4. You are given an unknown hydrated metal salt containing bromide ion, MBr2.nH2

Question

4. You are given an unknown hydrated metal salt containing bromide ion, MBr2.nH2O. a. You dissolve 0.500 g alibis salt in water and add excess Silver nitrate solution. AgNO3, to precipitate the bromide ion as insoluble Silver bromide, AgBr. After filtering, washing, drying, and weighing, the AgBr is found to weigh 0.609 g. What is the mass percent bromide in the metal salt? b. A second 0.500 g sample is dehydrated to remove water of hydration. After drying, the sample is found o weigh 0.325 g. What is the mass percent water in the metal salt? c. The metal cation has a charge of two. What is the molar mass of the metal? What is the identity of the metal? d. Determine the complete empirical formula of the hydrated metal salt.

Explanation / Answer

(a) Mass of AgBr, W1 = 0.609 g
Molar mass of AgBr, MW1 = 187.77 g/mol
moles of AgBr, n1 = W/MW1 = 0.609/187.77 = 0.00324 mol

Ag+(aq) + Br-(aq) = AgBr(s)
moles of Br-, n3 = moles of AgBr = 0.00324 mol
Molar mass of Br- ion, MW3 = 79.91 g/mol
Total mass of Br-, W3 = n3*MW3 = 0.00324*79.91 = 0.259 g
Total mass of salt, W2 = 0.5 g
Mass percent of bromide in the salt = W3/W2*100 = 0.259/0.5*100 = 51.78%

(b) Mass of hydrated sample, W2 = 0.5 g
Mass of dehydrated sample, W4 = 0.325 g
Mass of water, Ww = W2-W4 = 0.5 - 0.325 = 0.175 g
Mass percent of water in sample = Ww/W2*100 = 0.175/0.5*100 = 35%

(c) Metal cation has a charge of two. Molar mass of metal, M2+ = MWm
Ag+(aq) + Br-(aq) = AgBr(s)
moles of Br-, n3 = moles of AgBr = 0.00324 mol

2 moles of Br- ions are presnent in 1 mol of hydrated metal salt MBr2.nH2O
1 moles of Br- ions are presnent in 1/2 mol of hydrated metal salt MBr2.nH2O
0.00324 moles of Br- ions are presnent in 0.00324*1/2 = 0.00162 mol of hydrated metal salt

MBr2.nH2O

Moles of hydrated metal salt MBr2.nH2O, n2 = 0.00162 mole
Moles of metal ion M2+ = 0.00162 mole
Mass of metal ions, Wm = 0.00162*MWm
Mass of Br-, W3 = n3*MW3 = 0.00324*79.91 = 0.259 g

Total mass of dehydrated metal salt, MBr2 = Wm+W3= (0.00162*MWm + 0.259) g
Given, mass of dehydrated sample, W4 = 0.325 g
W4 =Wm + W3
0.325 = 0.00162*MWM + 0.259
Molar mass of metal, MWm =(0.325-0.259)/0.00162 = 40.74 g/mole
The molar mass 40.74 is nearer to that of Calcium(40.08), so the metal is calcium.

(e) Empirical formula of the hydrated metal salt, CaBr2.nH2O

Mass of water, Ww = W2-W4 = 0.5 - 0.325 = 0.175 g
Moles of water, nw = 0.175/18 = 0.0097 moles
Moles of hydrated metal salt CaBr2.nH2O, n2 = 0.00162 mole

0.00162 moles of hydrated salt has 0.0097 moles of H2O
1 mole of hydrated salt has n = 0.0097/0.00162 = 6 moles of H2O

Empirical formula of the hydrated metal salt, CaBr2.6H2O
Empirical formula mass, EMW = 40.08+2*79.91+6*18=307.9
Mass of n2 = 0.00162 moles of hydrated sample = 0.00162*307.9 = 0.5 g which is given in (a)

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