7.A – Types of Compounds and DA practice
Instructions: Determine the electronegativity difference (ΔEN) for the following chemical bonds in the indicated blanNa. No worNa need be shown. Then using the ΔEN
determine if the bond is considered a nonpolar covalent bond [NC], polar covalent [ PC ], ionic bond [ I ], or metallic bond [M].
ΔEN
Bond Type
ΔEN
Bond Type
ΔEN
Bond Type
1. Li – H
│0.98− 2.20│= 1.22
= [ PC ]
14. Cu – Zn
│1.90− 1.65│= 0.25
=[M ]
27. C – H
│2.55− 2.20│= 0.35
= [ NC ]
2. Li – Cl
│0.98− 3.16│= 2.18
=[I ]
15. Ca – Cl
│1.00− 3.16│= 2.16
=[I ]
28. Si – H
│1.90− 2.20│= 0.30
= [ NC ]
3. Na – H
│0.93− 2.20│= 1.27
= [ PC ]
16. Ca – S
│1.00− 2.58│= 1.58
= [ PC ]
29. C – O
│2.55− 3.34│= 0.89
= [ PC ]
4. Na – Cl
│0.93− 3.16│= 2.23
=[I ]
17. Sr – F
│0.95− 4.00│= 3.05
=[I ]
30. Al – Cl
│2.20− 3.16│= 0.96
= [ PC ]
5. K – H
│0.82− 2.20│= 1.38
= [ PC ]
18. Ba – F
│0.89− 4.00│= 3.11
=[I ]
31. C – Cl
│2.55− 3.16│= 0.61
= [ PC ]
6. K – Cl
│0.82− 3.16│= 2.34
=[I ]
19. Fr – F
│0.70− 4.00│= 3.30
=[I ]
32. H – H
│2.20 − 2.20│= 0.00
= [ NC ]
7. Fr – N
│0.70− 3.04│= 2.34
=[I ]
20. H – O
│2.20 – 3.34│= 1.24 = [ PC ]
33. S – O
│2.58− 3.34│= 0.86
= [ PC ]
8. Fr – Cl
│0.70− 3.16│= 2.46
=[I ]
21. H – S
│2.20−2.58 │= 0.38
= [ NC ]
34. P – O
│2.19− 3.34│= 1.25
= [ PC ]
9. Fe – Fe
│1.83− 1.83│= 0.00
=[M ]
22. H – Se
│2.20−2.55│= 0.35
= [ NC ]
35. Cs – F
│0.79− 4.00│= 3.21
=[I ]
10. Mg – F
│1.31− 4.00│= 2.69
=[I ]
23. H – Br
│2.20−2.96│= 0.76
= [ PC ]
36. O – O
│3.34− 3.34│= 0.00
= [ NC ]
11. Mg – O
│1.31− 3.34│= 2.13
=[I ]
24. H – I
│2.20−2.66│= 0.46
= [ NC ]
37. N – O
│3.04− 3.34│= 0.40
= [ NC ]
12. Cu – Cu
│1.90− 1.90│= 0.00
=[M ]
25. N – H
│3.04− 2.20│= 0.84
= [ PC ]
38. Fe – Co │1.83− 1.88│= 0.05
=[M ]
13. Ca – O
│1.00− 3.34│= 2.44
=[I ]
26. P – H
│2.19− 2.20│= 0.01
= [ NC ]
39. I – Br
= [ NC ]
│2.66− 2.96│= 0.30
Instructions: ON A SEPARATE SHEET OF PAPER, use dimensional analysis to perform the following conversions related to the compounds. You MUST use dimensional
analysis to preform each conversion. Remember, Sig Figs, NW = NC, Boxed Answers and N3 (this includes labeling and in your set-up!!!)
