 1 mol O = 4.63 mol O 16.0g O Divide each element by the lowest number of moles in the compound: 1.85 mol N = 1 mol N 1.85 4.63 mol O = 2.50 mol O Now find the lowest common denominator. 1 mol N x 2= N2 2.5 mol O x 2 = O5 Therefore, N2O5 is the empirical formula

Practice Problems Calculate the empirical formula for a compound that is 94.1% O and 5.90%H Calculate the empirical formula for a compound that is 79.8% C and 20.2% H

CH3 C2H6 C. Molecular Formula
“True Formula” - the actual number of atoms in a compound CH3 empirical formula ? C2H6 molecular formula

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.

2(CH2)  C2H4 C. Molecular Formula
The empirical formula for ethylene is CH2. Find the molecular formula if the molecular mass is g/mol? empirical mass = g/mol 28.1 g/mol 14.03 g/mol = 2.00 2(CH2)  C2H4

Example of Molecular Formula
Calculate the molecular formula of the compound whose Molar mass is 60.0g and empirical formula is CH4N. Molar Mass of Compound Empirical formula Empirical formula mass (efm) Molar Mass efm Molecular Formula 60.0g CH4N = 30.0g CH4N 60.0/30.0 = 2 CH4N x 2 = C2H8N2