General Chemistry Help

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CHM 130 HELP

Significant Digits

Conversions

Math Help / Problem Solving – downloadable lessons on the left margin

Study Tips and Helpful Advice

How to use your Calculator

Electron dot structures tutorial

Electron dot structures practice probs

Nomenclature

Stoichiometry (practice problems like grams to grams)

Oxidation states, ionic formulas and polyatomic ions 

Writing and balancing chemical reactions

Chem Tutor - help for many basic topics including math help

Chem Web - graphics intensive - great reviews, has interactive quizzes, fun

Interactive Periodic Table - info on most every element

CHM 151 HELP

Significant Digits Study Tips and Helpful Advice Oxidation states, ionic formulas and polyatomic ions  Molarity and dilution  (practice problems) Quantum numbers, Shells, Orbitals Chem Tutor - help for many basic topics including math help Chem Web - graphics intensive - great reviews, has interactive quizzes, fun Gen Chem Topic Review - Purdue has great info for 151 students Practice Problems by Topic very helpful

How to use your Calculator Conversions Avogadro's Number Writing and balancing chemical reactions Stoichiometry (practice problems) Oxidation and Reduction  basic info Nomenclature Graphing Interactive Periodic Table - info on most every element Math Help / Problem Solving – downloadable lessons on the left margin

CHM 152 HELP

Significant Digits Study Tips and Helpful Advice How to use your Calculator Conversions Chem Tutor - help for many basic topics including math help Chem Web - graphics intensive - great reviews, has interactive quizzes, fun Interactive Periodic Table - info on most every element Gen Chem Topic Review - Purdue has great info for 151 and 152 students Dr. Diebolt's help page - CHM 152 info for her sections Math Help / Problem Solving – downloadable lessons on the left margin

Graphing Kinetic Activities Equilibrium Acids, Bases and their equilibria Acids, Bases, Salts Table Buffer and Titration Activities Thermodynamics Redox Electrochemistry Nuclear chemistry

Table of Acid, Base and Salt Information

Chemical	Example reaction	Equili brium?	In the beaker (picture)
Strong acid	HCl(aq) + H2O(l) g Cl^- (aq) + H3O^+(aq)	no	Cl^-, H^+ (products only)
Weak acid	HF (aq) + H2O(l) D F^- (aq) + H3O^+(aq)	yes	HF, F^-, H^+ (both sides)
Strong base	NaOH(s) g Na^+ (aq) + OH^- (aq) (does not react with water, dissolves in it)	no	Na^+, OH^- (products only)
Weak base	NH3(aq) + H2O(l) D OH^- (aq) + NH4^+(aq)	yes	NH3, OH^-, NH4^+ (both sides)
Acidic salt	NH4Cl g NH4^+ (aq) + Cl^- (aq) (does not react with water, dissolves in it) then the non neutral ion (conjugate acid) reacts further with water NH4^+ (aq) + H2O(l) D NH3(aq) + H3O^+(aq)	yes	NH3, H^+ , NH4^+ Cl^-(both sides)
Basic salt	KF g K^+ (aq) + F^- (aq) (does not react with water, dissolves in it) then the non neutral ion (conjugate base ) reacts further with water F^- (aq) + H2O(l) D HF (aq) + OH^- (aq)	yes	HF, F^-, OH^-, K^+ (both sides)
Neutral salt	NaCl g Na^+ (aq) + Cl^- (aq) (does not react with water, dissolves in it) both ions are neutral so no further reaction	no	Na^+, Cl^- (products only)

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Significant Digits

Significant digits are not something we make up to terrorize you all semester long.  They represent the accuracy of a measurement.  For example, a cheap bathroom scale bought at the dollar store reads your weight as 152 pounds, not 152.45809 pounds.  It is not that accurate.  Significant digits are very important in all measurements.  Now remember that conversion factors such as (1 L / 1000 mL) DO NOT affect the significant digits because they are exact numbers. 

1. Rules - note that # means a non-zero digit (123456789)
1. Zeros in front of any # never count
2. Zeros after any # do not count unless they are also after a decimal place
3. Zeros in between any digits that count, count also
2. Examples
1. 2040 - 3 sig dig
2. 2040.0 - 5 sig dig
3. 00204.0 - 4 sig dig
4. 0.00204 - 3 sig dig
5. 0.020400 - 5 sig dig
6. 2.0400 - 5 sig dig
7. 100,000 - 1 sig dig
8. 100,000.0 7 sig dig
9. 0.0001 - 1 sig dig
10. 0.000100 - 3 sig dig
3. When multiplying or dividing, the answer must have the same number of sig dig as the least sig dig in the problem.  Your answer can not be more accurate than any measurement in the problem. 
4. When adding or subtracting the answer must have the same number of decimal places as the least number of decimal places in the problem. Your answer can not be more accurate than any measurement in the problem. 
5. Example:  Pretend we weigh something on a cheap scale and it is 3.5 grams.  Then we weigh something else and it is 4.2448 grams.  If we add them together the answer is 7.7 grams because we can only have one decimal place.  If we multiply them the answer is 15 grams² because the answer can only have 2 sig dig.
6. Now remember that conversion factors such as (1 L / 1000 mL) DO NOT affect the significant digits because they are exact numbers. 
7. More examples:
1. 3340 x 1.2 = 4.0 x 10³  The answer must have 2 sig dig thus 4000 is incorrect because it only has 1 sig dig.
2. 88359 / 3 = 30000  The answer can only have 1 sig dig.
3. 8.888 x 3.29853 = 29.32  The answer must have 4 sig dig.
4. 1.25 + 3.2 = 4.5  The answer can only have one decimal place.
5. 145 - 0.222 = 145  The answer can not have any decimal places.
6. 145 - 0.99 = 144  The answer can not have any decimal places.
7. 0.042 + 1.33 = 1.37  The answer can have only 2 decimal places.
8. More problems
1. How many significant digits are in 23,000?
2. How many significant digits are in 4000.00?
3. How many significant digits are in 0.00023040?
4. How many significant digits are in 400?
5. How many significant digits are in 7.2500?
6. How many significant digits are in 0.00333?
7. Write the numbers in 11-16 above in scientific notation.
8. Write these numbers in scientific notation:  42,000;  27.0040;  150,000,000;  0.00000003500;  50,000;  0.0205
9. Answers to the above
1. 2
2. 6
3. 5
4. 1
5. 5
6. 3
7. 2.3 x 10⁴, 4.00000 x 10³, 2.3040 x 10^-4, 4 x 10², 7.2500 x 10⁰, 3.33 x 10^-3 (Note that you must keep the correct number of significant digits when putting into scientific notation
8. 4.2 x 10⁴;  2.70040 x 10¹;  1.5 x 10⁸, 3.500 x 10^-8;  5 x 10⁴;  2.05 x 10^-2

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Conversions and Density!

