Combustion analysis of 0.213 g of a compound containing C, H, and O produces 0.468 g of CO2 and 0.255 g of H2O. Mass spectral analysis shows that the compound has a molar mass around 120 g mol-1. What is the compound's:
Empirical Formula:(-------) Molecular Formula:(--------)
g C = 0.468 x (12.01/44.01) = 0.1277 g C.
g H = 0.255 x (2/18.015) = 0.0283 g H.
g O = 0.213 - 0.1277 - 0.2831 =0.057 g O.
Now convert those to moles.
0.1277 g C/atomic mass C = ??
0.0283 g H/atomic mass H = ??
0.057/atomic mass O = ??
Now find the ratio of these to each other in small whole numbers. The easy way to do that is to divide the smallest number by itself which assures you of getting 1.000 for that value, then divide the other numbers by that same small number. Finally, round to whole numbers and you should have the empirical formula.
To find the molecular formula, you want to know how many units of the empirical formula are linked together to make the molecule. To determine this, just divide 120 (the molar mass) by the empirical formula mass, and round to a whole number. For example, just suppose the empirical formula turns out to be C3H8O, the empirical formula mass is 60 (3*C+8*H+1*O), Then 120/60 = 2 so you know the molecular formula is (C3H8O)2 or you can write it as C6H16O2.
To find the compound's empirical and molecular formulas, we need to follow several steps:
Step 1: Determine the number of moles of CO2 and H2O produced.
Moles of CO2 = Mass of CO2 / Molar mass of CO2
Moles of H2O = Mass of H2O / Molar mass of H2O
Given:
Mass of CO2 = 0.468 g
Molar mass of CO2 = 44.01 g/mol
Mass of H2O = 0.255 g
Molar mass of H2O = 18.02 g/mol
Moles of CO2 = 0.468 g / 44.01 g/mol ≈ 0.0106 mol
Moles of H2O = 0.255 g / 18.02 g/mol ≈ 0.0141 mol
Step 2: Determine the empirical formula by finding the mole ratios of the elements.
We need to find the simplest whole number ratio of carbon, hydrogen, and oxygen atoms in the compound.
Divide the moles of each element by the smallest number of moles obtained.
Smallest number of moles = 0.0106 (moles of CO2)
Moles of C = 0.0106 / 0.0106 = 1
Moles of H = (0.0141 / 0.0106) ≈ 1.33
Moles of O = 0.0106 / 0.0106 = 1
To obtain whole numbers, multiply the moles of each element by the same factor that will give the simplest whole number ratio.
Multiply the moles of H by 3 to get a whole number:
Moles of C = 1
Moles of H = 1.33 * 3 ≈ 4
Moles of O = 1
Simplified empirical formula: CH4O
Step 3: Determine the molecular formula by finding the ratio of the empirical formula mass to the molar mass of the compound.
Empirical formula mass = (12.01 g/mol) + (1.01 g/mol) + (16.00 g/mol) = 29.02 g/mol
Calculate the factor to multiply the empirical formula by to obtain the molar mass of the compound:
Factor = Molar mass of the compound / Empirical formula mass
= 120 g/mol / 29.02 g/mol
≈ 4.13
Multiply the empirical formula by the factor calculated:
Molecular formula: (CH4O) * 4.13 = C4H16O
Therefore, the compound's empirical formula is CH4O, and its molecular formula is C4H16O.