A compound contains only carbon and hydrogen. A 0.857 g sample is burned in oxygen to give 1.45 L of carbon dioxide, measured dry at 98.0 kPa and 20.0 degrees C. What is the compound's empirical formula?

Question

anonymous · Answer

This is the formula to calculate the moles of carbon dioxide:
$$n=\frac{PV}{RT}$$
These are the values of the variables:
$$R=8.31\frac{kPaL}{molK}$$
$$T=20C=293.15K$$
$$P=98.0kPa$$
$$V=1.45L$$
Plug them into the formula:
$$n=\frac{98.0*1.45}{8.31*293.15}=0.0583 moles CO_2$$
Calculating Carbon moles:
$$0.0583molesCO_2*\frac{1moleC}{1molesCO_2}=0.0583molesC$$
Now we can use the molecular weight to determine how much of the weight represents carbon:
$$0.0583molesC\frac{12.0107gC}{1moleC}=0.701gC$$

anonymous · Answer

Calculating the amount of hydrogen:
$$0.857g-0.701g=0.156g$$
$$0.156gH\frac{1.0079gH}{1moleH}=0.158molesH$$
Divide by lowest value to determine ratios:
$$\frac{0.158}{0.0583}\approx2.7$$
Rounding up we can determine that we have:
$$CH_3$$

anonymous · Answer

Woops. I made a booboo on the second set of calculations. Should be:
$$0.156gH\frac{1moleH}{1.0079gH}=0.155molesH$$
Note the grams have to cancel.
$$\frac{0.155}{0.0583}=2.65\approx3$$
The answer still ends up being $$CH_3$$
This is based upon feedback I received from my Chem teacher during summer. I was advised that you aim for the smallest set if variables as large molecules usually don't exist in nature.