40) 50.25 g C6H12
50.25 mol C6H12
→ mole C6H12
1 mol C6H12
84.18 g C6H12
41) 38.1 mole CuFeS2
38.1 mol CuFeS2
→ g CuFeS2
184.54 g CuFeS2
1 mol CuFeS2
= 0.5969 mol C6H12
= 7030 mol C6H12
42) 1.2x1024 FU C15H11NO4I4 → mol C15H11NO4I4
1.2 x 1024 FU C15H11NO4I4
1 mol C15H11NO4I4
6.022 x 1023 FU C15H11NO4I4
= 2.0 mol C15H11NO4I4
43) 13.95 mol Al2O3 ▪ 3 H2O → FU Al2O3 ▪ 3 H2O
13.95 mol Al2O3 ▪ 3 H2O
6.022 x 1023 FU Al2O3 ▪ 3 H2O
1 mol Al2O3 ▪ 3 H2O
= 8.401 x 1024 FU Al2O3 ▪ 3 H2O
44) 2.54 x 1021 FU Al4C3
→ g Al4C3
21
2.54 x 10 FU Al4C3
1 mol Al4C3
6.022 x 1023 FU Al4C3
45) 42.3 g Fe
42.3 g Fe
→ FU Fe
46) 34.512 g NaOH
34.512 g NaOH
→ FU OH-
143.95 g Al4C3
1 mol Al4C3
= 0.607 gl Al4C3
1 mol Fe
55.85 g Fe
6.022 x 1023 FU Fe
1 mol Fe
= 4.56 x 1023 FU Fe
1 mol NaOH
40.00 g NaOH
6.022 x 1023 FU NaOH
1 mol NaOH
1 FU OH−
1 FU NaOH
1 mol FeCl3
6.022 x 1023 FU FeCl3
162.20 g FeCl3
1 mol FeCl3
1 mol F
6.022 x 1023 FU F
19.00 g F
1 mol F
47) 1.23x1024 FU Fe3+
→ g FeCl3
1.23 x 1024 FU Fe3+
1 FU FeCl3
1 FU Fe3+
48) 1.2x1024 FU Ag2SiF6
→gF
1.2 x 1024 FU Ag2SiF6
6 FU F
1 FU Ag2SiF6
= 5.1958 x 1023 FU NaOH
= 230 g F
= 331 g FeCl3
7.B – Lewis Structures
Instructions: Draw Lewis Structures for each of the following atoms and then draw the Lewis Structures for its most common ion. Tell which has a larger radius: the atom or
the ion. Tell which has a larger radius: the atom or the ion.
Sodium
Atom
# Val e
Lead Atom
# Val e
-
_1_
# Val e
# Val e
-
_0_
-
_4_
Strontium
Atom
Sodium Ion
-
_2_
Lead (IV)
Ion
# Val e
Strontium
Ion
# Val e
-
_0_
_0_
-
Calcium
Atom
# Val e
Nitrogen
Atom
# Val e
Selenium
Atom
# Val e
-
_2_
-
Calcium
Ion
# Val e
Nitride Ion
# Val e
_5_
-
_0_
-
Aluminum
Atom
# Val e
Sulfur Atom
# Val e
_8_
-
_6_
Selenide
Ion
# Val e
_8_
-
_3_
-
Aluminum
Ion
# Val e
Sulfide Ion
# Val e
_6_
-
Bromine
Atom
# Val e
_7_
-
_0_
-
_8_
-
Bromide
Ion
# Val e
-
_8_
Instructions: For problems 19 – 30, given the following Lewis structures, provide the formula they represent. REMEMBER, the more EN atom is written last.
Formula
_HF_
_H2C2HOCO2H_
Formula
_N2_
Formula
_NH3_
Formula
_CH4_
Formula
_CF4_
Formula
_NO+_
Formula
C2H6O
Formula
_C6H6_
Formula
_PF5_
Formula
_NH4+_
Formula
_ClO2−_
7.C – VSEPR
Instructions: Given the following Lewis Structures, complete the table as indicated. Be sure to use your “Teal Sheet” as a reference.
Lewis Structure
Chemical
Formula
Central
Atom
# of
Ligands
# of
Lone
Pairs
AXmEn
Ligands
VSEPR
Geometry
BeCl2
Be
2
0
AX2
BH3
B
3
0
COH2
C
3
CO2
C
2
Lewis Structure
Chemical
Formula
Central
Atom
# of
Ligands
# of
Lone
Pairs
AXmEn
Notation
VSEPR
Geometry
Linear
C2H2
C
2
0
AX2
Linear
AX3
Trigonal
Planar
C2H4
C
3
0
AX3
Trigonal
Planar
0
AX3
Trigonal
Planar
C2H4
C
3
0
AX3
Trigonal
Planar
0
AX2
Linear
HF
F
1
3
AXE3
Linear
Instructions: For each write the correct Lewis electron-dot structure for the compound ON A SEPARATE SHEE OF PAPER. All 3 steps must be shown for CR.
Then complete the table
Formula
H2O
Correct Lewis Structure
Central Atom(s)
# of Ligands
on each central
# of Lone Pairs
on each central
AXmEn Notation
on each central
VSEPR Geometry
of each central
O
2
2
AX2E2
Bent
N
4
0
AX4
Tetrahedral
O
3
1
AX3E
Trigonal
Pyramidal
C1
4
0
AX4
Tetrahedron
O1
2
2
AX2E2
Bent
C
4
0
AX4
Tetrahedron
C1
3
0
AX3
Trigonal
Planar
O1
2
2
AX2E2
Bent
C1
4
0
AX4
Tetrahedron
C2
3
0
AX3
Trigonal
Planar
O1
2
2
AX2E2
Bent
NH4+
H3O+
CH3OH
C4H8 (Hint: ring)
CO2H2
CH3COOH
7.D – Nomenclature Review and % Composition
Instructions: ON A SEPARATE SHEET OF PAPER, write the formula or the name for the following compounds.
