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# Percent Composition - PowerPoint PPT Presentation

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.

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## PowerPoint Slideshow about ' Percent Composition' - nyssa-tanner

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Presentation Transcript

• Can be calculated if given:

masses of elements in compound

OR

the chemical formula

• 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

• 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.

• 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

• CH4O

• yes, cannot be reduced further

• C2H6

• no, empirical would be CH3

• C3H10O

• yes

• C6H6O2

• no. What would empirical be?

• C3H3O

• 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?

• Write the mass (g) of each element in the compound. So….

25.9% N = 25.9g

74.1% O = 74.1g

• N = 25.9g = 1.85 mol

14.0g/mol

• O = 74.1g = 4.63 mol

16.0g/mol

• Write the empirical formula. number of moles by the smallest number of moles.

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…….. number of moles by the smallest number of moles.

• Page 135, Problems #20 & 21

Molecular Formula number of moles by the smallest number of moles.

Given the empirical formula and the gram formula mass (gfm)

OR

Given the percent composition and the gram formula mass (gfm)

Example #1 number of moles by the smallest number of moles.

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 number of moles by the smallest number of moles.

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 number of moles by the smallest number of moles.

C = 60.0g/mol = 5

12.0g

H = 10.0g/mol = 10 1.0g

O = 32.0g/mol = 2

16.0g

number of moles by the smallest number of moles. 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!

And Now….. number of moles by the smallest number of moles.

• Oh Yeah! And there’s more…

• Page 136, Problems #22 & 23