**Backpackers often use canisters of white gas to fuel a cooking stove’s burner. If one canister contains 1.45 L of white gas, and the density of the gas is 0.710 g/cm^3, what is the mass of the fuel in kilograms?**

The above problem was addressing the topic of conversions in chapter 1. After reading the factors given, it was clear that the solution would come from the formula d = m/v (density = mass/volume). I first went to the conversion tables at 1.2 and 1.3 in the book to obtain the proper rates for the calculations. I have attached both tables below. Since the problem gave the values for volume (1.45 L) and density (0.710 g/cm^3), the equation would have to be manipulated to solve for mass.

The new equation would then be m = d x v. Now that I had the correctly formatted equation, I noticed that the measurements given were not directly convertible. Thus, I would first have to convert the 1.45 L of white gas to mL’s. I then had to convert the mL’s to cm^3 in order to find the find the density in g/cm^3. The result left me with 1,450 which I converted to g (grams) by multiplying it by 0.710 from the conversion tables noted above. This left me with a result of 1,029.5 g which I converted to kg (kilograms) by multiplying by 1 kg/1000g. my final answer to the mass was 1.03 kg. The equation described in this paragraph is notated below.

1.45 L = (1 mL/10^-3 L) x (1 cm^3/1 mL) x (0.710/1 cm^3) x (1 kg/1000g)

I have gone back and forth as to whether my answer as it stands (1.03 kg) is correctly displayed or if it should be written in powers of 10 (i.e. .010 x 10^-2 kg). Any guidance would be greatly appreciated.

My work is attached below.