1. copper(II) bromate
7. lead(IV) sulfite
13. PbO2
2+
−
4+
2−
= Ionic = Cu BrO3
= Ionic = Pb SO4
= Ionic = Pb4+O2−
= Cu(BrO3)2
= Pb(SO4)2
= Lead (IV) Oxide
2. dichlorine heptoxide
8. dihydrogen monosulfide
14. S2O5
= Molecular
= Molecular
= Molecular
= Cl2O7
= H2S
= Disulfur Pentoxide
3. copper(II) sulfide
9. lead(IV) acetate
15. PbO
= Ionic = Cu2+S2−
= Ionic = Pb4+C2H3O2−
= Ionic = Pb2+O2−
= CuS
= Pb(C2H3O2)4
= Lead (II) Oxide
4. dihydrogen dioxide
10. hydrogen chloride
16. N2O
= Molecular
= Ionic = H+Cl−
= Molecular
= H2O2
= HCl
= Dinitrogen monoxide
5. copper(II) phosphate
11. lead(IV) phosphide
17. (NH4)3PO3
= Ionic = Cu2+PO43−
= Ionic = Pb4+P3−
= Ionic = NH4+PO33−
= Cu3(PO4)2
= Pb3P4
= Ammonium Phosphite
6. nitrogen monoxide
12. sulfur trioxide
18. CCl4
= Molecular
= Molecular
= Molecular
= NO
= SO3
= Carbon Tetrachloride
19. (NH4)2S
= Ionic = NH4+S2−
= Ammonium Sulfide
20. ClO
= Molecular
= Chlorine monoxide
21. NH4C2H3O2
= Ionic = NH4+ C2H3O2−
= Ammonium Acetate
22. H2O
= Molecular
= Dihydrogen monoxide
23. BaO
= Ionic = Ba2+O2−
= Barium Oxide
24. PF3
= Molecular
= Phosphorous Trifluoride
Instructions: ON A SEPARATE SHEET OF PAPER, answer the following set of questions related to % composition. Remember, Sig Figs, NW = NC, Boxed Answers and
N3 (this includes labeling and in your set-up!!!)
25. First write the chemical formula, then determine the % composition of each
26. Calculate the theoretical mass % composition of water (That is to say….
element in each the following compounds.
determine the mass % of each element).
a. carbon tetrachloride
= Molecular
= H2O
Molar Mass = 18.02 g/mol H2O
= Molecular
= CCl4
Molar Mass = 153.81 g/mol CCl4
% H = (2 x 1.01 g/mol H) ÷ (18.02 g/mol) x 100% = 11.2 %H
% C = (1 x 12.01 g/mol C) ÷ (153.81 g/mol) x 100% = 7.808 %C
% O = (1 x 16.00 g/mol O) ÷ (18.02 g/mol) x 100% = 88.79 %O
% Cl = (4 x 35.45 g/mol C) ÷ (153.81 g/mol) x 100% = 92.19 %Cl
b. potassium sulfite
27. Perform a theoretical elemental analysis of aluminum nitrate. (That is to
say….. calculate the mass % composition of each element.)
= Ionic = K+SO32− = K2SO3
Molar Mass = 158.27 g/mol K2SO3
% K = (2 x 39.10 g/mol K) ÷ (158.27 g/mol) x 100% = 49.41 %K
= Ionic = Al3+NO3− = Al(NO3)3
Molar Mass = 213.01 g/mol Al(NO3)3
% S = (4 x 32.07 g/mol S) ÷ (158.27 g/mol) x 100% = 20.26 %S
% Al = (1 x 26.98 g/mol Al) ÷ (213.01 g/mol) x 100% = 12.67 %Al
% O = (4 x 48.00 g/mol O) ÷ (158.27 g/mol) x 100% = 30.33 %O
% N = (3 x 14.01 g/mol N) ÷ (213.01 g/mol) x 100% = 19.73 %N
c. barium hydroxide
% O = (9 x 16.00 g/mol O) ÷ (213.01 g/mol) x 100% = 67.602 %O
2+
−
= Ionic = Ba OH = Ba(OH)2
Molar Mass = 171.34 g/mol Ba(OH)2
28. If you experimentally analyzed magnesium oxide and found that it was made
% Ba = (1 x 137.32 g/mol Ba) ÷ (171.34 g/mol) x 100% = 80.145 %Ba
of 0.180 g of magnesium and 0.119 g of oxygen, calculate the % (by mass) of
% O = (2 x 16.00 g/mol O) ÷ (171.34 g/mol) x 100% = 18.68 %O
each compound.
% H = (2 x 1.01 g/mol H) ÷ (171.34 g/mol) x 100% = 1.179 %H
Total Mass = 0.180 g + 0.119 g = 0.299 g
d. gallium (III) chromate
% Mg = 0.180 g ÷ 0.219 g x 100% = 60.2 %N
= Ionic = Ga3+CrO42− = Ga2(CrO4)3 Molar Mass = 487.44 g/mol Ga2(CrO 4)3
% O = 0.119 g ÷ 0.219 g x 100% = 39.8%O
% Ga = (2 x 69.72 g/mol Ga) ÷ (487.44 g/mol) x 100% = 28.172 %Ga
% Cr = (3 x 52.00 g/mol O) ÷ (487.44 g/mol) x 100% = 32.004 %Cr
% O = (12 x 16.00 g/mol H) ÷ (487.44 g/mol) x 100% = 39.389 %O
29. What mass of aluminum would you be able to extract from 1.65 g of
aluminum oxide?
