Answer:
Empirical formula is C₄H₁₀O
Step-by-step explanation:
Values for C, H and O are determined as centesimal composition.
64.80 g of C in 100g of compound
13.62g of H in 100 g of compound
21.58 g of O in 100 g of compound.
We convert the mass to moles:
64.80 g . 1mol/ 12g = 5.4 moles of C
13.62 g . 1 mol /1g = 13.62 moles of H
21.58 g . 1 mol/16g = 1.35 moles of O
We pick the lowest value and we divide:
5.4 moles of C / 1.35 = 4 C
13.62 moles of H / 1.35 = 10 H
1.35 moles of O / 1.35 = 1 O
Empirical formula is C₄H₁₀O, it can be the diethyl ether.
We confirm, the excersise is well done.
Molar mass = 74g/mol
74 g of compound we have (12 . 4)g of C
In 100 g of compound we may have (100 . 48) / 74 = 64.8 g