atChem 111 Gases Show all your work. Sig. figs. must be correct (20 points) A sa
ID: 1084696 • Letter: A
Question
atChem 111 Gases Show all your work. Sig. figs. must be correct (20 points) A sample of a compound that contains carbon, sulfur, hydrogen and oxygen. A 1.200 g of sample is burned completely produced 896.0 mL of co2 at STP and 0.7208 g of water. In another experiment, a 1.500 g sample of the compound produced 100.0 mL 502 at 25.0 °C and 3.05 atm. Calculate the empirical formula. (R 0.0821 L.atm/mol.K) To find the molar mass another experiment was done: 3.57 L of the compound at 2.33 atm, and 25 c weighted 35.0g. Find the molecular formula of the compound.Explanation / Answer
No of mol of CO2 = PV/RT
= (1*0.896)/(0.0821*273.15)
= 0.04 mol
no of mol of C in sample = 0.04 mol
No of mol of H2o formed = 0.7208/18 = 0.04 mol
No of mol of H in sample = 0.04*2 = 0.08 mol
No of mol of SO2 formed = (3.05*0.1)/(0.0821*298.15) = 0.0125 mol
No of mol of S present in 1.2 g sample = 0.0125*1.2/1.5 = 0.01 mol
mass of O in sample = 1.2 - (0.04*12+0.08*1+0.01*32) = 0.32 g
mol of O = 0.32/16 = 0.02 mol
simplest ratio
C = 0.04/0.01 = 4
H = 0.08/0.01 = 8
s = 0.01/0.01 = 1
O = 0.02/0.01 = 2
empirical formula : C4H8SO2
molarmass(w) = wRT/PV
= 35*0.0821*298.15/(2.33*3.57)
= 103 g/mol
n = 103/120 = 0.858
Moleculear formula = (C4H8SO2)0.858
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