Question

What formula should I use given these variables? I have converted Celcius to Kelvin and mmhg to pressure.

A sample of octane, (C8H18, d = 0.702 g/mL), was completely reacted in excess oxygen, and it yielded 2.70 L of carbon dioxide at 701 mmHg and 35.0 Celcius. What is the volume of the original sample?

Answer 1

You know from the general formula for a complete combustion reaction that the number of CO2 molecules produced is the same as the number of carbon atoms in the hydrocarbon that was burned. So in this case, the ratio of C8H18 to CO2 is 1 to 8. If we assume temperature is the same throughout the reaction, then to keep pressure the same before and after the reaction, the volume must increase by a factor of 8 from reactants to products. In other words, the volume of the reactant is ⅛ that of the product, in this case, 2.70 L / 8 = 0.3375 L.

If you want to use a formula to confirm this, let's look at the ideal gas law, PV=nRT. I'm going to work it out step by step:

Before the reaction, \[P_{1}V_{1}=n_{1}R_{1}T_{1}\]
And after the reaction, \[P_{2}V_{2}=n_{2}R_{2}T_{2}\]
We know R1=R2 because it's the gas constant, so we can actually set these two equations equal to each other:
\[\frac{P_{1}V_{1}}{n_{1}T_{1}}=\frac{P_{2}V_{2}}{n_{2}T_{2}}\]
We're assuming temperature and pressure are constant throughout this reaction, meaning P1=P2 and T1=T2. That means we can take both these variables out of the equation:

\[\frac{V_{1}}{n_{1}}=\frac{V_{2}}{n_{2}}\]
We know from what I said before that n2=8*n1. So we can sub that into the equation:
\[\frac{V_{1}}{n_{1}}=\frac{V_{2}}{8n_{1}}\]
Notice though that we can now get rid of the n1 in the denominator, leaving:
\[V_{1}=\frac{V_{2}}{8}\]
The volume before the reaction is ⅛ the volume after the reaction! And we didn't have to sub in any numbers, or do any conversions! Keep in mind this only works because we're only looking at 1 reactant and 1 product (based on carbon atoms), so you can't use this strategy to talk about the whole reaction unless you tweak things a bit.

Anyways...hope that makes sense! I you have any questions let me know!

Answer 2

Hm that's not quite right. You didn't take into account the density.

First find moles of CO2,

\(\sf n_{CO_2}=\dfrac{PV}{RT}=\dfrac{[2.7~L*\dfrac{701}{760}atm]}{[(35+273)K*0.08205736 L*atm/K*mol]}=0.09854~mol\)

From the balanced equation (not shown), the ratio relating octane to carbon dioxide is:

\(\sf \dfrac{n_{octane}}{1}=\dfrac{n_{CO_2}}{8}\rightarrow \sf n_{octane}=\dfrac{0.09854~mol}{8}=0.01232~mol\)

Find mass:

\(\sf m_{octane}=M_{octane}*n_{octane}=114.23 g/mol*0.01232~mol=1.41~g\)

Find volume:

\(\sf \rho=\dfrac{m}{V}\rightarrow V=\dfrac{m}{\rho}=\dfrac{1.41~g}{0.702~g/mL}=2 ~mL\)

Answer 3

I guess your mistake was to assume that octane was a gas at 35 celsius but it's not, it's a liquid.

Answer 4

You're right - that was my mistake. Thanks for correcting it!

Answer 5

no probs!