1. Unit 2 Formulas and Equations
AP Chemistry
Unit 2
Formulas and Equations

2. The mole is the unit for counting atoms and molecules
6.02 × 1023 atoms of (insert name of element here) has a mass of (insert mass of element from periodic table here)

8. How many atoms are in 5.00 g of oxygen gas
5.00 g O 2 1 × 1 mol O g O 2 × 2 mol O 1 mol O 2 × 6.02 × atoms 1 mol O = × 1023

9. 3.64 g NaCl 50.0 mL × 1 mol NaCl 58.5 g NaCl × 1000 mL 1 L =1.24 M
Determine the concentration (in molarity) when 3.64 g of NaCl is added to enough water to make 50.0 mL of solution.
3.64 g NaCl 50.0 mL × 1 mol NaCl 58.5 g NaCl × 1000 mL 1 L =1.24 M

10. 3.19 mmol Ca NO 3 2 25.0 mL × 1 𝑚𝑜𝑙 1000 𝑚𝑚𝑜𝑙 × 1000 𝑚𝐿 1 𝐿 =0.128 𝑀
Determine the concentration (in molarity) when 3.19 mmol of Ca(NO3)2 is added to enough water to make 25.0 mL of solution.
3.19 mmol Ca NO mL × 1 𝑚𝑜𝑙 1000 𝑚𝑚𝑜𝑙 × 1000 𝑚𝐿 1 𝐿 =0.128 𝑀
Note that the milli- in mmol and mL effectively canceled!

11. A student has 12. 0 M HCl available but needs 50. 0 mL of 0
A student has 12.0 M HCl available but needs 50.0 mL of M for an experiment. How much of the concentrated acid should she use?
0.500 mol HCl 1 L HCl (𝑑) × L HCl (𝑑) 1 × 1 L conc. HCl (𝑠) 12.0 mol HCl L
or 2.08 mL

12. Recognizing that molarity times volume gives moles…
20.0 mL of M HNO3 is diluted to mL. Determine the new concentration of nitric acid
Recognizing that molarity times volume gives moles…
0.280 mol H NO 3 1 L × 20.0 mL mL = M

14. Composition stoichiometry deals with the makeup of a single compound
How many grams of iron is in 3.85 g of Fe2(SO4)3?
3.85 g Fe 2 SO × 1 mol Fe 2 SO g Fe 2 SO × 2 mol Fe 1 mol Fe 2 SO × 55.8 g Fe 1 mol Fe =1.07 g

15. F.M. of CuCO3 is 187.0 g Mass of copper is 63.5 × 2 = 127.0 g
What percent of Cu2CO3 is copper?
F.M. of CuCO3 is g
Mass of copper is 63.5 × 2 = g
%Cu= g Cu g Cu 2 CO 3 ×100=67.9%

17. A balanced chemical equation is necessary All problems follow
Reaction stoichiometry deals with relationships in chemical reactions
A balanced chemical equation is necessary
All problems follow
g A → mol A → mol B → g B
All fractions, other than the first (given), must equal one!

18. 2 C 4 H O 2 →10 H 2 O+8 CO 2
Butane burns in oxygen according to the equation shown above. When g of butane burns in excess oxygen, how much water should be formed?
15.85 g C 4 H × 1 mol C 4 H g C 4 H 10 × 10 mol H 2 O 2 mol C 4 H 10 × 18.0 g H 2 O 1 mol H 2 O =24.6 g

19. We have 11.85 g of Na 3 PO 4 and we need 11.61 g
3Ca NO Na 3 PO 4 → Ca 3 PO NaNO 3
17.42 g of calcium nitrate reacts with g of sodium phosphate according to the reaction shown above. Determine which one is the limiting reactant.
17.42 g Ca NO × 1 mol Ca NO g Ca NO × 2 mol Na 3 PO 4 3 mol Ca NO × g Na 3 PO 4 1 mol Na 3 PO 4 = g Na 3 PO 4
We have g of Na 3 PO 4 and we need g
Therefore Na 3 PO 4 is the excess reactant
Which means Ca NO is the limiting reactant!

20. Fe 2 O 3 +2Al→ Al 2 O 3 +2Fe
12.91 g of iron(III) oxide reacts with aluminum as shown above. If 8.87 g of molten iron is formed, what is the percent yield?
12.91 g Fe 2 O 3 1 × 1 mol Fe 2 O g Fe 2 O 3 × 2 mol Fe 1 mol Fe 2 O 3 × 55.8 g Fe 1 mol Fe =9.03 g Fe
%𝑌𝑖𝑒𝑙𝑑= ×100=98.2%

22. 2) determine moles of each element
An empirical formula is a chemical formula in lowest terms.
Steps in determining an empirical formula from percent composition.
1) assume 100 g if given in %
2) determine moles of each element
3) divide each molar amount by the smallest one
4) check for ½’s 1/3’s ¼’s etc.

