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Empirical and Molecular Formulas and Percent Composition

Empirical and Molecular Formulas and Percent Composition. Unit 7: Stoichiometry – Part III Mrs. Callender. What is percent composition and how do I calculate it?. What is the difference between an Empirical and Molecular Formula?. How do I calculate Empirical and Molecular Formulas?.

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Empirical and Molecular Formulas and Percent Composition

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  1. Empirical and Molecular Formulas and Percent Composition Unit 7: Stoichiometry – Part III Mrs. Callender

  2. What is percent composition and how do I calculate it? What is the difference between an Empirical and Molecular Formula? How do I calculate Empirical and Molecular Formulas? Lesson Essential Questions…..

  3. Calculate the Percent Composition of AlAsO4? 1. Find the formula mass of the compound. What does the term “Percent” mean? 26.98 g + 74.92 g + 4(16.00 g) = 165.90 g Percent means a “part of the whole”. x 100 = 16.26% Al = 26.98 165.90 74.92 165.90 64.00 165.90 As = x 100 = 45.16% x 100 = 38.58% O = 100.00 %

  4. Calculate the Percent Composition of Glucose C12H22O11? 12(12.01) g + 22(1.01) g + 11(16.00 g) = 342.34 g 1. Find the formula mass of the compound. x 100 = 42.10% C = 144.12 342.34 22.22 342.34 176.00 342.34 H = x 100 = 6.49% x 100 = 51.41% O = 100.00 %

  5. There are two types of chemical formulas: F O R M U L A S EMPIRICAL FORMULA The lowest whole number ratio of elements in a compound. MOLECULAR FORMULA The true number of atoms of each element in a compound or formula. Molecular formula = (Empirical Formula)n Molecular Formula of Benzene - C6H6 or (CH)6 Empirical Formula of Benzene - CH

  6. Practice…. What are the empirical formulas for the following? 1. C2H4 4. C2H6O2 CH2 CH3O 2. C8H18 C4H9 5. X39Y13 X3Y C14H18N2O5 6. C28H36N4O10 3. WO2 WO2 IONIC compounds should already be empirical formulas because of how the charge is balanced. 4. Na2SO4 Na2SO4

  7. Law of Definite Proportions law stating that every pure substance always contains the same elements combined in the same proportions by weight. Ratios of subscripts will always be in small whole numbers. Ex. CO2 always has a ratio of one carbon to two oxygen's.

  8. Lets Start with an easy problem…. A compound with an empirical formula of C2OH4 and a molar mass of 88 grams per mole. What is the molecular formula of this compound? Step 1: Find the molar mass (formula mass) of the empirical formula. Step 2: Divide the molecular formula molar mass by the empirical formula molar mass. Step 3: Take that number and multiply it by the empirical formula. Step 4: Check molar mass of molecular formula. 2(12.01 g) + 16.00 g + 4(1.01 g) = 44.06 g 4(12.01 g) + 2(16.00 g) + 8(1.01 g) = 88.12 g Is the empirical formula mass the same as the molecular formulas molar mass? 88 = 2 44.06 This mass matches closely to the molecular formula mass so you have successfully solved the problem. NO C2OH4 2 X C4O2H8

  9. The local CSI found and unknown compound to contain 13.5 g Ca, 10.8 g O and 0.675 g of H. Calculate the empirical formula for this compound. 13.5 g Cax 1 1 mole Ca 40.08 g Ca = 0.337 mole Ca = 0.675 mole O 10.8 g Ox 1 1 mole O 16.00 g O 0.675 g Hx 1 1 mole H 1.01 g H = 0.668 mole H Step 3: Divide all moles by the smallest number of moles. Step 4: How can we write this formula better? Step 2: Write down the elements into the form of a compound with the mole amount. Step 1: Start with the grams of each element and convert to moles by using molar mass. Ca O H 2 2 1 (O H) .668 Ca .675 .337 2 .337 .337 .337

  10. NutraSweet is 57.14% C, 6.16% H, 9.52% N, and 27.18% O. Calculate the empirical formula of NutraSweet and find the molecular formula. (The molar mass of NutraSweet is 294.34 g/mol) 1 mole C 12.01 g C 57.14 g Cx 1 = 4.76 mole C 14(12.01 g) + 18(1.01 g) + 2(14.01 g) + 5(16.00 g) = 294.34 g = 6.10 mole H 1 mole H 1.01 g H 6.16 g Hx 1 Step 6: Since the molar mass of the molecular formula is the same as the molar mass of the empirical formula, they are one in the same and you have successfully answered the problem. Step 5: Find the molar mass of your empirical formula. Compare this to the molar mass given in the problem. = 0.680 mole N 1 mole N 14.01 g N 9.52 g Nx 1 27.18 g Ox 1 1 mole O 16.00 g O = 1.70 mole O Step 4: The 2.5 can NOT be rounded so it must be multiplied by 2 to get rid of the decimal. Step 1: Start by changing % to grams by thinking you have 100 g of the sample. Change the grams to moles. Step 3: Divide all moles by the smallest number of moles. C N O H Step 2: Write down the elements into the form of a compound with the mole amount. C14H18N2O5 2.5 1 9 2 x 7 .680 1.70 14 18 6.10 2 5 4.76 .680 .680 .680 .680

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