Chemistry 2100
Problem Set #4: Chapters 5 and 10
1.
(a)
(b)
(c)
(d)
(e)
2.
(a)
(b)
(c)
Hydrogen gas can be reacted with chlorine gas to give hydrogen chloride.
Write a balanced equation for this reaction.
H2
+
Cl2
→
2 HCl
If 0.623 g of chlorine gas is reacted with excess hydrogen gas, what mass of hydrogen
chloride is produced?
MCl2 = 70.906 g/mol
MHCl = 36.461 g/mol
(i)
moles Cl2 = 0.623 g × 1 mol .
70.906 g
moles Cl2 = 0.00879 mol
(ii)
moles HCl = 0.00879 mol Cl2 × 2 mol HCl
1 mol Cl2
moles HCl = 0.0176 mol HCl
(iii)
mass HCl = 0.0176 mol × 36.461 g
1 mol
mass HCl = 0.641 g
***3 sig. fig.***
If all of the hydrogen chloride produced in part (b) is dissolved in water to make 1.000 L of
hydrogen chloride solution, what is the pH of the solution?
(i)
concentration HCl = 0.0176 mol
1.000 L
concentration HCl = 0.0176 mol/L
(ii)
concentration H+ = concentration HCl (since HCl is a strong acid)
concentration H+ = 0.0176 mol/L
***3 sig. fig.***
+
(iii) pH = –log[H ]
= –log(0.0176)
pH = 1.754
***3 decimal places (since there are 3 sig. fig. in [H+])***
Is the solution described in part (c) acidic or basic?
acidic (pH is less than 7 and HCl is a strong acid)
If the reaction described in part (b) actually produced 0.625 g of hydrogen chloride,
calculate the percent yield.
actual yield .× 100%
% yield =
theoretical yield
= (0.625 g) × 100%
(0.641 g)
***theoretical yield is answer to part (b)***
% yield = 97.5%
For each of the following reactions, identify the acid, the
conjugate base.
HCO2H
+
HCO3→
HCO2acid
base
conj. base
C2H5OH
+
H2SO4
→
C2H5OH2+
base
acid
conj. acid
+
H2O
→
H3O+
H2 O
base
acid
conj. acid
base, the conjugate acid and the
+
+
+
H2CO3
conj. acid
HSO4conj. base
OHconj. base
3.
Write two different acid-base reaction equations where the only reactants are water and the
bisulfite ion (HSO3-).
+
HSO3→
HO+
H2SO3
H2O
acid
base
H2O
base
4.
(a)
(b)
(c)
(d)
+
HSO3acid
→
H3O+
+
SO32-
A 2.50 mg sample of an unknown compound was analyzed by mass spectrometry and found
to have a molecular formula of C25H35N5.
What is the empirical formula of the unknown compound?
C5H7N
Calculate the percent composition of the unknown compound.
(i)
MC5H7N = 81.117 g/mol
(ii)
%C =
MC5 × 100%
MC5H7N
= (5 × 12.011 g/mol) × 100%
(81.117 g/mol)
% C = 74.035%
***5 sig. fig. (100% is an exact number)***
(iii) %H = MH7 × 100%
MC5H7N
= (7 × 1.0079 g/mol) × 100%
(81.117 g/mol)
% H = 8.6977%
***5 sig. fig. (100% is an exact number)***
(iv)
%N =
MN × 100%
MC5H7N
= (14.007 g/mol) × 100%
(81.117 g/mol)
% N = 17.268%
***5 sig. fig. (100% is an exact number)***
Calculate the number of molecules of unknown compound in the sample analyzed.
(i)
MC25H35N5 = 405.59 g/mol
(ii)
moles C25H35N5 = 2.50 mg ×
1g
× 1 mol .
1000 mg
405.59 g
moles C25H35N = 0.00000616 mol (or 6.16 × 10-6 mol)
(iii)
molecules C25H35N = 0.00000616 mol × 6.022 × 1023 molecules
1 mol
18
molecules C25H35N = 3.71 × 10 molecules
***3 sig. fig.***
Calculate the number of hydrogen atoms in the sample analyzed.
35 atoms H
.
atoms H = 3.71 × 1018 molecules C25H35N ×
1 molecule C25H35N
atoms H = 1.30 × 1020 atoms H
***3 sig. fig.***
5.
(a)
(b)
(c)
6.
An unknown compound was subjected to elemental analysis and found to contain 36.70% K,
33.27% Cl and 30.03% O.
Calculate the empirical formula of the unknown compound.
(i)
100 g unknown contains 36.70 g K, 33.27 g Cl and 30.03 g O.
(ii)
moles K = 36.70 g × 1 mol .
39.098 g
moles K = 0.9387 mol
(iii)
moles Cl = 33.27 g × 1 mol .
35.453 g
moles Cl = 0.9384 mol
(iv)
moles O = 30.03 g × 1 mol .
15.999 g
moles O = 1.877 mol
(v)
0.9387 mol K : 0.9384 mol Cl : 1.877 mol O = 1 mol K : 1 mol Cl : 2 mol O
0.9384
0.9384
0.9384
(vi)
Empirical formula is KClO2
If its molecular formula is the same as its empirical formula, name the unknown compound.
potassium chlorite
The unknown compound is either a Brønsted-Lowry acid or base. Which is it?
It must be a base. It has no H+ to give up.
A balanced reaction equation for the complete combustion of propane (C3H8) is shown:
C3H8
(a)
(b)
+
5 O2
→
3 CO2
+
4 H2O
If this reaction proceeds with a 95.4% yield and burning a cylinder of propane produced
34.6 g of CO2, calculate the theoretical yield of CO2.
actual yield .× 100%
% yield =
theoretical yield
theoretical yield = actual yield.× 100%
% yield
= 34.6 g × 100%
95.4%
theoretical yield = 36.3 g
***3 sig. fig.***
Calculate the mass of propane burned.
***Use theoretical yield not actual yield!***
(i)
MC3H8 = 44.096 g/mol
MCO2 = 44.009 g/mol
(ii)
moles CO2 = 36.3 g × 1 mol .
44.009 g
moles CO2 = 0.824 mol
(iii)
moles C3H8 = 0.824 mol CO2 × 1 mol C3H8
3 mol CO2
moles C3H8 = 0.275 mol C3H8
(iv)
mass C3H8 = 0.275 mol × 44.096 g
1 mol
mass C3H8 = 12.1 g
***3 sig. fig.***
(c)
Calculate the mass of oxygen used to burn the propane.
moles CO2 = 0.824 mol (part b)
(i)
MO2 = 31.998 g/mol
(ii)
moles O2 = 0.824 mol CO2 × 5 mol O2 .
3 mol CO2
moles O2 = 1.37 mol O2
(iii)
mass O2 = 1.37 mol × 31.998 g
1 mol
mass O2 = 43.9 g
***3 sig. fig.***
7.
(a)
(b)
(c)
Predict the products for each of the following reactions:
LiOH
+
HClO2
→
LiClO2
Mg
+
H2SO4
→
MgSO4
HBr
+
NH4OH
→
NH4Br
+
+
+
H2O
H2
H2O