Answer: The molecular formula will be ![C_6H_6O_6](https://tex.z-dn.net/?f=C_6H_6O_6)
Explanation:
Mass of
= 17.95 g
Mass of
= 4.87 g
Molar mass of carbon dioxide = 44 g/mol
Molar mass of water = 18 g/mol
For calculating the mass of carbon:
In 44g of carbon dioxide, 12 g of carbon is contained.
So, in 17.95 g of carbon dioxide, =
of carbon will be contained.
For calculating the mass of hydrogen:
In 18g of water, 2 g of hydrogen is contained.
So, in 4.87 g of water, =
of hydrogen will be contained.
Mass of oxygen in the compound = (12.00) - (4.89+0.541) = 6.57 g
Mass of C = 4.89 g
Mass of H = 0.541 g
Mass of O = 6.57 g
Step 1 : convert given masses into moles.
Moles of C =![\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{4.89g}{12g/mole}=0.407moles](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5Ctext%7B%20given%20mass%20of%20C%7D%7D%7B%5Ctext%7B%20molar%20mass%20of%20C%7D%7D%3D%20%5Cfrac%7B4.89g%7D%7B12g%2Fmole%7D%3D0.407moles)
Moles of H=![\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.541g}{1g/mole}=0.541moles](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%7B0.541g%7D%7B1g%2Fmole%7D%3D0.541moles)
Moles of O=![\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{6.57g}{16g/mole}=0.410moles](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%7B6.57g%7D%7B16g%2Fmole%7D%3D0.410moles)
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C =![\frac{0.407}{0.407}=1](https://tex.z-dn.net/?f=%5Cfrac%7B0.407%7D%7B0.407%7D%3D1)
For H =![\frac{0.541}{0.407}=1](https://tex.z-dn.net/?f=%5Cfrac%7B0.541%7D%7B0.407%7D%3D1)
For O = ![\frac{0.410}{0.407}=1](https://tex.z-dn.net/?f=%5Cfrac%7B0.410%7D%7B0.407%7D%3D1)
The ratio of C : H : O = 1: 1 : 1
Hence the empirical formula is
.
Hence the empirical formula is ![CHO](https://tex.z-dn.net/?f=CHO)
The empirical weight of
= 1(12)+1(1)+1(16)= 29 g.
If 0.25 moles has mass of 44.0 g
Thus 1 mole has mass of = ![\frac{44.0}{0.25}\times 1=176g](https://tex.z-dn.net/?f=%5Cfrac%7B44.0%7D%7B0.25%7D%5Ctimes%201%3D176g)
Thus molecular mass is 176 g
Now we have to calculate the molecular formula.
![n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{176g}{29g}=6](https://tex.z-dn.net/?f=n%3D%5Cfrac%7B%5Ctext%7BMolecular%20weight%20%7D%7D%7B%5Ctext%7BEquivalent%20weight%7D%7D%3D%5Cfrac%7B176g%7D%7B29g%7D%3D6)
The molecular formula will be=![6\times CHO=C_6H_6O_6](https://tex.z-dn.net/?f=6%5Ctimes%20CHO%3DC_6H_6O_6)