userdata-paziras-Chem101-Chap_03A

Empiricalformulamassofch2 is 1c1x120amu 120amu

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Unformatted text preview: s = ¾¾¾¾¾¾¾¾ = 1.000 0.008351 moles Sn Empirical Formula: Sn1.000O2.043 (includes experimental error) 10 SnO2 Chemistry 101 Chapter 3 Sample Problem # 1 Arsenic (As) reacts with oxygen (O) to form a compound that is 75.7 % As and 24.3 % O by mass. What is the empirical formula of this oxide ? Step 1: PERCENT MASS 75.7 % As 24.3% O Assume 100 g compound Step 2: MASS 75.7 g As 24.3 g O MOLE 1 mole As ? moles As = 75.7 g As x ¾¾¾¾¾ = 74.92 g As 1.01 moles As atoms 1 mole O ? moles O = 24.3 g O x ¾¾¾¾¾ = 16.00 g O 1.52 moles O atoms Step 3: DIVIDE BY SMALL 1.52 moles O Relative number of O atoms = ¾¾¾¾¾¾¾¾ = 1.50 1.01 moles As 1.01 moles As Relative number of As atoms = ¾¾¾¾¾¾¾¾ = 1.00 1.01 moles As Empirical Formula: As1.00O1.50 ??????????? (subscipts must be whole numbers) 11 Chemistry 101 Chapter 3 Step 4: MULTIPLY ‘TIL WHOLE: (As1.00O1.50) x 2 = As2O3 A Simple Rhyme for a Simple Formula (to be remembered) 1. 2. 3. 4. Percent to Mass Mass to Mole Divide by small Multiply ‘til Whole 12 Chemistry 101 Chapter 3 Sample Problem # 1 A sample of Freon contains 0.423 g C, 2.50 g Cl, and 1.34 g F. What is the Empirical Formula o f Freon? 1. Percent to Mass This step...
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