Combustion analysis of 0.6943 g of an unknown compound containing carbon, hydrogen, and oxygen, produced 1.471 g CO2 and 0.226 g H2O. Determine the molecular formula, given the molar mass is 166 g/ mol.
Answer – We are given, mass of compound = 0.6943 g
Mass of CO2 = 1.471 g , mas of H2O = 0.226 g , molecular mass = 166 g/mol
First we need to calculate moles of CO2 and H2O from the given mass
Moles of CO2 = 1.471 g / 44.0 g.mol-1 = 0.0334 mole
Moles of H2O = 0.226 g / 18.015 g.mol-1 = 0.0125 moles
Moles of C from the moles of CO2
1 moles of CO2 = 1 moles of C
So, 0.0344 moles of CO2 = ?
= 0.0334 moles of C
Moles of H from moles of H2O
1 moles of H2O = 2 moles of H
So, 0.0125 moles of H2O = ?
= 0.02521 moles of H
Mass of C = 0.0334 moles * 12.011 g/mol
=0.402 g of C
Mass of H = 0.0251 moles * 1.0079 g/mol
= 0.0253 g of H
Total mass of compound = mass of C + mass of H + mass of O
0.6943 g = 0.402 g + 0.0253 g + mass of O
Mass of O = 0.6943 g – 0.402 g – 0.0253 g
= 0.267 g of O
Moles of O = 0.267 g / 15.998 g.mol-1
= 0.0167 moles of O
So moles of O both are smallest, so we need to divide each mole by this mole
So , C = 0.0334 /0.0167 = 2
H = 0.0251 / 0.0167 = 1.5
O = 0.0167 /0.0167 = 1
Now we need to multiply each by 2 for whole number of H
C = 2*2 = 4
H = 1.5 *2 = 3
O = 1*2 = 2
Empirical formula is C4H3O2
Molecular formula = n * empirical formula
n = molecular formula mass / empirical formula mass
= 166 / 83
= 2
So molecular formula = 2 * C4H3O2
= C8H6O4
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