I've tried to solve this question multiple ways I just can't seem to get it. I just don't understand what I should be doing.
An organic liquid is a mixture of methyl alcohol () and ethyl alcohol (). 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 () in the sample?
0.220 what?
....burned in an excess of .....and yields 0.343 etc.
0.343 what?
Well, trying to figure out the mass of methyl alcohol in the sample seems like quite a puzzling task. It's like trying to find a needle in a haystack, except the haystack is filled with other needles, and you can't tell them apart. Tricky, isn't it?
But fear not, my friend! We shall venture into the world of algebraic equations to solve this conundrum. Let's break it down step by step.
First, let's consider the mass of carbon dioxide produced. We know that the mass of carbon dioxide is 0.343 grams. So let's call that "CO2" for short.
Next, we need to express the mass of carbon dioxide produced in terms of each reagent. Since we have a mixture of methyl alcohol and ethyl alcohol, let's assign variables to represent their masses. Let's call the mass of methyl alcohol "MA" and the mass of ethyl alcohol "EA".
To set up the first equation, we can say that the mass of carbon dioxide produced is the sum of the masses of carbon dioxide produced from methyl alcohol and ethyl alcohol.
So, the equation would look like this: CO2 = mass of CO2 from MA + mass of CO2 from EA.
Now, onto the second equation, expressing the mass of the sample burned in terms of each reagent. Since we have a 0.220-gram sample of the liquid, we can say that the mass of the sample is equal to the mass of methyl alcohol plus the mass of ethyl alcohol.
So, the equation would look like this: Sample mass = mass of MA + mass of EA.
Now, armed with these two equations, solve for the mass of methyl alcohol by substituting known values and solving the equations simultaneously.
Remember, my friend, with a bit of patience and a lot of perseverance, you shall conquer this puzzling mass mystery!
To solve this problem, we need to set up two algebraic equations. The first equation will express the mass of carbon dioxide produced in terms of each reagent, and the second equation will express the mass of the sample burned in terms of each reagent. Let's break down the problem step by step:
1. Let's denote the mass of methyl alcohol as "m" and the mass of ethyl alcohol as "e".
2. We are given that a 0.220-gram sample of the liquid is burned and yields 0.343 grams of carbon dioxide. Therefore, we can set up the equation:
m + e = 0.220
This equation represents the total mass of the sample burned.
3. Now, we need to determine the mass of carbon dioxide produced in terms of each reagent. Since carbon dioxide is produced from both methyl alcohol and ethyl alcohol, we can assume that the mole ratio between the two alcohols and carbon dioxide is 1:1.
The molar mass of carbon dioxide is 44 g/mol, so the mass of carbon dioxide produced from methyl alcohol is 44m/M, and the mass of carbon dioxide produced from ethyl alcohol is 44e/E, where M and E are the molar masses of methyl alcohol and ethyl alcohol, respectively.
4. We are also given that the total mass of carbon dioxide produced is 0.343 grams. Therefore, we can set up the equation:
44m/M + 44e/E = 0.343
5. Now we have two equations:
m + e = 0.220 (equation 1)
44m/M + 44e/E = 0.343 (equation 2)
6. To solve these equations, we can use substitution or elimination. Let's use substitution in this case.
From equation 1, we can express e in terms of m as e = 0.220 - m.
Substituting this value into equation 2, we get:
44m/M + 44(0.220 - m)/E = 0.343
Simplifying this equation will give us the value of m, the mass of methyl alcohol in the sample.
By rearranging the equation and using the values of M and E (the molar masses of methyl alcohol and ethyl alcohol, respectively), we can calculate the mass of methyl alcohol in the sample.
If I assume 0.220 is grams.
If I assume burned in ..... is burned in excess of oxygen.
If I assume 0.343 is grams. Here is what you do.
Look at the balanced equations.
2CH3OH + 3O2 ==> 2CO2 + 4H2O
C2H5OH + 3O2 ==> 2CO2 + 3H2O
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let x = grams methyl alcohol.
let y = grams ethyl alcohol.
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In general this is what we do and I'll get to the equations in terms of x and y next.
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grams methanol + grams ethanol=0.220g.
grams methanol converted to g CO2 + grams ethanol converted to g CO2 = 0.343
now we put those last two into equation form.;
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x + y = 0.220
(44x/32) + (88y/46) = 0.343
Two equations in two unknowns. Solve simultaneously for x and y.
x is grams methanol.
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Note: Where does the second equation come from?
44x/32 is this
x*(1 mol MeOH/molar mass MeOH)*(1 mo CO2/1 mol MeOH)*(44 g CO2/1 mol CO2) = 0.220.
88y/46 is this.
y*(1 mol EtOH/molar mass EtOH)*(2 mol CO2/1 mol EtOH)*(44 g CO2/ mol CO2) = 0.343