When 0.500 g of butyric acid is combusted completely in excess oxygen, and it produces 0.999 g of carbon dioxide and 0.409 g of water, how do you find the empirical formula of butyric acid?

I know the empirical formula now, but how do you get to it with only this information?

2 Answers
Oct 21, 2017

C2H4O

Explanation:

The idea is to calculate the ratio of moles of each element. In order to do so we must first calculate the % of the mass of the C & H to arrive at the amount of O in a molecule of butyric acid:

%C in CO2 = 12g/m / 44g/m = 27.3% C
%H in H2O = 2g/m / 18g/m = 11.1% H

The %O must be what's left over: 100% - (27.3% + 11.1%) = 61.6% O

g C: 0.999g x 27.3% = 27.3g
g H: 0.409g x 11.1% = 0.0454g
g O: 0.500 - (27.3g + 0.0454g) = 0.182g

Converting to moles (using atomic weights):

moles C: 27.3g / 12g/m = 0.0227mole C
moles H: 0.0454g / 1g/m = 0.0454mole H
moles O: 0.182g / 16g/m = 0.0114mole O

Dividing each by the least number of moles to get the lowest whole number ratio of atoms:

0.0227mole C / 0.0114mole = 2 carbon atoms
0.0454mole H / 0.0114mole = 4 hydrogen atoms
0.0114mole O / 0.0114mole = 1 oxygen atom

Hence the empirical formula of butyric acid is C2H4O.

Butyric acid's molecular formula is twice that, or: C4H8O2:

H3C-CH2-CH2-CO2H is it's structure. It is the active principle in spoiled milk & butter.

Oct 24, 2017

Butyric acid contains carbon, hydrogen and oxygen, as seen from the fact that the products of the combustion are CO2 and H2O.

In any empirical formula, the first step is to convert all given masses into moles:

moles CO2=0.999g÷44.0gmol=0.0227mol

Since CO2 molecules contains one C atom, the number of moles of carbon in this carbon dioxide and also originally in the butyric acid must 0.0227 mol.

Similarly,

moles H2O=0.409g÷18.0gmol=0.0227mol

However, since each molecule of H2O contains two H atoms, the amount of H in this water, and also in the butyric acid is 2×0.0227=0.0454 mol.

So, we have determined that butyric acid contains 0.0227 mol C and 0.0454 mol H.

Now, determine the mass of these elements

#0.0227 "mol" xx 12 g/"mol" = 0.2724 g of C

#0.0454 "mol" xx 1.0 g/ "mol" = 0.0454 g of H

This accounts for 0.3178 g of the butyric acid sample. The remaining 0.5000.3178=0.1822g must be oxygen in the original compound.

moles O = 0.1822g÷16.0gmol=0.0114 mol

Now, we can establish a ratio:

moles C : moles H : moles O = 0.0227 : 0.0454 : 0.0114

Divide each value by the smallest (0.0114) to get a more recognizable ratio:

C : H : O = 2 : 4 : 1

So, the empirical formula is C2H4O