# Question 0b136

Mar 29, 2017

Probably glucose (I assume other than C and H, only oxygen is in the molecule)

#### Explanation:

Since the question does not provide any other element, I assume it contains oxygen (in addition to given hydrogen and carbon). I also assume that the compound components are provided in weight ratio (40% by weight for Carbon and 6.67% by weight for Hydrogen).

My solution:
Carbon 12 grams/mol, Hydrogen 1 gram/mol, and Oxygen 16 grams/mol.

What is the oxygen weight? 100-40-6.67 = 53.33%

Then,
for Carbon, $180 \times \left(\frac{0.40}{12}\right) = \frac{72}{12} = 6$
for Hydrogen, $180 \times \left(\frac{0.0667}{1}\right) = \frac{12}{1} = 12$
For Oxygen, $180 \times \left(\frac{0.5333}{16}\right) = \frac{96}{16} = 6$

The formula will be ${C}_{6} {H}_{12} {O}_{6}$, which is glucose. Glucose molecular weight is exactly 180 grams.

Mar 29, 2017

Given

C->40%;H->6.67%

So the rest will be O->(100-40-6.67)%=53.33%#

Dividing the given percentage of C , H and O with their respective atomic masses and taking their ratio we get the ratio of number of atoms of the constituent elements present in the compound.

So the ratio of number of atoms of C , H and O in the compound will be

${n}_{C} : {n}_{H} : {n}_{O} = \frac{40}{12} : \frac{6.67}{1} : \frac{53.33}{16}$

$= 3.33 : 6.67 : 3.33$

$= \frac{3.33}{3.33} : \frac{6.67}{3.33} : \frac{3.33}{3.33} = 1 : 2 : 1$

Hence the empirical formula of the compound is $C {H}_{2} O$

let its molecular formula be ${\left(C {H}_{2} O\right)}_{n}$

So calculated value of molar mass will be $\left(12 \times 1 + 1 \times 2 + 1 \times 16\right) n = 30 n g \text{/} m o l$

But given molar mass is $180 g \text{/} m o l$

Hence $30 n = 180 \implies n = 6$

So molecular formula of the compound is ${C}_{6} {H}_{12} {O}_{6}$