= Ionic = Al3+O2− = Al2O3
Molar Mass = 101.96 g/mol Al2O3
% Al = ( 2 x 26.98 g/mol Al) ÷ 101.96 g/mol x 100% = 52.92 %Al
1.65 g Al2O3 x 0.5292 % Al = 0.873 g Al
7.E – Empirical Formula
Instructions: ON A SEPARATE SHEET OF PAPER, answer the following set of questions related to empirical formulas. Remember, NW = NC, Boxed Answers and N3
(this includes labeling and in your set-up!!!)
1. One compound of platinum and chlorine is known to consist of 42.1 % chlorine. Another consists of 26.7% chlorine.
a. Determine the empirical formulas for each compound.
%→g
mole ratio
42.1% Cl
42.1g Cl ÷ 35.45 g/mol Cl
= 1.18871 mol Cl
÷ 0.29680 mol
= 4 Cl
57.9 % Pt = 57.9 g Pt
57.9g Pt ÷ 195.08 g/mol Pt
= 0.29680 mol Pt
÷ 0.29680 mol
= 1 Pt
%→g
mole ratio
= 0.75317 mol Cl
÷ 0.37574 mol
= 2 Cl
= 0.37574 mol Pt
÷ 0.37574 mol
= 1 Pt
26.7% Cl
= 42.1 g Cl
= 26.7 g Cl
26.7g Cl ÷ 35.45 g/mol Cl
73.3% Pt = 73.3 g Pt
73.3g Pt ÷ 195.08 g/mol Pt
b. Name each compound.
PtCl4 = Plantium (IV) Chloride
PtCl4
PtCl2
PtCl2 = Plantium (II) Chloride
2. Consider compound made of silver and fluorine that is 85 % silver?
a. What is the empirical formula?
%→g
mole ratio
85% Ag
= 85g Ag
85g Ag
÷ 107.87 g/mol Ag
= 0.78799 mol Ag
÷ 0.78947 mol
= 1 Ag
15% F
= 15 g F
15g F
÷ 19.00 g/mol F
= 0.78947 mol F
÷ 0.78947 mol
= 1 Cl
AgF
b. Name this compound.
AgCl = Silver Chloride (You need to remember that there are 3 transition metals that do NOT get a Roman number because they only
ever have one charge option. Ag+ Zn2+ Cd2+ You should memorize these ion charges.)
3. An ionic compound was found to be 24.58 % potassium, 34.81 % manganese, and the remainder is oxygen.
a. Determine the empirical formula of this compound.
%→g
mole ratio
24.58% K = 24.58g K
24.58g K ÷ 39.10 g/mol K
34.81% Si = 34.81g Si
34.81g Si ÷ 54.94 g/mol Si
40.61% O = 40.61g O
40.61g O ÷ 16.00 g/mol O
b. Name this compound.
KSiO4 = Potassium Permanganate
= 0.6286 mol K
= 0.6336 mol Si
= 2.5381 mol O
÷ 0.6286 mol
÷ 0.6286 mol
÷ 0.6286 mol
=1K
= 1 Si
=4O
KSiO4
4. 6.840 g of an aluminum compound was analyzed. The polyatomic ion portion of the compound was made of 1.922 g of sulfur and 3.837 g of oxygen. The remainder of the
compound was made of aluminum.
a. Determine the empirical formula of this compound.
mole ratio
1.081g Al
1.922g S
3.837g O
÷ 26.98 g/mol Al
÷ 32.07 g/mol S
÷ 16.00 g/mol O
= 0.040067 mol Al
= 0.059931 mol S
= 0.23981 mol O
÷ 0.040067mol
÷ 0.040067mol
÷ 0.040067mol
= 1 Al
= 1.5 S
=6O
x2
x2
x2
= 2 Al
=3S
= 12 O
Al2S3O12
Should be…
Al2(SO4)3
b. Name this compound.
Al2(SO4)3 = Aluminum Sulfate
5. An ionic compound was analyzed and determined to be 37.70 % sodium, 22.95 % silicon, and 39.34 % oxygen.
a. Determine the empirical formula of this compound.
%→g
mole ratio
37.70% Na = 37.70g Na
22.95% Si = 22.95g Si
39.34% O = 39.34g O
b. Name this compound.
