1 / 8

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.

sophie
Télécharger la présentation

CHEMISTRY

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related