An organic liquid is a mixture of methyl alcohol (CH3OH) and ethyl alcohol (C2H2OH). A 0.220-sample of the liquid is burned in an excess of and yields 0.343 (carbon dioxide).
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?
I've tried to solve this problem multiple ways and I'm just not getting the right answer. I don't think I really understand what I'm supposed to be doing at all. I want to actually understand what I'm doing.
Masses of Components in a Mixture
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GrahamKemp
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Solving Two Unknowns From Two Linear Equations.
Note: Ethyl Alcohol (Ethanol) is C2H5OH or CH3CH2OH.
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Given Values:
Mass of Sample: ms = 0.220 g
Mass of CO2: m0 = 0.343 g
Other Known Values:
Molar Masses of compounds of interest:
Carbon dioxide, M0 = 44.00964 ± 0.00003 g/mol.
Methanol, M1 = 32.04200 ± 0.00002 g/mol.
Ethanol, M2 = 46.06867 ± 0.00004 g/mol.
Relationship Between Unknown Amounts:
Alkanols burned in excess of O2 go to carbon dioxide and water. Balance individual reactions then combine in terms of unknown amounts, where n1=m1/M1 etc.:
n1 CH3OH + 1½n1 O2 = n1 CO2 + 2n1 H2O
n2 CH3CH2OH + 3n2 O2 = 2n2 CO2 + 3n2 H2O
================================================================
n1 CH3OH + n2 CH3CH2OH + (1½n1+3n2) O2 = (n1+2n2) CO2 + (2n1+3n2) H2O
1. Expressing the mass of carbon dioxide produced in terms of each reagent
From the above relationship, the amount of CO2 is related to the amount of the alcohols: n0=(n1+2n2). So:
m0 = M0((m1/M1)+(2m2/M2))
2. Expressing the mass of sample burned in terms of each reagent.
It is simply the sum of their mass:
ms = m1+m2
What is the mass of methyl alcohol (CH3OH) in the sample?
We have two linear equations of two unknowns (m1 and m2). Rearranging the second equation to give m2=(ms-m1), use this to substitute into the first equation, and rearrange that to solve for m1.
m0 = M0((m1/M1) + (2(ms-m1)/M2) )
<=> m1 = ((m0/M0)-(2ms/M2))÷((1/M1)-(2/M2))
Thus: m1 = ((0.343/44.00964)-(2*0.220/46.06867))÷((1/32.04200)-(2/46.06867))
.: m1 = 0.144 g
Additionally: Since m2=(ms-m1).
.: m2 = 0.076 g
SO:
Mass of Methanol: 0.144g, Mass of Ethanol: 0.076g.
------------------------
Given Values:
Mass of Sample: ms = 0.220 g
Mass of CO2: m0 = 0.343 g
Other Known Values:
Molar Masses of compounds of interest:
Carbon dioxide, M0 = 44.00964 ± 0.00003 g/mol.
Methanol, M1 = 32.04200 ± 0.00002 g/mol.
Ethanol, M2 = 46.06867 ± 0.00004 g/mol.
Relationship Between Unknown Amounts:
Alkanols burned in excess of O2 go to carbon dioxide and water. Balance individual reactions then combine in terms of unknown amounts, where n1=m1/M1 etc.:
n1 CH3OH + 1½n1 O2 = n1 CO2 + 2n1 H2O
n2 CH3CH2OH + 3n2 O2 = 2n2 CO2 + 3n2 H2O
================================================================
n1 CH3OH + n2 CH3CH2OH + (1½n1+3n2) O2 = (n1+2n2) CO2 + (2n1+3n2) H2O
1. Expressing the mass of carbon dioxide produced in terms of each reagent
From the above relationship, the amount of CO2 is related to the amount of the alcohols: n0=(n1+2n2). So:
m0 = M0((m1/M1)+(2m2/M2))
2. Expressing the mass of sample burned in terms of each reagent.
It is simply the sum of their mass:
ms = m1+m2
What is the mass of methyl alcohol (CH3OH) in the sample?
We have two linear equations of two unknowns (m1 and m2). Rearranging the second equation to give m2=(ms-m1), use this to substitute into the first equation, and rearrange that to solve for m1.
m0 = M0((m1/M1) + (2(ms-m1)/M2) )
<=> m1 = ((m0/M0)-(2ms/M2))÷((1/M1)-(2/M2))
Thus: m1 = ((0.343/44.00964)-(2*0.220/46.06867))÷((1/32.04200)-(2/46.06867))
.: m1 = 0.144 g
Additionally: Since m2=(ms-m1).
.: m2 = 0.076 g
SO:
Mass of Methanol: 0.144g, Mass of Ethanol: 0.076g.