Download 2 • Elements and Compounds

South Pasadena • Honors Chemistry
Name
2 • Elements and Compounds
Period
2.5
PROBLEMS
1. Write the balanced formation equation for the
following compounds. Indicate the states.
(a) N2O (g)
2 N2 (g) + O2 (g) → 2 N2O (g)
(b) NaCl (s)
2 Na (s) + Cl2 (g) → 2 NaCl (s)
(c) PBr3 (ℓ)
OR
2 P (s) + 3 Br2 (ℓ) → 2 PBr3 (ℓ)
P4 (s) + 6 Br2 (ℓ) → 4 PBr3 (ℓ)
(d) SO3 (g)
OR
2 S (s) + 3 O2 (g) → 2 SO3 (g)
S8 (s) + 12 O2 (g) → 8 SO3 (g)
2. Write the balanced decomposition equation for the
following compounds. Indicate the states.
(a) Al2O3 (s)
2 Al2O3 (s) → 4 Al (s) + 3 O2 (g)
(b) CaO (s)
2 CaO (s) → 2 Ca (s) + O2 (g)
(c) OF2 (g)
2 OF2 (g) → O2 (g) + 2 F2 (g)
(d) HI (g)
2 HI (g) → H2 (g) + I2 (s)
3. A sample of the common amino acid lysine was
found to have 226.4 mg of carbon, 44.3 mg of
hydrogen, 88.0 mg of nitrogen, and 100.5 mg of
oxygen.
(a) Find the % composition of lysine.
Total mass = 226.4 mg C + 44.3 mg H
+ 88.0 mg N + 100.5 mg O = 459.2 mg
226.4 mg C
%C =
× 100% = 49.30% C
459.2 mg
44.3 mg H
%H =
× 100% = 9.65% H
459.2 mg
88.0 mg N
%N =
× 100% = 19.2% N
459.2 mg
100.5 mg O
%O =
× 100% = 21.89% O
459.2 mg
(b) Find the empirical compound for lysine.
Assume 100 g sample.
1 mol C 
nC = 49.30 g C 
12.01 g C = 4.105 mol C
1 mol H 
nH = 9.65 g H 
1.008 g H = 9.57 mol H
1 mol N 
nN = 19.2 g N 
14.01 g N = 1.37 mol N
1 mol O 
nO = 21.89 g O 
16.00 g O = 1.368 mol O
–
Date
CHEMICAL ANALYSIS
Empirical Formula:
C4.105 H 9.57 N 1.37 O1.368
1.368
1.368
1.368
1.368
 C3H7NO
(c) If the molar mass of lysine is 146.19 g/mol,
find the molecular formula for lysine.
Molar Mass of Lysine
=
Empirical Formula Mass
146.19 g/mol
=
3(12.01) + 7(1.008) + 1(14.01) + 1(16.00)
146.19 g/mol 2
=
73.10 g/mol
1
Molecular Formula = (Empirical Formula) ×
2 = (C3H7NO) × 2 = C6H14N2O2
4. 0.85 mol of a sample of benzene, composed of only
carbon and hydrogen, has a mass of 66.4 g.
(a) What is the molar mass of benzene?
Molar Mass =
total mass
66.4 g
=
total moles 0.85 mol
= 78 g/mol
(b) How any molecules of benzene are in this
sample?
66.4 g benzene 
1 mol benzene
 78 g benzene 
6.022 × 10 molecules benzene
1 mol benzene


23
= 5.1 × 1023 molecules benzene
(c) If benzene contains 7.74% H, find the empirical
formula and molecular formula of the
compound.
1 mol C 
nC = 92.26 g C 
12.01 g C = 7.681 mol C
1 mol H 
nH = 7.74 g H 
= 7.68 mol H
1.008
g H

