CHEMISTRY
Question 2
2. Answer the following questions related to hydrocarbons.
(a) Determine the empirical formula of a hydrocarbon that contains 85.7 percent carbon by mass.
1 mol C
= 7.14 mol C
12.01 g C
1 mol H
= 14.3 g H ×
= 14.2 mol H
1.008 g H
nC = 85.7 g C ×
nH
1 point for moles of C and moles of H
7.14 mol C 14.2 mol H
:
7.14
7.14
1 mol C : 1.99 mol H
1 point for ratio of moles of C to moles of H
The empirical formula is CH2
1 point for correct formula
(b) The density of the hydrocarbon in part (a) is 2.0 g L–1 at 50°C and 0.948 atm.
(i) Calculate the molar mass of the hydrocarbon.
mass
RT
molar mass
mass
RT
RT
= density ×
molar mass =
×
V
P
P
L atm
0.0821
× 323 K
mol K
molar mass = 2.0 g L−1 ×
0.948 atm
−1
molar mass = 56 g mol
PV = nRT =
1 point for correct substitution
1 point for the answer
(ii) Determine the molecular formula of the hydrocarbon.
empirical mass × n = molar mass
empirical mass for CH2 = 14 g mol−1
14 g mol−1 × n = 56 g mol−1
n = 4
The molecular formula is C4H8.
1 point for correct formula CHEMISTRY
Question 2 (cont’d.)
(c) Two flasks are connected by a stopcock as shown below. The 5.0 L flask contains CH4 at a pressure of
3.0 atm, and the 1.0 L flask contains C2H6 at a pressure of 0.55 atm. Calculate the total pressure of the
system after the stopcock is opened. Assume that the temperature remains constant.
Pf of CH4 =
Pf of C2H6 =
PV
(3.0 atm) (5.0 L)
i i
=
= 2.5 atm CH4
Vf
6.0 L
1 point for final pressure of CH4 or C2H6
PV
(0.55 atm) (1.0 L)
i i
=
= 0.092 atm C2H6
Vf
6.0 L
PT = Pf CH4 + Pf C2H6 = 2.5 atm + 0.092 atm = 2.6 atm
1 point for the total pressure
(d) Octane, C8H18(l), has a density of 0.703 g mL–1 at 20°C. A 255 mL sample of C8H18(l) measured at
20°C reacts completely with excess oxygen as represented by the equation below.
2 C8H18(l) + 25 O2(g) → 16 CO2(g) + 18 H2O(g)
Calculate the total number of moles of gaseous products formed.
n products = 255 mL C8H18 ×
0.703 g C8 H18
1 mol C8 H18
×
×
1 mL C8 H18
114 g C8 H18
34 mol products
= 26.7 mol products
2 mol C8 H18
1 point for substitution of any of these
conversion factors
1 point for the correct answer
Chem Gases Key
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