Question

28. A compound contains Carbon, Hydrogen and Sulfur. When 153.90 grams of the compound is burned in a combustion apparatus, 222.00 grams of CO2 and 113.52 grams of water are collected. What is the empirical formula of this compound? (5pts) 29. How many molecules of hydrogen are there in 3.00 g of ethanol, CH3CH2OH? (3pts)

Answer 1

Solution:

28 Solution: First, determine the mass of tog carbon in 222 g of co2 by converting to mol of Carbon, - mol of carbon - 2222/00, x 1 mol g102 x 1 mol of carbon 44-010 ges inol of log 222 mol of carbon mol.wt of co= 44.010 g = 5:044 mol of a atomic wt of t= 12.011 g Now, gram of carbon, gram of carbon : 5.044 mol of cx 120g oft Imolofc : 60.589 of carbon 44.010 mo How simillar way we will calculate mass of H in 113.52g water mole of H = 113.52 of 140 x 1 mol tho da molt I 18.02 8 1 mol H2O ? 113.52x2 mol B 18.02 = 6.299x2 mol H = 12.598 Now gram of H in gram of t = 6.299 molt x 1.008 g molt - 6-3.5 g Now given compound contains, Carbon, Hychrogen and Sulfur only. And mass of CxHy is given as 153.90g As above we have calculated mass of c and it in grams as co 60.589 1 6.299 6.35g

. Now we need to calculate mass of s in the 153.90g of sample .. mass of sample s = 153.90 g of sample - 60.589 of C - 6-35g of H 86.979 Now we need to calculate & composition of each element in sample thus, % C = 60.58 g x 100 = 0.3936 x 100 = 39.36% 153.90 153.90g % H = 6.35 g 153.909 x 100 = 0.0412 x 100 % s = 86.97 9 x 100 153.90g = 0.5651 x 100 = 56.517 Now, we have carbon and Hydrogen in moles but not yet Calculated moles of sulfur. thus, atomic wt of s is mol of us = 86.97 gofs t I mots s. - 32.065g T 32.0658/mol 86.97 mol ofs. 32.065 2.71 moln Now we have mol of c = 5.044 mol of t = 6.20 6.299 12.598 mol of s = 2.71 from the number of mol obtained at a tentative empirical formula is 05.044 412.598 52.71 stesston Ch. 044 6.299 2.7 bolon motes. next we need to round off all the digts, thus we will dinde eath motes q elements by Smallst number.

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