Question

Compound j burns in excess Oxygen to give carbondioxide and water only . when 3.00 g sample of compound j is burnt in excess of oxygen , 4.40 g of carbondioxide and 1.80 g of water are formed
What is the empirical formula
A) CH B) CHO C) CH2 D) CH2O

Answer 1

Answer:- D) $CH_2O$

Solution:- The moles of carbon are calculated from the given grams of carbon dioxide as carbon dioxide contains only one carbon so the mol ratio of carbon to carbon dioxide will be 1:1.

$4.40gCO_2(\frac{1molCO_2}{44gCO_2})(\frac{1molC}{1molCO_2})$

= 0.1 mol C

Similarly, the moles of hydrogen are calculated from given grams of water. There is 2:1 mol ratio between hydrogen and water as water molecule has two hydrogen atoms.

$1.80gH_2O(\frac{1molH_2O}{18gH_2O})(\frac{2molH}{1molH_2O})$

= 0.2 mol H

We need the grams of C and H also to make sure if the compound has only C and H or some oxygen is also there.

Moles of C and H are multiplied by their atomic masses to get the grams.

$0.1molC(\frac{12g}{1mol})$

= 1.2g C

$02.molH(\frac{1g}{1mol})$

= 0.2g H

sum of masses of C and H = 1.2 g + 0.2 g = 1.4 g

mass of oxygen in the compound is calculated on subtracting the sub of C and H masses from the given mass of the compound.

mass of oxygen = 3.00 g - 1.4 g = 1.6 g

Let's convert the grams of oxygen to moles.

$1.6gO(\frac{1mol}{16g})$

= 0.1 mol O

From above calculations, we have 0.1 mol of C, 0.2 moles of H and 0.1 mol of O in the compound.

So, the mol ratio of C:H:O is 1:2:1.

Hence, the empirical formula of the compound is $CH_2O$ and the right choice is D.