For conversions within the metric system, you must memorize the conversion (for example:  1000 mL = 1 L, or 1000 g = 1 kg should be memorized)  Remember that metric conversions are exact ratios and thus will not limit your significant digits for the answer.  First start with what you are given.  Figure out what units you need.  You will need to create a ratio (conversion factor) between the units given and the units needed.  Ask yourself which unit is bigger.  Put a "1" by that unit.  Then ask how many of the smaller units are in the bigger unit.  Put that number in front of the smaller unit.  There is your conversion factor.  Make sure the units cancel and you get the units you need.  Always write your units down.  Practice as many of the following as you need - the answers are below.

1 mile = 5280 feet, 12 inches = 1 foot, 3 feet = 1 yard, 1.609 km = 1 mile, 2.54 cm = 1 inch, 1 pound = 0.4536 kg, 1 quart = 946.4 mL.

1. How many pounds does 2.00 kg of cheese weigh?

2. How many mL are in 0.50 quarts?

3. How many inches are in 1.00 km?

4. How many feet are in 3.45 km?

5. How many km are in 5.00 miles?

6. How many yards are in 72.5 miles?

7. How many cc's are in 979 mL?
8. How many Kelvins are in -15.5^oC?
9. How many degrees Celsius are in 315 K?
10. How many degrees Fahrenheit are in 30.0^oC?
11. How many Gg are in 4.32 x 10¹¹ g?
12. How many meters are in 0.423 Mm?
13. How many kL are in 8383 L?
14. How many cm are in 0.783 m?
15. How many meters are in 252 mm?
16. How many grams are in 5232 mg?
17. How many liters are in 7.14 x 10⁶ nL?
18. How many liters are in 2.52 x 10⁴ mL?
19. How many mm are in 0.123 m?
20. If the mass of a lead ball is 23.5 g and the volume is 3.5 mL, what is the density of the lead ball?
21. If the density of carbon tetrachloride is 0.793 g/mL, and a sample has a volume of 9.29 mL, what is the mass?
22. If the density of ethanol is 0.828 g/mL and a sample has a mass of 14.5 g what is the volume?
23. How many km are in 2.88 m?
24. How many cL are in 4.56 x 10^-3 L?
25. How many ng are in 6.36 x 10^-7 g?
26. How many Mg are in 4.14 Gg?
27. How many mm are in 3.50 x 10⁵ pm?
28. How many nL are in 434 µL?
29. A water sample of mass 0.0204 kg is how many liters? d (H₂O) = 1.00 g/mL
30. If gold's density is 19.32 g/mL, how much would a 0.0333 L sample weigh in grams?
31. Table salt has a density of 2.16 g/mL.  If you used 2.00 mL on your food, how much in mg is that?
32. The density of ethanol is 0.802 g/mL.  How much in grams does 9.85 x 10^-2L mass?
33. How many GL are in 2.55 x 10⁷ L?
34. How many Mm are in 345 km?
35. How many cg are in 0.497 g?
36. How many cg are in 2.49 x 10³ mg?
37. How many mL are in 0.258 L?
38. How many µm are in 6.36 x 10^-7 m?
39. How many nm are in 345 µm?
40. What is the density in g/mL of a substance that masses 0.987 kg and has a volume of 4.52 x 10²mL?
41. How much water in mL would 5.25 mg of copper displace?  d (Cu) = 11.53 g/mL

Answers:

1. 2.00 kg (pound / 0.4536 kg) = 4.41 pounds

2. 0.50 quarts (946.4 mL / quart) = 470 mL

3. 1.00 km (mile / 1.609 km)(5280 feet / mile)(12 inch / ft) = 3.94 x 10⁴ or 39,400 inches

4. 3.45 km (mile / 1.609 km)(5280 ft / mile) = 11,320 or 1.13 x 10⁴ feet

5. 5.00 mile (1.609 km / mile) = 8.05 km

6. 72.5 miles (5280 ft / mile)(yd / 3 ft) = 1.28 x 10⁵ or 128,000 yards

7. 979 cc  (cc = mL)
8. -15.5^oC + 273 =  258K
9. 315K - 273 = 42^oC
10. 86^oF
11. 4.32 x 10¹¹ g (Gg / 10⁹ g) =  432 Gg
12. 0.423 Mm (10⁶ m / Mm) = 4.23 x 10⁵ m
13. 8383 L ( kL / 1000 L) = 8.383 kL
14. 0.783 m (100 cm / m) = 78.3 cm
15. 252 mm (m / 1000 mm) = 0.252 m
16. 5232 mg ( g / 10⁶mg) = 5.232 x 10^-3 g
17. 7.14 x 10⁶ nL (L / 10⁹ nL) = 7.14 x 10^-3 L
18. 2.52 x 10⁴ mL (L / 1000 mL) = 25.2 L
19. 0.123 m ( 1000mm / m) = 123 mm
20. 23.5 g / 3.5 mL =  6.7 g/mL
21. 9.29 mL (0.793 g/mL) = 7.37 g
22. 14.5 g ( mL / 0.828 g) = 17.5 mL
23. 2.88 m ( 1 km / 1000 m) = 2.88 x 10^-3
24. 4.56 x 10^-3 L ( 100 cL / 1 L) = 4.56 x 10^-1 or 0.456 cL
25. 6.36 x 10^-7 g ( 10⁹ ng / 1 g) = 6.37 x 10² or 637 ng
26. 4.14 Gg ( 10⁹ g / 1 Gg )( 1 Mg / 10⁶ g) = 4.14 x 10³ Mg
27. 3.50 x 10⁵ pm ( 1 m / 10¹² pm)( 1000 mm / m) = 3.50 x 10^-4 mm
28. 434 µL ( L / 10⁶ µL)(10⁹ nL / L) = 4.34 x 10⁵ nL
29. Since density of water = 1.000 g/mL:  0.0204 kg ( 1000g / kg)(1 mL / 1 g)(L / 1000 mL) = 0.0204 L
30. 0.0333 L ( 1000 mL / L )(19.32 g / mL) = 643 g
31. 2.00 mL (2.16 g / mL)(1000 mg / g ) = 4320 mg
32. 9.85 x 10^-2L ( 1000mL / L)(0.802 g / mL) = 79.0 g
33. 2.55 x 10⁷ L (  GL / 10⁹ L) = 2.55 x 10^-2 or 0.0255 GL
34. 345 km (1000m / km)(Mm / 10⁶ m) = 0.345 Mm
35. 0.497 g (100 cg / g) = 49.7 cg
36. 2.49 x 10³ mg (  g / 1000 mg)(100 cg / g) = 249 cg or 2.49 x 10² cg
37. 0.258 L (1000 mL / L) = 2.58 x 10² or 258 mL
38. 6.36 x 10^-7 m ( 10⁶µm / m) = 0.636 µm
39. 345 µm (  m / 10⁶ µm) ( 10⁹nm / m) = 3.45 x 10⁵ nm
40. 0.987 kg  (1000 g / kg) = 987 g  then divide by 4.52 x 10²mL = 2.18 g/mL
41. 5.25 mg ( g / 1000 mg)(  1 mL / 11.53 g Cu) = 4.55 x 10^-4 mL

