Question

A student has 530.0 mL of a 0.1082 M aqueous solution of MnSO4 to use in an experiment. He accidentally leaves the container uncovered and comes back the next week to find only a solid residue. The mass of the residue is 12.79 g. Determine the chemical formula of this residue. If the substance is a hydrate, use a period instead of a dot in its formula.

Answer 1

Parol ů to calculate the chemical formula Answer: - . The main objective of given residue. Given Date: Given molarity of volume of MnSO4 solution = 0.1082 M . 530.0 ml We know M = number of moles x 1000 v in ml number of moles of Mn Sou (n.) a Holarity & vin me 1000 n 0.1082 x 530.0 1000 57.346 1000 ni = 0.057346 moles But na weight molecular weight weight of Masou - nix molecular weight 0.057346 x 151.00 Sind = 8.6592 g pou given weight of Solid residue = 12.799

Pg.o2 Hence weight of H₂O = 12.799 - 8.65929 - 4.1307 g number of moles of Hoo (na) - weight Molecular weight 4.13079 - 18 glmol (na) = 0.2294 moles. mole ratio of mnsou to Hao I get the ratio divide A. ? The other values of by n,&nz , choose least one 0.057346 This least one to get ratio (n,: 02) and Masou a ..:057346 0.057346 461 H₂o - 0.2294 au 0.057346 24 The molecular formala for residue is - Ansouhhzo