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EMPIRICAL FORMULA

EMPIRICAL FORMULA. The empirical formula represents the smallest ratio of atoms present in a compound. The molecular formula gives the total number of atoms of each element present in one molecule of a compound.

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EMPIRICAL FORMULA

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  1. EMPIRICAL FORMULA • The empirical formula represents the smallest ratio of atoms present in a compound. • The molecular formula gives the total number of atoms of each element present in one molecule of a compound. The empirical formula is the simplest formula and the molecular formula is the “true” formula.

  2. Hydrates Compounds containing Water molecules MgSO4 7H2O Magnesium Sulfate heptahydrate CaSO4 2H2O Calcium Sulfate dihydrate Ba(OH)2 8H2O Barium Hydroxide octahydrate CuSO4 5H2O Copper II Sulfate pentahydrate Na2CO3 10H2O Sodium Carbonate decahydrate

  3. Examples of Names and Formulas • KNO2 Potassium Nitrite BaSO3 Barium Sulfite • Mg(NO3)2 Magnesium Nitrate Na2SO4 Sodium Sulfate • LiClO4 Lithium Perchlorate Ca(BrO)2 Calcium Hypobromite • NaClO3 Sodium ChlorateAl(IO2)3 Aluminum Iodite • RbClO2 Rubidium Chlorite KBrO3 Potassium Bromate • CsClO Cesium Hypochlotite LiIO4 Lithium Periodate

  4. Examples of Names and Formulas ofOxoanions and their Compounds - II • Calcium Nitrate Ammonium Sulfite • Strontium Sulfate Lithium Nitrite • Potassium HypochloriteLithium Perbromate • Rubidium Chlorate Calcium Iodite • Ammonium Chlorite Boron Bromate • Sodium Perchlorate Magnesium Hypoiodite

  5. Examples of Names and Formulas ofOxoanions and their Compounds - II • Calcium Nitrate Ca(NO3)2Ammonium Sulfite(NH4)2SO3 • Strontium Sulfate SrSO4 Lithium NitriteLiNO2 • Potassium Hypochlorite KClO Lithium PerbromateLiBrO4 • Rubidium Chlorate RbClO3 Calcium Iodite Ca(IO2)2 • Ammonium Chlorite NH4ClO2 Boron BromateB(BrO3)3 • Sodium Perchlorate NaClO4 Magnesium Hypoiodite Mg(IO)2

  6. Determining Names and Formulas of Ionic Compounds Containing Polyatomic Ions a) BaCl2 5 H2O b) Magnesium Perchlorate c) (NH4)2SO3 d) Calcium Nitrate

  7. Determining Names and Formulas of Ionic Compounds Containing Polyatomic Ions Ba+2 is the cation Barium, Cl- is the Chloride anion. There are five water molecules therefore the name is: Barium Chloride Pentahydrate a) BaCl2 5 H2O b) Magnesium Perchlorate Magnesium is the Mg+2 cation, and perchlorate is the ClO4- anion, therefore we need two perchlorate anions for each Mg cation therefore the formula is: Mg( ClO4)2 c) (NH4)2SO3NH4+ is the ammonium ion, and SO3-2 is the sulfite anion, therefore the name is: Ammonium Sulfite d) Calcium NitrateCalcium is the Ca+2 cation, and nitrate is the NO3- anion, therefore the formula is: Ca(NO3)2

  8. Names and Formulas of Binary Covalent Compounds 1) The element with the lower group number in the periodic table is the first word in the name; the element with the higher group number is the second word. (Important exception: When the compound contains oxygen and a halogen, the halogen is named first.) 2) If both elements are in the same group, the one with the higher period number is named first. 3) The second element is named with its root and the suffix “-ide.” 4) Covalent compounds have Greek numerical prefixes (table 2.6)to indicate the number of atoms of each element in the compound. The first word has a prefix only when more than one atom of the element is present; the second word always has a numerical prefix.

