Question

A compound that is composed of carbon, hydrogen, and oxygen contains 70.6% C, 5.9% H, and 23.5% O by mass. The molecular weight of the compound is 136 amu. What is the molecular formula? A) B) C) D) E)
A) c8 H8 O2
B)C8 H4 O
C)C4 H4 O
D)C9 H12 O

Answer 1

Answer: The molecular formula will be $C_8H_8O_2$

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= 70.6 g

Mass of H = 5.9 g

Mass of O = 23.5 g

Step 1 : convert given masses into moles.

Moles of C =$\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{70.6g}{12g/mole}=5.9moles$

Moles of H =$\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{5.9g}{1g/mole}=5.9moles$

Moles of O =$\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{23.5g}{16g/mole}=1.5moles$

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

For C = $\frac{5.9}{1.5}=4$

For H = $\frac{5.9}{1.5}=4$

For O =$\frac{1.5}{1.5}=1$

The ratio of C : H: O= 4: 4:1

Hence the empirical formula is $C_4H_4O$

The empirical weight of $C_4H_4O$ = 4(12)+4(1)+1(16)= 68g.

The molecular weight = 136 g/mole

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{136}{68}=2$

The molecular formula will be=$2\times C_4H_4O=C_8H_8O_2$