A sulphide of iron contains 46.5% of iron by weight. The empirical formula of the sulphide is:

F e S

No worries! We‘ve got your back. Try BYJU‘S free classes today!

F e_{2} S

No worries! We‘ve got your back. Try BYJU‘S free classes today!

F e S_{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

F e_{2} S_{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D $F e S_{2}$
A sulphide of iron contains 46.5% of iron by weight.
100 g of sample will contain 46.5 g of iron and $100 - 46.5 = 53.5$ g of sulphur.
The atomic masses of iron and sulphur are 55.8 g/mol and 32.1 g/mol.
The number of moles of iron $= \frac{46.5 g}{55.8 g / m o l} = 0.8333$ .
The number of moles of sulphur $= \frac{53.5 g}{32.1 g / m o l} = 1.667$
The ratio of the number of moles of iron to the number of moles of sulphur $= 0.8333 : 1.667 = 1 : 2$
Hence, the empirical formula of the sulphide is $F e S_{2}$ .

Suggest Corrections

Similar questions

Q. Fe +S =Fes. Calculate the mass in g of iron sulphide formed when 28g of iron combines with 16g of sulphur.

Q. The molecular formula of iron(II) sulphide is:

0.7 g of iron reacts directly with 0.4 g of sulphur to form ferrous sulphide. If 2.8 g f iron dissolves in dilute HCl and excess of sodium sulphide solution is added, 4.4 g of iron sulphide is precipitated. If it obeys the law of constant composition, what is the ratio of weights of Fe : S ?

0.7 g of iron reacts directly with 0.4 g of sulphur to form ferrous sulphide. If 2.8 g of iron dissolves in dilute HCl and excess of sodium sulphide solution is added, 4.4 g of iron sulphide is precipitated. If it obeys the law of constant composition, what is the ratio of weights of Fe : S ?

In the formation of iron sulphide, by heating iron and sulphur, the mass of iron(II) sulphide __________ when compared with a total mass of iron and sulphur.

Join BYJU'S Learning Program