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Nomenclature Rules

1. Binary Ionic Compounds (ionic compounds with two elements)
1. use the metal's name + the nonmetal's name + the suffix "ide"
2. NaCl is sodium chloride, Li₂O is lithium oxide, KBr is potassium bromide, MgI₂ is magnesium iodide
3. sodium nitride is Na₃N, calcium bromide is CaBr₂, aluminum fluoride is AlF₃
4. note that ionic compounds are always empirical formulas - we would never say that table salt is Na₆Cl₆
2. Ionic compounds with polyatomic ions
1. treat polyatomic ions as groups
2. never change the name of the polyatomic ion
3. NaOH is sodium hydroxide, Ca(NO₃)₂ is calcium nitrate, MgSO₄ is magnesium sulfate
4. ammonium sulfide is (NH₄)₂ S, calcium carbonate is CaCO₃, magnesium phosphate is Mg₃(PO₄)₂
3. Binary Covalent Compounds (covalent compounds with two elements)
1. prefix + nonmetal + prefix + nonmetal + suffix "ide"
2. prefixes are mono, di, tri, tetra, penta, hexa, hepta, octa
3. if there is only one of the first nonmetal, don't use the mono prefix
4. carbon tetraiodide is CI₄, sulfur dioxide is SO₂, phosphorus pentachloride is PCl₅
5. P₂O₅ is diphosphorus pentoxide, N₂H₄ is dinitrogen tetrahydride, SF₂ is sulfur difluoride
4. Transition metal compounds:  name of transition metal, (charge in Roman numerals) + name of nonmetal + ide
1. Fe₂O₃ - iron(III) oxide
2. CuS - copper(II) sulfide
3. Co(NO₃)₃ - cobalt(III) nitrate
4. FeSO₄ - iron(II) sulfate
5. iron(III) flouride - FeF₃
6. iron(II) phosphate - Fe₃(PO₄)₂

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Oxidation States and Ionic Formulas

 Here are some rules about oxidation states, which is a fancy name for charges.

• Alkali metals are always charged +1 when in a compound. Their goal is to lose one electron, thus becoming +1 cations.
• Alkaline Earth metals are always charged +2 when in a compound. Their goal is to lose two electrons, thus becoming +2 cations.
• B column is usually charged +3 or -5 depending on what they are bonded to.  They can lose 3 electrons becoming +3 OR gain 5 electrons becoming -5. Actually Al, Ga and In are always +3.   
• C column is usually charged +4 or -4 depending on what it is bonded to. They can lose 4 electrons becoming +4 OR gain 4 electrons becoming -4. (They can also be +2 or -2.)  Example  CH₄  C is -4 and H is +1.  Example CO₂  C is +4 and O is -2.
• N column is usually charged -3 in compounds but sometimes +5. They can lose 5 electrons becoming +5 OR gain 3 electrons becoming -3.  Example NH₃  N is -3 and H is +1.
• Chalcogens are usually charged -2 in compounds (Oxygen is always charged -2 when in a compound except for peroxides like H₂O₂ where it is -1) unless they are bonded to something more electronegative than they are, in this case they can be +6 +4 or +2.  Example sulfate ion.  SO₄^2-  Here the total charge is -2 so S is +6, each O is -2 (there are 4 of them so that adds up to -8 for all four O's) so that adds up to -2 overall, which is why sulfate is an ion and not a molecule - it has an overall charge.
• Halogens are usually charged -1 in compounds (F always is -1) unless they are bonded to something more electronegative than they are, in this case they can be +7  +5 or +3.   Example IF₇.  Here F is -1 since it always is -1.  That makes I +7.  Note that IF₅ and IF₃ also exist.  
• Noble gases are stable and unreactive, they don't often have a charge - they are neutral atoms.
• Transition metals have many charges, too many to predict.  Exceptions are Zn and Cd which are always +2 and silver which is always +1.  My research focused on Re which can be charged from +1 all the way up to +9.  
• H is usually +1.  When H is bonded to a metal however, it can be -1.  Example NaH where Na is +1 since it is always +1 so H must be -1. 
• Remember the atoms in a pure element have a charge of ZERO!!!  Atoms don't have charges unless they are bonded to something else.

OK, armed with this information,  Let's predict chemical formulas for compounds from the following and name them:

1. lithium and oxygen
2. magnesium and phosphate
3. ammonium and sulfur
4. calcium and fluorine
5. aluminum and chlorine
6. potassium and nitrate
7. sodium and sulfate
8. phosphorus and oxygen
9. silicon and hydrogen
10. calcium and hydroxide
11. silver and carbonate
12. sodium and bromine

Answers

1. Li₂O  (Li is +1 so we need 2 of them since O is -2)  lithium oxide
2. Mg₃(PO₄)₂  (Mg is +2 so we need 3 of them, phosphate is -3 so we need 2 of them) magnesium phosphate
3. (NH₄)₂S   (ammonium is +1 so we need 2 of them since S is -2) ammonium sulfide
4. CaF₂  (Ca is +2 and F is -1 so we need two F)  calcium fluoride
5. AlCl₃  (Al is +3 and Cl is -1 so we need 3 of them)  aluminum chloride
6. KNO₃  ( K is +1 and nitrate is -1)  potassium nitrate
7. Na₂SO₄ (Na is +1 so we need 2 of them since sulfate is -2) sodium sulfate
8. P₂O₅  (O is -2 according to the rules, so P must be +5 and not -3 since negatives repel, we need 2 P's to make +10 and 5 O's to make -10)  diphosphorus pentoxide
9. SiH₄  (Since H is +1, Si must be -4, and so we need 4 H's) silicon tetrahydride
10. Ca(OH)₂  (Ca is +2 and OH is -1 so we need 2 of them)  calcium hydroxide
11. Ag₂CO₃  (Ag is +1 so we need 2 of them since carbonate is -2)  silver carbonate or I'd also accept silver(I) carbonate
12. NaBr  (Na is +1 and Br is -1)  sodium bromide

You should know these polyatomic ions by the way!

• sulfate SO₄^2-
• phosphate PO₄^3-
• carbonate  CO₃^2-
• hydroxide  OH^-
• ammonium  NH₄⁺
• acetate  CH₃COO^-
• nitrate  NO₃^-
• bicarbonate HCO₃^-
• monohydrogen phosphate HPO₄^2-
• dihydrogen phosphate H₂PO₄^-1

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avogadro's number

The following practice problems deal with converting between moles and atoms or molecules using Avagadro's Number.  Then we'll also convert between grams and atoms or molecules using Avogadro's Number and Molar Mass.  Remember if we start with an entire molecule but want to end with just atoms (or vice versa), we need to use molar ratios such as ( 2 H atoms / 1 H₂O molecule). 

