Chemistry 201/211
Worksheet 2
Fall 2004
October 7, 2004
Oregon State University
1.) Complete the table:
sodium chloride
ammonium iodide
potassium phosphate
potassium chromate
iron (III) chloride
lithium hydride
nitric acid
magnesium cyanide
NaCl
NH4I
K3PO4
K2CrO4
FeCl3
LiH
HNO3
Mg(CN)2
2.) Given a cube of lead that is 1.000 cm on each side and knowing that the density of
lead is 11.35 g/cm3, calculate how many atoms of lead are contained in the sample?
Atoms are spherical; therefore, the lead atoms in the sample cannot fill all the available
space. As an approximation, assume that 60% of the space of the cube is filled with
spherical lead atoms. Calculate the volume of one lead atom from this information.
(Hint: first calculate the volume of a “lead cube”). From the calculated volume (V), and
the formula V=4/3r3, estimate the radius (r) of a lead atom.
V  l 3  (1.000 cm) 3  1.000cm 3
M  D  V  11.35 g 3  1.000cm 3  11.35g
cm
1mol Pb 6.022  10 23 atoms
# atoms  11.35g 

 3.299  10 22 atoms
207.2g Pb
1mol Pb
1.000cm 3
3.299  10 atoms
22
 3.031  10 23 cm
3
V  4 r 3  1.819  10 23 cm
3
atom
3
atom
r3 3
4
 60%  1.819  10 23 cm
1.819  10 23 cm
3
atom
3
atom
 1.631  10 8 cm
3.) The number of molecules in 22.4 L of air at the surface of the earth is around one
mole. Assume that the average volume of air a person takes in and expels in breathing is
4 L and that the air around the earth has a volume of 41018 m3 (Calculated from V=4r2h
where h is ~8 km of atmosphere above the surface). Mozart lived for 35 years and died
sufficiently long ago that the molecules of air that he breathed are sufficiently dispersed
throughout the atmosphere. Under all these conditions, how many molecules of air
breathed by Mozart are in your lungs right now? Assume that a person takes 20 breaths
of air per minute.
atom
35 years 
365days 24hours 60 min 20breaths 4 L 1mol 6.022  10 23 molecules


 4.0  10 31 molecules
1year
1day
1hour 1 min 1breath 22.4 L
1mol
1L 22.4mol 6.022  10 23 molecules
 100cm  1ml
4  1018 m 3 
 5  10 46 molecules in the atmostpher e

