Chemistry Final Exam Review

1.Calculate the molar mass of: (Ca(C2H3O2)2)

Ca = 40.1 g/mol

4 * C = + 48.0 g/mol

6 * H = + 6.0 g/mol

4 * O2 = + 64.0 g/molmolar mass (Ca(C2H3O2)2) = 158.1 g/mol

Chapter 10: Molar Mass

Molar Mass

2.Cu3(BO3)2

Ans: 308.3 g/mol

3. Pb3(PO4)2

Ans: 811.5 g/mol

2. Number of atoms in a molecule1. 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

3. (NH4)2HPO4

Answer: 9

4. C4H10O

Answer: 10

mole

Representative particlesMass

Volume of gas at STP

1.0 mole

6.02 x 10 23 particles

6.02 x 10 23 particles

1.0 mole

1.0 mole

molar mass

molar mass

1.0 mol

22.4 L

1 . 0 mol e

1.0 mole

22 .4 L

Mole conversions

Figure 10.12 (pg 303)

1. Calculate the mass, in grams, of 2 molecules of ribose C5H10O5.

a. Calculate molar mass: C5H10O5 molar mass = C: 5 * 12.0 g/mol = 60.0 g/mol

H: 10* 1.0 g/mol = +10.0 g/mol O: 5* 16.0 g/mol = +80.0 g/mol

C5H10O5 molar mass = 150.0 g/mol

b. Solve using mole conversion:2 molecules * 1mole * 150.0 g = 4.98 x 10-22 g

6.02 x 1023 mole

Problem: Calculate the mass, in grams, of a molecule of aspirin (C9H8O4)

Molar mass C9H8O4 = 180.0 g/moleAnswer: 2.99 x 10-22 g

Mole conversions: Finding mass in grams

Mole conversions: Finding volume in liters

2. Calculate the vol., in liters, of 1.50 mole Cl2 at STP

a. Use molar volume to convert moles to liters:

1.50 mole Cl2 * 22.4 L = 33.6 L mole

Problems: Calculate the vol., in liters, of the following gases at STP:

i. 7.6 mole Ar ii. 0.44 mole C2H6 iii. 835g SO3

Answer: i. 1.7 x 102 L Ar ii. 9.9 L C2H6

iii. 234 L SO3

Mole conversions: Finding the # of atoms

3. Calculate the number of atoms in 5.78 mol NH4NO3.

Answer: 3.13 x 1025 atoms

% CompositionNicotine has a molecular formula of C10H14N2, calculate the % composition of nitrogen (ONLY) in this compound.

1. Calculate the molar mass

10 * C = 120.0 g/mol

14 * H = + 14.0 g/mol

2 * N2 = + 28.0 g/mol molar mass C10H14N2 = 162.0 g/mol

% Composition

2. Divide molar mass of nitrogen by total mass of compound

% Composition = 28.0 g/mol * 100 % 162.0 g/mol

= 17.3 %

C/E: 5/25/10

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

1. Change % composition (ratio of the mass) to the ratio of moles by using molar mass2. Reduce the ratio to the lowest whole-number ratio

Empirical Formula from % Composition and molar mass

Solving:25.9 g N * 1mol N = 1.85 mol N

14.0 g N

74.1 g O * 1mol O = 4.63 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; 4.63 mol O = 2.50 mol O 1.85 1.85

This becomes: N1O2.5

Multiply by 2: N2O5

Empirical Formula from % Composition and molar mass

Calculate empirical formula of each compound:

1. 94.1% O; 5.9% H

Answer: HO

2. 67.6% Hg; 10.8% S; 21.6% O

Answer: HgSO4

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: 58.8 g * 1 mol = 4.9 mol 12.0 g

H: 9.8 g * 1 mol = 9.8 mol 1.0 g

O: 31.4 g * 1 mol = 1.96 mol 16.0 g

Molecular Formula from % Composition and molar mass

2. 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 1.96 1.96 1.96

This becomes: C2.5H5O1

Multiply by 2: C5H10O2 (Empirical formula)

3. Compute molar mass of empirical formula

Molecular Formula from % Composition and molar mass

molar mass C5H10O2: C = 5 * 12.0 g/mol = 60.0 g/mol H = 10 * 1.0 g/mol = 10.0 g/mol

O = 2 * 16.0 g/mol = 32.0 g/mol 102.0 g/mol

3. Compute the ratio of molar mass of compound and empirical formula

Molar mass methyl butanoate = 102 g/mol = 1 Empirical formula molar mass 102 g/mol

4. Multiply empirical formula with the ratio in step 3

C5H10O2 * 1 = C5H10O2

So, empirical formula = molecular formula

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 97 g/mol

4. Molecular formula = C4H5N2O1 * 2 = C8H10N4O2

Balancing equations, types of chemical reactions, and Reactions in

aqueous solutions

Chapter 11

Balancing Chemical Equations

______C8H18 + _______ O2 ______CO2 +______H2O

_____(NH4)3PO4 + _____ Pb(NO3)4 _____ Pb3(PO4)4 +____(NH4)NO3

2 25 16 18

4 3 12

Types of Chemical Reactions

A. HCl(aq) + Na(s) NaCl(aq) + H2(g)

B. O2(g) + P4 P2O5

C. FeCl3(aq) + NH4OH(aq) ? Fe(OH)3(s) + NH4Cl(aq) D. C6H22O11 + O2(g) CO2(g) + H2O(l) E. FeBr3 Br2(g) + Fe(s)

single replacement

synthesis

double replacement

combustion

decomposition

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

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)