1. How many atoms are in 4.00 moles of carbon?
2. How many moles are in 1.25 x 10²⁵ atoms of neon?
3. How many molecules are in 2.99 moles of ammonia?
4. How many moles are in 5.25 x 10²⁴ molecules of lithium sulfide?
5. How many molecules are in 4.24 moles of sodium chloride?
6. How many H atoms are in 8.35 moles of water?
7. How many O atoms are in 1.44 moles of magnesium sulfate?
8. If I have 3.45 x 10²⁵ H atoms in a pure CH₄ sample, how many moles of CH₄ are present?
9. How many molecules are in 5.25 grams of LiBr?
10. How much does 6.46 x 10²⁵ molecules of CaS mass?
11. How many atoms are in 24.7 grams of argon?
12. How much does 3.33 x 10²⁴ nitrogen atoms mass?
13. How many molecules are in 34.5 grams of calcium phosphate?
14. How many C atoms are in 2.64 grams of C₂H₄?
15. How many Br atoms are in 14.3 grams of bromine?
16. A sample of silicon dioxide contains 5.25 x 10²⁶ oxygen atoms.  How much does this sample mass?

Answers

1. 4.00 mol C (6.02 x 10²³ atoms / mol) = 2.41 x 10²⁴ C atoms
2. 1.25 x 10²⁵ atoms Ne ( mol / 6.02 x 10²³ atoms) = 20.8 mol Ne
3. 2.99 mol NH₃ (6.02 x 10²³ molecules / mol) = 1.80 x 10²⁴ molecules NH₃
4. 5.25 x 10²⁴ molecules Li₂S (mol / 6.02 x 10²³ molecules) = 8.72 mol Li₂S
5. 4.24 mol NaCl (6.02 x 10²³ molecules / mol) = 2.55 x 10²⁴ molecules NaCl
6. 8.35 mol H₂O ( 2 H / 1 H₂O) (6.02 x 10²³ atoms / mol) = 1.01 x 10²⁵ H atoms
7. 1.44 mol MgSO₄ ( 4 O / MgSO₄) (6.02 x 10²³ atoms / mol) = 3.47 x 10²⁴ O atoms
8. 3.45 x 10²⁵ H atoms ( 1 CH₄ molecule/ 4 H atoms) (mol / 6.02 x 10²³ molecules) = 14.3 mol CH₄
9. 5.25 g LiBr (mol / 86.8 g)(6.02 x 10²³ molecules / mol) = 3.64 x 10²² molecules LiBr
10. 6.46 x 10²⁵ molecules CaS (mol / 6.02 x 10²³ molecules)(72.2 g / mol) = 7750 g CaS
11. 24.7 g Ar ( mol / 39.9 g) (6.02 x 10²³ atoms / mol) = 3.73 x 10²³ atoms Ar
12. 3.33 x 10²⁴ N atoms (mol / 6.02 x 10²³ atoms) (14.0 g / mol ) = 77.4 g N
13. 34.5 g Ca₃(PO₄)₂ (mol / 310.3 g) (6.02 x 10²³ molecules / mol) = 6.69 x 10²² molecules Ca₃(PO₄)₂
14. 2.64 g C₂H₄ ( mol / 28.0 g) (2 C / 1 C₂H₄) (6.02 x 10²³ atoms / mol) = 1.14 x 10²³ C atoms
15. 14.3 g Br₂ ( mol / 159.8 g) (2 Br / Br₂) (6.02 x 10²³ atoms / mol) = 1.08 x 10²³ Br atoms
16. 5.25 x 10²⁶ O atoms (mol / 6.02 x 10²³ atoms) (1 SiO₂ / 2 O ) ( 60.1 g / mol) = 2.62 x 10⁴ g SiO₂

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Balancing Chemical Reactions

Remember when writing reactions, you must have all the correct formulas for the reactant(s) and product(s) first.  Formulas depend on the charges of the elements and polyatomic ions involved.  Once you have determined the formulas, you can NOT change them later - don't change the subscripts in the formulas.  For example sodium and chlorine make NaCl.  You cannot later while balancing change this to NaCl₂.  You can only add coefficients in front of the compounds to balance the reaction - like 2 NaCl.

Formation rxn:  reactants are elements in their natural state forming a product compound

Combustion rxn:  hydrocarbon (molecule with only C and H) adding to oxygen producing water and carbon dioxide

Precipitation rxn:  (a type of double replacement) when the cations swap anions and one product is NOT soluble

Acide Base rxn: Where the H⁺ from an acid comes off and the base (often hydroxide ion) gains the H+

1. Balance the following formation reaction:  Ca (s) + I₂ (s) --> CaI₂
2. Balance the following formation reaction:  Na (s) + Br₂ (l) --> NaBr
3. Write and balance the formation reaction for magnesium and chlorine.
4. Write and balance the formation reaction for potassium and oxygen.
5. Write and balance the formation reaction for calcium and oxygen.
6. Write and balance the formation reaction for phosphorus and chlorine.
7. Balance the following combustion reaction:  C₄H₈ + O₂ (g) --> CO₂ (g) + H₂O (g)
8. Balance the following combustion reaction:  C₆H₁₄ + O₂ (g) --> CO₂ (g) + H₂O (g)
9. Write and balance the combustion reaction for C₆H₆.
10. Write and balance the combustion reaction for C₃H₆.
11. Balance the following double displacement reaction:  CaI₂ + Na₂SO₄ --> CaSO₄ + NaI
12. Balance the following double displacement reaction:  MgSO₄ + KNO₃ --> Mg(NO₃)₂ + K₂SO₄
13. Write and balance the double displacement reaction for magnesium chloride and potassium oxide.
14. Write and balance the double displacement reaction for calcium nitride and lithium bromide.
15. Write and balance the double displacement reaction for silver nitrate and iron(III) chloride.

Answers

1. already balanced
2. 2 Na (s) + Br₂ (l) --> 2 NaBr
3. Mg (s) + Cl₂ (g) --> MgCl₂
4. 4 K (s) + O₂ (g) --> 2 K₂O
5. 2 Ca (s) + O₂ (g) --> 2 CaO
6. 2 P (s) + 5 Cl₂ (g) --> 2 PCl₅
7. C₄H₈ + 6 O₂ (g) --> 4 CO₂ (g) + 4 H₂O (g)
8. 2 C₆H₁₄ + 19 O₂ (g) --> 12 CO₂ (g) + 14 H₂O (g)
9. 2 C₆H₆ + 15 O₂ (g) --> 12 CO₂ (g) + 6 H₂O (g)
10. 2 C₃H₆ + 9 O₂ (g) --> 6 CO₂ (g) + 6 H₂O (g)
11. CaI₂ + Na₂SO₄ --> CaSO₄ + 2 NaI
12. MgSO₄ + 2 KNO₃ --> Mg(NO₃)₂ + K₂SO₄
13. MgCl₂ + K₂O --> MgO + 2 KCl
14. Ca₃N₂ + 6 LiBr --> 3 CaBr₂ + 2 Li₃N
15. 3 AgNO₃ + FeCl₃ --> 3 AgCl + Fe(NO₃)₃

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Molarity and dilution

Molarity (M) = moles per liter (mol/L)  Note you may need to change grams to moles using the molar mass first.

