Empirical Formula

# Empirical Formula - Empirical Formula Empirical • Formulas represent the symbolic language Formulas of chemistry of • Empirical Formula the

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Unformatted text preview: Empirical Formula Empirical • Formulas represent the symbolic language Formulas of chemistry of • Empirical Formula: the SIMPLEST WHOLE Empirical NUMBER ratio of the various atoms in a NUMBER compound. compound. • Molecular Formula: specifies the EXACT Molecular FORMULA of the molecule FORMULA Empirical Formula Empirical • H2F10 Molecular Formula Empirical Formula Empirical • Benzene – C6H6 (molecular formula) (molecular – CH (empirical formula) Empirical Formula Empirical Two-handed person person Three-handed person H2F10 H3F15 Empirical Formula Empirical • Benzene Acetylene – C6 H6 CH CH C2 H2 Empirical Formula Calc. Empirical • Determine the empirical and molecular Determine formula for a compound that gives the following analysis (in mass percents) following • 71.65% Cl 24.27% C 4.07% H • The molecular weight (molar mass) is The known to be 98.96 known Empirical Formula Calc. Empirical • 71.65% Cl 24.27% C 4.07% H • The molecular weight is known to be The 98.96 98.96 • Steps to solving: – Percent to mass – Mass to mol – Divide by small – Multiply by whole Empirical Formula Calc. Empirical • 71.65% Cl 24.27% C 4.07% H • The molecular weight is known to be The 98.96 98.96 • Steps to solving: – Percent to mass • • • • Assume 100 g sample 71.65 g Cl 24.27 g C 4.07 g H Empirical Formula Calc. Empirical • • • 71.65% Cl 24.27% C 4.07% H The molecular weight is known to be 98.96 Steps to solving: – Mass to mol mol 71.65 g Cl = 2.021 mol Cl 35.4527 g Cl mol 24.27 g C = 2.021 mol C 12.011 g C mol 4.07 g H = 4.04 mol H 1.008 g H Empirical Formula Calc. Empirical • • • 71.65% Cl 24.27% C 4.07% H 24.27% The molecular weight is known to be 98.96 Steps to solving: – Divide by small 1 2.021 mol Cl = 1 2.021 mol 1 2.021 mol C = 1 2.021 mol 1 4.04 mol H = 2 2.021 mol Empirical Formula Calc. Empirical • 71.65% Cl 24.27% C 4.07% H • The molecular weight is known to be The 98.96 98.96 • Steps to solving: – Multiply by whole • Cl = 1 •C =1 •H =2 CH2Cl Empirical Formula Calc Empirical • Empirical Formula • CH2Cl • Molecular Formula – Compare weight of empirical formula to Compare molecular formula molecular – Empirical formula weight 1( 35.453) + 1( 12.011) + 2 ( 1.008 ) = 49.48 g / mol Empirical Formula Calc Empirical • Empirical Formula 49.48 g/mol • CH2Cl • Molecular Formula molecular weight 98.96 = =2 empirical formula weight 49.48 C2H4Cl2 Empirical Formula Calc Empirical • Empirical Formula 49.48 g/mol • CH2Cl • Molecular Formula C2H4Cl2 Empirical Formula Calc. #2 Empirical • A white powder is analyzed and found to white contain 43.64% phosphorus and 56.36% oxygen by mass. • The compound has a molecular weight The (molar mass) of 283.88. • What are the compound’s empirical and What molecular formula? molecular Empirical Formula Calc. #2 Empirical • A white powder is analyzed and found to white contain 43.64% phosphorus and 56.36% oxygen by mass. • Step 1: Percent to mass Empirical Formula Calc. #2 Empirical • A white powder is analyzed and found to white contain 43.64% phosphorus and 56.36% oxygen by mass. • Step 1: Percent to mass – 43.64 g P – 56.36 g O Empirical Formula Calc. #2 Empirical • A white powder is analyzed and found to white contain 43.64% phosphorus and 56.36% oxygen by mass. • Step 2: Mass to mol mol 43.64 g P = 1.409 mol P 30.97 g P mol 56.36 g O = 3.523 mol O 15.999 g O Empirical Formula Calc. #2 Empirical • A white powder is analyzed and found to white contain 43.64% phosphorus and 56.36% oxygen by mass. • Step 3: Divide by small 1 1.409 mol P =1 P 1.409 1 3.523 mol O = 2.5 O 1.409 Empirical Formula Calc. #2 Empirical • A white powder is analyzed and found to white contain 43.64% phosphorus and 56.36% oxygen by mass. • Step 4: Multiply till whole 1P x2=2P 2.5 O x 2 = 5 O P 2 O5 Empirical Formula Calc. #2 Empirical • P 2 O5 • The compound has a molecular weight of The 283.88. • What are the compound’s empirical and What molecular formula? molecular Empirical Formula Calc. #2 Empirical • P2O5 2(30.974 g/mol P) + 5(15.999 g/mol O) = 141.94 g/mol • Calculate empirical formula weight Empirical Formula Calc. #2 Empirical • P2O5 • Calculate empirical formula weight • Compare the empirical formula weight to Compare molecular weight (molar mass) molecular molecular weight 283.88 = =2 empirical formula weight 141.94 Empirical Formula Calc. #2 Empirical • P2O5 • Calculate empirical formula weight • Compare the empirical formula weight to Compare molecular weight molecular • (P2O5) x 2 = P4O10 Self Learning Self • Click here to view a video on calculating Click here the empirical formula the • Click here to take a practice quiz Click here ...
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## This note was uploaded on 01/28/2011 for the course CHM 101 taught by Professor Kershisnick during the Summer '10 term at Oakton.

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