Question

Find the empirical formula of a compound that is 48.38 % carbon, 8.12 % hydrogen, and 43.5 % oxygen

Answer 1

n_C = 48.38g C x $\frac{1molC}{12gC}$ = 4.03 ÷ 2.71 = 1.5molC

n_H = 8.12g H x $\frac{1molH}{1gH}$ = 8.12 ÷ 2.71 = 3molH

n_O = 43.5g O x $\frac{1molO}{16gO}$ = 2.71 ÷ 2.71 = 1molO

now we got a number 1.5 so we multiply by 2 to all

n_C = 1.5 ( 2 ) = 3molC

n_H = 3 ( 2 ) = 6molH

n_O = 1 ( 2 ) = 2molO

now we got empirical formula is

C₂H₆O₂