Na2SiO3 = Sodium Silicate
37.70g Na ÷ 22.99 g/mol Na
22.95g Si ÷ 28.09 g/mol Si
39.34g O ÷ 16.00 g/mol O
= 1.63984 mol Na
= 0.81702 mol Si
= 2.45875 mol O
÷ 0.81702 mol
÷ 0.81702 mol
÷ 0.81702 mol
= 2 Na
= 1 Si
=3O
Na2SiO3
6. Analysis of an ionic compound was found to be 21.20 % nitrogen, 6.06 % hydrogen, 24.3 % sulfur, and 48.45 % oxygen.
a. Determine the empirical formula of this compound.
%→g
mole ratio
21.20% N = 21.20g N
21.20g N ÷ 14.01 g/mol N
= 1.5132 mol N
÷ 0.7577 mol
=2N
N2H8SO4
6.06% H = 6.06g H
6.06g H ÷ 1.01 g/mol H
= 6 mol H
24.3% S = 24.3g S
24.3g S ÷ 32.07 g/mol S
= 0.7577 mol S
48.45% O = 48.45g O
48.45g O ÷ 16.00 g/mol O
= 3.0281 mol O
b. Name this compound. (Hint: This compound is made of two polyatomic ions.)
(NH4)2SO4 = Ammonium Sulfate
÷ 0.7577 mol
÷ 0.7577 mol
÷ 0.7577 mol
=8H
=1S
=4O
Should be..
(NH4)2SO4
7. A 10.00 g sample of an ionic compound is composed of 4.94 g of potassium, 2.03 g of sulfur, and the remainder is oxygen.
a. Determine the empirical formula of this compound.
mole ratio
4.94g K
2.03g S
3.03g O
÷ 39.10 g/mol K
÷ 32.07 g/mol S
÷ 16.00 g/mol O
= 0.18938 mol K
= 0.06329 mol S
= 0.12634 mol O
÷ 0.06329 mol
÷ 0.06329 mol
÷ 0.06329 mol
=2K
=1S
=3O
K2SO3
b. Name this compound.
K2SO3 = Potassium Sulfite
8. Analysis of an iron, carbon, oxygen ionic compound was determined to be 48.2 % iron, 10.4 % carbon, and 41.4 % oxygen.
a. Determine the empirical formula of this compound.
%→g
mole ratio
48.2% Fe
10.4% C
41.4% O
= 48.2g Fe
= 10.4g C
= 41.4g O
48.2g Fe ÷ 55.85 g/mol Fe
10.4g C ÷ 28.09 g/mol C
41.4g O ÷ 16.00 g/mol O
= 0.86302mol Fe
= 0.85950 mol C
= 2.5875 mol O
÷ 0.85950 mol
÷ 0.85950 mol
÷ 0.85950 mol
= 1 Fe
=1C
=3O
69.9g Fe ÷ 55.85 g/mol Fe
= 1.25157 mol Fe
÷ 1.25157 mol
= 1 Fe
x 2 = 2 Fe
30.1 g O ÷ 16.00 g/mol O
= 1.88125 mol O
÷ 1.25157 mol
= 1.5 O
x2 =3O
FeCO3
b. Name this compound.
FeCO3 = Iron (II) Carbonate
9. An iron oxide compound was analyzed and determined to be 69.9 % iron.
a. Determine the empirical formula of this compound.
%→g
mole ratio
69.9% Fe
= 69.9g Fe
30.1 % O = 30.1 g O
b. Name this compound.
Fe2O3 = Iron (II) Oxide
Fe2O3
10. An ionic compound made of iron and a polyatomic ion was determined to be 14.70 % iron, 41.07 % chromium, and 44.23 % oxygen.
a. Determine the empirical formula of this compound.
%→g
mole ratio
14.70% Fe =14.70g Fe
41.07% Cr = 41.07g Cr
44.23% O = 44.23g O
14.70g Fe ÷ 55.85 g/mol Fe
41.07g Cr ÷ 52.00 g/mol Cr
44.23g O ÷ 16.00 g/mol O
= 0.26321 mol Fe
= 0.78981 mol Cr
= 2.76438 mol O
÷ 0.26321 mol
÷ 0.26321 mol
÷ 0.26321 mol
= 1 Fe
x 2 = 2 Fe
= 3 Cr
x 2 = 6 Cr
= 10.5 O x 2 = 21 O
b. Name this compound. (Hint: To get the correct ratios, be careful about not rounding off your molar masses or the calculated mole values.)
Fe2(Cr2O7)3 = Iron (II) Dichromate
Fe2Cr6O21
Should be..
Fe2(Cr2O7)3
7.F – Empirical Formula - Hydrate
Instructions: ON A SEPARATE SHEET OF PAPER, answer the following set of questions related to hydrates and empirical formulas. Remember, NW = NC, Boxed
Answers and N3 (this includes labeling and in your set-up!!!)
1. Calculate the % water in MgSO4 • 7 H2O
Molar Mass = 246.52 g/mol MgSO4 • 7 H2O
% H2O = ( 7 x 18.02 g/mol H2O) ÷ (246.52 g/mol) x 100% = 51.17% H2O
2. Determine the formula for a cobalt (II) chloride hydrate that had an original mass of 1.62 g and was heated to remove the water then had a mass of 0.88 g.
