Answer: The molecular formula of the compound is ![C_3H_{12}O_3](https://tex.z-dn.net/?f=C_3H_%7B12%7DO_3)
Explanation:
If percentage are given then we are taking total mass is 100 grams.
So, the mass of each element is equal to the percentage given.
Mass of C = 37.5 g
Mass of H = 12.5 g
Mass of O = 50.0 g
Step 1 : convert given masses into moles.
Moles of C =![\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{37.5g}{12g/mole}=3.125moles](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%7B%20given%20mass%20of%20C%7D%7D%7B%5Ctext%7B%20molar%20mass%20of%20C%7D%7D%3D%20%5Cfrac%7B37.5g%7D%7B12g%2Fmole%7D%3D3.125moles)
Moles of H = ![\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{12.5g}{1g/mole}=12.5moles](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%7B%20given%20mass%20of%20H%7D%7D%7B%5Ctext%7B%20molar%20mass%20of%20H%7D%7D%3D%20%5Cfrac%7B12.5g%7D%7B1g%2Fmole%7D%3D12.5moles)
Moles of O = ![\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{50.0g}{16g/mole}=3.125moles](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%7B%20given%20mass%20of%20O%7D%7D%7B%5Ctext%7B%20molar%20mass%20of%20O%7D%7D%3D%20%5Cfrac%7B50.0g%7D%7B16g%2Fmole%7D%3D3.125moles)
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C =![\frac{3.125}{3.125}=1](https://tex.z-dn.net/?f=%5Cfrac%7B3.125%7D%7B3.125%7D%3D1)
For H= ![\frac{12.5}{3.125}=4](https://tex.z-dn.net/?f=%5Cfrac%7B12.5%7D%7B3.125%7D%3D4)
For O =![\frac{3.125}{3.125}=1](https://tex.z-dn.net/?f=%5Cfrac%7B3.125%7D%7B3.125%7D%3D1)
The ratio of C: H: O = 1: 4: 1
Hence the empirical formula is
.
The empirical weight of
= 1(12)+4(1)+1(16)= 32g.
The molecular weight = 93.0 g/mole
Now we have to calculate the molecular formula.
![n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{93}{32}=3](https://tex.z-dn.net/?f=n%3D%5Cfrac%7B%5Ctext%7BMolecular%20weight%20%7D%7D%7B%5Ctext%7BEquivalent%20weight%7D%7D%3D%5Cfrac%7B93%7D%7B32%7D%3D3)
The molecular formula will be=![3\times CH_4O=C_3H_{12}O_3](https://tex.z-dn.net/?f=3%5Ctimes%20CH_4O%3DC_3H_%7B12%7DO_3)