Question

A compound gave following
C=57.82%
O=38.58% and the rest is hydrogen
Relative molecular mass = 166
Find empirical formula and molecular formula?
(C=12 O=16 H=1)

Answer 1

Answer: The empirical formula is $C_{4}O_{2}H_3$   and molecular formula is C_8O_4H_6[/tex]

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 = 57.82 g

Mass of O = 38.58 g

Mass of H = (100-(57.82+38.58) = 3.6 g

Step 1 : convert given masses into moles.

Moles of C =$\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{57.82g}{12g/mole}=4.81moles$

Moles of O =$\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{38.58g}{16g/mole}=2.41moles$

Moles of H =$\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{3.6g}{1g/mole}=3.6moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{4.81}{2.41}=2$

For O = $\frac{2.41}{2.41}=1$

For H = $\frac{3.6}{2.41}=1.5$

Converting them into whole number ratios by multiplying by 2.

The ratio of C : O : H= 4: 2 : 3

Hence the empirical formula is $C_{4}O_{2}H_3$  

The empirical weight of $C_{4}O_{2}H_3$  = 4(12) + 2(16) + 1(3)=  83 g

The molecular weight = 166 g/mole

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight of metal}}{\text{Equivalent weight of metal}}=\frac{166}{83}=2$

Thus Molecular formula=$n\times Empirical formula=2\times C_{4}O_{2}H_3=C_8O_4H_6$

Answer 2

So, the mass of each element is equal to the percentage given.

Mass of C = 57.82 g

Mass of O = 38.58 g

Mass of H = (100-(57.82+38.58) = 3.6 g

Moles of C =\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{57.82g}{12g/mole}=4.81moles molar mass of C given mass of C=12g/mole57.82g=4.81moles

Moles of O =\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{38.58g}{16g/mole}=2.41moles molar mass of O given mass of O=16g/mole38.58g=2.41moles

Moles of H =\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{3.6g}{1g/mole}=3.6moles molar mass of H given mass of H=1g/mole3.6g=3.6moles

For C = \frac{4.81}{2.41}=22.414.81=2

For O = \frac{2.41}{2.41}=12.412.41=1

For H = \frac{3.6}{2.41}=1.52.413.6=1.5

The ratio of C : O : H= 4: 2 : 3

Hence the empirical formula is C_{4}O_{2}H_3C4O2H3  

The empirical weight of C_{4}O_{2}H_3C4O2H3  = 4(12) + 2(16) + 1(3)=  83 g

The molecular weight = 166 g/mole

Now we have to calculate the molecular formula.

n=\frac{\text{Molecular weight of metal}}{\text{Equivalent weight of metal}}=\frac{166}{83}=2n=Equivalent weight of metalMolecular weight of metal=83166=2

Thus Molecular formula=n\times Empirical formula=2\times C_{4}O_{2}H_3=C_8O_4H_6n×Empiricalformula=2×C4O2H3=C8O4H6