3
1
m
1000
ml
1
L
1
mol
1
cm


31
molecules breathed by Mozart
4.0  10 molecules
 8  10 16 is the ratio of
46
5  10 molecules
molecules in the atmosphere
3
8  10 16 * 4 L 
1mol 6.022  10 23 molecules
 2.2  10 7 molecules
22.4 L
1mol
4.) Which of the following statements is/are true? If the statement is false, make it true.
A 558.5 gram sample of iron contains
a.) 558.5 moles of iron False, it equals 10 moles of Fe
b.) 6.022  1023 atoms of iron False, it equals 6.022  1024
c.) ten times as many atoms as 0.5200 grams of chromium False, it equals 100X
d.) twice as many atoms as 60.06 grams of carbon True
558.5 gFe 
1molFe
6.022  10 23 atoms
 10molFe 
 6.022  10 24 atoms
55.8 gFe
1molFe
.5200 gCr 
1molCr
6.022  10 23 atoms
 0.01000molCr 
 6.022  10 21 atoms
52.0 gCr
1molCr
60.06 gC 
1molC
6.022  10 23 atoms
 5.005molC 
 3.013  10 24 atoms
12.0 gC
1molC
5.) Lithium has two isotopes. 7Li has a mass of 7.016 amu. Does the second isotope
have more, less or the same number of
a.) protons
b.) neutrons
c.) electrons
The abundance of 7Li is 92.23%. What is the mass in amu of the second isotope?
The second isotope has the same number of protons and electrons (if it were an ion
the electrons might differ). The average mass of Li from the book is 6.941 amu.
The other isotope must then be lighter. It can’t have fewer protons or else it would
be a different element. It must have fewer neutrons than 7Li.
0.9223  7.016  (1  0.9223)  X  6.941
X  6.051 6 Li
6.) Label each statement as true (T) or false (F).
a.) There are 6.0221023 atoms of sodium in 22.99 lbs of sodium. False
b.) In N2O4, the mass of oxygen is twice that of nitrogen. False
c.) One mole of chlorine atoms has a mass of 35.45 g. True
d.) Boron has two isotopes, B-10 (10.01 amu) and B-11 (11.01amu).
There is more naturally occurring B-10 than B-11. False
7.) Many phosphorus compounds are quite toxic, and they have been used as insecticides
and poison gases. The nerve gas Sarin, which was released in a Tokyo subway station in
1995, has a molecular formula of C4H10PO2F.
a.) Determine the composition (percent by mass) of each element in Sarin
b.) An unknown compound is believed to be Sarin. When a 10.0 gram sample of
this compound is completely combusted, 12.5 g CO2 and 6.4 g H2O are
produced, along with other combustion products. Using numerical
calculations show that this unknown compound could be Sarin.
c.) Do you have enough evidence to say that the compound is not C4H10P2O6Cl
a.)
4C = 412.0 g = 48.0 g
48.0/140.1 = 0.343 = 34.3%
10 = 10 g = 10.1 g
10.1/140.1 = .0721 = 7.21%
1P = 1 g = 31.0 g
31.0/140.1 = 0.221 = 22.1%
2O = 216.0 g = 32.0 g
32.0/140.1 = 0.228 = 22.8%
1F= 119.0 g = 19.0 g
19.0/140.1 = 0.136 = 13.6%
140.1 g
C 4 H 10 P O 2 F  XO 2  4CO 2  5H 2 O  etc
10 g C 4 H 10 P O 2 F 
b.) 10 g C 4 H 10 P O 2 F 
12.5 g CO 2 
6.4 g H 2 O 
34.3 g C
 3.43 g C
100 g C 4 H 10 P O 2 F
7.21 g H
 0.721 g H
100 g C 4 H 10 P O 2 F
12.0 g C
 3.41 g C
44.0 g CO 2
2.01 g H
 0.72 g H
18.0 g H 2 O
c.) The grams of carbon and hydrogen produced match within the significant
figures reported. If no other combustion product contain any carbon or hydrogen
this might indeed have been Sarin gas. This analysis only validates the carbon and
hydrogen. Any of the other elements may be in different proportions, or may have
different constituents. To see if the compound is C4H10P2O6Cl one would go through
the same procedure and find that the mass of the carbon and hydrogen do not
match.
8.) One form of phosphorus, known as white phosphorus, consists of molecules of P4.
Because white phosphorus is spontaneously flammable in air it is usually stored under
water.
a.) A graduated cylinder contains 120.0 mL of water. A piece of white
phosphorus weighing 10.412 g is submerged in a graduated cylinder and the
water level rises to 125.7 mL. What is the density of white phosphorus
b.) White phosphorus reacts with oxygen in the air to produce tetraphosphorus
decaoxide. Write and balance the chemical equation for this process.
c.) The same 10.412 g sample of white phosphorus is removed from water and
completely combusted in an excess of oxygen. What mass of product will be
formed?
a.) D  M  10.412 g
 0.08677 g
V
120.0 mL
mL
b.) P4  5O 2  P4 O10
c.) 10.412 g P4 
1 mol P4 O10
1 mol P4
 8.4050  10  2 mol P4 
 8.4050  10  2 mol P4 O10
123.88 g P4
1mol P4
8.4050  10  2 mol P4 O10 
283.88 g P4 O10
 23.860 g P4 O10
1 mol P4 O10
9.) Methanol, CH3OH, can be made by the reaction of carbon monoxide and hydrogen.
CO(g) + 2H2(g)  CH3OH(l)
Suppose 356g of CO are mixed with 65.0 g of H2.
a.) Which is the limiting reactant?
b.) What is the maximum mass of methanol that can be formed?
c.) What mass of the excess reactant remains after the limiting reactant has been
consumed?
d.) If only 310 g of CH3OH is produced, what is the percent yield?
If both elements are used to completion:
1 mol CO 1 mol CH 3 OH
a.) 356 g CO 

 12.7 mol CH 3 OH
28.0 g CO
1 mol CO
1 mol H 2 1 mol CH 3 OH
65.0 g H 2 

 16.1 mol CH 3 OH
2.02 g H 2
2 mol H 2
Since the CO produces the least amount of Methanol, it must be the limiting
reagent.
32.0 g CH 3 OH
b.) 12.7 mol CH 3 OH 
 406. g CH 3 OH
1 mol CH 3 OH
c.) 12.7 mol CH 3 OH 
2 mol H 2
2.02 g H 2

 51.3 g H 2 used
1 mol CH 3 OH 1 mol H 2
65.0 g - 51.3 g  13.7 g H 2 left
d .)
310 g CH 3 OH
406. g CH 3 OH
 100%  76%
Given the balanced equation: A + 3B  2C
The molar mass of C is 20.0 g/mol. If one mole of A produces 10.0 grams of C,
what is the percent yield of the reaction?
a.) 400%
b.) 50%
c.) 25%
d.) 10%
e.) 1.0%
10.)
One mole of A will produce 2 mols of C. One mole of A should then produce 40 g of
C (2 mol20.0g/mol). Only 10.0 g of C was produced so the % yield is
10 g
 100  25%
40 g