= Ionic = Co2+Cl−
= CoCl2
Mass of Water = 1.62g – 0.88g = 0.74 g H2O
mole ratio
0.88g CoCl2 ÷ 129.83 g/mol CoCl2
= 0.006778 mol CoCl2
÷ 0.006778 mol
= 1 CoCl2
0.74g H2O ÷ 18.02 g/mol H2O
= 0.041065 mol H2O
÷ 0.006778 mol
= 6 H2O
CoCl2 · 6H2O
3. Determine the formula for a calcium sulfate hydrate that was determined to be 44 % water.
= Ionic = Ca2+SO42− = CaSO4
%→g
mole ratio
66% CaSO4 = 66g CaSO4
66g CaSO4 ÷ 136.15 g/mol CaSO4 = 0.006778 mol CaSO4 ÷ 0.006778 mol
= 1 CaSO4
44% H2O
44g H2O
= 6 H2O
= 48.2g H2O
÷ 18.02 g/mol H2O
= 0.041065 mol H2O
÷ 0.006778 mol
CaSO4 · 6H2O
4. Determine the formula for a lead(II) acetate hydrate that had an original mass of 1.21 g and when heated, its mass decreased by 0.18 g
= Ionic = Pb2+C2H3O2−
= Pb(C2H3O2)2
Mass of lead (II) acetate = 1.21g – 0.18g = 1.03 g Pb(C2H3O2)2
mole ratio
1.03g Pb(C2H3O2) ÷ 532.52 g/mol Pb(C2H3O2)2 = 0.001934 mol Pb(C2H3O2)2
÷ 0.001934 mol
= 1 Pb(C2H3O2)2
0.18g H2O ÷ 18.02 g/mol H2O
÷ 0.001934 mol
= 3 H2O
= 0.009989 mol H2O
Pb(C2H3O2)2 · 3H2O
5. A hydrate was determined to be 14.7 % water. The anhydrate was analyzed to be 65.9% barium and 34% chlorine. Determine the formula of this hydrate
%→g
mole ratio
65.9% Ba = 65.9g Ba
65.9g Ba
÷ 137.33 g/mol Ba
= 0.470867 mol Ba
÷ 0. 470867mol
= 1 Ba
34% Cl
34g Cl
÷ 35.45 g/mol Cl
= 0.95910 mol Cl
÷ 0. 470867 mol
= 2 Cl
14.7g H2O
÷ 18.02 g/mol H2O
= 0.8158mol H2O
÷ 0. 470867 mol
= 2 H2O
= 34 g Cl
14.7% H2O = 14.7g H2O
BaCl2 · 3H2O
6. The magnesium sulfite hydrate is heated to remove its water, the water removed is nearly the same mass as the anhydrate itself. What is the formula and name of this
hydrate?
= Ionic = Mg2+SO32− = MgSO3
%→g
mole ratio
50% MgSO3= 50g MgSO3
50g MgSO3 ÷ 104.38 g/mol MgSO3 = 0.47902 mol MgSO3÷ 0.47902 mol
= 1 MgSO3
50% H2O
50g H2O
= 6 H2O
= 50g H2O
÷ 18.02 g/mol H2O
= 2.77469 mol H2O
÷ 0.47902 mol
MgSO3 · 6H2O
7. Anhydrous lithium perchlorate (4.78 g) was dissolved in water and re-crystallized. Care was taken to isolate all the lithium perchlorate as its hydrate. The mass of the
hydrated salt obtained was 7.21 g. What is the formula and name of this hydrate?
= Ionic = Li+ClO4−
= LiClO4
Mass of Water = 7.21g – 4.78g = 2.43 g H2O
mole ratio
4.78g LiClO4÷ 106.39 g/mol LiClO4
= 0.044929 mol LiClO4
÷ 0.044929 mol
= 1 LiClO4
2.43g H2O ÷ 18.02 g/mol H2O
= 0.134850 mol H2O
÷ 0.044929 mol
= 3 H2O
LiClO4 · 3H2O
8. A hydrate was found to be 48.8 % water. The anhydrate was analyzed and found to be 17.8% Chromium, 32.9% sulfur, and the rest oxygen. Determine the formula of this
hydrate and then name it.
%→g
mole ratio
17.8% Cr
= 17.8g Cr
17.8g Cr
÷ 52.00 g/mol Cr
= 0.342407 mol Cr
÷ 0.342407 mol
= 1 Cr
CrS2O3
32.9% S
= 32.9 g S
32.9g S
÷ 32.07 g/mol S
= 1.02588 mol S
÷ 0.342407mol
=3S
Should be.
