Question

1. Using the mass percent of iron in iron (III) chloride, calculate the number of moles of iron in 139.87g of FeCl₃. ( The Molar mass of iron is 55.85 grams.)

2. If 2.317 g of Iron (III) bromide is transferred to a volumetric flask, what is the molarity (M), if the volume of solution is 250 ml?

If 2.317 g of Iron (III) bromide is transferred to a volumetric flask, what is the molarity (M), if the volume of solution is 250 ml?

Answer 1

1)

Molar mass of FeCl3,
MM = 1*MM(Fe) + 3*MM(Cl)
= 1*55.85 + 3*35.45
= 162.2 g/mol


MM of FeCl3 = 162.2
Lets calculate the mass of element

mass of Fe = 1*MM(Fe)
= 55.85 g

Mass % of Fe = (55.85*100)/162.2
= 34.43 %

SO,
mass of Fe = 34.43 % of 139.87
= 34.43 * 139.87 / 100
= 48.16 g

Molar mass of Fe = 55.85 g/mol


mass(Fe)= 48.16 g

use:
number of mol of Fe,
n = mass of Fe/molar mass of Fe
=(48.16 g)/(55.85 g/mol)
= 0.8623 mol
Answer: 0.8623 mol

2)

Molar mass of FeBr3,
MM = 1*MM(Fe) + 3*MM(Br)
= 1*55.85 + 3*79.9
= 295.55 g/mol


mass(FeBr3)= 2.317 g

use:
number of mol of FeBr3,
n = mass of FeBr3/molar mass of FeBr3
=(2.317 g)/(295.55 g/mol)
= 7.84*10^-3 mol
volume , V = 250 mL
= 0.25 L


use:
Molarity,
M = number of mol / volume in L
= 7.84*10^-3/0.25
= 3.136*10^-2 M
Answer: 3.136*10^-2 M