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Percent Composition. The relative amounts of elements in a compound. % composition = . mass of element. x 100. mass of compound. 24g C. 85.7% C. x 100. =. 28g CH 4. Percent Composition from Formula. C 2 H 4. 2 moles of C x 12 g/mol = 24 g.

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## Percent Composition

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**Percent Composition**• The relative amounts of elements in a compound. • % composition = mass of element x 100 mass of compound 24g C 85.7% C x 100 = 28g CH4**Percent Composition from Formula**C2H4 2 moles of C x 12g/mol = 24 g 4 moles of H x 1 g/mol = 4 g 28 g 24g/28g x 100 = 85.7% C 4g/ 28g x 100 = 14.3% H**You try one:**NaC2H3O2**Empirical Formula**• The lowest whole number ratio of elements in a compound**Calculate empirical formula from percent composition:**1st - Begin by pretending to have 100 g sample & convert the % to grams. 2nd - Convert grams to the moles by dividing by molar mass. 3rd - Obtain whole number ratios of moles by: 1) dividing each number by the smallest number 2) multiplying to get rid of any fractions 4th - Write the empirical formula using the number of moles as subscripts.**Example 1**• Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N. • Assume 100 g so. . . . . • 38.67 g C / 12 g/mol C = 3.22 mole C • 16.22 g H / 1.07g/mol H = 16.09 mole H • 45.11 g N / 14 g/mol N = 3.22 mole N • Divide each by the smallest value. . . . . 3.22**Example 1 cont.**• 3.220 mole C / 3.22 = 1 • 16.09 mole H / 3.22 = 5 • 3.22 mole N / 3.22 = 1 • The empirical formula is CH5N**Example 2**A compound is 43.64 % P and 56.36 % O. What is the empirical formula? • 43.6 g P /31 g/mol = 1.4 mole P /1.4 = 1 • 56.4 g O /16 g/mol = 3.5 mole O /1.4 = 2.5 Divide by smallest number • Multiply the result to get rid of any fractions. P1O2.5 = P2O5 2 X**Molecular Formulas**Are whole number multiples of empirical formulas. • Divide molar mass of compound by molar mass of empirical formula. • Multiply subscripts in compound by the number from step 1. Ex. An empirical formula is CH5N. The molar mass is 93 g/mol. 93 g/mol / 31 g/mol = 3 3 x CH5N = C3H15N3

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