mead Chem. 111 NAME e (LAST) FIRST) CLS Chem 111 Gases Show all your work. Sig.

Question

mead Chem. 111 NAME e (LAST) FIRST) CLS Chem 111 Gases Show all your work. Sig. figs. must be correct (20 points) A sample of a compound that contains carbon, sulfur, hydrogen and oaxygen. 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 S02 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. Molar ma35.

Explanation / Answer

Organic compound + O2 -------> CO2 + H2O

number of moles of CO2 n = PV/RT (from ideal gas equation)

n = 1 * 0.896/(0.0821*273.15)

n = 0.0399 moles

1 mole of Co2 requires 1 mole of C atoms

0.0399 moles of CO2 requires 0.0399 moles of C atoms

mass of carbon in the given compound = 0.0399*12 = 0.4788 g/mol

percentage of Carbon in the given organic compound = 0.4788/1.2 *100 = 39.9%

1 mole of H2O requires 2 moles of H atoms

18 g of H2O requires 2 g of H atoms

0.7208 g of H2O requires 2/18*0.7208 g of H atoms = 0.08 g/mol

Percentage of Hydrogen = 0.08/1.2*100 = 6.67%

number of moles of sulpher n = PV/RT (from ideal gas equation)

n = 3.05*0.1/(0.0821*298)

n = 0.0125 moles

mass of sulpher = 0.0125 * 32 g/mol = 0.4 g

percentage of sulpher = 0.4/1.5*100 = 26.67%

percentage of oxygen = 100 - (39.9+6.67+26.67) = 26.76%

Empirical formula = C4H8SO2

Element %/atomic mass No.of moles simple no.of moles Whole no. ratio C 39.9/12 = 3.325 3.325/0.8335=3.989 4 4 H 6.67/1=6.67 6.67/0.8335=8.002 8 8 S 26.67/32= 0.8335 0.8335/0.8335=1 1 1 O 26.76/16=1.6725 1.6725/0.8335=2.006 2 2

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