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Percent Composition Percent = part / whole Example: MgO Find molar mass of whole compound: MgO = 24.3 + 16.00 = 40.3 grams % Mg (by mass) = mass of Mg.

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Presentation on theme: "Percent Composition Percent = part / whole Example: MgO Find molar mass of whole compound: MgO = 24.3 + 16.00 = 40.3 grams % Mg (by mass) = mass of Mg."— Presentation transcript:

1 Percent Composition Percent = part / whole Example: MgO Find molar mass of whole compound: MgO = = 40.3 grams % Mg (by mass) = mass of Mg = 24.3 g x 100 = 60.3% molar mass of MgO 40.3 g % O (by mass) = mass of O = 16.0 g x 100 = 39.7% molar mass of MgO 40.3 g

2 Empirical Formulas Gives the lowest whole-number ratio of elements and compounds in a formula Gives the lowest whole-number ratio of elements and compounds in a formula 4 Steps: 1.Change the percent to grams 2.Convert grams to moles 3.Divide each of the moles by the smallest number to find the ratio 4.Round ratio to a whole number. If the ratio is not a whole number (ex. 1.5) multiply each element by 2 to get a whole number

3 Examples A compound contains 94.1% Oxygen and 5.9% Hydrogen. What is its empirical formula? 1) 94.1% = 94.1g O 5.9% = 5.9g H 2) 94.1g O 16 g O 1 mol O = 5.88 mol O 5.9g H 1 g H 1 mol H = 5.9 mol H 3) = = (can round to 1.0) 4) Ratio = 1:1Formula = OH

4 Examples A compound contains 67.6% Mercury and 10.8% Sulfur and 21.6% Oxygen. What is its empirical formula? 1) 67.6% = 67.6g Hg 10.8% = 10.8g S 21.6% = 21.6g O 3) = 1.00 Hg = 1.00 S 4) Ratio = 1:1:4Formula = HgSO g Hg = mol Hg 10.8g S = mol S 21.6g O = 1.35 mol O = 4.02 O

5 Examples What is the empirical formula for a compound containing 70.0% Fe and 30.0% O? 1) 70.0% = 70.0g Fe 30.0% = 30.0g O 3) = 1.00 Fe 4) Ratio = 1 : 1.5 5) Mult ratio by 2 = 2 : g Fe = 1.25 mol Fe g O = mol O = 1.5 O Formula = Fe 2 O 3

6 Molecular Formulas Same as empirical formula, or a simple whole-number multiple of it Steps: 1)Calculate the mass in grams of the empirical formula provided 2)Divide the molar mass by the mass of the empirical formula 3)Multiply this whole number ratio by the empirical formula

7 Examples Calculate the molecular formula of the compound whose molar mass is 60.0g and empirical formula is CH 4 N. 1) CH 4 N = (1) + 14 = 30.0 g 2) 60.0g 30.0g = 2.0 3) 2.0 (CH 4 N)= C 2 H 8 N 2

8 Examples What is the molecular formula of ethylene glycol (CH 3 O), used in antifreeze. It has a molar mass of 62 g/mol. 1) CH 3 O = 31.0 g 2) 62.0g 31.0g = 2.0 3) 2.0 (CH 3 O)= C 2 H 6 O 2 Find the molecular formula of C 3 H 2 Cl, which is mothballs. Its molar mass is 147 g/mol. 1) C 3 H 2 Cl = 73.0 g 2) 147.0g 73.0g 3) 2.0 (C 3 H 2 Cl) = C 6 H 4 Cl 2 = 2.0


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