Percent Composition Can be calculated if given: masses of elements in compound OR the chemical formula

Percent Composition Can be used to:  calculate the mass of elements in a compound  determine the empirical formula of a compound  determine the molecular formula of a compound

Empirical Formula shows the simplest mole ratio of the elements. CO is a 1:1 ratio of carbon to oxygen H2O is a 2:1 ratio CO2 is a 1:2 ratio Empirical formulas can’t be reduced.

Molecular Formula shows the actual number of atoms in a molecule. The molecular formula for hydrogen peroxide is H2O2. Its empirical formula would be HO. Often the molecular formula is the same as the empirical formula: H2O, CO2

Empirical? CH4O C2H6 C3H10O C6H6O2 yes, cannot be reduced further no, empirical would be CH3 C3H10O yes C6H6O2 no. What would empirical be? C3H3O

Calculating Empirical Formulas A chemist with an unknown compound can easily figure out its percent composition, but it is much more meaningful to know its formula. EXAMPLE: What is the empirical formula for a compound that is 25.9% nitrogen and 74.1% oxygen?

Method Write the mass (g) of each element in the compound. So…. 25.9% N = 25.9g 74.1% O = 74.1g

2. Convert the mass of each element to moles. N = 25.9g = 1.85 mol 14.0g/mol O = 74.1g = 4.63 mol 16.0g/mol

Calculate the simplest whole number ratio by dividing the number of moles by the smallest number of moles. 1.85 : 4.63 = 1 : 2.5 1.85 1.85 (If the result is not within 0.1 of a whole number, multiply all numbers by a whole number) 2 ( 1 : 2.5) = 2 : 5

Write the empirical formula. N2 O5 For inorganic compounds, write the most positive element first. For organic compounds, write C first, H second and all others alphabetically.

A special present just for you…….. Page 135, Problems #20 & 21

Molecular Formula Given the empirical formula and the gram formula mass (gfm) OR Given the percent composition and the gram formula mass (gfm)

Example #1 Calculate the molecular formula for NaO having a gfm of 78g.  Determine the efm (empirical formula mass). NaO = 23.0g + 16.0g = 39.0 Divide the efm into the gfm. 78.0 = 2 39.0 This is the conversion factor used to determine the molecular formula. Na2O2

Example #2 Find the molecular formula for a compound having a composition of 58.8% C, 9.8% H and 31.4% O and a gmm of 102g/mol. Determine the mass of each component. C = 102g/mol x 58.8% = 60.0g/mol H = 102g/mol x 9.8% = 10.0g/mol O = 102g/mol x 31.4% = 32.0g/mol

 convert to moles C = 60.0g/mol = 5 12.0g H = 10.0g/mol = 10 1.0g O = 32.0g/mol = 2 16.0g

 Use moles as subscripts for components of compound C5H10O2 Check the gmm of this compound…does it equal 102.0g/mol? 5(12.0) + 10(1.0) + 2(16.0) = 102.0g/mol YES!

Oh Yeah! And there’s more… And Now….. Oh Yeah! And there’s more… Page 136, Problems #22 & 23

Now Try page 139, #41 44