Chemical reaction of octane burning: 2C_8H_18 + 25O_2 -> 18H_2O + 16CO_2. Octane has a density of 703.00 kg/m^3, or 703.00 g/L. A gallon is 3.78541 L. What is the mass (in g) of 1.0000 gallon of octane?

Jan 9, 2017

The density of octane turns out to be 2661 g/gal (based on a US gallon), but see the method used to find the answer, below... (this method is what really matters!)

Explanation:

Keeping track of the units as you go through this calculation will guarantee that you get it right.

$703.00 \frac{k g}{m} ^ 3 \times 1000 \frac{g}{k g} \times \frac{1}{1000} {m}^{3} / L \times 3.78541 \frac{L}{g a l}$

Note that after the first value, each remaining number is just a conversion factor, selected to change the units from kg and ${m}^{3}$ to g and gal. The only one you might wonder about is the $\frac{1}{1000} {m}^{3} / L$ term. This was used to change from ${m}^{3}$ to L, since there are 1000 L in a cubic meter.

Once the units work out the way you need them to, the number s will as well. In the end you get #2661 g per gallon.

Jan 9, 2017

The $\text{mass}$ is simply the $\text{volume"xx"density} \ldots \ldots \ldots \ldots$

Explanation:

And you have included all the information you need:

$\text{volume"xx"density} =$

$= 1.0000 \cdot {\cancel{\text{gallon"xx3.78541*cancelL*cancel"gallon}}}^{-} 1 \times 703.00 \cdot g \cdot \cancel{{L}^{-} 1} \cong 2700 \cdot g$

Note (i) that here we are using $\text{US gallons} \equiv 3.785 \cdot L$, whereas an $\text{Imperial Gallon} \equiv 4.546 \cdot L$.

Note (ii) that I included the units in the calculation because it is a check on our calculation. It is all too easy to divide when you should multiply or vice versa. We have all done this. That the units of the calculation gives final units of $\text{mass}$ persuades me that I have performed the operation in the correct way (FOR ONCE!!!).