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Empirical and Molecular Formulas
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Empirical Formula Empirical Formula
A formula that gives the simplest whole-number ratio of the atoms of each element in a compound. Molecular Formula Empirical Formula H2O2 HO CH3O
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2. Divide each mole by the smallest mole.
Determine the empirical formula for a compound containing g Cl and g Ca. Steps 1. Find mole amounts. 2. Divide each mole by the smallest mole.
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1. Find mole amounts. 2.128 g Cl x 1 mol Cl = mol Cl 35.45 g Cl 1.203 g Ca x 1 mol Ca = mol Ca 40.08 g Ca
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2. Divide each mole by the smallest mole.
Cl = mol Cl = 2.00 mol Cl 0.0300 Ca = mol Ca = 1.00 mol Ca Ratio – 1 Ca: 2 Cl Empirical Formula = CaCl2
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“Percent to mass Mass to mole Divide by small Multiply ‘til whole”
A compound weighing g consists of 72.2% magnesium and 27.8% nitrogen by mass. What is the empirical formula? Hint “Percent to mass Mass to mole Divide by small Multiply ‘til whole”
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A compound weighing 298. 12 g consists of 72. 2% magnesium and 27
A compound weighing g consists of 72.2% magnesium and 27.8% nitrogen by mass. What is the empirical formula? Percent to mass: Mg – (72.2%/100)* g = g N – (27.8%/100)* g = g Mass to mole: Mg – g*( 1 mole/ 24.3 g)=8.86 mol N – g*( 1 mole/14.01 g)=5.92 mol
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Divide by small: Mg mole/5.92 mole = 1.50 N mole/5.92 mole = 1.00 mole Multiply ‘til whole: Mg – 1.50 x 2 = 3.00 N – 1.00 x 2 = 2.00 Mg3N2
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Molecular Formula The molecular formula gives the actual number of atoms of each element in a molecular compound. Steps 1. Find the empirical formula. 2. Calculate the Empirical Formula Mass. 3. Divide the molar mass by the “EFM”. 4. Multiply empirical formula by factor.
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Find the molecular formula for a compound whose molar mass is ~124
Find the molecular formula for a compound whose molar mass is ~ and empirical formula is CH2O3. 2. “EFM” = g /62.03 = 2 4. 2(CH2O3) = C2H4O6
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1. Find the empirical formula.
Find the molecular formula for a compound that contains 4.90 g N and 11.2 g O. The molar mass of the compound is 92.0 g/mol. Steps 1. Find the empirical formula. 2. Calculate the Empirical Formula Mass. 3. Divide the molar mass by the “EFM”. 4. Multiply empirical formula by factor.
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Empirical formula. A. Find mole amounts.
4.90 g N x 1 mol N = mol N 14.01 g N 11.2 g O x 1 mol O = mol O 16.00 g O
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B. Divide each mole by the smallest mole.
N = = 1.00 mol N 0.350 O = = 2.00 mol O Empirical Formula = NO2 Empirical Formula Mass = g/mol
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Molecular formula Molar Mass = 92.0 g/mol = 2.00
Emp. Formula Mass g/mol Molecular Formula = 2 x Emp. Formula = N2O4
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A g compound containing only carbon, hydrogen, and oxygen is found to be 48.38% carbon and 8.12% hydrogen by mass. The molar mass of this compound is known to be ~ g/mol. What is its molecular formula?
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A g compound containing only carbon, hydrogen, and oxygen is found to be 48.38% carbon and 8.12% hydrogen by mass. The molar mass of this compound is known to be ~ g/mol. What is its molecular formula? g C – (48.38/100)* g = g g H – (8.12/100)* g = g g O – (43.5/100)* g = g mole C g * ( 1 mole ) = mol 12.01 g mole H – g * ( 1 mole ) = mol 1.01 g mole O – g * ( 1 mole ) = mol 16.00 g
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From last slide: 21.29 mol C, 42.49 mol H, 14.27 mol O
A g compound containing only carbon, hydrogen, and oxygen is found to be 48.38% carbon and 8.12% hydrogen by mass. The molar mass of this compound is known to be ~ g/mol. What is its molecular formula? From last slide: mol C, mol H, mol O C – 21.29/14.27 = 1.49 H – 42.49/14.27 = 2.98 (essentially 3) O – 14.27/14.27 = 1.00 C – 1.49 x 2 = 3 H – 3 x 2 = 6 O – 1 x 2 = 2 C3H6O2
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From last slide: Empirical formula = C3H6O2
A g compound containing only carbon, hydrogen, and oxygen is found to be 48.38% carbon and 8.12% hydrogen by mass. The molar mass of this compound is known to be ~ g/mol. What is its molecular formula? From last slide: Empirical formula = C3H6O2 “EFM” = 74.09 Molar mass = = ~3 EFM 3(C3H6O2) = C9H18O6
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