Question

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?

Answer 1

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.

Answer 2

Butyric acid contains carbon, hydrogen and oxygen, as seen from the fact that the products of the combustion are `CO_2` and `H_2O`.

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

moles `CO_2= 0.999 g -: 44.0 g/"mol"=0.0227 "mol"`

Since `CO_2` 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 `H_2O= 0.409 g-:18.0 g/"mol" = 0.0227 "mol"`

However, since each molecule of `H_2O` contains two H atoms, the amount of H in this water, and also in the butyric acid is `2xx0.0227=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 `0.500 - 0.3178 = 0.1822 g` must be oxygen in the original compound.

moles O = `0.1822 g -: 16.0 g/"mol" = 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 `C_2H_4O`