Question

An organic liquid is a mixture of methyl alcohol (CH3OH) and ethyl alcohol (C2H5OH). A 0.220-g sample of the liquid is burned in an excess of O2(g) and yields 0.345 g CO2(g) (carbon dioxide). Part A Set up two algebraic equations, one expressing the mass of carbon dioxide produced in terms of each reagent and the other expressing the mass of sample burned in terms of each reagent.What is the mass of methyl alcohol (CH3OH) in the sample?

Answer 1

Answer:

Mass of Methyl alcohol (or methanol) = 0.1403g

Explanation:

Step 1: Given data

Mass of Sample: ms = 0.220 g

Mass of CO2: m0 = 0.345 g

Mass of CH3OH = m1

Mass of C2H5OH = m2

Molar mass CO2 = 44.01 g/mole

Molar mass CH3OH = 32 g/mole

Molar mass C2H5OH 46.07 g/mole

Step 2: Tjhe balanced equation

2 CH3OH + 3 O2 = 2 CO2 + 4 H2O

C2H5OH + 3 O2 =2 CO2 + 3 H2O

------------------------------------------------------

2 CH3OH + C2H5OH + 6 O2 = 4 CO2 + 7 H2O

Step 3: Calculate mass

In the balanced equation we notice that , the amount of CO2 is related to the amount of the alcohols: Moles CO2 = 4

mass of CO2 = Molar mass CO2((mass CH3OH/ Molar mass CH3OH)+(mass C2H5OH/Molar mass C2H5OH))

mass of CO2 = 44.01 ((mass CH3OH/ 32.04) + (mass C2H5OH/ 46.07)

The sample has a mass of 0.220 g = mass CH3OH + mass C2H5OH

mass C2H5OH = 0.220G - mass CH3OH

This we will insert in the equation, so we will only have 1 unknown mass

0.345g = 44.01((m1/32.04) + (2(0.220-m1)/M2) )

<=> m1 = ((mCO2/44.01)-(2*0.22/46.07))÷((1/32.04)-(2/46.07))

Thus: m1 = ((0.345/44.01)-(2*0.220/46.07))÷((1/32.04)-(2/46.07))

Thus m1 = 0.1403 g  = Mass of CH3OH

Since m2=(ms-m1).

m2 = 0.220-0.1403 = 0.0797 g = Mass of C2H5OH

Mass of methyl alcohol (Methanol) = 0.1403g

Mass of ethyl alcohol (Ethanol) = 0.0797g.