1. Chemistry Final Exam Review

2. Calculate the molar mass of: (Ca(C2H3O2)2) Ca = 40.1 g/mol
Chapter 10: Molar Mass
Calculate the molar mass of:
(Ca(C2H3O2)2)
Ca = g/mol
4 * C = g/mol
6 * H = g/mol
4 * O2 = g/mol
molar mass (Ca(C2H3O2)2) = g/mol

4. Molar Mass
3. Pb3(PO4)2
Ans: g/mol
Cu3(BO3)2
Ans: g/mol

5. 2. Number of atoms in a molecule
How many atoms of hydrogen are present in four molecules of
C3H7O?
Answer: 28
How many atoms of hydrogen are present in:
2. Al(OH)3
Answer: 3
(NH4)2HPO4
Answer: 9
C4H10O
Answer: 10

6. Mole conversions
Figure (pg 303)

7. Mole conversions: Finding mass in grams
Calculate the mass, in grams, of 2 molecules of ribose C5H10O5.
a. Calculate molar mass:
C5H10O5 molar mass = C: 5 * 12.0 g/mol = g/mol
H: 10* 1.0 g/mol = g/mol
O: 5* 16.0 g/mol = g/mol
C5H10O5 molar mass = g/mol
Solve using mole conversion:
2 molecules * 1mole * g = x g
6.02 x mole
Problem: Calculate the mass, in grams, of a molecule of aspirin (C9H8O4)
Molar mass C9H8O4 = g/mole
Answer: x g

8. Mole conversions: Finding volume in liters
2. Calculate the vol., in liters, of 1.50 mole Cl2 at STP
Use molar volume to convert moles to liters:
1.50 mole Cl2 * L = L
mole
Problems: Calculate the vol., in liters, of the following gases at STP:
i. 7.6 mole Ar
ii mole C2H6
iii. 835g SO3
Answer: i x 102 L Ar
ii L C2H6
iii L SO3

9. Mole conversions: Finding the # of atoms
3. Calculate the number of atoms in 5.78 mol NH4NO3.
Answer: x 1025 atoms

10. % Composition 1. Calculate the molar mass 10 * C = 120.0 g/mol
Nicotine has a molecular formula of C10H14N2,
calculate the % composition of nitrogen (ONLY) in this compound.
1. Calculate the molar mass
10 * C = g/mol
14 * H = g/mol
2 * N2 = g/mol
molar mass C10H14N2 = g/mol

11. % Composition Divide molar mass of nitrogen by total mass of compound
% Composition = g/mol * 100 %
162.0 g/mol
= %
C/E: 5/25/10

12. Empirical Formula from % composition
Calculate the empirical formula of a compound that contains 25.9%
nitrogen and 74.1% oxygen.
unknown:
Empirical formula = N?O?
Known:
% nitrogen = 25.9
% oxygen = 74.1
Change % composition (ratio of the mass) to the ratio of moles by
using molar mass
2. Reduce the ratio to the lowest whole-number ratio

13. Empirical Formula from % Composition and molar mass
Solving:
25.9 g N * 1mol N = mol N
14.0 g N
74.1 g O * 1mol O = mol O
16.0 g O
Mole ratio of nitrogen to oxygen is N1.85O4.63.
Divide by the smaller number:
1.85 mol N = 1 mol N; mol O = mol O
This becomes: N1O2.5
Multiply by 2: N2O5

14. Empirical Formula from % Composition and molar mass
Calculate empirical formula of each compound:
94.1% O; 5.9% H
Answer: HO
67.6% Hg; 10.8% S; 21.6% O
Answer: HgSO4

15. Molecular Formula from % Composition and molar mass
The % composition of methyl butanoate is 58.8% C, 9.8% H, and
31.4% O and its molar mass is 102 g/mol. What is the empirical formula?
What is the molecular formula?
Solve:
1. Compute moles using molar mass
C: g * 1 mol = 4.9 mol
12.0 g
H: g * 1 mol = 9.8 mol
1.0 g
O: g * 1 mol = 1.96 mol
16.0 g

16. Molecular Formula from % Composition and molar mass
Divide by the smallest number of moles
1.96 mol O = 1 mol O; 9.8 mol H = 5 mol H; 4.9 mol C = 2.5 mol
This becomes: C2.5H5O1
Multiply by 2:
C5H10O2 (Empirical formula)
3. Compute molar mass of empirical formula

17. Molecular Formula from % Composition and molar mass
molar mass C5H10O2: C = 5 * g/mol = g/mol
H = 10 * 1.0 g/mol = g/mol
O = 2 * g/mol = g/mol
102.0 g/mol
3. Compute the ratio of molar mass of compound and empirical formula
Molar mass methyl butanoate = g/mol = 1
Empirical formula molar mass g/mol
Multiply empirical formula with the ratio in step 3
C5H10O2 * 1 = C5H10O2
So, empirical formula = molecular formula

18. Molecular Formula from % Composition and molar mass
Problem: Caffeine is analyzed and found to have the following
% composition: 49.5% carbon, 5.2 % hydrogen, 28.9% nitrogen and
16.4% oxygen. The molar mass of caffeine is 195 g/mol. Find the
molecular formula of this compound.
1. C4H5N2O1
2. Molar mass of empirical formula = 97 g/mol
3. Molar mass caffeine = 195 g/mol = 2
Molar mass empirical formula g/mol
4. Molecular formula = C4H5N2O1 * 2 = C8H10N4O2

19. Balancing equations, types of chemical reactions, and Reactions in aqueous solutions
Chapter 11

20. Balancing Chemical Equations16
______C8H18 + _______ O2  ______CO2 +______H2O 2 25 18 4 3 12
_____(NH4)3PO4 + _____ Pb(NO3)4  _____ Pb3(PO4)4 +____(NH4)NO3

21. Types of Chemical Reactions
single replacement
synthesis
double replacement
combustion
decomposition

22. Ba(NO3)2 (aq) + K2CO3 (aq)  BaCO3 (s) + 2KNO3 (aq) Total Ionic:
Net ionic equations, formation of a precipitate, and identification of spectator ions
Ba(NO3)2 (aq) + K2CO3 (aq)  ?
Ba(NO3)2 (aq) + K2CO3 (aq)  BaCO3 (s) + 2KNO3 (aq)
Total Ionic:
Ba+2 (aq) + NO3-1 (aq) + K +1 (aq) + CO3-2 (aq)  BaCO3 (s) + K+1 (aq) + NO3- (aq)
Spectator ions: NO3-1 (aq), K +1 (aq)
Net Ionic: Ba+2 (aq) + CO3-2 (aq)  BaCO3 (s)
D:5/26/10

23. Net ionic equations, formation of a precipitate, and identification of spectator ions (continued)
AgNO3 (aq) + K3PO4 (aq)  ?
AgNO3 (aq) + K3PO4 (aq)  Ag3PO4 (s) + KNO3 (aq)
Total Ionic:
Ag+ (aq)+ NO3- (aq)+ K+ (aq)+ PO43- (aq) Ag3PO4 (s) + NO3- (aq) + K+ (aq)
Spectator Ions: NO3- (aq), K+ (aq)
Net Ionic: 3Ag+ (aq) + PO43- (aq)  Ag3PO4 (s)