Question

Experiment 10: Empirical Formulas Post-lab: 1. In the experiment, how did you figure out the mass of any in your compound After heating minus before heating 2.8360 2.1.65g of 2. 1.651g of silver is combined with 0.1224g of oxygen. Find the empirical formula mole -0. 1.659 0.00 0.06 0.007 0.0 107.89-0 0.122490x Imole = 0.007 16.009 Empirical formula: 2.14 U O I 3. 7.00g of iron is reacted with 3.00g of oxygen. Find the empirical formula a exlmole -0.125 0.15 0.187-1.49 56.85 A25 0.1a5 1.23 oog Ox Imole 0.189 16.00g apirical formuli. Fei 03 4. 0.880g of iron is reacted with chlorine to form 2.000 of an iron chloride compound Find the empirical formula. 380g Fex I mole=0.016. 55.85 2009 CI X Imole -0.000 Empirical formula: Fei (1z 35.45 0.0% 0.056 = 3.5+2 0.0% -7 106

Answer 1

2) mass of silver : 1.6519 .. mole of silver Given mass 1.6519 = 107-879/mo 0.0153 mol Mw mass of 0. = 0.12249 0.12249 & 0.00765 mol. -- mole of Or 16 g/mol so, ratio of mole of Ag and 'o' 0.00765 . empirical formaila - Ago 3) mass of iron = 79 i mole of iron gran = 0.125 mol mass of 6 = 39 i mole of 6 39 0.1875 mol 16 g/mol - ratio of mole of fe and o 0.125 : 2 011875 formula = fe2 Oz so, empirical 4) mole of fe= 55,859 1 mol - mass. Of fe = 0.880 g 0.8809 ; - 0.015756 mol mass of cd = mass of compound- mass of fe = 2.09 - 0.889 = 1129 & mole of de 0.03159 mol 4129 35.459/mol

So, ratio of mol of fe and a 0.015756 : Ź 0.03159 fe Cla so, empirical formular - r ana 5) mass of Ti = 0.7159 - mole of Ti = 0.7159 a 0.015 mol 47.87 g/mol mass of Cl = mass of compound mass of Ti = 2.8369-0.8159 - 2.1219 .: mole of a : 2.1219 0.06 mol 35-45 g/mol ratio of mol of Ti and ca => 0.015 - Ž 0.06 so, empirical formula = Ticla 6) % mass of 'C' : 40.lil. 1. mass of t = 6.7% 1. mars of co = 53.3% let mass of compound be 100g then mass of 'C' = 40.49 mass of 'H' = 6.79 mass of 'o = 53.39 mole of C 40.49 3.367 mol 12 g/mol mole of H = 6.70 6.7 mol mole of or 53.39 = 3.33 mol Iglmel CS T6 g (mol

So, ratio of mol of 'C' H and 'o' is 3.367: 6.7: 3.33 12:1 so, empirical formula & CH₂O empirical mass 30g and molecular mass 60g/mol . molecular mas : n x empirical mass 4 constant in : Gog/mol. 2 30g .: molecular formula = nx empirical formula 2 x (CH2O) : molecular formula - C H₂O2 7) mass of N = 0.39559 mole of N = 0.39559 - T4 g/mol = 0.02825 mol mass of 'o' = 0.gourg i mol ofo = 0.90459 -0.0565 mol T6 g/mol - ratio of 'N' and 'o' 0.02825 = - 0.0565 empirical formula No₂ empirical mass = 469 ne molecular mass . 929 . empirical mass 469 = 2 in Molecular formula= 2 x empirical formada • 2 X (NO) = N204 CS