Empirical Formula: C7.681 H7.68  CH
7.68
7.68
Molecular Formula:
Molar Mass
=
Empirical Formula Mass
78 g/mol
78 g/mol
6
=
=
1(12.01) + 1(1.008) 13.02 g/mol 1
Molecular Formula
= (Empirical Formula) × 6 = (CH) × 6
= C6H6
(d) If the density of benzene is 0.876 g/mL, find
the volume of this sample.
66.4 g C6H6 
1 mL C6H6 
0.876 g C6H6 = 75.8 mL
C6H6(Note: do not use 1 mol = 22.4 L since
it’s not a gas at STP)
5. A compound was analyzed and found to contain
53.30% carbon, 11.19% hydrogen, and 35.51%
oxygen by mass.
(a) Calculate the empirical formula of the
compound.
Assume 100 g sample.
1 mol C 
nC = 53.30 g C 
12.01 g C = 4.438 mol C
1 mol H 
nH = 11.19 g H 
= 11.10 mol H
1.008 g H
1 mol O 
nO = 35.51 g O 
16.00 g O = 2.219 mol O
Empirical Formula: C4.438 H11.10 O2.219 
2.219
2.219
P + Se → PxSey
nP:
45.2 g P 
1 mol P 
30.97 g P = 1.46 mol P
nSe:
(131.6 g – 45.2 g)
= 86.4 g Se 
Empirical Formula:
1 mol Se 
78.96 g Se = 1.09 mol Se
P1.46 Se1.09  P1.33Se 
1.09
1.09
P4Se3
8. An 80.50 g sample of an unknown oxide of
chromium (CrxOy) is heated until all of the oxygen
is vaporized, leaving 55.08 g of solid chromium
metal. The molar mass of this unknown compound
is between 150 and 160 g/mol.
(a) Find the % composition of each element of the
unknown compound.
2.219
C2H5O
(b) If the molar mass of the compound is 90.12
g/mol, what is the molecular formula of the
compound?
Molar Mass
=
Empirical Formula Mass
90.12 g/mol
90.12 g/mol
=
2(12.01) + 6(1.008) + 1(16.00) 46.07 g/mol
2
=
1
Molecular Formula
= (Empirical Formula) × 2 = (C2H5O) × 2
= C4H10O2
6. A 0.77 g sample of nitrogen reacts with chlorine to
form 6.61 g of the chloride. Determine the
empirical formula of nitrogen chloride.
N2 + Cl2  NxCly
1 mol N 
0.77 g N 
14.01 g N = 0.0550 mol N
1 mol Cl 
nCl:
(6.61 g – 0.77 g) = 5.84 g Cl 
35.45 g Cl
= 0.165 mol Cl
Empirical Formula: N0.0550 Cl 0.165  NCl3
(b) Find the empirical formula and molecular
formula of the unknown compound.
(c) Write the balanced chemical equation for this
decomposition. Include states.
nN:
0.0550
(d) Find the volume at STP of the oxygen gas that
is vaporized.
0.0550
7. A 45.2 g sample of phosphorous reacts with
selenium to form 131.6 g of the selenide. Find the
empirical formula of phosphorous selenide.
(e) Find the number of chromium atoms in this
sample.
9. Tartaric acid is the white, powdery substance that
coats tart candies such as Sour Patch Kids.
Combustion analysis of a 12.01 g sample of tartaric
acid – which contains only carbon, hydrogen, and
oxygen – produced 14.08 g CO2 and 4.32 g H2O.
Determine the empirical formula of tartaric acid.
CxHyOz (s) + O2 (g) → CO2 (g) + H2O (ℓ)
(a) Determine the empirical formula of tartaric
acid.
CxHyOz (s) + O2 (g) → CO2 (g) + H2O (ℓ)
MM:
1(12.01) + 2(16.00)
= 44.01 g/mol
2(1.008) + 1(16.00)
= 18.016 g/mol
Carbon:
14.08 g CO2 
1 mol CO2   1 mol C 
44.01
g CO2 1 mol CO2

12.01 g C
= 0.3199 mol C 
 1 mol C  = 3.843 g C
Hydrogen:
1 mol H2O   2 mol H 
4.32 g H2O 
18.02 g H2O 1 mol H2O
1.008 g H
= 0.480 mol H 
 1 mol H  = 0.483 g H
Oxygen:
12.01 g CxHyOz – (3.843 g C + 0.483 g H)
= 7.68 g O
1 mol O 
7.68 g O 
16.00 g O = 0.480 mol O
Empirical Formula: C0.3199 H 0.480 O 0.480
0.3199
 CH1.5O1.5
0.3199
0.3199
 C2H3O3
(b) If the molar mass of tartaric acid is 150 g/mol,
find its molecular formula.
Molecular Formula:
Molar Mass
Empirical Formula Mass
150 g/mol
=
2(12.01) + 3(1.008) + 3(16.00)
150 g/mol
2
=
=
75.04 g/mol 1
Molecular Formula
= (Empirical Formula) × 2
= (C2H3O3) × 2 = C4H6O6