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Chemistry

Chemistry. Percent Mass & Empirical Formulas. Mass percent of an element = Atomic mass of element x subscript x 100% Molecular mass of compound Formula: %Ele = At.M x Sub x 100% mm

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Chemistry

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  1. Chemistry Percent Mass & Empirical Formulas

  2. Mass percent of an element = • Atomic mass of element x subscript x 100% • Molecular mass of compound • Formula: %Ele = At.M x Sub x 100% • mm • Example: Determine the percentage of each element in the compound Calcium perchlorate. • Solution: Write the correct formula Ca+2(ClO4) –1 Ca(ClO4)240.0 + 2[35.5 + 4(16.0)] 40.0 + 70.0 + 128238 amu

  3. Ca(ClO4)2 • %Ele = At.M x Sub x 100% • mm • % Ca = 40.0 x 1 x 100% = 16.8 % • 238 • % Cl = 35.5 x 2 x 100% = 29.4% • 238 • % O = 16.0 x 2 x 4 x 100% = 53.8 % • 238

  4. What is the mass of Chlorine in a 5.36 g sample of Calcium perchlorate? • Mass ele = %ele x mass of compound • 100 • The percent of Chlorine in the compound was 29.4% • % Cl = 29.4 x 5.36 g = 1.58 g • 100

  5. Empirical Formula: The simplest whole number ration of atoms in a compound. • Once a compound is analyzed the results can be expressed in either percentages or actual masses of the individual elements. The method for determining the Empirical formula is the same, that is convert mass to moles and fromthat determine a mole ratio. • MASS MOLE RATIO

  6. MOLE: A chemical unit that has a mass, a volume and numbers of particles ( atoms, molecules or formula units.) • 1.0O mole of H2O = 18.0 g the molar mass • 1.00 mole of NaCl = 58.5 g the molar mass • How many moles are in 45.0 g of water? • Unit Factor Method • 45.0 g x 1.00 mol = 2.50 mol water • 18.0 g

  7. EXAMPLE: • Analysis: A compound was analyzed and it was determined that it contained 32.39% Na, 22.54% S, and 45.07% O. Determine the empirical formula. • Solution: When given percentages assume there is 100 g of specimen and multiply the percent of each element times the 100g. • Na 0.3239% x 100g = 32.39g • atomic mass = 23.0 g/mol • S 0.2254% x 100g = 22.54g • Atomic mass = 32.1 g/mol • O 0.4507% x 100g = 45.07g • Atomic mass = 16.0 g/mol

  8. /0.702 = 2 • CALCULATE THE NUMBER OF MOLES • Na = 32.39 g x 1.00 mol = 1.408 mol • 23.0 g • S = 22.54 g x 1.00 mol = 0.702 mol • 32.1 g • O = 45.07g x 1.00 mol = 2.817 mol • 16.0 g /0.702 = 1 Smallest number /0.702 = 4 Divide by the smallest number Empirical Formula Na2SO4

  9. EXAMPLE TWO: • An analysis of a compound determined that a 7.155 g sample of a compound contained 3.933 g of Cobalt and the identity of the other substance was sulfur. Determine the empirical formula. • Solution: The mass of sulfur can be calculated from subtracting the mass of cobalt from the mass of the compound.

  10. 7.155 g - 3.933 g = 3.222 g of Sulfur • Co = 3.933g x 1.00 mole = 0.0667 mol/0.0667 = 1.00 • 59.0 g • S = 3.222g x 1.00 mole = 0.1004 mol/0.0667 = 1.50 • 32.1 g • Since the 1.00 : 1.50 cannot be rounded off, convert to nearest fraction 1.50 = 1 ½ • 1 : 3/2 multiply by the denominator of 2 • (2) 1 : (2) 3/2 = 2 : 3 • Co2S3

  11. Fraction Equivalent to decimals • 0.25 = ¼ 0.5 = ½ 0.75 = ¾ • 0.33 = 1/3 0.67 = 2/3 • What if the ratio was • 1.25 : 0.5 • Change to fraction • 1 1 : 1 • 4 2 • Make both fractions • 5 : 1 • Multiply by 4 4 2 • 5 : 2 Ratio 4 x x 4

  12. MOLECULAR FORMULA: The empirical formula is the simplest whole number ratio, but the molecular formula is the actual formula. Most of the time they are the same formula. In order to calculate the molecular formula the molar mass must be known (given in the problem). • An unknown hydrocarbon with a mass of 15.589g containing carbon, hydrogen and oxygen is combusted in a combustion furnace where oxygen is passed through the sample. The CO2 produced from the reaction is trapped in the CO2 absorber and the mass wasdetermined to be 7.509g. The other product of combustion was water and it was condensed and weighed. The mass of water was 3.585 g. Determine the Empirical Formula? If it was determined the molecular masswas 118.14 amu, what is the Molecular Formula?

  13. Experimental Set Up

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