<p>A 8.8g mixture of methanol and ethanol is completely combusted with oxygen and yields 14.020 g of carbon dioxide gas. </p>
<p>2CH3OH + 3O2 -> 2CO2 + 4H2O
CH3CH2OH + 3O2 -> 2CO2 + 3H2O</p>
<p>Calculate the percent composition by mass of methanol in the mixture</p>
<p>Molar mass of methanol = 32.0426 g
Molar mass of ethanol = 46.0694 g</p>
<p>Simple question forgot how to do it, can anyone show me the setup and how to solve this?
Please be detailed, but dont need to show exact numerical answer if u dont wanna</p>
<p>Is it system of equations? This is definitely not simple…</p>
<p>Edit: OK, here’s what I got. All numbers are approximate but I did all rounding after final calculations.</p>
<p>0.319 mol Co2 produced. 0.319 = moles methanol + moles ethanol * 2
.319 = m + 2e</p>
<p>32.0426 g * moles methanol + 46.0694 g * moles ethanol = 8.8 g
32.0426m + 46.0694e = 8.8</p>
<p>Solve the system of equations you get moles ethanol = 0.0781 and moles methanol = 0.1623.
grams ethanol = 3.600
grams methanol = 5.200</p>
<p>5.200 / 8.8 * 100%=<br>
59% (2 significant digits)</p>
<p>Is there a better way to do this? I checked my answer, I believe it’s right, but I don’t recall ever having to solve systems of equations for chem…</p>
<p><em>bumps someone else’s thread</em></p>
<p>Is there a better way to do this problem? I checked my answer, I believe it’s right, but I don’t recall ever having to solve systems of equations for chemistry…</p>
<p>From what I can tell, you know from the equations that 2 mol methanol –> 2 mol CO2, and 1 mol ethanol –> 2 mol CO2. We have 14.02g CO2 gas, which is roughly 1/3 mol CO2. Therefore, we need some combination of methanol and ethanol that will yield 1/3 mol CO2. So we can have 1/6 mol methanol and 1/12 mol ethanol to –> 1/3 mol CO2. 1/6 mol methanol = ~32/6, or 16/3, or 5.2. 5.2/8.8 is similar to bobtheboy’s answer, from what I can tell, and simpler. Hopefully all the rounding would still be accurate enough on an MC…?</p>
<p>Lol I love chemistry so much</p>
<p>@heinochus, but that method would not cut it on the FRQ would it? You may get the answer, but you have to show proper work. Is the only way to properly do it without guess and check by using system of equations?</p>
<p>@bob’s method i believe is how i learned it.</p>
<p>I was assuming that this was a MC :# guess not</p>
<p>Well, I just wanted to ensure correct methodology.
Although if it was an MC, your way might be better.</p>
<p>Haha, I just did and it comes out to 59.09%, which is basically exactly what you got.</p>
<p>[the haha isn’t a gloating one, I just thought it was kind of funny.]</p>