  9. Determining Names and Formulas of Binary Covalent Compounds Problem: What are the name or chemical formulas of the following chemical compounds: a)Carbon dioxideb)PCl3c) Give the name and chemical formula of the compound formed from two P atoms and five O atoms. Solution: a)Carbon dioxide b) PCl3 c)The compound formed from two P atoms and five O atoms

  10. Determining Names and Formulas of Binary Covalent Compounds Problem: What are the name or Chemical formulas of the following Chemical compounds: a)Carbon dioxideb)PCl3c) Give the name and chemical formula of the compound formed from two P atoms and five O atoms. Solution: a) The prefix “di-” means “two.” The formula is CO2 b) P is the symbol for phosphorous; there are three chlorine atoms which require the prefix “tri-.” The name of the compound is: phosphorous trichloride c)P comes first in the name (lower group number). The compound is diphosphorous pentaoxide ( commonly called “phosphorous pentaoxide”)

  11. EMPIRICAL FORMULA Assume 100g sample Calculate mole ratio Use Atomic Masses Mass % of elements Empirical Formula Grams of each element Moles of each element

  12. EMPIRICAL FORMULA Step 1: If given the % composition, assume a 100g sample then convert % to grams. Step 2: Use the atomic masses to convert grams to moles. Step 3: Divide the moles of each element by the SMALLEST mole fraction. Step 4: The results from step 3 should be a whole number, if not, make it so by multiplying by a common factor.

  13. EMPIRICAL FORMULA 1. Calculate the empirical formula from a sample containing 43.4% Na, 11.3% C, and 45.3% O. smallest 43.4%  43.4 g Na1 mole Na 23 g/mol Na =1.887 moles Na 11.3%  11.3 g C (1 moleC / 12 g/mol C) = 0.9417 moles C 45.3%  45.3 g O (1 mole O/ 16 g/mol O) = 2.831 moles O 1.887/0.9417 =2.00 Na 2.831/0.9417 = 3.00 O . 9417/0.9417 = 1.00 C Empirical Formula = Na2CO3

  14. EMPIRICAL FORMULA 2. When 8.00 g of calcium metal is heated in air, 11.20 g of metal oxide is formed. Calculate the empirical formula. According to the Law of Conservation of mass, 11.20 g Product - 8.00 g Ca = 3.20 g Oxygen (reactive part of air) smallest 8.00 g Ca (1 mole / 40 g/mol) = 0.200 moles Ca 3.20 g O (1 mole / 16 g/mol) = 0.200 moles O 0.200 / 0.200 = 1 Empirical Formula = CaO

  15. EMPIRICAL FORMULA 3. A compound was found to have a composition of 33.0 % Sr, 26.8 % Cl, and 40.2 % water. Calculate the empirical formula of this hydrate. smallest 33.0%  33.0 g Sr (1 mole/87.6 g/mol) = 0.3767 moles Sr 26.8%  26.8 g Cl (1 mole/35.45 g/mol) = 0.7560 moles Cl 40.2%  40.2 g H2O (1 mole/18.0g/mol) = 2.233 moles H2O 0.7560 / 0.3767 = 2 Cl2.233 / 0.3767 = 5.9 = 6 H2O Empirical Formula = SrCl2 . 6 H2O

  16. EMPIRICAL FORMULA & Molecular Formula 4. Propylene contains 14.3 % H, 85.7% C, and has a molar mass of 42.0 g/mol. What is its molecular formula? smallest 14.3%  14.3 g H (1 mole/1.01 g/mol) = 14.19 moles H 85.7%  85.7 g C (1 mole/12.01 g/mol) = 7.142 moles C 14.19 / 7.142 = 1.987 = 2 H Empirical Formula = CH2 Molar mass / empirical mass = multipier (42.0 g/mol / 14.0 g/mol) = 3 3 x CH2 becomes the molecular formula C3H6

  17. PRACTICE PROBLEM #12 A ______ 1. Which contains the larger number of MOLES of atoms? a) 125.0 g KCl b) 25.0 g CaSO4 c) 17.0 g of N2 ______ 2. What is the empirical formula of the compound whose composition is 39.7% K, 27.8% Mn, and 32.5% O? ______ 3. Determine the empirical formula of a compound that contains 89.7 % bismuth and 10.3 % oxygen. ______ 4. Write the molecular formula for a compound that contains 54.5 % C, 9.1% H, and 36.4 % O and has a molar mass of 132 amu? K2MnO4 Bi2O3 C6H12O3

  18. GROUP STUDY PROBLEM #12 ______ 1. Which contains the larger number of MOLES of atoms? a) 125.0 g HBr b) 25.0 g C6H11O6 c) 17.0 g of Br2 ______ 2. A sample of a compound weighing 4.18 g contains 1.67 g of sulfur and the rest is oxygen. What is the empirical formula? ______ 3. What is the empirical formula of the compound whose composition is 28.7% K, 1.4% H, 22.8 % P, and 47.1% O? ______ 4. A compound contains 92.3% C and 7.7% H and has a molar mass of 78.0 g/mol. Determine the molecular formula.

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