Question

(a) the characteristic odor of pineapple is due to ethyl butyrate, a compound containing carbon, hydrogen, and oxygen. combustion of 4.17 mg of ethyl butyrate produces 9.48 mg of co2 and 3.87 mg of h2o. what is the empirical formula of the compound?

Answer 1

By stoichiometry and assume that:

CxH2xOy + zO2 -> xCO2 + xH2O 

CO2: 9.48/44 = 0.215 mmol 
H2O: 3.87/18 = 0.215 mmol 
mass of C = 0.215 * 12 = 2.58 mg 
mass of H = 0.215 * 2 * 1 = 0.43 mg 
mass of O in ethylbutyrate = 4.17 - 2.58 - 0.43 = 1.11 mg 
So C/O = 2.58/1.11 ≈ 3 

Thus we have C3H6O

 

Answer 2

Answer: The empirical formula for the given compound is $C_3H_6O$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=9.48mg=9.48\times 10^{-3}g$

Mass of $H_2O=3.87mg=3.87\times 10^{-3}g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in $9.48\times 10^{-3}g$ of carbon dioxide, $\frac{12}{44}\times 9.48\times 10^{-3}=2.58\times 10^{-3}g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in $3.87\times 10^{-3}g$ of water, $\frac{2}{18}\times 3.87\times 10^{-3}=4.30\times 10^{-4}g$ of hydrogen will be contained.

For calculating the mass of oxygen:

Mass of oxygen in the compound = $(4.17\times 10^{-3})-[(2.58\times 10^{-3})+(4.30\times 10^{-4})]=1.16\times 10^{-3}g$

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =$\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{2.58\times 10^{-3}g}{12g/mole}=2.15\times 10^{-4}moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{4.30\times 10^{-4}g}{1g/mole}=4.30\times 10^{-4}moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.16\times 10^{-3}g}{16g/mole}=7.25\times 10^{-5}moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is $7.25\times 10^{-5}$ moles.

For Carbon = $\frac{2.15\times 10^{-4}}{7.25\times 10^{-5}}=2.96\approx 3$

For Hydrogen  = $\frac{4.30\times 10^{-4}}{7.25\times 10^{-5}}=5.93\approx 6$

For Oxygen  = $\frac{7.25\times 10^{-5}}{7.25\times 10^{-5}}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 6 : 1

Hence, the empirical formula for the given compound is $C_3H_6O_1=C_3H_6O$