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Molecular Formulas A molecular formula is either the same as an empirical formula or it is a 
simple whole number multiple of its empirical formula.

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Presentation on theme: "Molecular Formulas A molecular formula is either the same as an empirical formula or it is a 
simple whole number multiple of its empirical formula."— Presentation transcript:

1 Molecular Formulas A molecular formula is either the same as an empirical formula or it is a 
simple whole number multiple of its empirical formula.

2 Determining Molecular Formulas
Calculate the molecular formula of a compound whose molar mass is 60.0 g/mol and has an empirical formula of CH4N. What you need: Empirical Formula Molar Mass Calculate the molar mass of the empirical formula C: 12.0g x 1 H: 1.0g x 4 N: 14.0g x 1 = 12.0g = 4.0g = 14.0g 30.0 g/mol Molecular Formula Molar Mass Empirical Formula Molar Mass 60.0 g/mol = Multiple = 2 30.0 g/mol Empirical Formula X Multiple = Molecular Formula CH4N x 2 = C2H8N2

3 Determining Molecular Formulas
Find the molecular formula of ethylene glycol which has a molar mass of 62 g/mol and an empirical formula of CH3O. What you need: Empirical Formula Molar Mass Calculate the molar mass of the empirical formula C: 12.0g x 1 H: 1.0g x 3 O: 16.0g x 1 = 12.0g = 4.0g = 16.0g 31.0 g/mol Molecular Formula Molar Mass Empirical Formula Molar Mass 62 g/mol = Multiple = 2 31.0 g/mol Empirical Formula X Multiple = Molecular Formula CH3O x 2 = C2H6O2

4 Determining Molecular Formulas
The compound methyl butanoate has a percent composition of 58.8% C, 9.8% H, and 31.4% O and has a molar mass of 102 g/mol. What is its molecular formula? What you need: Empirical Formula Empirical Formula = C5H10O2 Molar Mass Calculate the molar mass 
of the empirical formula 58.8g C 12.0g C 1 mol C 9.8g H 1.0g H 1 mol H 31.4g O 16.0g O 1 mol O = 4.90 mol C = 9.8 mol H = 1.96 mol O C: 12.0g x 5 H: 1.0g x 10 O: 16.0g x 2 = 60.0g = 10.0g = 32.0g = 2.50 = 5.0 = 1.00 x 2 = 5 = 10 = 2 1.96 mol 102.0 g/mol 102 g/mol = 1 102.0 g/mol C5H10O2 x 1 = C5H10O2

5 Determining Molecular Formulas
What is the molecular formula of a compound that has a percent composition of 94.1% O and 5.9% H and has a molar mass of 34 g/mol. What you need: Empirical Formula Empirical Formula = HO or OH Molar Mass Calculate the molar mass 
of the empirical formula 5.9g H 1.0g H 1 mol H 94.1g O 16.0g O 1 mol C H: 1.0g O: 16.0g = 5.9 mol H = 5.88 mol O = 1.0 = 1.00 5.88 mol 17.0 g/mol 34 g/mol = 2 17.0 g/mol HO x 2 = H2O2 or O2H2


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