Question

10.023g of a sample containing sodium chloride and MgSO4 x 5H2O has amass of 8.987g after being thoroughly heated. What is the percent mass of NaCl in the sample?

Answer 1

Given

Intial Total mass of sample = 10.023 g

Final total mass of sample = 8.897 g

Mass of water evaporated = Initial - final = 10.023 g - 8.897 g = 1.126 g

Molar mass of water = 18 g/mol

No. of moles of Water = Mass / Molar mass = 1.126 g / 18 g/mol = 0.0625 moles

As 1 mole of MgSO4 .5H2O contains 1 mole of MgSO4 and 5 moles of H2O

No. of moles of MgSO4.5H2O = No. of moles of water / 5 = 0.0625 /5 = 0.0125 moles

Molar mass of MgSO4.5H2O = 210.44 g/mol

Mass of MgSO4.5H2O = No. of moles * Molar mass = 0.0125 moles * 210.44 g/mol = 2.633 g

Mass of NaCl = intial mass - mass of MgSO4.5H2O = 10.023 - 2.633 = 7.39 g

Percent mass = Mass of NaCl / Intial total mass = (7.39 g / 10.023 g) * 100 % = 73.73 % Answer