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

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

Step-by-step 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* 10^(-3)g`

Mass of
`H_2O=3.87mg=3.87* 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* 10^(-3)g` of carbon dioxide,
`(12)/(44)* 9.48* 10^(-3)=2.58* 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* 10^(-3)g` of water,
`(2)/(18)* 3.87* 10^(-3)=4.30* 10^(-4)g` of hydrogen will be contained.

For calculating the mass of oxygen:

Mass of oxygen in the compound =
`(4.17* 10^(-3))-[(2.58* 10^(-3))+(4.30* 10^(-4))]=1.16* 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}}=(2.58* 10^(-3)g)/(12g/mole)=2.15* 10^(-4)moles`

Moles of Hydrogen =
`\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=(4.30* 10^(-4)g)/(1g/mole)=4.30* 10^(-4)moles`

Moles of Oxygen =
`\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=(1.16* 10^(-3)g)/(16g/mole)=7.25* 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* 10^(-5)` moles.

For Carbon =
`(2.15* 10^(-4))/(7.25* 10^(-5))=2.96\approx 3`

For Hydrogen =
`(4.30* 10^(-4))/(7.25* 10^(-5))=5.93\approx 6`

For Oxygen =
`(7.25* 10^(-5))/(7.25* 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`

Answer 2

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