49.3% O
= 49.3g O
49.3g O
÷ 16.00 g/mol O
= 3.08125 mol O
÷ 0.342407mol
=6O
Cr(SO3)3
51.2% Cr(SO3)3= 51.2 Cr(SO3)3
51.2g Cr(SO3)3÷ 292.21 g/mol Cr(SO3)3= 0.175216 mol Cr(SO3)3÷ 0.175216 mol = 1 Cr(SO3)3
48.8% H2O = 48.8 H2O
48.8g H2O
÷ 18.02 g/mol H2O
= 2.7081 mol H2O
÷ 0.175216 mol
= 7 H2O
Cr(SO3)3 · 7H2O
7.G – Empirical & Molecular Formula
Instructions: ON A SEPARATE SHEET OF PAPER, answer the following set of questions related to hydrates and empirical formulas. Remember, NW = NC, Boxed
Answers and N3 (this includes labeling and in your set-up!!!)
1. Acetylene is a gas used in welding torches. As a fossil fuel it contains carbon and hydrogen. It contains 92.3 % carbon, and 7.7 % hydrogen. The molecular mass is 26
g/mole. Determine the empirical and molecular formula.
%→g
mole ratio
92.3% C = 92.3g C
7.7% H
= 7.7g H
Empirical Formula Mass =
Molecular Formula Multiple =
92.3g C ÷ 12.01 g/mol C
= 7.6238 mol C
7.7g H
÷ 1.01 g/mol H
= 7.6238 mol H
13.01 g/mol CH
26 g/mol (CH)x ÷ 13.01 g/mol CH = 2
C2H2
÷ 7.6238 mol
÷ 7.6238 mol
=1C
=1H
CH
2. All simple saccharides contain 40.0 % carbon, 6.7 % hydrogen, and 53.3 % oxygen. The molecular mass of these saccharides (also Known as carbohydrates) is 180
g/mole. Determine the empirical and molecular formula.
%→g
mole ratio
40.0% C = 40.0g C
6.7% H
= 6.7g H
53.3% O = 53.3g O
Empirical Formula Mass =
Molecular Formula Multiple =
40.0g C ÷ 12.01 g/mol C
= 3.3306 mol C
6.7g H
÷ 1.01 g/mol H
= 6.6337 mol H
53.3g O ÷ 16.00 g/mol O
= 3.3313 mol O
30.03 g/mol CH2O
180 g/mol (CH2O)x ÷ 30.03 g/mol CH2O = 6
÷ 3.3306 mol
÷ 3.3306 mol
÷ 3.3306 mol
=1C
=2H
=1O
CH2O
C6H12O12
3. The hydrocarbon butane - the fuel in bic lighters - has the following composition: 82.7 % carbon and 17.3% hydrogen. The molecular mass of butane is 58.2 g/mole.
Determine the empirical and molecular formulas.
%→g
mole ratio
82.7% C = 82.7g C
17.3% H = 17.3g H
Empirical Formula Mass =
Molecular Formula Multiple =
82.7g C ÷ 12.01 g/mol C
= 6.8859 mol C
÷ 6.8859 mol
17.3g H ÷ 1.01 g/mol H
= 17.1287 mol H
÷ 6.8859 mol
29.07 g/mol C2H5
58.2 g/mol (C2H5)x ÷ 29.07 g/mol C2H5= 2
C4H10
=1C
= 2.5 H
x2= 2C
x2=5H
C2H5
4. Chlorofluorocarbons (CFCs), the propellant that was widely used in aerosol cans until it was found to have a detrimental effect on the upper atmosphere ozone layer. One
particular CFC is made of 37.3 % carbon, 6.2 % hydrogen, 19.7 % fluorine, and 36.8 % chlorine. The molecular mass of this compound is 96.5 g/mole. Determine the
molecular and empirical formulas.
%→g
mole ratio
37.3 % C = 37.3 g C
6.2% H
= 6.2g H
19.7% F = 19.7g F
36.8% Cl = 36.8 g Cl
Empirical Formula Mass =
Molecular Formula Multiple =
37.3 g C ÷ 12.01 g/mol C
= 3.1057 mol C
6.2g H
÷ 1.01 g/mol H
= 6.1386 mol H
19.7g F ÷ 19.00 g/mol F
= 1.0368 mol F
36.8g Cl ÷ 35.45 g/mol Cl
= 1.0381 mol Cl
96.54 g/mol C3H6FCl
96.5 g/mol (C3H6FCl)x ÷ 96.54 g/mol C3H6FCl = 1
÷ 1.0368 mol
÷ 1.0368 mol
÷ 1.0368 mol
÷ 1.0368 mol
=3C
=6H
=1F
= 1 Cl
C3H6FCl
C3H6FCl
5. Hydrazine is used as a rocket fuel. Its molecular mass is 92.0 g/mole. Analysis of this compound shows that for every 1.0 g of nitrogen there will be 2.28 g of oxygen.
