Answer:
C₆H₈O
Explanation:
Empirical Formula: This is the simplest formula of a compound.
From the question,
Percentage by mass of the elements: C = 74.97%, H = 8.39%, O = 16.64%
Molar mass of the elements : C = 12 g/ mole, H = 1 g/mole, O = 16 g/mole
Step 1: Divide the percentage by mass by the molar mass
C : H : O
74.97/12 8.39/1 16.64 /16
6.2475 : 8.39 : 1.04
Step 2: Dived by the smallest ratio ( 1.04)
C : H : O
6.2475/1.04 : 8.39/1.04 : 1.04/1.04
6 : 8 : 1
Hence, The empirical formula of the compound = C₆H₈O
Answer:
The empirical formula of the unknown compound is C₆H₈O
Explanation:
The simplest ratio (whole number) of the atoms in a compound is the empirical formula of the compound.
To determine the empirical formula of the compound given, the following steps are followed;
(i). Since percentage compositions of the elements in the compound are given, it can be assumed that all calculated from 100 grams of the compound. Therefore, the percentages can be changed directly to grams as follows;
74.97% Carbon (C) => 74.97g of Carbon (C)
8.39% Hydrogen (H) => 8.39g of Hydrogen (H)
16.64% Oxygen (O) => 16.64g of Oxygen (O)
(ii) Convert the grams to corresponding number of moles;
Recall that;
1 mole of Carbon contains 12 grams of the atom
=> 74.97 grams will be [tex]\frac{74.97}{12}[/tex] moles of Carbon
=> 74.97 grams will be 6.2475 moles of Carbon
Also,
1 mole of Hydrogen atoms contains 1 gram of the atom
=> 8.39 grams will be [tex]\frac{8.39}{1}[/tex] moles of Hydrogen
=> 8.39 grams will be 8.39 moles of Hydrogen
And,
1 mole of Oxygen atoms contains 16 grams of the atom
=> 16.64 grams will be [tex]\frac{16.64}{16}[/tex] moles of Hydrogen
=> 16.64 grams will be 1.04 moles of Hydrogen
(iii) Normalize the results calculated in step (ii) above by dividing each by the smallest (which is 1.04)
| Carbon | Hydrogen | Oxygen
Moles | 6.2475 | 8.39 | 1.04
Divide by smallest | [tex]\frac{6.2475}{1.04}[/tex] = 6.007 | [tex]\frac{8.39}{1.04}[/tex] = 8.07 | [tex]\frac{1.04}{1.04}[/tex] = 1
(iv) Round off the results in (iii) above to the nearest unit
| Carbon | Hydrogen | Oxygen
Moles | 6.2475 | 8.39 | 1.04
Divide by smallest | [tex]\frac{6.2475}{1.04}[/tex] = 6.007 | [tex]\frac{8.39}{1.04}[/tex] = 8.07 | [tex]\frac{1.04}{1.04}[/tex] = 1
Rounding off | 6 | 8 | 1
(v) Derive the empirical formula
By using the numbers resulting from step (iv) as subscript, place them beside each of their corresponding elements as follows;
=> C₆H₈O₁
The number 1 (which is the subscript of Oxygen can be removed)
=> C₆H₈O
Therefore, the empirical formula of the unknown compound is C₆H₈O
