Question

What is the empirical formula for a compound if it contains 0.558g of iron and the mass of the compound (made of iron and oxygen) is 0.718g? In a gas mixture containing oxygen, nitrogen, and helium, the total pressure is 925 torr. The partial pressure of oxygen is 425torr and helium is 75torr. What is the partial pressure of nitrogen

Answer 1

PART 1

Given : 1) Mass of Iron = 0.558 g & 2) Mass of Iron + Mass of Oxygen = 0.718 g

Mass of Iron + Mass of Oxygen = 0.718 g

$\therefore$ Mass of Oxygen = 0.718 g - mass of Iron = 0.718 g - 0.558 g = 0.160 g

Now, calculate no. of moles of Fe and O.

We have, No. of moles = Mass / Molar Mass

$\therefore$ No. of moles of Fe = 0.558 g / (55.85 g /mol ) = 0.00999 mol Fe

$\therefore$ No. of moles of O = 0.160 g / ( 16.00 g / mol ) = 0.0100 mol O

Ratio of number of moles of Fe : O is

0.00999 / 0.00999 = 1.00 mol Fe

0.0100 / 0.00999 = 1.00 mol O

In given oxide, Fe and O are present in the 1 : 1 ratio. Hence, empirical formula of oxide is FeO.

ANSWER : Empirical formula of oxide = FeO.

PART 2

Given : Total pressure of mixture of a gas = 925 torr

Partial pressure of O ₂ gas = 425 torr

Partial pressure of Hegas = 75 torr

According to Dalton's law , the total pressure exerted by a mixture of non reacting gases is equal to the sum of partial pressures of individual gases.

Mathematically ,we can write P _total = P ₁ + P ₂+ P ₃ + .................

$\therefore$P _total = P _oxygen + P _Helium + P _Nitrogen

$\therefore$ P _Nitrogen = P _total - ( P _oxygen + P _Helium )

$\therefore$ P _Nitrogen = 925 torr - ( 425 torr + 75 torr )

P _Nitrogen = 925 torr - 500 torr

P _Nitrogen = 425 torr

ANSWER : Partial pressure of nitrogen gas in the mixture = 425 torr