Question

A carbohydrate contains 40% carbon and 6.66% hydrogen. Determine its molecular formula, if its molecular weight is 180.

Answer 1

Answer : The molecular formula will be, $C_6H_{12}O_6$

Solution : Given,

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 g

Mass of H = 6.66 g

Mass of O = 53.34 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{40g}{12g/mole}=3.33moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{6.66g}{1g/mole}=6.66moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{53.34g}{16g/mole}=3.33moles$

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 = $\frac{6.66}{3.33}=2$

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

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

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $C_1H_2O_1$

The empirical weight of $C_1H_2O_1$  = 1(12) + 2(1) + 1(16) = 30 g/eq

The molecular weight = 180 g/mole

Now we have to calculate the molecular formula.

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

where, n is the valency.

$n=\frac{180g/mole}{30g/eq}=6$

The molecular formula is always the multiple of empirical formula.

$\text{(Empirical formula)}_n=\text{Molecular formula}$

$(C_1H_2O_1)_6=C_6H_{12}O_6$

So, the molecular formula will be, $C_6H_{12}O_6$

Answer 2

The above answer is absolutely correct.