Question

Vinegar contains carbon, hydrogen, and oxygen with percent masses of 40.01% C and 6.70% H. If the molar mass of vinegar is about 60 g/mol, what is its molecular formula

Answer 1

Answer: The molecular formula will be $C_2H_4O_2$

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= 40.01 g

Mass of H = 6.70 g

Mass of O = 100-(40.01+ 6.70) = 53.29 g

Step 1 : convert given masses into moles.

Moles of C =$\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{40.01g}{12g/mole}=3.33moles$

Moles of H =$\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{6.70g}{1g/mole}=6.70moles$

Moles of O =$\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{53.29g}{16g/mole}=3.33moles$

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

For C = $\frac{3.33}{3.33}=1$

For H = $\frac{6.70}{3.33}=2$

For O =$\frac{3.33}{3.33}=1$

The ratio of C : H: O= 1: 2: 1

Hence the empirical formula is $CH_2O$

The empirical weight of $CH_2O$ = 1(12)+2(1)+1(16)= 30 g.

The molecular weight = 60 g/mole

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{60}{30}=2$

The molecular formula will be=$2\times CH_2O=C_2H_4O_2$