University:
College
Course:
College Chemistry
Academic year:

2022
Views:

408

Pages:

13
Author:

Dipin K.

Worksheet 111 - Answer Key SET A: STOICHIOMETRY AND Chemical Formulas Calculations 1) A compound with the formula, BxHyO3, contains 36.14% by mass oxygen. What is the value of the integer, x? Ans: x = 6 Setup: 36.14 g O * (1 mol O / 16.00 g O) * (2 mol H / 1 mol O) * (1.01 g H / 1 mol H) = 15.21 g H Find mass of B in 100g sample: 100g sample - 36.14g O - 15.21g H = 48.65 g B B H O 48.65g 15.21g 36.14g 10.80g/mole 1.01g/mole 16.00g/mole 4.505 mole 15.06 mole 2.259 mole 4.505/2.259 15.06/2.259 2.259/2.259 2 6.666 = 6 2/3 1 6 20 3 Empirical Formula B6H20O3 x = 6 SET A 2) A mixture of cobalt (II) oxide and cobalt (III) oxide contains 32.50% by mass cobalt (II) oxide. What is the total number of oxide ions in a 122 g of the mixture? Setup: 32.50 g CoO * (1 mole CoO / 74.90 g CoO) * (1 mole O²⁻ ion / 1 mole CoO) = 0.4345 mole O²⁻ ion 67.50 g Co₂O₃ * (1 mole Co₂O₃ / 165.89 g Co₂O₃) * (3 moles O²⁻ ions / 1 mole Co₂O₃) = 1.221 mole O²⁻ ion Total moles O²⁻ = 0.4345 mole + 1.221 mole = 1.659 mole O²⁻ ion 1.659 mole O²⁻ ion * (6.02 x 10²³ ions / 1 mole O²⁻ ion) = 9.99 x 10²³ ions 3) A sulfur containing compound is treated chemically to convert all its sulfur into barium sulfate. A 8.19 mg sample of the compound gave 5.46 mg barium sulfate. a) What is the percentage of sulfur in the compound? Setup: 5.46 x 10⁻³ g BaSO₄ * (1 mole BaSO₄ / 233.39 g BaSO₄) * (1 mole S / 1 mole BaSO₄) = 7.52 x 10⁻⁵ mole S % S = (7.52 x 10⁻⁵ g S / 8.19 x 10⁻³ g compound) x 100 = 9.18% b) If there is one sulfur atom in the molecule, what is the molar mass of the compound? Setup: 5.46 x 10⁻³ g BaSO₄ * (1 mole BaSO₄ / 233.39 g BaSO₄) * (1 mole S / 1 mole BaSO₄) = 0.0235 x 10⁻³ mole S 0.0235 x 10⁻³ mole S * (1 mole compound / 1 mole S) = 0.0235 x 10⁻³ mole compound molar mass of compound = 8.19 x 10⁻³ g compound / 0.0235 x 10⁻³ mole compound = 349 g/mole SET C 1) A phosphorus containing compound is treated chemically to convert all its phosphorus into Mg3(PO4)2. A 7.88 g sample of the compound gave 4.75 g Mg3(PO4)2. What is the percentage by mass of phosphorus in the compound? 1 Setup: 4.75 g Mg3(PO4)2 * (1 mole Mg3(PO4)2 / 262.89 g Mg3(PO4)2) * (2 moles P / 1 mole Mg3(PO4)2) * (31.0 g P / 1 mole P) = 1.12 g P %P = (1.12 g P / 7.88 g compound) * 100 = 14.2% P 2) The percent by mass of boron in the compound, B7X3, is 42.1%. What is the molar mass of X? Setup: 42.1 g B * (1 mole B / 10.8 g B) * (3 moles X / 7 moles B) = 1.67 moles X mass of X = 100 g compound - 42.1 g B = 57.9 g molar mass of X = 57.9 g X / 1.67 moles X = 34.7 g/mole