Question

A-CH3SH reacts with CO according to the following unbalanced chemical equation:

CH3SH + CO → CH3CO(SCH3) + H2S

Calculate the mass in grams of the excess reagent remaining after the complete reaction of 2.17 g of CH3SH with 1.45 g of CO. ( )

B- a compound containing only C, H and O was subjected to combustion analysis. A sample of 4.270×10¹ g produced 9.200×10¹ g of CO₂ and 3.766×10¹ g of H₂O. Determine the empirical formula of the compound and enter the appropriate subscript after each element.

C( ) H( ) O( )

If the molar mass of the compound is 204.266 g/mol, determine the molecular formula of the compound and enter the appropriate subscript after each element.
C( ) H( ) O( )

Answer 1

(A): mass of CH3SH = 2.17 g

molecular mass of CH3SH = 48.1 gmol^-1

Hence moles of CH3SH = 2.17g / 48.1 gmol^-1​ = 0.0451 mol

mass of CO = 1.45 g

molecular mass of CO = 28.0 gmol^-1

Hence moles of CH3SH = 1.45g / 28.0 gmol^-1​ = 0.0518 mol

The balanced equation for the reaction of CH3SH with CO is

2CH3SH + CO -- > CH3CO(SCH3) + H2S

stoichiometric mol: 2 mol 1 mol 1 mol 1 mol

Here first we need to find the limiting reactant.

2 mol of CH3SH reacts with the 1 mol of CO.

Hence 0.0451 mol of CH3SH that will react with the moles of CO

= (0.0451 mol CH3SH) x ( 1 mol CO / 2 mol CH3SH)

= 0.02255 mol CO

Moles of CO initially taken = 0.0518 mol

Hence moles of CO remain unreacted = 0.0518 - 0.02255 = 0.02925 mol CO

Hence mass of CO remain unreacted = 0.02925 mol x 28.0 g/mol = 0.819 g (answer)

(Note that CH3SH is the limiting reactant here)

B:Let the organic cmpound be CxHyOz

Moles of the organic compound = mass/molecular mass = 4.270*10¹ g / 204.266 g/mol = 0.2090 mol

Moles of CO2 produced =  mass/molecular mass of CO2 = 9.200*10¹ g / 44.0 g/mol = 2.09 mol

Moles of H2O produced =  mass/molecular mass of H2O = 3.766*10¹ g / 18.0 g/mol = 2.09 mol

0.2090 mol of CxHyOz produces the moles of CO2 = 2.09 mol

Hence 1 mol of CxHyOz that will produce the moles of CO2 = 2.09 mol / 0.2090 mol = 10

Hence number of C atoms in CxHyOz, x = 10

Also

0.2090 mol of CxHyOz produces the moles of H2O = 2.09 mol

1 mol of H2O contains 2 mol of H atom.

Hence 1 mol of CxHyOz that will produce the moles of H-atom = 2 x 2.09 mol / 0.2090 mol = 20

Hence total number of Hydrogen atoms in CxHyOz = 20

Given molecular mass of the compound = 204.266 g/mol

Also the molecular mass of the compound can be calculated as M = 12x + 20y + 16z = 204.266 g/mol

=> 12*10+20*1+ 16z = 204.266

=> z = 64.266/16 = 4

Hence number of O-atoms = 4

Hence Empirical formulae = C10H20O4

Emprircal formullae mass = 10*12+20*1+16*4 = 204.266 g/mol

Since molecular mass= empirical formulae mass. Hence molecular formulae of the compound is also C10H20O4