Dilution equation:  M₁V₁ = M₂V₂   where 1 is before and 2 is after, thus V₂ is the total final volume

1. How many moles are in 234 mL of a 1.29 M solution?
2. What is the molarity if 9.25 moles of sodium sulfide is dissolved to make a 787 mL solution?
3. What is the molarity if 43.2 grams of calcium chloride is dissolved to make a 1145 mL solution?
4. How many grams do you need to prepare 0.500 L of a 4.00 M solution of sodium bromide?
5. The salt is blood serum is 0.14 M.  How many grams of sodium chloride are in 5.00 mL of blood serum?
6. I want to make a 0.555 M potassium iodide solution.  I have 12.7 grams of potassium iodide and I want to use it all.  What will be the volume of my solution?
7. I used 54.7 grams of sodium phosphate to make a 0.250 M solution.  What is the volume in mL?
8. I have 20.0 mL of a 2.34 M solution.  If I add 15.0 mL to the solution, what is the final molarity?
9. I had 25.0 mL of a 5.00 M solution.  How much water did I add if I diluted the solution to 2.30 M?
10. What is the final volume of a 0.833 M solution if the original was 10.0 mL of a 1.00 M solution?
11. What was the original molarity of 25.0 mL of a 0.424 M solution if the original volume was 10.0 mL?
12. Concentrated HCl is 6.00 M.  In lab we used 1.00 M HCl.  If we need 500.0 mL in lab, how much of the concentrated do I need to dilute?

Answers

1. 1.29 mol / L = x moles / 0.234 L, x = (1.29 mol/L)(0.234 L) = 0.302 moles
2. M = (9.25 moles / 0.787 L) = 11.8 M
3. 43.2 g CaCl₂ (mol / 111.1 g) = 0.389 moles, M = (0.389 mol / 1.145 L) = 0.340 M
4. (0.500 L )(4.00 mol/L) = 2.00 moles NaBr needed, 2.00 mol NaBr (102.9 g/mol) = 206 grams
5. (0.14 mol/L)(0.00500 L) = 7.00 x 10^-4 moles, 7.00 x 10^-4 moles NaCl (58.5 g/mol) = 0.0410 g
6. 12.7 g KI (mol / 166.0 g) = 0.0765 moles KI,   0.0765 mol ( 1 L / 0.555 mol) = 0.138 L
7. 54.7 g Na₃PO₄ (mol / 164 g) = 0.334 moles,   0.334 mol ( 1 L / 0.250 mol) = 1.34 L = 1340 mL
8. (2.34 M)(20.0 mL) = M₂ (35.0 mL), M₂ = 1.34 M
9. (5.00 M)(25.0 mL) = (2.30 M) V₂,  V₂ = 54.3 mL so I added 29.3 mL
10. (10.0 mL)(1.00 M) = (0.833 M) V₂,    V₂ = 12.0 mL
11. M₁ (10.0 mL) = (0.424 M)(25.0 mL) ,     M₁ = 1.06 M
12. (6.00 M) V₁ = (1.00 M)(500.0 mL),   V₁ = 83.3 mL

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Stoichiometry

Use the following balanced chemical reaction to answer practice questions 1 - 11. 

H₂SO₄  +  2 NaOH  g   Na₂SO₄  +  2 H₂O

1. How many moles of water can be produced from 0.234 moles H₂SO₄ ?

2. How many moles of sodium hydroxide are needed to completely react with 0.878 moles H₂SO₄ ?

3. How many moles of sodium hydroxide were used to produced 1.32 moles of sodium sulfate?

4. How many moles of H₂SO₄  are needed to completely react with 3.13 grams of sodium hydroxide?

5. How many grams of sodium sulfate could be produced from 0.525 mol sodium hydroxide?

6. How many moles of water could be produced from 25.0 grams H₂SO₄ ?

7. How many grams of sodium sulfate could be produced from 13.8 grams H₂SO₄ ?

8. How many grams of sodium hydroxide were needed to produce 65.0 grams of water?

9. How many grams of sodium hydroxide are needed to completely react with 21.3 grams H₂SO₄ ?

10. How many grams of water were produced if 47.0 grams of sodium sulfate were also produced?

11. How many grams of sodium sulfate can be produced from 23.7 grams H₂SO₄ and 23.7 grams sodium hydroxide?

Answers

1. 0.234 mol H₂SO₄ (2 H₂O / 1 H₂SO₄ ) = 0.468 mol H₂O

2. 0.878 mol H₂SO₄ (2 NaOH / 1 H₂SO₄ ) = 1.76 mol NaOH

3. 1.32 mol Na₂SO₄  (2 NaOH / 1 Na₂SO₄ ) = 2.64 mol NaOH

4. 3.13 g NaOH (mol / 40.0 g)(1 H₂SO₄ / 2 NaOH) = 0.0391 mol H₂SO₄

5. 0.525 mol NaOH (1 Na₂SO₄ / 2 NaOH)(142.1 g / mol) = 37.3 g Na₂SO₄

6. 25.0 g H₂SO₄ (mol / 98.1 g)(2 H₂O / 1 H₂SO₄ ) = 0.510 mol H₂O

7. 13.8 g H₂SO₄  (mol / 98.1 g)(1 Na₂SO₄ / 1 H₂SO₄ )(142.1 g / mol) = 20.0 g Na₂SO₄

8. 65.0 g H₂O (mol / 18.0 g)(2 NaOH / 2 H₂O)(40.0 g / mol) = 144 g NaOH

9. 21.3 g H₂SO₄ (mol / 98.1 g)(2 NaOH / 1 H₂SO₄ )(40.0 g / mol) = 17.4 g NaOH

10. 47.0 g Na₂SO₄ (mol / 142.1 g)(2 H₂O / 1 Na₂SO₄ )(18.0 g / mol) = 11.9 g H₂O

11. This is a limiting reagent problem, so we work both out:  23.7 g H₂SO₄  (mol / 98.1 g)(1 Na₂SO₄ / 1 H₂SO₄ )(142.1 g / mol) = 34.3 g Na₂SO₄ possible    OR     23.7 g NaOH (mol / 40.0 g)(1 Na₂SO₄ / 2 NaOH )(142.1 g / mol) = 42.1 g Na₂SO₄ possible .  Since 34.3 grams is the smaller of the two possible amounts, it is the real answer which means H₂SO₄ is the limiting reagent and NaOH is in excess.

