Question

The characteristic odor of pineapple is due to an ester known as ethyl butanoate. It contains C, H, and O. Combustion of 2.78 g of this compound produced 6.32 g of CO2 and 2.58 g of H2O. What is the empirical formula of this compound? If the molecular mass of this compound us 166 amu, what is the molecular formula for this compound?Let's determine the number of grams of carbon, hydrogen, and oxygen by performing a combustion analysis:First, determine the percent of carbon dioxide:CO2= % C= mass of carbon/mass of carbon dioxide x 100% =Second, determine the percent of hydrogen in carbon dioxide:H2O= % H = mass of hydrogen/mass of water x 100%=Using the above percentages, determine the mass of each element in the original 2.78 grams sample:Grams of C=Grams of H=Grams of O (subtract the grams of C and grams of H from the 2.78 g total)=Now, let's convert the number of grams of each element into moles below:-------------------------------------------------------|||______________________________Now, let's divide the number of moles of each element by the smallest number obtained:-------------------------------------------------------|||______________________________What is the empirical formula for this compound?_____________Using the actual molecular mass of 116 amu, determine the molecular formula for this compound:_____________

Answer 1

`\begin{gathered} empirical\text{ formula}=C_3H_6O \\ molecular\text{ formula}=C_6H_(12)O_2 \end{gathered}`

Explanations:

Given the following parameters

Mass of ethyl butanoate = 2.78grams

Mass of CO2 = 6.32grams

Mass of H2O = 2.58grams

Determine the mole of CO2 and H2O

Mole of CO2 = mass/molar mass

Mole of CO2 = 6.32/44.01

Mole of CO2 = 0.144moles

Since there are 1 mole of carbon in CO2, the mass of carbon present will be:

Mass of C = mole * molar mass

Mass of C = 0.144 * 12

Mass of C = 1.728grams

Mole of H2O = mass/molar mass

Mole of H2O = 2.58/18

Mole of H2O = 0.143moles

Since there are 2 atoms of hydrogen in H2O, the required mass of hydrogen will be:

Mass of H = 2(0.143) * 1

Mass of H = 0.287grams

Determine the mass of oxygen

Mass of oxygen =Total mass - (grams of C + grams of H)

Mass of oxygen = 2.78 - (1.728 +0.287)

Mass of oxygen = 0.765grams

Convert the number of grams to moles

Mole of Carbon = mass/molar mass

mole of carbon = 1.728/12 = 0.143moles

Mole of Hydrogen = 0.287/1 = 0.287moles

Mole of Oxygen = 0.77/16 = 0.04813moles

Divide the number of moles of each element by the smallest number obtained

Ratio of Carbon = 0.143/0.04813 = 2.97 ≈ 3

Ratio of Hydrogen = 0.287/0.04813 = 5.963 ≈ 6

Ratio of Oxygen = 0.04813/0.04813 = 1

Hence the empirical formula for the compound is C₃H₆O

Determine the molecular mass of the compound

`\begin{gathered} (C_3H_6O)_n=116 \\ [3(12)+1(6)+16]n=116 \\ (36+6+16)n=116 \\ 58n=116 \\ n=(116)/(58) \\ n=2 \end{gathered}`

The molecular formula will be expressed as:

`(C_3H_6O)_2=C_6H_(12)O_2`