University:
Campbell University
Course:
CHEM 111 | General Chemistry I
Academic year:

2023
Views:

85

Pages:

6
Author:

Chanel Morton

Percent Composition Key Name: ______ 1. Calculate the percentage composition of the following: a) C₂H₆ C: 12.01 g/mol x 2 = 24.02 g/mol H: 1.01 g/mol x 6 = 6.06 g/mol Answer: C = 79.85%, H = 20.15% b) FeCl₂ Fe: 55.85 g/mol Cl: 35.45 g/mol x 2 = 70.90 g/mol Answer: Fe = 44.06%, Cl = 55.94% c) FeCl₃ Fe: 55.85 g/mol Cl: 35.45 g/mol x 3 = 106.35 g/mol Answer: Fe = 34.43%, Cl = 65.57% d) C₂H₄O C: 12.01 g/mol x 2 = 24.02 g/mol H: 1.01 g/mol x 4 = 4.04 g/mol O: 16.00 g/mol Answer: C = 52.98%, H = 8.90%, O = 38.12% e) CaCO₃ Ca: 40.08 g/mol C: 12.01 g/mol O: 16.00 g/mol x 3 = 48.00 g/mol Answer: Ca = 40.08%, C = 12.00%, O = 48.00% f) NaOH Na: 22.99 g/mol O: 16.00 g/mol H: 1.01 g/mol Answer: Na = 57.48%, O = 40.00%, H = 2.53% g) CaCl₂•2H₂O Ca: 40.08 g/mol Cl: 35.45 g/mol x 2 = 70.90 g/mol H: 1.01 g/mol x 4 = 4.04 g/mol O: 16.00 g/mol x 2 = 32.00 g/mol Answer: Ca = 27.26%, Cl = 47.67%, H = 2.75%, O = 21.77% h) (NH₄)₃PO₄ N: 14.01 g/mol x 3 = 42.03 g/mol H: 1.01 g/mol x 12 = 12.12 g/mol P: 30.97 g/mol O: 16.00 g/mol x 4 = 64.00 g/mol Answer: N = 28.19%, H = 8.13%, P = 20.77%, O = 42.92% i) AgCl(NH₃)₂ Ag: 107.87 g/mol Cl: 35.45 g/mol N: 14.01 g/mol x 2 = 28.02 g/mol H: 1.01 g/mol x 6 = 6.06 g/mol Answer: Ag = 60.80%, Cl = 19.95%, N = 15.79%, H = 3.42% j) C₁₇H₁₅N₃O₂Cl Answer: C = 62.10%, Cl = 10.78%, H = 4.61%, N = 12.79%, O = 9.73% k) Sn(SO₄)₂•2H₂O Answer: Sn = 34.22%, S = 18.49%, O = 46.13%, H = 1.16% l) (NH₄)₂Sn(OH)₆ Answer: N = 20.02%, H = 14.14%, O = 56.00%, Sn = 10.83% 2. Calculate the percentage of the bold species in each of the following: a) CaCl₂•2H₂O Answer: Ca = 24.51% b) NiSO₄•7H₂O Answer: Ni = 44.91% c) Ce(CO)₉H₂O Answer: Ce = 22.96% d) Al₂(SO₄)₃•18H₂O Answer: Al = 48.67% e) Cr(NH₃)₆Cl₃•H₂O Answer: Cr = 36.76% f) Cr(NH₃)₆Cl₃•H₂O Answer: Cr = 36.76%