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Study Buddy: Determining Formulas & Nomenclature Terms Description Percent Composition Representation The percent by mass of each element present in a compound. Empirical Formula The simplest ratio of atoms present in a compound Molecular Formula The actual number of atoms of each element that occur in one molecule of the compound Law of Definite Proportions Law of Multiple Proportions A chemical compound always contains the same proportion of elements by mass regardless of amount of sample Whenever two elements form more than one compound, the elements will combine in whole number ratios and the compounds will have different properties Mass of element X 100 Mass of compound Molecular C6H12O6 H2 O 2 CH4 Empirical CH2O HO CH4 Water is always 88.8% O and 11.2% H by mass CO or CO2 exist, but not CO1.5 Representations: 1. Draw ionic models for the ions when they react, name them and write the formula: a. Magnesium and chlorine MgCl2 Mg+2 Magnesium chloride Cl-1 b. Iron(III) and oxygen Fe2O3 Fe+3 O-2 Iron(III) oxide Li2SO4 c. Lithium and sulfate Li+1 SO4-2 Lithium sulfate 2. Explain when each of these would be used in chemical names: a. Roman numerals transition metals with more than one valency b. Prefixes covalent compounds (no mono- on 1st element) c. –ide ending the anion of ionic cmpds and the 2nd name of covalent cmpds d. Ending unchanged the cation of ionic cmpds, the 1st name of covalent cmpds & polyatomics 3. Write the name or formula for each of the following compounds: Name Carbon tetrachloride Tin(IV) oxide Dinitrogen tetroxide Iron(II) phosphate Phosphorus pentafluoride Methane Sodium nitrate Calcium sulfide 12/8/2015 Ionic or Covalent Covalent Ionic Covalent Ionic Covalent Covalent Ionic Ionic Formula CCl4 SnO2 N2O4 Fe3(PO4)2 PF5 CH4 NaNO3 CaS SCIE_CHEM_FORM_MAT_STUDYBUDDYTE_AL copyright © CFISD 2015 1 4. What is the name and formula of the compound that forms when aluminum reacts with the sulfate ion? Aluminum sulfate, Al2(SO4)3 a. How many atoms of each element are in one formula unit of this compound? Al _2__ S _3__ O _12__ b. How many moles of each element are in one mole of this compound? Al _2__ S _3__ O _12__ c. What is the percent composition by mass for each of the elements in this compound? Molar mass of Al2(SO4)3 = 2(27.0)+3(32.1)+12(16.0) = 342.3 g/mol % Al = 2(27.0)/342.3 x 100 = 15.8% Al % S = 3(32.1)/342.3 x 100 = 28.1% S % O = 12(16.0)/342.3 x 100 = 56.1% O 5. Lead (II) chromate, PbCrO4, was used as a pigment in paints. How many moles of the compound are in 75.0 g of lead (II) chromate? Molar mass of PbCrO4 = 323 g/mol X mol PbCrO4 75.0 g PbCrO4 = 1 mol PbCrO4 323 g PbCrO4 0.232 moles PbCrO4 a. How many atoms of oxygen are present in the 75.0 g of PbCrO4? X atoms oxygen (4)(6.02x1023) atoms oxygen 5.59x1023 atoms oxygen = 1 mol PbCrO4 0.232 mol PbCrO4 6. What is the difference between a molecular formula and an empirical formula? A molecular formula shows the actual number of atoms in one molecule of a compound, whereas an empirical formula shows the simplest ratio of atoms in the compound. The molecular formula represents a compound, but the empirical may not represent an actual compound that exists. 7. Describe how a chemist would use an empirical formula to determine the molecular formula? Once the chemist has determined empirical formula by using the percent composition of each of the elements, the chemist would use the actual molar mass of the compound to determine the “multiplier” by comparing it to the molar mass of the empirical formula. The chemist would then multiply each of the subscripts by the multiplier. 8. A compound was analyzed and found to contain 22.2% nitrogen, 1.6% hydrogen, and 76.2% oxygen. What is the empirical formula of the compound? Nitrogen – 22.2% x 100g = 22.2g / 14.0 g/mol = 1.59 mol 1.59 mol/1.59 mol = 1 HNO3 Hydrogen – 1.6% x 100g = 1.6g /1.0 g/mol = 1.6 mol 1.6 mol/1.59 mol = 1 Oxygen – 76.2% x 100 = 76.2g / 16.0 g/mol = 4.76 mol 4.76 mol/1.59 mol = 3 9. A compound composed of hydrogen and oxygen is found to contain 0.59 g of hydrogen and 9.44 g of oxygen. The molar mass of this compound is 34.0 g/mol. Find the empirical and molecular formulas. Hydrogen – 0.59g / 1.0 g/mol = 0.59 mol 0.59/0.59 = 1 empirical formula HO Oxygen – 9.44g / 16.0 g/mol = 0.590 mol 0.590/0.59 = 1 Molar mass of HO 1.00 + 16.0= 17.0 g/mol Molar mass of compound 34.0 g/mol 12/8/2015 Multiplier = 34.0/17.0 = 2 Empirical Formula H2O2 SCIE_CHEM_FORM_MAT_STUDYBUDDYTE_AL copyright © CFISD 2015 2