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CHEMISTRY

This guide explains the concepts of empirical and molecular formulas in chemistry, emphasizing how to calculate the percent composition of elements in a compound. By using magnesium nitrate (Mg(NO3)2) as an example, the step-by-step calculation of the percent mass of magnesium, nitrogen, and oxygen is demonstrated. Additionally, we explore how to determine the empirical formula from percentage composition data and provide examples that illustrate the process of converting mass to moles to find empirical formulas.

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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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