Question

Determine the empirical formula for a compound that contains c, h and o. it contains 52.14% c and 34.73% o by mass.

Answer 1

Answer: C2H6O

Explanation:

Empirical formula is the formula showing proportions of the elements present in a compound. Empirical formula will not show the arrangements of atoms in a compound.

From the question above, we are given that the compound contains 52.14% of carbon, 34.73% of oxygen and the amount of Hydrogen is unknown.

Therefore, to determine the amount of Hydrogen, we say;

100% of the whole compound - (52.14% + 34.73%).

Percentage of hydrogen in the compound= 13.13% .

Therefore, Mole of Hydrogen atom= 13.13/1 = 13.13.

Mole of carbon atom = 52.14/12= 4.345.

Mole of Oxygen atom= 34.73/16 = 2.171.

The next thing to do is to divide the each mole by the smallest mole(that is, the mole of oxygen atom).

Hence, Hydrogen,H = 13.13/2.171 =6.

Carbon,C= 4.345/2.171 = 2.

Oxygen = 2.171/2.171 = 1.

Therefore the empirical formula is;

C2H6O.

Answer 2

H %= 100- (52.14 + 34. 73) equals 13.13 %
Assuming 100 g of this compound
Mass H= 13.13 g
Moles H= 13.13 g ÷ 1.008g/ moles= 13

Mass C= 52.14 g
Moles C= 52.14 g ÷ 12.011 g/ moles= 4

The empirical formula is C4H1302