23. Determine the empirical formula for a compound that is found to be 52
Determine the empirical formula for a compound that is found to be 52.14% C, 13.13% H and 34.73% O
C = g
H = g
O = g
C = g / g/mol = mol
H = g / g/mol = mol
O = g /16.00 g/mol = mol
Divide all by the smallest, mol
C = mol / mol =
H = mol / mol = 6.000
O = mol / mol = 1.000
E.F. = C2H6O

24. Determine the empirical formula for a compound that is found to be 24
Determine the empirical formula for a compound that is found to be % S, % Cr, and 48.96% O
S = g
Cr = g
O = g
S = g / g/mol = mol
Cr = g / g/mol = mol
O = g / g/mol = mol
Divide all by the smallest, moles
S = mol / mol =
Cr = mol / mol = 1.000
O = mol / mol = 6.000
E.F. = Cr2S3O12 or
Cr2(SO4)3

25. A molecular formula can be determined from an empirical formula if additional information is given, such as molar mass.

26. Determine the molecular formula for a compound that is found to be 9
Determine the molecular formula for a compound that is found to be 9.15% hydrogen, 36.32% oxygen, and 54.53% carbon. The molar mass was found to be 120 g/mol +/-25 g/mol.
C = g
H = 9.15 g
O = g
C = g / g/mol = mol
H = 9.15 g / g/mol = mol
O = g /16.00 g/mol = mol
Divide all by the smallest, mol
C = mol / mol = 2.000
H = mol / mol = 3.999
O = mol / mol = 1.000
E.F. = C2H4O

27. Determine the molecular formula for a compound that is found to be 9
Determine the molecular formula for a compound that is found to be 9.15% hydrogen, 36.32% oxygen, and 54.53% carbon. The molar mass was found to be 120 g/mol +/-25 g/mol.
E.F. = C2H4O
Mass of E.F. is 2 × 12.0
4 × 1.0
1 × 16.0
44.0 g/mol
1 × E.F. = 44.0 g/mol
2 × E.F. = 88.0 g/mol
3 × E.F. = g/mol
4 × E.F. = g/mol
5 × E.F. = g/mol
Range is 95 – 145 g/mol
M.F. = 3 × E.F.
C6H12O3

29. Many ionic substances that crystallize in water trap water molecules in the structure.
These “hydrates” have very specific formulas
For example: MgSO4·7H2O

30. A hydrate of magnesium chloride is placed into a crucible of mass 22
A hydrate of magnesium chloride is placed into a crucible of mass g. Before heating the crucible and contents have a mass of g. After heating the mass is g. Determine the formula for the hydrate.
Mass of MgCl2 is – = g
Mass of H2O is – = g
Moles of MgCl2 is / = mol
Moles of H2O is / = mol
Dividing by gives
MgCl2·7H2O

31. Hydrates
MgCO3∙5H2O
Na2CO3∙2H2O
Na2CO3∙10H2O 2 10
LiClO4∙3H2O
ZnSO4∙7H2O
BaCl2∙2H2O
CoCl2∙2H2O

33. Oxidation Numbers Ion’s charges are their oxidation numbers
The most electronegative element (usually the last one) is treated like an ion
Everything else must be calculated

34. Determine the oxidation number of chlorine in chlorate
ClO3-
Cl = x
O = -2 × 3 = -6
x + (-6) = -1
x = 5

35. Determine the oxidation number of chromium in Al2(Cr2O7)3
Cr = 6x
O = -2 × 21 = -42
6 + 6x + (-42) = 0
x = 6 or
Cr2O72-
Cr = 2x
O = -2 × 7 = -14
2x + (-14) = -2
x = 6

37. Strong acids ionize completely in water There are seven strong acids
They are formed when hydrogen bonds to soluble anions
NO3-
ClO4-
ClO3-
Cl-
Br- I-
SO42-
The only exception is
C2H3O2-

38. C2H4Cl2 can be put together in two different ways
Isomers exist among some organic compounds. They have the same formula but a different structure
C2H4Cl2 can be put together in two different ways
H− Cl | C | H − Cl | C | H −H or
Cl H C=C Cl H or
H− Cl | C | Cl − H | C | H −H
Structural
isomers
Cl H C=C H Cl
Optical
isomers

39. Any carbon with four different things bonded to it will give rise to an isomer
Cl− H | C | Br −I vs Cl− I | C | Br −H

40. Most metals oxidize but in different ways
Iron will oxidize (rust) until the whole sample is consumed
Aluminum forms an oxide layer (tarnish) that protects the underlying metal
Stainless steel contains chromium which forms a protective oxide layer to keep the iron from oxidizing
Steel can be coated with a protective layer of zinc in a process called galvanization

43. Spectrophotometry uses light to determine the concentration of colored solutions

44. Absorbance varies with the wavelength of light used

45. Once the optimal wavelength is chosen, a series of standard solutions is prepared and tested

46. Spectrophotometry makes use of the Beer-Lambert law

47. It boils down to one equation:
A = abc or A = εlc
Where: A is the absorption of light
a (ε) is the molar absorptivity
b (l) is the path length
c is the concentration
A = -log T
𝑇= 𝐼 𝑡 𝐼 0
Where: T is transmittance
It = Intensity of light transmitted
I0 = Intensity of original light

49. Quiz
1) 3.42 × 1024
9) C9H20
2) 1.11 × 1025
10) g
3) 2.66× 1020
11) Ni3(PO4)2·12H2O
4) 2.73× 1024
12) CaCl2·2H2O
5) M
13) (a) 5
6 (a) M
(b) 1
(b) M
(c) -½
(c) M
(d) 6
7) 30.34%
14) M
8) 97.4%
15) 0.49 M