Question

Compound has a molar mass of and the following composition: elementmass % carbon47.09% hydrogen6.59% chlorine46.33% Write the molecular formula of .

Answer 1

The given question is incomplete. The complete question is:

Compound X has a molar mass of 153.05 g/mol and the following composition:

element mass %

carbon 47.09%

hydrogen 6.59%

chlorine 46.33%

Write the molecular formula of X.

Answer: The molecular formula of X is $C_6H_{10}Cl_2$

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= 47.09 g

Mass of H = 6.59 g

Mass of Cl = 46.33 g

Step 1 : convert given masses into moles.

Moles of C =$\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{47.09g}{12g/mole}=3.92moles$

Moles of H =$\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{6.59g}{1g/mole}=6.59moles$

Moles of Cl =$\frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac{46.33g}{35.5g/mole}=1.30moles$

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

For C = $\frac{3.92}{1.30}=3$

For H = $\frac{6.59}{1.30}=5$

For Cl =$\frac{1.30}{1.30}=1$

The ratio of C : H: Cl= 3: 5 :1

Hence the empirical formula is $C_3H_5Cl$

The empirical weight of $C_3H_5Cl$ = 3(12)+5(1)+1(35.5)= 76.5g.

The molecular weight = 153.05 g/mole

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{153.05}{76.5}=2$

The molecular formula will be=$2\times C_3H_5Cl=C_6H_{10}Cl_2$