What is empirical formula of an oxide of iron which has 69.9 % iron and 30 % oxygen by mass?

Question

What is empirical formula of an oxide of iron which has 69.9 % iron and 30 % oxygen by mass? (A) FeO (B) Fe 2 O 3 (C) Fe 3 O 4 (D) None

Tardigrade Admin · Accepted Answer

% of mass Atomic mass Relative no.of moles Simple ratio Simple whole no Iron Fe 69.9 55.85 69.9 / 55.85=1.25 (1.25/1.25)=1 2 Oxygen O 30.1 16.0 30.1 / 16=1.88 (1.88/1.25)=1.5 3 Formula = Fe 2 O 3

		$%$ of mass	Atomic mass	Relative no.of moles	Simple ratio	Simple whole no
Iron	$F e$	$69.9$	$55.85$	$69.9/55.85 = 1.25$	$\frac{1.25}{1.25} = 1$	$2$
Oxygen	$O$	$30.1$	$16.0$	$30.1/16 = 1.88$	$\frac{1.88}{1.25} = 1.5$	$3$

Q. What is empirical formula of an oxide of iron which has $69.9%$ iron and $30%$ oxygen by mass?

FeO

$F e_{2} O_{3}$

$F e_{3} O_{4}$

None

$%$ of mass Atomic mass Relative no.of moles Simple ratio Simple whole no

Iron $F e$ $69.9$ $55.85$ $69.9/55.85 = 1.25$ $\frac{1.25}{1.25} = 1$ $2$

Oxygen $O$ $30.1$ $16.0$ $30.1/16 = 1.88$ $\frac{1.88}{1.25} = 1.5$ $3$

Formula $= F e_{2} O_{3}$

Q. What is empirical formula of an oxide of iron which has 69.9% iron and 30% oxygen by mass?

FeO

Fe2​O3​

Fe3​O4​

None

% of mass Atomic mass Relative no.of moles Simple ratio Simple whole no Iron Fe 69.9 55.85 69.9/55.85=1.25 1.251.25​=1 2 Oxygen O 30.1 16.0 30.1/16=1.88 1.251.88​=1.5 3 Formula =Fe2​O3​

Q. What is empirical formula of an oxide of iron which has $69.9%$ iron and $30%$ oxygen by mass?

$F e_{2} O_{3}$

$F e_{3} O_{4}$

$%$ of mass Atomic mass Relative no.of moles Simple ratio Simple whole no

Iron $F e$ $69.9$ $55.85$ $69.9/55.85 = 1.25$ $\frac{1.25}{1.25} = 1$ $2$

Oxygen $O$ $30.1$ $16.0$ $30.1/16 = 1.88$ $\frac{1.88}{1.25} = 1.5$ $3$

Formula $= F e_{2} O_{3}$