<u>Answer:</u> The empirical formula for the given compound is ![C_3H_6O](https://tex.z-dn.net/?f=C_3H_6O)
<u>Explanation:</u>
The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:
![C_xH_yO_z+O_2\rightarrow CO_2+H_2O](https://tex.z-dn.net/?f=C_xH_yO_z%2BO_2%5Crightarrow%20CO_2%2BH_2O)
where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.
We are given:
Conversion factor: 1 g = 1000 mg
Mass of ![CO_2=6.32mg=0.00632g](https://tex.z-dn.net/?f=CO_2%3D6.32mg%3D0.00632g)
Mass of ![H_2O=2.58g=0.00258g](https://tex.z-dn.net/?f=H_2O%3D2.58g%3D0.00258g)
Mass of compound = 2.78 mg = 0.00278 g
We know that:
Molar mass of carbon dioxide = 44 g/mol
Molar mass of water = 18 g/mol
- <u>For calculating the mass of carbon:</u>
In 44g of carbon dioxide, 12 g of carbon is contained.
So, in 0.00632 g of carbon dioxide,
of carbon will be contained.
- <u>For calculating the mass of hydrogen:</u>
In 18g of water, 2 g of hydrogen is contained.
So, in 0.00258 g of water,
of hydrogen will be contained.
- Mass of oxygen in the compound = (0.00278) - (0.00172 + 0.000286) = 0.000774 g
To formulate the empirical formula, we need to follow some steps:
- <u>Step 1:</u> Converting the given masses into moles.
Moles of Carbon =![\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.00172g}{12g/mole}=1.43\times 10^{-4}moles](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%7BGiven%20mass%20of%20Carbon%7D%7D%7B%5Ctext%7BMolar%20mass%20of%20Carbon%7D%7D%3D%5Cfrac%7B0.00172g%7D%7B12g%2Fmole%7D%3D1.43%5Ctimes%2010%5E%7B-4%7Dmoles)
Moles of Hydrogen = ![\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.000286g}{1g/mole}=2.86\times 10^{-4}moles](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%7BGiven%20mass%20of%20Hydrogen%7D%7D%7B%5Ctext%7BMolar%20mass%20of%20Hydrogen%7D%7D%3D%5Cfrac%7B0.000286g%7D%7B1g%2Fmole%7D%3D2.86%5Ctimes%2010%5E%7B-4%7Dmoles)
Moles of Oxygen = ![\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.000774g}{16g/mole}=4.83\times 10^{-5}moles](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%7BGiven%20mass%20of%20oxygen%7D%7D%7B%5Ctext%7BMolar%20mass%20of%20oxygen%7D%7D%3D%5Cfrac%7B0.000774g%7D%7B16g%2Fmole%7D%3D4.83%5Ctimes%2010%5E%7B-5%7Dmoles)
- <u>Step 2:</u> Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is ![4.83\times 10^{-5}mol](https://tex.z-dn.net/?f=4.83%5Ctimes%2010%5E%7B-5%7Dmol)
For Carbon = ![\frac{1.43\times 10^{-4}}{4.83\times 10^{-5}}=2.96\approx 3](https://tex.z-dn.net/?f=%5Cfrac%7B1.43%5Ctimes%2010%5E%7B-4%7D%7D%7B4.83%5Ctimes%2010%5E%7B-5%7D%7D%3D2.96%5Capprox%203)
For Hydrogen = ![\frac{2.86\times 10^{-4}}{4.83\times 10^{-5}}=5.92\approx 6](https://tex.z-dn.net/?f=%5Cfrac%7B2.86%5Ctimes%2010%5E%7B-4%7D%7D%7B4.83%5Ctimes%2010%5E%7B-5%7D%7D%3D5.92%5Capprox%206)
For Oxygen = ![\frac{4.83\times 10^{-5}}{4.83\times 10^{-5}}=1](https://tex.z-dn.net/?f=%5Cfrac%7B4.83%5Ctimes%2010%5E%7B-5%7D%7D%7B4.83%5Ctimes%2010%5E%7B-5%7D%7D%3D1)
- <u>Step 3:</u> Taking the mole ratio as their subscripts.
The ratio of C : H : O = 3 : 6 : 1
Hence, the empirical formula for the given compound is ![C_3H_{6}O_1=C_3H_6O](https://tex.z-dn.net/?f=C_3H_%7B6%7DO_1%3DC_3H_6O)