University:
Dakota State University
Course:
CHEM 112 | General Chemistry I
Academic year:

2021
Views:

131

Pages:

2
Author:

slypery5hy

**Mole Calculations** **Name:** 1. Calculate the mass of the following: a) 1.00 mol of NH4Cl 1.00 mol * 53.5 g/mol = 53.5 g b) 4.50 mol of NH4Cl 4.50 mol * 53.5 g/mol = 241 g c) 3.25 mol of PCl3 3.25 mol * 137.32 g/mol = 446 g d) 0.00355 mol of Na3PO4 0.00355 mol * 163.94 g/mol = 0.584 g e) 0.0125 mol of XeF4 0.0125 mol * 207.29 g/mol = 2.59 g f) 2.60 mol of C2H6 2.60 mol * 30.07 g/mol = 78.2 g g) 3.25 x 10^2 mol of NH3 3.25 x 10^2 mol * 17.03 g/mol = 5540 g h) 7.90 x 10^-4 mol of H2SO4 7.90 x 10^-4 mol * 98.08 g/mol = 0.0684 g i) 1.00 x 10^-3 mol of NaOH 1.00 x 10^-3 mol * 40.00 g/mol = 0.0400 g j) 1.75 x 10^-4 mol of Fe 1.75 x 10^-4 mol * 55.85 g/mol = 0.00977 g 2. Calculate the number of moles in the following: a) 17.9 g of H2SO4 17.9 g * (1 mol / 98.08 g) = 0.173 mol b) 91.5 g of H2O 91.5 g * (1 mol / 18.02 g) = 5.08 mol c) 53.0 g of C 53.0 g * (1 mol / 12.01 g) = 4.41 mol d) 0.125 mg of CuS 0.125 mg * (1 g / 1000 mg) * (1 mol / 95.61 g) = 1.31 x 10^-6 mol e) 450 kg of CH4 450000 g * (1 mol / 16.04 g) = 28000 mol f) 225 g of (NH4)2SO4 225 g * (1 mol / 132.14 g) = 1.70 mol g) 55.2 mg of Cl2 0.0552 g * (1 mol / 70.90 g) = 7.79 x 10^-4 mol h) 128.2 g of SO2 128.2 g * (1 mol / 64.06 g) = 2.001 mol **KEY** 1. 2955 g of Ag Answer: 2955 g * (1 mol / 107.87 g) = 27.39 x 10^6 mol 2. 0.0845 g of KMnO4 Answer: 0.0845 g * (1 mol / 158.04 g) = 5.35 x 10^-4 mol 3. Calculate the molar mass of each of the substances mentioned in the following. a) A 0.250 mol sample of methane (CH4) has a mass of 4.00 g. Answer: 4.00 g / 0.250 mol = 16.0 g/mol b) A 0.00248 mol sample of cholesterol has a mass of 0.947 g. Answer: 0.947 g / 0.00248 mol = 382 g/mol c) The mass of 6.47 x 10^-4 mol of diamond is 7.76 mg. Answer: 0.00776 g / 6.47 x 10^-4 mol = 12.0 g/mol d) A 3.44 x 10^-5 mol sample of a particular protein has a mass of 74.8 g. Answer: 74.8 g / 3.44 x 10^-5 mol = 2.17 x 10^6 g/mol 4. Calculate the volume at STP (standard temperature and pressure) occupied by the following. a) 12.5 mol of NH3(g) Answer: 12.5 mol * (22.4 L/mol) = 280 L b) 0.350 mol of O2(g) Answer: 0.350 mol * (22.4 L/mol) = 7.84 L c) 4.25 mol of HCl(g) Answer: 4.25 mol * (22.4 L/mol) = 95.2 L 5. Calculate the number of moles of the following gases at STP. a) 85.9 L of H2(g) Answer: 85.9 L * (1 mol / 22.4 L) = 3.83 mol b) 375 mL of SO3(g) Answer: 0.375 L * (1 mol / 22.4 L) = 0.0167 mol c) 5.00 mL of OCl2(g) Answer: 0.00500 L * (1 mol / 22.4 L) = 2.23 x 10^-4 mol