Now answer the following stoichiometric problems.

1. Write and balance the formation reaction between nitrogen and hydrogen.  If I start with 9.33 grams of nitrogen, how much product can I make?  If I actually made 9.74 grams of product, what was my percent yield?

2. Write and balance the formation reaction that produces phosphorus pentachloride.  How many grams of chlorine do I need in order to produce 1.42 grams of phosphorus pentachloride?

3. Write and balance the combustion reaction for the hydrocarbon C₆H₁₂.  How many grams of oxygen will I need to completely combust 534 grams of C₆H₁₂?

4. Write and balance the metathesis reaction between calcium phosphate and potassium bromide.  If I produced 42.4 grams of calcium bromide, how many grams of calcium phosphate did I start with?

5. Write and balance the metathesis reaction between sodium chloride and potassium fluoride.  If I start with 66.2 grams potassium fluoride, how many grams of sodium fluoride can I produce?  If I actually produce 35.7 grams of sodium fluoride, what is my percent yield?

6. Write and balance the formation reaction for carbon dioxide.  If I start with 1.77 grams carbon, how much product can I make?  If I actually produce 2.10 grams of product, what is my percent yield?

7. If my theoretical yield is 10.0 grams and my actual yield is 7.24 grams, what is my percent yield?

8. If my theoretical yield is 10.0 grams and my actual yield is 11.24 grams, what is my percent yield?

9. How many grams of potassium sulfate can be produced by the metathesis reaction of 8.88 grams potassium chloride and 4.44 grams magnesium sulfate?

Answers

1. N₂ (g) +  3 H₂ (g) g 2 NH₃ (g)      9.33 g N₂ (mol / 28 g)(2 NH₃ / 1 N₂)(17.0 g / mol) = 11.3 g NH₃ is what I could make.  However, my percent yield is (9.74 / 11.3) x 100 = 86.2%

2. 2 P (s) + 5 Cl₂ (g)  g  2 PCl₅       1.42 g PCl₅ (mol / 208.5 g)( 5 Cl₂ / 2 PCl₅)(71.0 g / mol) = 1.21 g Cl₂

3. C₆H₁₂  +  9 O₂ (g)  g  6 CO₂ (g)  +  6 H₂O (g)     534 g C₆H₁₂ (mol / 84.0 g)(9 O₂ / 1 C₆H₁₂ )(32.0 g / mol) = 1830 g O₂

4. Ca₃(PO₄)₂ + 6 KBr g 2 K₃PO₄ + 3 CaBr₂      42.4 g CaBr₂(mol / 199.9 g)(1 Ca₃(PO₄)₂ / 3 CaBr₂ )(310.3 g / mol) = 21.9 g Ca₃(PO₄)₂

5. NaCl + KF g NaF + KCl     66.2 g KF (mol / 58.1 g)(1 NaF / 1 KF)(42.0 g / mol) = 47.9 g NaF    my percent yield is (35.7 / 47.9) x 100 = 74.5%

6. C (s) + O₂ (g) g CO₂ (g)   1.77 g C (mol / 12.0 g)(1 CO₂ / 1 C)(44.0 g / mol) = 6.49 g CO₂     my percent yield is (2.10 / 6.49) x 100 = 32.4%

7. (7.24 / 10.0) x 100 = 72.4%

8. (11.24 / 10.0) x 100 = 112.4%  - Something is wrong here.  I can't create matter.  When  percent yield is higher than 100% there is a human error.  Perhaps I weighed more reactant out than I realized.  Perhaps water condensed inside my reaction vessel.  Perhaps my theoretical yield calculation was wrong. 

9. First write the reaction:  2 KCl + MgSO₄ g  K₂SO₄ + MgCl₂  Now this is a limiting reagent problem.  Calculate the answer based on 8.88 g KCl:  8.88 g KCl ( mol / 74.6 g)(K₂SO₄ / 2KCl)(174.3 g/mol) = 10.4 g K₂SO₄ .  Now calculate the answer based on 4.44 g MgSO₄ :  4.44 g MgSO₄ ( mol / 120.4g)( K₂SO₄ / MgSO₄ )(174.3 g/mol) = 6.43 g K₂SO₄ .  Compare and the answer is really 6.43 g, magnesium sulfate is the limiting reagent and potassium chloride is in excess. 

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Oxidation and Reduction

reduction = gaining electrons, since electrons are negative the charge goes down, when a chemical is reduced it is called the oxidizing agent

Oxidation = losing electrons, when electrons go away the atom is more positive so the charge goes up, when a chemical is oxidized it is called the reducing agent

exothermic = heat is produced, heat is a product, the reaction gets warm as heat is released

endothermic = heat is need to react, heat is a reactant, the reaction feels cold as it absorbs surrounding heat

1. Is the reaction occurring in an instant hot pack exothermic or endothermic?
2. What is the charge on sulfur in H₂SO₄?
3. What is the charge on iron in FePO₄?
4. What is the charge on Cl in Cl₂ (g)?
5. What is the charge on a sodium atom in Na (s)?
6. For the following reactions, identify what is oxidized, what is reduced, what is the oxidizing agent, and what is the reducing agent:
1. 3 Ni (s) + 2 Al³⁺ (aq)  -->  3 Ni²⁺ (aq) + 2 Al (s)  (Note the reaction is balanced so that the charge on each side of the arrow is equal)
2. Cu²⁺ (aq) + Cd (s) --> Cu (s) + Cd²⁺ (aq)
3. 2 Na (s) + 2 H₂O (l) --> 2 NaOH (aq) + H₂ (g)
4. 2 Li (s) + I₂ (s) --> 2 LiI (s)
5. 4 Al (s) + 3 O₂ (g) --> 2 Al₂O₃ (s)
6. 2 CuNO₃ + PdCl₄ --> 2 CuCl₂ + Pd(NO₃)₂

Answers

1. exothermic - it is producing heat
2. +6 (O is -2 and there are four of them, H is +1 so that adds up to -6 thus S must be +6)
3. +3 (phosphate ion is -3 so iron must be +3)
4. zero (all elements are zero)
5. zero (it's an element)
6. Answers:
1. oxidized = reducing agent = Ni (s), reduced = oxidizing agent = Al³⁺ (aq)
2. oxidized = reducing agent = Cd (s), reduced = oxidizing agent = Cu²⁺ (aq)
3. oxidized = reducing agent = Na (s), reduced = oxidizing agent = H₂O (l)
4. oxidized = reducing agent = Li (s), reduced = oxidizing agent = I₂ (s)
5. oxidized = reducing agent = Al (s), reduced = oxidizing agent = O₂ (g)
6. oxidized = reducing agent = CuNO₃ , reduced = oxidizing agent = PdCl₄