Determine the empirical and molecular formulas.
mole ratio
1.0g N
÷ 14.01 g/mol N
2.28g O
÷ 16.00 g/mol O
Empirical Formula Mass =
Molecular Formula Multiple =
= 0.071378 mol N
÷ 0.071378 mol
= 0.1425 mol O
÷ 0.071378 mol
46.01 g/mol NO2
92.0 g/mol (NO2)x ÷ 46.01 g/mol NO2= 2
=1N
=2O
NO2
N2O4
6. Tobacco leaves contain between 2 to 8 % nicotine. Nicotine is made of 74.0 % carbon, 8.7 % hydrogen, and 17.3 % nitrogen. The molecular mass is 162 g/mole.
Determine the empirical and molecular formulas.
%→g
mole ratio
74.0% C = 74.0 g C
8.7% H
= 8.7g H
17.3% N = 17.3g N
Empirical Formula Mass =
Molecular Formula Multiple =
74.0 g C ÷ 12.01 g/mol C
= 6.16153 mol C
8.7g H
÷ 1.01 g/mol H
= 8.61386 mol H
17.3g N ÷ 14.01 g/mol N
= 1.23483 mol N
81.13 g/mol C5H7N
162 g/mol (C5H7N)x ÷ 81.13 g/mol C5H7N = 2
÷ 1.23483 mol
÷ 1.23483 mol
÷ 1.23483 mol
C10H14N2
=5C
=7H
=1N
C5H7N
7. Sulfadiazine, a drug used for the treatment of bacterial infections, analyzes to 48 % carbon, 4.0 % hydrogen, 22.4 % nitrogen, 12.8 % sulfur, and 12.8 % oxygen. The
molecular mass is 250.0 g/mole. Determine the empirical and molecular formulas.
%→g
mole ratio
48% C
= 48g C
4.0% H
= 4.0g H
22.4% N = 22.4 g N
12.8% S = 12.8g S
12.8% O = 12.8g O
Empirical Formula Mass =
Molecular Formula Multiple =
48 g C
÷ 12.01 g/mol C
= 3.9967 mol C
4.0g H
÷ 1.01 g/mol H
= 3.9604 mol H
22.4g N ÷ 14.01 g/mol N
= 1.5989 mol N
12.8 S
÷ 32.06 g/mol S
= 0.3991 mol S
12.8g O ÷ 16.00 g/mol O
= 0.8 mol O
250.44 g/mol C10H10N4SO2
250.0 g/mol (C10H10N4SO2)x ÷ 250.44 g/mol C10H10N4SO2
÷ 0.3991 mol
÷ 0.3991 mol
÷ 0.3991 mol
÷ 0.3991 mol
÷ 0.3991 mol
= 10 C
= 10 H
=4N
=1S
=2O
C10H10N4SO2
C10H10N4SO2
8. What is the empirical formula for a compound which contains 0.0134 g of iron, 0.00769 g of sulfur and 0.0115 g of oxygen?
mole ratio
0.0134g Fe ÷ 55.85 g/mol Fe
0.00769g S ÷ 32.07 g/mol S
0.0115g O ÷ 16.00 g/mol O
= 0.0002399 mol Fe
= 0.0002398 mol S
= 0.0071875 mol O
÷ 0.0002398mol
÷ 0.0002398mol
÷ 0.0002398mol
= 1 Fe
=1S
=3O
FeSO3
9. Some molecular compound made of phosphorus and oxygen with a molar mass of 284 g/mole is made of 43.7% phosphorus. Determine the empirical and molecular
formulas of this compound.
mole ratio
43.7g P
÷ 30.97 g/mol N
56.3g O
÷ 16.00 g/mol O
Empirical Formula Mass =
Molecular Formula Multiple =
= 1.41104 mol N
÷ 1.41104 mol
= 3.51875 mol O
÷ 1.41104 mol
141.94 g/mol P2O5
284 g/mol (P2O5)x ÷ 141.94 g/mol P2O5= 2
=1P
= 2.5 O
x2=2P
x2=5O
P2O5
P4O10
10. Barry Um has a sample of a compound which weighs 200 grams and contains only carbon, hydrogen, oxygen and nitrogen. By analysis, he finds that it contains 97.56
grams of carbon, 4.878 g of hydrogen, 52.03 g of oxygen and the rest nitrogen. Find its empirical formula.
Mass of nitrogen = 200 g – 97.56 g – 4.878 g – 52.03 g = 45.532 g N
mole ratio
97.56g C
4.878g H
52.03g O
45.532 g N
÷ 12.01 g/mol C
÷ 1.01 g/mol H
÷ 16.00 g/mol O
÷ 14.01 g/mol N
= 8.1232 mol C
= 4.8297 mol H
= 3.2519 mol O
= 3.2500 mol N
÷ 3.2500 mol
÷ 3.2500 mol
÷ 3.2500 mol
÷ 3.2500 mol
=2.5 C
= 1.5 H
=1O
=1N
x2=5C
x2=3H
x2=2O
x2=2N
C5H3O2N2