Chem 142
Name ________________________________________
Summer 2011
Section ________________
Worksheet 3: Stoichiometry
1. Calculate the average atomic mass of Zirconium (Zr). 90Zr (51.45%), 91Zr
(11.27%), 92Zr (17.17%), 94Zr (17.33%),96Zr (2.78%).
"( 90amu ! 0.5145 ) + ( 91amu ! 0.1127 ) %
$
'
average atomic mass = $ + ( 92amu ! 0.1717 ) + ( 94amu ! 0.1733) ' = 91.32amu
$ + ( 96amu ! 0.0278 )
'
#
&
2. Sucrose is the common sugar used in all homes, and chemical analysis
tells us that the chemical composition is 42.14% carbon, 6.48%
hydrogen and 51.46% oxygen. What is the molecular formula of sucrose
if its molar mass is approximately 340 g/mol?
Assume 100 g of sucrose.
1mol C
moles C = 42.14g C !
= 3.508mol C
12.011g C
1mol H
moles H = 6.48g H !
= 6.429mol H
1.008g H
1mol O
moles O = 51.46g O !
= 3.216mol O
16.00g O
empirical formula  CH2O
empirical equivalents =
340 g mol sucrose
= 11
30 g mol empirical
molecular formula  C11H22O11
3. Determine the empirical and molecular formulas for a deadly nerve gas
that gives the following mass percent analysis.
C = 39.10%
H = 7.67% O = 26.11% P = 16.82% F = 10.30%
The molar mass is known to be 184.1 g/mol.
Assume 100 g of compound.
1mol C
moles C = 39.10g C !
= 3.255mol C
12.011g C
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Chem 142
Summer 2011
1mol H
= 7.61mol H
1.008g H
1mol O
moles O = 26.11g O !
= 1.632mol O
16.00g O
1mol P
moles P = 16.82g P !
= 0.5431mol P
30.97g P
1mol F
moles F = 10.30g F !
= 0.54211mol F
19.00g F
moles H = 7.76g H !
empirical formula  C6H14O3PF
empirical equivalents =
184.1g mol compound
=1
184.1g mol empirical
molecular formula  C6H14O3PF
4. Erythrose (M = 120 g/mol) is an important chemical compound used
often as a starting material in chemical synthesis and contains carbon,
hydrogen, and oxygen. Combustion analysis of a 700.0 mg sample
yielded: 1.027 g CO2 and 0.4194 g H2O. Determine the molecular
formula.
12.011g C
1mol C
!
= 0.02334mol C
44.011g CO2 12.011g C
2.016g H
1mol H
moles H = 6.48g H 2O !
!
= 0.04656mol H
18.016g H 2O 1.008g H
1mol O
moles O = ( 0.700g erythrose ! 0.2803g C ! 0.04693g H ) "
= 0.02330mol O
16.00g O
moles C = 1.027g CO2 !
empirical formula  CH2O
empirical equivalents =
120 g mol sucrose
=4
30 g mol empirical
molecular formula  C4H8O4
5. A combustion device was used to determine the empirical formula of a
compound containing ONLY carbon, hydrogen and oxygen. A 0.6349 g
sample of the unknown produced 1.603 g of CO2 and 0.2810 g H2O.
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Chem 142
Summer 2011
Determine the empirical formula of the compound. Assuming the
empirical and molecular formulas are the same, what is the balanced
chemical reaction for the combustion of this liquid?
12.011g C
1mol C
!
= 0.03642mol C
44.011g CO2 12.011g C
2.016g H
1mol H
moles H = 0.2810g H 2O !
!
= 0.03119mol H
18.016g H 2O 1.008g H
1mol O
moles O = ( 0.700g erythrose ! 0.4374g C ! 0.03144g H ) "
= 0.01038mol O
16.00g O
moles C = 1.603g CO2 !
empirical formula  C3.5H3O
Need whole numbers so multiply everything by 2 to get: C7H6O2
2C7H6O2 (l)+ 15O2 (g)  6H2O (g) + 14CO2 (g)
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