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Quantum Numbers, shells, orbitals

1. Shells, Subshells and Orbitals
1. Shells are three dimensional spheres around the nucleus.  They are numbered 1, 2, 3, 4, 5, 6, 7...  Make sure you understand they are spheres not circles - imagine the nucleus as a tiny steel ball bearing.  The shells would be like a ping pong ball around the steel ball, then a baseball, then a soccer ball, then a beach ball, etc... The shells are quantized because nothing exists between them.  Electrons must be on a shell, not in between.  Stairs are also quantized, ramps are not.
2. Subshells are on the shells.  We have s, p, d, f, g... subshells.  Shell one has one subshell - s.  Shell 2 has 2 subshells - s and p.  Shell 3 has 3 subshells - s, p, d.  Shell 4 has 4 subshells - s, p, d, f.  And so forth. 
3. Subshells consist of orbitals.  The s subshell has one orbital.  The p subshell has 3 orbitals.  The d subshell has 5 orbitals.  The f subshell has 7 orbitals.  And so forth with odd numbers of orbitals.
4. Two electrons can fit into one orbital, but they must have opposite spins.  Electrons are always spinning.  
5. Putting it together:
1. Shell one - one subshell (s), one orbital, 2 electrons maximum.
2. Shell two - two subshells (s, p), 4 orbitals, 8 electrons maximum.
3. Shell three - three subshells (s, p, d), 9 orbitals, 18 electrons maximum.
4. Shell four - four subshells (s, p, d, f), 16 orbitals, 32 electrons maximum.
6. Example questions
1. How many subshells are on the fourth shell?  4
2. How many orbitals are on the third shell?   9
3. How many orbitals are in a p subshell?  3
4. How many electrons can fit into a d subshell?  10
5. How many orbitals are in an f subshell?  7
6. How many electrons can fit into a p orbital?  2

7. Shells exist around the nucleus at certain distances.  There is nothing between the shells and electrons can not exist anywhere but on the shells.  We use this word to describe the shells and how they exist.  answer:  quantized

8. How many subshells are on the third shell?   3

9. How many electrons can fit into an orbital?   2

10. How many orbitals are on the first shell?    1

11. How many electrons maximum can fit on the fourth shell?    32

2. Quantum Numbers
1. A quantum number is an address for an electron.  No two electrons have the same address. A quantum number looks like this:  (n, l, m_l, m_s) where n = shell number, l = subshell number, m_l = orbital number and m_s = spin number.
2. Shell number n.  Simply 1, 2, 3, 4, 5, 6, 7, ...
3. Subshell number l.  We have to assign numbers to the subshells.  We let 0 = s, 1 = p, 2 = d, 3 = f, etc.  Note that l must be less than n. For example, if n = 3, then l must be 2, 1, or 0 because the third shell only has s, p, and d subshells, not an f subshell - so l can not be 3 or higher.
4. Orbital number m_l.  We let the middle orbital be number 0 and then number the other orbitals if there are any. 
1. When l = 0, it is the s subshell, there is only one orbital, it is numbered 0.
2. When l = 1, it is the p subshell, there are three orbitals, they are numbered -1, 0, 1.
3. When l = 2, it is the d subshell, there are five orbitals, they are numbered -2, -1, 0, 1, 2.
4. When l = 3, it is the f subshell, there are seven orbitals, they are numbered -3, -2, -1, 0, 1, 2, 3.
5. So if an electron is in a p subshell, what are the possible m_l numbers?  -1, 0, 1
6. So if an electron is in a f subshell, what are the possible m_l numbers?  -3, -2, -1, 0, 1, 2, 3.
7. Note that m_l possible values are from -l to +l.  For example, if l = 2 the possible m_l values are -2, -1, 0, 1, 2.
5. Spin number m_s.  We assign +1/2 to a clockwise spin and -1/2 to a counter clockwise spin.  So the possible m_s numbers are just -1/2 and +1/2.  Nothing else.

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How to use your Calculator! 

1. Entering a number in scientific notation: Let's enter 3.53 x 10⁸ in our calculator.  Push in [3.53] then push your [EE] or [EXP] button which stands for "times ten to the power of" all in that one little button.  Then push in [8].  There you have it.  Now try 2.52 x 10^-23.  Push in [2.52] then [EE] or [exp] then [+/-] to get the negative sign, then [23].  There you have it.  
2. Making your calculator display numbers in scientific notation:  Punch in the number 234000.  Now there are 2 main ways to make your calculator display this in scientific notation depending on your calculator.  (1) Push the [2nd] function button.  Now find the button that says SCI over the top of it.  It is usually the [8 or 5] key.  Now your calculator should show 2.34 EE 5 or something similar.  Graphing calculators do it like this (2) Find the mode button on your calculator and push it.  Move the cursor by pushing the arrow buttons until the cursor highlights the SCI mode.  Push [enter]. Now quit the mode selection process by pushing [quit].  Now your calculator should show the number in scientific notation like 2.34 EE 5 or something similar.   
3. NEVER EVER ENTER SCIENTIFIC NUMBERS BY ACTUALLY PUSHING IN [x] [10] [^] or [y^X] BUTTONS.  Always use the [EE] or [EXP] buttons. 
4. Calculations with scientific notation: Try the following calculations.  
1. (2.55 x 10⁸ )(8.21 x 10⁴) = 2.09 x 10¹³
2. (4.24 x 10^-4)(7.51 x 10^-5) = 3.18 x 10^-8
3. (4.77 x 10³) / (2.24 x 10^-7) = 2.13 x 10¹⁰
4. 525 / [(2.24 x 10³)(8.35 x 10^-3)] = 28.1
5. 17.3 (7.13 x 10^-23) / [(77.34)(6.26 x 10^-18)] = 2.55 x 10^-6

Did any of the answers have less digits than you had?  That's due to SIGNIFICANT DIGITS.  When multiplying and dividing the answer can only have the same amount of digits as the number in the problem with the least amount of significant digits. Your answer can never be more significant than your given information.  Thus 6a and 6c has 2 digits while 6b and 6d has 3 digits in the answer.  Now remember that conversion factors such as (1 L / 1000 mL) DO NOT affect the significant digits because they are exact numbers.  Note than when adding and subtracting you focus on decimal places.  Your answer can not have more decimal places than the number with the least amount of decimal places. Try some more. 

