Key for Week 3 Worksheet
Problem #1
a). Use the results of the analysis to find the fractional mass composition of each element in the
compound. Convert these results to the smallest possible molar ratios. This will yield the
empirical formula, since there is a 1:1 correspondence between moles of atoms in a compound and
the number of atoms in one unit of compound.
b). First get the empirical formula, then use the mass spectrometer results to obtain the molecular
formula.
N: This is already given as a percent composition, so we simply assume 1 g of substance and
have 0.2885 g N.
C: 13.56 mg CO2 x
H: 3.47 mg H2O x
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= 3.701 mg C à
= 0.388 mg H à
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= 49.48% C
= 5.19% H
O: 100% - (28.85% from N + 49.48% from C + 5.19% from H) = 16.48% O
So now assume 1 g of substance and we have 0.2885 g N, 0.4948 g C, 0.0519 g H, and 0.1648 g
O. Calculate the moles of each atom:
0.2885 g N x
0.4948 g C x
0.0519 g H x
0.1648 g O x
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= 0.0206 mole N
= 0.0412 mole C
= 0.0515 mole H
= 0.0103 mole O
For smallest possible ratios we get (round to nearest integer):
N:
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= 2; C:
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= 4; H:
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= 5; O:
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=1
So, empirical formula is C4H5N2O and which has a molecular mass of 97.10 u.
The mass spectrometer results indicate a molecular mass of 190–210 u. The range of integer
values would thus be 190/97.10 = 1.96 to 210/97.10 = 2.16. In any case, 2 is the only close
integer, so the molecular formula must be C8H10N4O2.
Problem #2
a). To balance the oxidation of caffeine, you could start
with any of carbon, hydrogen, or nitrogen, since all
those elements occur in one species on each side of
the equation. Oxygen must be balanced last, since it
occurs in all the product species. See below:
Balance C:
C8H10N4O2 + O2 à 8CO2 + H2O + NO2
Balance H:
C8H10N4O2 + O2 à 8CO2 + 5H2O + NO2
Balance N:
C8H10N4O2 + O2 à 8CO2 + 5H2O + 4NO2
Caffeine
Balance O:
C8H10N4O2 +
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O2 à 8CO2 + 5H2O + 4NO2
Eliminate fraction:
2C8H10N4O2 + 27O2 à 16CO2 + 10H2O + 8NO2
b). 0.0385 mole caffeine x
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= 0.192 mole H2O
c). Strategy: convert mass to moles and apply relationship above to get moles of water, and then
convert back to mass.
7.48 g caffeine x
d). 7.48 g caffeine x
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x
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= 3.47 g H2O
= 16.6 g O2
Problem #3
a). Let’s say we only had 10.0 g of oxygen:
10.0 g O2 x
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x
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= 2.09 g H2O
b). Note that while not explicitly displayed, 7.48 g caffeine does correspond to the 0.0385 moles of
caffeine given in problem 2b. Now we simply get the yield as:
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x 100% = 60.2% yield
Student’s answers will vary depending upon their chosen mass of oxygen.