Answer:
C5 H10 O3
Step-by-step explanation:
To find the empirical formula of something, you need to find out how many moles of each element there are in it.
We know that there are 0.01962 moles of CO2 and all of the carbon in the original compound winds up in the CO2, so let's use that to help us find the moles of carbon in the original compound. We will also need to calculate the mass of carbon to find out how much oxygen is in the original compound.
The set-up will look like this: moles of CO2 ⇒ moles of carbon ⇒ grams of carbon
For every mole of CO2, there is one mole of carbon, so there are 0.01962 moles of carbon.
(0.01962 moles C) * (12 g/ 1 mole) = 0.23544 g of carbon in original sample
Let's do the same thing with the hydrogen using the H2O. However, remember that there are 2 moles of H for every 1 mole of water.
(0.01961 moles H2O) * (2 moles H / 1 mole H2O) = 0.03922 moles H
Now find the mass of hydrogen that was in the original compound.
(0.03922 moles H) * (1.008 g / 1 mole H) = 0.039534 g of hydrogen
Using the original mass (0.4647 g), let's subtract the mass of carbon and hydrogen to find how much of the mass of the compound was oxygen.
0.4647 g - (0.039534 g H + 0.23544 g C) = 0.189726 g O
Using the molar mass of oxygen we can convert the grams of O to moles of O.
(0.189726 g O) * (1 mole / 16 g) = 0.011858 moles O
So, we have 0.01962 moles C, 0.03922 moles H, and 0.011858 moles O
To find the empirical formula, divide all the moles by the smallest number of moles.
0.01962/0.011858 = 1.65
0.03922/0.011858 = 3.3
0.011858/0.011858 = 1
From this we get C1.65 H3.3 O1 and if we multiply it all by 3 we get C5H10O3
Combustion analysis can be pretty challenging, but just remember to find the moles of carbon from the moles of CO2, the moles of H from the moles of water, and the moles of O by subtracting the masses of C and H from the original mass and then converting that mass to moles.
Once you have the moles of everything, you can divided each of the moles of the smallest number of moles (in this case it was oxygen) and then multiplying it by a number that gets you close to whole numbers.