Northern Arizona University
CHM 151, General Chemistry I
Section 4, Fall 2004
Quiz #3
Dr. Brandon Cruickshank
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)
a) 19
2.
+
6 O2(g)
b) 25
c)
→
6 CO2(g)
38
+
d) 4
How many F atoms are in 8.0 moles of SF6? Circle the correct answer.
a)
6.02 ×10
23
b) 4.82 × 10
24
c)
2.89 × 10
6 H2O(l)
25
e)
14
e)
8.03 ×10
[3 pts]
d) 7.97 ×10
−23
8.0 mol SF6 ×
6 mol F
6.022 × 1023 F atoms
×
= 2.89 × 1025 F atoms
1 mol SF6
1 mol F
8.0 mol SF6 ×
6.022 × 1023 SF6 molecules
6 atoms F
×
= 2.89 × 1025 F atoms
1 mol SF6
1 molecule SF6
23
or
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]
1 mol C
= 6.302 mol C
12.01 g C
1 mol H
8.80 g H ×
= 8.730 mol H
1.008 g H
1 mol O
15.51 g O ×
= 0.9694 mol O
16.00 g O
75.69 g C ×
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
[Molar masses: O3, 48.00 g/mol; NaI, 149.89 g/mol] [4 pts]
1.25 × 10−3 g O3 ×
6.
g of ozone (O3)?
1 mol O3
2 mol NaI 149.89 g NaI
×
×
= 0.00781 g NaI
48.00 g O3 1 mol O3
1 mol NaI
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)
reagent.
Which reactant is the limiting reagent? You MUST show work to receive credit.
3.50 mol N 2 ×
2 mol Li3 N
= 7.00 mol Li3 N
1 mol N 2
16.00 mol Li ×
2 mol Li3 N
= 5.33 mol Li3 N
6 mol Li
[3 pts]
Li limits the amount of product that can be produced (5.33 mol Li3N) and is therefore the limiting
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:
5.33 mol Li3 N ×
23
34.83 g Li3 N
= 186 g Li 3 N
1 mol Li3 N
1 mole = 6.022 × 10 particles
Molar mass C = 12.01 g/mol
Molar mass Li = 6.941 g/mol
Molar mass O = 16.00 g/mol
Molar mass N = 14.01 g/mol
Molar mass H = 1.008 g/mol