1. Problem 3.14
How many moles of cobalt (Co) atoms are there in 6.00 X109 (6 billion) Co atoms?

2. PROBLEM 3.24-F
CALCULATE THE MOLAR MASS OF: (f) Mg3N2

3. PROBLEM 3.27
CALCULATE THE NUMBER OF C, H, AND O ATOMS IN 1.50 G OF OF GLUCOSE (C6H12O6)

4. PROBLEM 3.30
The density of water is 1.00 g/ml at 4 Celsius degrees. How many water molecules are in 2.56 ml of water at this temperature?

5. Percent composition of an element in a compound =
n x molar mass of element
molar mass of compound
x 100%
n is the number of moles of the element in 1 mole of the compound
%C =
2 x (12.01 g)
46.07 g
x 100% = 52.14%
C2H6O
%H =
6 x (1.008 g)
46.07 g
x 100% = 13.13%
%O =
1 x (16.00 g)
46.07 g
x 100% = 34.73%
52.14% % % = 100.0%

6. MASS RELATIONSHIPS: PERCENT COMPOSITION AND FORMULAS
What is the percent of Hydrogen in H2O?

7. TYPES OF FORMULAS
MOLECULAR (C6H1206)
EMPIRICAL (CH2O)

8. Problem 3.39
Tin (Sn) exists in Earth’s crust as SnO2. Calculate the percent composition by mass of Sn and O in SnO2.

9. Determine the empirical formula of a compound that has the following percent composition by mass: K 24.75, Mn 34.77, O percent.

10. PROCEDURE Find grams of each element. CONVERT GRAMS TO MOLES
DIVIDE BY SMALLEST NUMBER OF MOLES
EXPRESS MOLE RATIO

11. Determine the empirical formula of a compound that has the following percent composition by mass: K 24.75, Mn 34.77, O percent.

12. PROBLEM 3.49 A, B
WHAT ARE THE EMPIRICAL FORMULAS OF THE COMPOUNDS WITH THE FOLLOWING COMPOSITIONS? (a) 2.1 PERCENT H, 65.3 PERCENT O, 32.6 PERCENT S; (b) PERCENT Al, 79.8 PERCENT Cl

13. Percent Composition and Empirical Formulas
Determine the empirical formula of a compound that has the following percent composition by mass: K 24.75, Mn 34.77, O percent.
nK = g K x
= mol K
1 mol K
39.10 g K
nMn = g Mn x
= mol Mn
1 mol Mn
54.94 g Mn
nO = g O x
= mol O
1 mol O
16.00 g O

14. PROBLEM 3.49
WHAT IS THE EMPIRICAL FORMULA OF A COMPOUND CONTAINING (A) 2.1 Percent H; 65.3 Percent O; 32.6 percent S

15. 3 ways of representing the reaction of H2 with O2 to form H2O
A process in which one or more substances is changed into one or more new substances is a chemical reaction
A chemical equation uses chemical symbols to show what happens during a chemical reaction
reactants
products
3 ways of representing the reaction of H2 with O2 to form H2O

16. Balancing Chemical Equations
Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation.
Ethane reacts with oxygen to form carbon dioxide and water
C2H6 + O2
CO2 + H2O
Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts.
2C2H6
NOT
C4H12

17. Balancing Chemical Equations
Start by balancing those elements that appear in only one reactant and one product.
C2H6 + O2
CO2 + H2O
start with C or H but not O
1 carbon
on right
2 carbon
on left
multiply CO2 by 2
C2H6 + O2
2CO2 + H2O
6 hydrogen
on left
2 hydrogen
on right
multiply H2O by 3
C2H6 + O2
2CO2 + 3H2O

18. Balancing Chemical Equations
Balance those elements that appear in two or more reactants or products.
multiply O2 by 7 2
C2H6 + O2
2CO2 + 3H2O
2 oxygen
on left
4 oxygen
(2x2)
+ 3 oxygen
(3x1)
= 7 oxygen
on right
C2H O2
2CO2 + 3H2O 7 2
remove fraction
multiply both sides by 2
2C2H6 + 7O2
4CO2 + 6H2O

19. Balancing Chemical Equations
Check to make sure that you have the same number of each type of atom on both sides of the equation.
2C2H6 + 7O2
4CO2 + 6H2O
4 C (2 x 2)
4 C
12 H (2 x 6)
12 H (6 x 2)
14 O (7 x 2)
14 O (4 x 2 + 6)
Reactants
Products
4 C
12 H
14 O

20. PROBLEM 3.59 BALANCE THE FOLLOWING EQUATIONS. C + O2  CO
H2 + Br2  HBr
K + H2O  KOH + H2
H2O2  H2O + O2
S8 + O2  SO2
KOH + H3PO4  K3PO4 + H2O
CH4 + Br2  CBr4 + HBr

