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CHEMISTRY

CHEMISTRY. Empirical Formula (simplest whole number ratio). Percent mass. The percentage of an element in a compound is determined by the mass of the element and the mass of the compound. % Ele = atomic mass Ele x number atoms x 100% molecular mass.

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CHEMISTRY

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  1. CHEMISTRY Empirical Formula (simplest whole number ratio)

  2. Percent mass • The percentage of an element in a compound is determined by the mass of the element and the mass of the compound. • % Ele = atomic mass Ele x number atoms x 100% • molecular mass

  3. Calculations • Find the percent of each element in magnesium nitrate • Mg(NO3)2 has a molecular mass of 148 au • % Mg = At. Mass Mg x # atoms (subscript) • molecular mass Mg(NO3)2 • % Mg = 24.3 x 1 X 100 % = 16.4 % • 148 • % N = 14.0 x 2 X 100 % = 18.9 % • 148 • % O = 3 x 16.0 x 2 X100 % = 64.9 % • 148

  4. The formula of a compound • The ratio of atoms in a compound depends on the positive and negative charges. • The ratio of moles of atoms is the same because of Avogadro's hypothesis that one mole has the same number of atoms.

  5. Percentage of Elements • What is the empirical formula of a compound that is 66.0% Ca and 34.0% P? • 1. Assume that there is 100 g of sample • 2. Therefore there is 66.0 g of Ca and 34.0 g of P • 66.0 g Ca x 1.00 mole Ca = 1.65 mol Ca • 40.1 g Ca • 34.0 g P x 1.00 mole P = 1.10 mol P • 31.0 g P • Divide by the smallest number • 1.65/1.10 = 1.5 x 2 = 3 • 1.10/1.10 = 1.0 x 2 = 2 • Ca3P2

  6. Mass of elements • What is the empirical formula of a compound if a 2.50 g sample contains 0.900 g of Calcium and 1.60 g of chlorine? • Convert the masses to moles. • 0.900 g Ca x 1.00 mol Ca = 0.0224 mol Ca • 40.1 g Ca • 1.60 g Cl x 1.00 mol Cl = 0.0451 mol Cl • 35.5 g Cl • Divide by the smaller number • 0.0451/0.0224 = 2.01 • 0.0224/0.0224 = 1.00 • CaCl2

  7. Problem • A compound has a percentage composition of 40.0% carbon, 6.71% hydrogen, and 53.3% oxygen. What is the empirical formula? • 40.0 g C x 1.00 mol C = 3.33 mol C • 12.0 g C • 6.71 g H x 1.00 mol H = 6.64 mol H • 1.01 g H • 53.3 g O x 1.00 mol O = 3.33 mol O • 16.0 g O • 3.33/3.33 = 1.00, 6.64/3.33 = 1.99, 3.33/3.33 = 1.00 • CH2O

  8. Molecular Formula • If the molar mass of CH2O is 90.1 g/mol what is the molecular formula? • ( Mass of Empirical Formula) X = Molar mass • (CH2O) X = 90.1 g/mole • ( 12.0 + 2(1.01) + 16.0 ) X = 90.1 g/mole • (30.02) X = 90.1 g/mol • X = 3 • (CH2O) 3 = C3H6O3

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