Northern Arizona University Quiz #3 CHM 151, General Chemistry I Dr. Brandon Cruickshank Section 4, Fall 2004 September 23, 2004

Name ______

1. Balance the following equation with the smallest set of whole numbers. The sum of all the coefficients is? Don't forget to count coefficients of one. Circle the correct answer. [3 pts]

1 C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l)

a) 19 b) 25 c) 38 d) 4 e) 14

2. How many F atoms are in 8.0 moles of SF6? Circle the correct answer. [3 pts]

a) 6.02 ×1023 b) 4.82 × 1024 c) 2.89 × 1025 d) 7.97 ×10−23 e) 8.03 ×1023

23 6molF 6.022× 10 Fatoms 25 8.0 mol SF6 ×× =2.89 × 10 F atoms 1molSF6 1molF or 23 6.022 × 10 SF6 molecules 6atomsF 25 8.0 mol SF6 ××=2.89 × 10 F atoms 1 mol SF661 molecule SF

3. Mesitylene, a hydrocarbon that occurs in small amounts in crude oil, has an empirical formula of C3H4. The experimentally determined molar mass of this substance is 121 g/mol. What is the molecular formula of mesitylene? [3 pts]

molar mass 121 g =≈3 empirical mass 40.06 g

The molecular formula is (C3H4)3 or C9H12.

4. Determine the empirical formula of ibuprofen, a headache remedy that contains 75.69% C, 8.80% H, and 15.51% O by mass. [5 pts]

1molC 75.69 g C ×=6.302 mol C 12.01 g C 1molH 8.80 g H ×=8.730 mol H 1.008 g H 1molO 15.51 g O ×=0.9694 mol O 16.00 g O

This gives the formula C6.302H8.730O0.9694. Dividing by the smallest number of moles (0.9694 mol) gives the formula C6.500H9.006O. This is not an empirical formula because the subscript of C is 6.500. Multiplying all the subscripts by 2 gives the empirical formula C13H18O2. 5. A method used by the Environmental Protection Agency (EPA) for determining the concentration of ozone (O3) in air is to pass the air sample through a “bubbler” containing sodium iodide (NaI), which removes the ozone according to the following equation:

O3 (g) + 2 NaI (aq) + H2O (l) → O2 (g) + I2 (s) + 2 NaOH (aq)

−3 How many grams of sodium iodide (NaI) are needed to remove 1.25 × 10 g of ozone (O3)? [Molar masses: O3, 48.00 g/mol; NaI, 149.89 g/mol] [4 pts]

−3 1molO3 2molNaI 149.89gNaI 1.25 ××10 g O3 × × =0.00781 g NaI 48.00 g O331 mol O 1 mol NaI

6. Lithium and nitrogen react to produce lithium nitride

6 Li (s) + N2 (g) → 2 Li3N (s)

Given that 3.50 moles of N2 are reacted with 16.00 moles of Li, answer the following questions:

a) Which reactant is the limiting reagent? You MUST show work to receive credit. [3 pts]

2molLi N 3.50 mol N ×=3 7.00 mol Li N 231molN 2 2molLi N 16.00 mol Li ×=3 5.33 mol Li N 6molLi 3

Li limits the amount of product that can be produced (5.33 mol Li3N) and is therefore the limiting reagent.

b) How many grams of Li3N are produced assuming complete reaction? [3 pts]

The amount of product Li3N was calculated in part (a), 5.33 mol Li3N. Converting to grams:

34.83 g Li3N 5.33 mol Li3N ×=186 g Li3N 1molLi3N

1 mole = 6.022 × 1023 particles Molar mass Li = 6.941 g/mol Molar mass N = 14.01 g/mol Molar mass C = 12.01 g/mol Molar mass O = 16.00 g/mol Molar mass H = 1.008 g/mol