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Determining Chemical Formulas Experimentally

Determining Chemical Formulas Experimentally. % composition, empirical and molecular formula. Percentage Composition. The mass of each element in a compound compared to the entire mass of the compound times 100 is the percent composition of that element.

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Determining Chemical Formulas Experimentally

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  1. Determining Chemical Formulas Experimentally % composition, empirical and molecular formula

  2. Percentage Composition • The mass of each element in a compound compared to the entire mass of the compound times 100 is the percent composition of that element. Example: What is the % composition of H and O in 1 mole of H2O? %H = 1.01 g H x 2 x 100 = 11% 18.02 g H2O =100% %O = 16.00 g O x 100 = 89% 18.02 g H2O

  3. Percentage Composition Example: A sample of an unknown compound with amass of 0.2370 g contains 0.0948 g of carbon, 0.126 g of oxygen, 0.016 g of hydrogen. What is the % composition of the compound? %C = 0.0948g X 100 = 40% 0.2370 g %O = 0.126 g x 100 = 53 % 0.2370 g %H = 0.016 g x 100 = 7% 0.2370

  4. Empirical Formula • The simplest whole number ratio of the elements in a compound is the empirical formula. Steps: • Find moles of each element in the compound. • Divide each mole value by the smallest mole value to find the mole ratio. • Write the formula using the mole ratios determined in step 2.

  5. A Poem Percent to grams Grams to moles Divide by smallest Multiply till whole

  6. Empirical Formula Example: A sample is analyzed and determined to contain 80 g C and 20 g H. What is the empirical formula? • 80 g C x 1 mole = 6.7 mol C 12.0 g 20 g H x 1 mole = 20 mol H 1.01 g

  7. Empirical Formula • Mole ratios: C: 6.7 = 1 H: 20 = 2.98 ~ 3 6.7 6.7 • CH3 is the empirical formula • You may round the mole ratio if it is within 0.05 of a whole number. If it is not a whole number you must multiply all the mole ratios by a factor to get a whole number.

  8. Empirical Formula Example: What is the empirical formula of a compound that is 70% Fe and 30%O? • 70% Fe = 70 g Fe x 1 mole = 1.25 mol 55.85 g 30% O = 30 g O x 1 mole = 1.88 mol 16.00 g • Fe = 1.25/1.25 = 1 O = 1.88/ 1.25 = 1.5

  9. Empirical Formula • Empirical formula must be in whole number ratios. Fe = 1 and O = 1.5, multiply both ratios by 2 Fe = 2 and O = 3, so the formula is: Fe2O3

  10. Molecular Formula • The formula of the actual molecular compound is the molecular formula. Steps: • Calculate the empirical mass. • Divide the molecular mass by the empirical mass. • Multiply each subscript in the empirical formula; by the number found in step 2.

  11. Molecular Formula • Example: The molecular mass of a compound is found to be 180g/mol. If the empirical formula is CH2O, determine the molecular formula. • CH2O = 12.01g/mol + (1.01g/mol x 2) + 16.00 g/mol = 30.02 g/mol • I80g/mole = 6 30.02 g/mole 3. CH2O x 6 = C6H12O6

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