1. ([24.5)(0.084852)] / [(12.2)(2.234)] = 0.0763 is the answer.  If you got 0.381 you put 2.234 in the numerator instead of the denominator.  Here is what you actually push into the calculator:  (24.5) times (0.084852) divided by (12.2) divided by (2.234).  If you push times before 2.234 the calculator puts it in the numerator .
2. (1.24 x 10^-19)(3.15 x 10¹³) = 3.91 x 10^-6 is the answer.  If you got the 3.91 correct but the wrong power of ten you used scientific notation incorrectly in your calculator.  Actually push in this:  1.24 EE (+/-)19 times 3.15 EE 13 = and you should get the correct answer.  To make your calculator show you the answer in scientific notation, for most calculators push 2nd function then SCI. 
3. [(3.2 x 10^-29)(3.776 x 10¹⁴)] / [(8.3 x 10⁹)(5.25 x 10²¹)] = 2.8 x 10^-46 (Only 2 significant digits because of all the numbers the least amount of sig dig was 2)
4. (231.5) / [(1.2345 x 10³)(9.8765 x 10^-4)] = 189.9
5. 3340 x 1.2 = 4.0 x 10³  The answer must have 2 sig dig thus 4000 is incorrect because it only has 1 sig dig.
6. 88359 / 3 = 30000  The answer can only have 1 sig dig.
7. 8.888 x 3.29853 = 29.32  The answer must have 4 sig dig.
8. 1.25 + 3.2 = 4.5  The answer can only have one decimal place.
9. 145 - 0.222 = 145  The answer can not have any decimal places.
10. 145 - 0.99 = 144  The answer can not have any decimal places.
11. 0.042 + 1.33 = 1.37  The answer can have only 2 decimal places.

For 152 students.  

1. Try the following problems using logs.  However many significant digits are in the number for which you take the log are how many decimal places your answer should have.  So log 4.11 should have an answer with 3 decimal places = .614.  Log 46 should have an answer with only 2 decimal places = 1.66.  Remember log is base 10, ln is natural log. You should know the difference.
1. log 5.2 = 0.72
2. log (4.27 x 10^-4) = -3.370
3. 9.2 = log x,    solve for x, x = 2 x 10⁹    (you have to use the [2nd] function and [log] buttons to raise both sides to the 10.)
4. 4.25 = -log x, solve for x, x = 5.6 x 10^-5
5. 13.45 = -log x, solve for x, x = 3.5 x 10^-14
6. ln 3.55 x 10⁸⁴ = 1.947 x 10²
7. 9.25 x 10^-3 = ln x, solve for x, x = 1.009
2. How do you solve 259.403 = ln x.  Solving for x > 1.0 x 10⁹⁹.  Sometimes x is so huge our calculators can't display it.  Here's what you do:
1. Example: ln x = 200, Solve by taking e both sides, x = 7.23 x 10⁸⁶
2. Eventually the exponent will be greater than 99 and our calculators don't show more than 2 digits in a calculator - so must solve like this:  ln x = 300, If you take e of both sides you get ERROR
   So ln x = 150 + 150 (divide the large number of 300 by 2 so you have smaller numbers)
   take each number to e
   x = (1.39 x 10⁶⁵)(1.39 x 10⁶⁵) and solve by doing 1.39 squared then adding the exponents
   x = 1.93 x 10¹³⁰    COOL!!!
3. So the answer for the above question where ln x = 259.403   is x = 4.54 x 10¹¹²

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Study and Exam Tips

There are 4 main reasons student don't like or do poorly in chemistry:

1. They get behind and never catch up

2. They don't know how to use their scientific calculator

3. They don't spend enough time on the material

4. They don't practice problems

How to Pass Chemistry:

1.  Notetaking: Take notes during lecture. If you miss a class, the lecture closely follows the review, also try to get notes from a classmate.

2.  You should read over your lecture notes later that SAME day. Before you go to bed or after dinner, read over the notes you took that very day. This increases your chance of remembering them by 50%!!!

3.  You should take your own notes as you read each chapter. You will then have not only lecture notes and the review, but your own personal notes for each chapter. Group them together in your notebook.

4.  Do NOT wait until exam time to look at your notes for the first time since taking them. By that time you will not even remember writing those notes that first week of school! If you wait until test time, you will look back at your notes and not even remember writing them! When the professor finishes a chapter, you should read ALL the notes on that chapter again, and maybe even again!

5.  Try to read ahead of the professor. Lectures are often fast paced and may sometimes be boring. It really helps if you have already seen the material, even quickly, before going to lecture. This makes it much easier to follow along and stay interested. Read the material before lecture trying to understand it as well as you can on your own. Then the professor may make it all come together. Otherwise the professor may go so quick that you are completely lost.

6.  NEVER GET BEHIND! Catching up in chemistry is near impossible. You should...

7.  Study in 30 minute intervals several days every week. Studies prove you remember the first and last 15 minutes you study well, so study 30 minutes! Take many breaks like call someone, write an email, listen to a song, or change to a non-scientific subject like English. Telling yourself that you will sit down on the weekend and study chemistry for 5 hours straight is a lie! Study a little bit almost every day, do not cram. Do not study chemistry only on Sunday for example. You will get burned out on Sunday and forget too much in between.

8.  Work end of the chapter problems that are answered in the back of the text so you can check yourself. Try online tutorials and interactive problems. Don't work problems that you have no answer for because you may do it wrong and thus learn incorrectly.

9. Ask for help always!!! Bug your professor. Don't let questions go unanswered!

Test Taking Strategies

1.  Stay on top of the course so you don't have to cram. Don't cram for an exam.

2.  Do not pull all nighter's on coffee or vivarin. You'll be dead walking by the exam and will not perform well. Also you quickly forget what you cram, so when the next test builds on previous knowledge (and the final is cumulative) you will be starting from square one again.

3.  You can only really focus on one subject for about 1.5 hours. So if you study the hour before the exam, your brain will shut down on you before the exam is finished. One hour before the exam, close your book and notes. You're done studying! If you don't understand it now, it is too late anyway. Chill out for this hour, let your brain rest so it is recharged for the exam. Watch TV, eat dinner, chat with friends. Do anything but study!

4.  Study with a group. Divide up the material and have each person present their part to the group.

5.  Study alone as well. Read all your notes. Read the reviews.  Do practice problems.  Look over quizzes and homework. I primarily write the exams from the chapter reviews posted on this web page. 

6.  Study in 30 minute intervals with breaks. Do not attempt to study for 3-4 hours straight because you will burn out and get bored.

7.  Try to study for an exam several nights, not just the night before. If you have a question, it will be too late to get help the night before most likely.  Plan ahead and study several days before each exam. 

Why should I practice?  I followed my professor in lecture just fine!

Doing a problem on your own is NOT the same as following your professor along in class.  It is easy to understand a problem when you simply watch and listen along while your professor does all the work.  Trying a similar problem may not be as easy as you think - which is why you must practice! 

I hear this all the time: "You made it seem so easy in class.  I totally understood it when you did this in class." Yet the student can not answer the problem on his/her own. 

We professors often don't have time in class to let you practice many problems - we typically work one example or two, then maybe give you a chance to work one problem on your own.  That is not enough - please work many practice problems on your own outside of class.  You will learn and perform much better.

Good Luck!