20) Determine the molecular formula of an oxide of iron, in which the mass per cent of iron and
oxygen are 69.9 and 30.1, respectively.

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Solution

$D e a r S t u d e n t M a s s p e r c e n t o f i r o n a n d o x y g e n a r e 69.1 % a n d 30.1 % r e s p e c t i v e l y . M a s s o f i r o n = 69.1 g M a s s o f o x y g e n = 30.1 g M o l e o f i r o n = \frac{69.1}{56} = 1.23 M o l e o f o x y g e n = \frac{30.1}{16} = 1.88 M o l e r a t i o i s - F e : O = 1.23 : 1.88 = \frac{1.23}{1.23} : \frac{1.88}{1.23} = 1 : 1.5 = 2 : 3 S o, t h e f o r m u l a o f i r o n o x i d e i s F e_{2} O_{3} . R e g a r d s$

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20) Determine the molecular formula of an oxide of iron, in which the mass per cent of iron and oxygen are 69.9 and 30.1, respectively.

Dear StudentMass percent of iron and oxygen are 69.1% and 30.1% respectively.Mass of iron=69.1 gMass of oxygen=30.1 gMole of iron=69.156=1.23Mole of oxygen=30.116=1.88Mole ratio is-Fe:O=1.23:1.88=1.231.23:1.881.23=1:1.5=2:3So, the formula of iron oxide is Fe2O3.Regards

20) Determine the molecular formula of an oxide of iron, in which the mass per cent of iron and
oxygen are 69.9 and 30.1, respectively.