21. Problem 3.59 BALANCE THE FOLLOWING EQUATIONS C + O2  CO
H2 + Br2  HBr
Zn + AgCl  ZnCl2 +Ag
(k) NaOH + H2SO4  Na2SO4 + H2O

22. How to “Read” Chemical Equations
2 Mg + O MgO
2 atoms Mg + 1 molecule O2 makes 2 formula units MgO
2 moles Mg + 1 mole O2 makes 2 moles MgO
48.6 grams Mg grams O2 makes 80.6 g MgO
NOT
2 grams Mg + 1 gram O2 makes 2 g MgO

23. Amounts of Reactants and Products
Chemical Equations:
Amounts of Reactants and Products
Write balanced chemical equation
Convert quantities of known substances into moles
Use coefficients in balanced equation to calculate the number of moles of the sought quantity
Convert moles of sought quantity into desired units

24. Problem 3.66
Silicon tetrachloride (SiCl4) can be prepared by heating Si in chlorine gas:
Si(s) + Cl2(g) SiCl4(l)
In one reaction mole of SiCl4 is produced. How many moles of molecular chlorine were used in the reaction.

25. Problem 3.68
Consider the combustion of butane (C4H10 ): 2C4H (O2)  8CO2 + 10H2O In a particular reaction, 5.0 moles of C4H10 are reacted with an excess of O2. Calculate the moles of CO2 formed.

26. Problem 3.71
When Potassium Cyanide (KCN) reacts with acids, a deadly poisonous gas, hydrogen cyanide (HCN) is given off. Here is the equation.
KCN(aq)+ HCl(aq)  KCl(aq) + HCN(g)
If a sample of grams of KCN is treated with an excess of HCl, calculate the amount of HCN formed in grams.

27. Problem 3.75
Limestone (CaCO3) is decomposed by heating to quicklime (CaO) and carbon dioxide.
Calculate how many grams of quicklime can be produced from 1.0 kg of limestone.
CaCO3  CaO + CO2

28. Limiting Reagent: Reactant used up first in the reaction.
2NO + O NO2
NO is the limiting reagent
O2 is the excess reagent

29. In one process, 124 g of Al are reacted with 601 g of Fe2O3.
2Al + Fe2O3  Al2O3 + 2Fe
Calculate the mass of Al2O3 formed.

30. Have more Fe2O3 (601 g) so Al is limiting reagent
In one process, 124 g of Al are reacted with 601 g of Fe2O3
2Al + Fe2O Al2O3 + 2Fe
Calculate the mass of Al2O3 formed.
g Al
mol Al
mol Fe2O3 needed
g Fe2O3 needed OR
g Fe2O3
mol Fe2O3
mol Al needed
g Al needed
1 mol Al
27.0 g Al x
1 mol Fe2O3
2 mol Al x
160. g Fe2O3
1 mol Fe2O3 x =
124 g Al
367 g Fe2O3
Start with 124 g Al
need 367 g Fe2O3
Have more Fe2O3 (601 g) so Al is limiting reagent

31. 1 mol Al 27.0 g Al x 1 mol Al2O3 2 mol Al x 102. g Al2O3 1 mol Al2O3 x
Use limiting reagent (Al) to calculate amount of product that
can be formed.
g Al
mol Al
mol Al2O3
g Al2O3
2Al + Fe2O Al2O3 + 2Fe
1 mol Al
27.0 g Al x
1 mol Al2O3
2 mol Al x
102. g Al2O3
1 mol Al2O3 x =
124 g Al
234 g Al2O3
At this point, all the Al is consumed and Fe2O3 remains in excess.

32. PROBLEM 3.83
Nitric Oxide (NO) reacts with oxygen gas to form nitrogen dioxide (NO2), a dark brown gas:
2NO(g) + O2(g)  2NO2(g)
In one experiment, moles of NO are mixed with moles of O2(g). Calculate which of the two is the limiting reagent. Calculate the moles of NO2 produced.

33. Reaction Yield Theoretical Yield is the amount of product that would
result if all the limiting reagent reacted.
Actual Yield is the amount of product actually obtained
from a reaction.
% Yield =
Actual Yield
Theoretical Yield
x 100%

34. Percent Yield
In an organic chemistry lab, 12.4 grams of an organic compound was produced.
According to calculations, the reaction should have produced 17.0 grams.
Calculate the percent yield.

35. Chemistry In Action: Chemical Fertilizers
Plants need: N, P, K, Ca, S, & Mg
3H2 (g) + N2 (g) NH3 (g)
NH3 (aq) + HNO3 (aq) NH4NO3 (aq)
2Ca5(PO4)3F (s) + 7H2SO4 (aq)
3Ca(H2PO4)2 (aq) + 7CaSO4 (aq) + 2HF (g)
fluorapatite