Download Chemistry I Review

Chemistry I Review
Writing Ionic Compounds
1. Cation + Anion
2. Charge determined by position in periodic table.
3. Criss-Cross charges (make them subscripts).
4. Polyatomic is placed in parenthesis if charge is different from cation.
5. One subscripts are understood.
6. Reduce subscripts to lowest terms.
Examples:
1. Formula between sodium & chloride
Na Na+
Cl ClCriss-cross charges & reduce
NaCl
2. Formula between magnesium & bromine
Mg Mg2+
Br BrCriss-cross charges & reduce
MgBr2
3. Formula between ammonium & phosphate
NH4 NH4+
PO4PO43Criss-cross charges & reduce
(NH4)3PO4
4. Formula between iron with a charge of 3 & sulfate
Fe Fe3+
SO4SO42Criss-cross charges & reduce
Fe2(SO4)3
Naming Ionic Compounds
1. Name cation.
2. Change anion ending to –ide.
3. Place roman numerals in parenthesis after a transition metal to indicate
charge.
4. For polyatomic ions, use the name of the ion.
Examples:
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KI
K = potassium I= iodine
Change –ine to –ide
Potassium iodide
Examples:
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MgBr2
Mg = magnesium Br= bromine
Change –ine to –ide
Magnesium bromide
Examples:
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Cu2S
Cu = copper S= sulfur
Change ending of anion to –ide
Use roman numeral to indicate charge of transition metal
Copper (I) sulfide
Examples:
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Fe(NO3)2
Fe = iron NO3 = nitrate
Name NO3 as is since polyatomic
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Use roman numeral to indicate charge of transition metal
Iron (II) nitrate
Naming Covalent Compounds
1. Nonmetal + nonmetal
2. Use prefixes
3. Never use mono with the first element
4. Do not change the ending of first element
5. Change ending of the second element to -ide
Prefixes
Mono: 1
Tri: 3
Penta: 5
Hepta: 7
Nona: 9
Di: 2
Tetra: 4
Hexa: 6
Octa: 8
Deca: 10
Examples:
Cl5Br6
Cl = chlorine Br: bromine
5=penta
6=hexa
Pentachlorine hexabromide
Examples:
I 3F
I = iodine F: fluorine
3=tri
1=mono
Triodine monofluoride
Examples:
NO8
N = nitrogen O: oxygen
1=mono
8=octa
Nitrogen octoxide
Writing Covalent Compounds
1. Identify the elements and prefixes
2. Name as is
Examples:
Disulfur Trioxide
Di sulfur = S2
Trioxide = O3
Formula is: S2O3
Examples:
Tetranitrogen Heptachloride
Tetranitrogen = N4
Heptachloride = Cl7
Formula is: N4Cl7
Naming Acids
1. If the anion ends in –ide, the acid will be named…
2. Hydro (root) – ic acid
3. This is usually for H plus one element
Examples
HCl
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Hydrochloric Acid
HI
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Hydroidic Acid
H2S
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Hydrosulfuric Acid
1. If you have a H plus an anion ending in –ate, the acid will be named…
(root) – ic acid
Examples
H2SO4
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Sulfuric Acid
HNO3
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Nitric Acid
H3PO4
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Phosphoric Acid
1. If you have a H plus an anion ending in –ite, the acid will be named…
(root) – ous acid
Examples
H2SO4
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Sulfurous Acid
HNO2
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Nitrous Acid
H3PO3
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Phosphorous Acid
Remember…
ate  ic
ite  ous
Balancing Equations
1. Chemical Equations:
Depict the kind of reactants and products and their relative amounts in a
reaction.
4 Al (s) + 3 O2 (g) ---> 2 Al2O3 (s)
The numbers in the front are called coefficients
2. When balancing a chemical reaction you may add coefficients in front of
the compounds to balance the reaction, but you may NOT change the
subscripts.
There are four steps:
1. Write the correct formula for the reactants & products.
2. Find the number of atoms for each element on both sides.
3. Determine the coefficients for each compound.
4. Check your answer.
Examples:
2 Al + 3 CuO  Al2O3 + 3 Cu
2 C3H2 + 7 O2  6 CO2 + 2 H2O
Al(OH)3 + 3 HBr  AlBr3 + 3 H2O
2 Na3PO4 + Fe2O3 3 Na2O + 2 FePO4
Stoichiometry
1. Write the chemical equation
2. Balance the chemical equation
3. Start with your given
4. Cross out until you get what you want
Examples:
How many moles of Fe2O3 will I form from 5.0 mol of Fe?
5.0 mole Fe x 2 mol Fe2O3
4 mol Fe
= 2.5 mol Fe2O3
How many g of NaCl will be produced from 1.25 mol of chlorine gas reacting
with sodium?
Write & balance the equation first.
2Na + Cl2  2NaCl
1.25 mol Cl2 x 2 mol NaCl x 58.44 g
1 mol Cl2
1 mol NaCl
= 146 g NaCl
Ammonium nitrate decomposes into dinitrogen monoxide gas and water.
Determine that amount of water produced if 25.0 g of ammonium nitrate
decomposes.
First write the balanced equation.
NH4NO3  N2O + 2 H2O
25.0 g NH4NO3 x 1 mol NH4NO3 x 2 mol H2O
80.04 g NH4NO3
1 mol NH4NO3
x
18.02 g H2O
1 mol H2O
= 11.2 g H2O
Molarity
1. Defined as the number of moles of solute per liter of solution.
2. Molarity = moles solutes / liter of solution
Examples:
Calculate the molarity of a solution made by dissolving 23.4 g sodium sulfate
in 125 mL of solution.
First convert grams into moles:
23.4 g Na2SO4 = 0.165 mol
Convert mL to Liters
125 mL = 0.125 L
M = mol / liters
0.165 mol / 0.125 L
= 1.32 M
How many grams of Na2SO4 are required to make 0.350 L of a 0.500 M
solution of Na2SO4?
Isolate moles from molarity
0.500 mol / L x 0.350 L = 0.175 mol
Convert moles to grams using GFM
0.175 mol x (142 g/ mol) = 24.85 g
What volume of 18.0 M sulfuric acid is needed to contain 2.45 g?
Convert grams to moles:
2.45 g x ( 1 mol / 98 g) = 0.025 mol
Convert mol to L
0.025 mol x (L / 18 mol) = 0.00139 L