# 1 Percent Composition: Identifies the elements present in a compound as a mass percent of the total compound mass. The mass percent is obtained by dividing.

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1 Percent Composition: Identifies the elements present in a compound as a mass percent of the total compound mass. The mass percent is obtained by dividing the mass of each element by the total mass of a compound and converting to percentage. Percentage Composition

2 Empirical Formula The empirical formula gives the ratio of the number of atoms of each element in a compound. CompoundFormulaEmpirical Formula Hydrogen peroxideH 2 O 2 OH BenzeneC 6 H 6 CH EthyleneC 2 H 4 CH 2 PropaneC 3 H 8 C 3 H 8

3 Percentage Composition Glucose has the molecular formula C 6 H 12 O 6. What is its empirical formula, and what is the percentage composition of glucose? Empirical Formula = smallest whole number ratio CH 2 O

4 Percentage Composition CH 2 O Total mass = 12.01 + 2.02 + 16.00 = 30.03 %C = 12.01/30.03 x 100% = 39.99% %H = 2.02/30.03 x 100% = 6.73% %O = 16.00/30.03 x 100% = 53.28%

5 Empirical Formula A compound’s empirical formula can be determined from its percent composition. A compound’s molecular formula is determined from the molar mass and empirical formula.

6 Empirical Formula A compound was analyzed to be 82.67% carbon and 17.33% hydrogen by mass. What is the empirical formula for the compound? Assume 100 g of sample, then 82.67 g are C and 17.33 g are H Convert masses to moles: 82.67 g C x mole/12.011 g = 6.88 moles C 17.33 g H x mole/1.008 g = 17.19 mole H Find relative # of moles (divide by smallest number)

7 Empirical Formula Convert moles to ratios: 6.88/6.88 = 1 C 17.19/6.88 = 2.50 H Or 2 carbons for every 5 hydrogens C 2 H 5

8 Empirical Formula Empirical Formula is: C 2 H 5 Formula weight is: 29.06 g/mole - The molecular weight is known to be 58.12 g/mole. Since the empirical formula is the lowest ratio the actual molecule would weigh more. By a whole number multiple. Divide the actual molar mass by the mass of one mole of the empirical formula. Then the molecular formula must be: C 4 H 10

Example A compound is known to be composed of 71.65 % Cl, 24.27% C and 4.07% H. Its molar mass is known (from gas density) is known to be 98.96 g. What is its molecular formula?

Example 71.65 Cl 24.27C 4.07 H. = 2.0mol = 4.0mol

11 Cl 2 C 2 H 4 Its molar mass is known (from gas density) is known to be 98.96 g. What is its molecular formula? We divide by lowest (2mol ) Cl 1 C 1 H 2 would give an empirical wt of 48.5g/mol

12 Cl 2 C 2 H 4 Its molar mass is known (from gas density) is known to be 98.96 g. What is its molecular formula? We divide by lowest (2mol ) Cl 1 C 1 H 2 would give an empirical wt of 48.5g/mol 2 X Cl 1 C 1 H 2 = Cl 2 C 2 H 4

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