Portage Chemistry 103 Module 2 Exam Study

The percentage of an element present in a compound can be calculated as shown below:

% of an element = weight of element / molecular weight of compound x 100

moles

= grams / molecular weight

Ca3(PO4)2

3 Ca = 3 x 40.08= 120.24 2 P = 2 x 30.97= 61.94 8 O = 8 x 16.00 = 128.00 310.18 % Ca = (120.24 ÷ 310.18) x 100 = 38.76% %P = (61.94 ÷ 310.18) x 100 = 19.97% %O = (128.00 ÷ 310.18) x 100 = 41.27%

1) Combination or Synthesis Reactions

A reaction in which a single product is formed from multiple reactants 2 Na + Cl2 → 2 NaCl 2 H2 + O2 → 2 H2O

Will the following reaction occur?

F2 + 2 Br-1 → 2 F-1 + Br2 F2 is higher on the nonmetal activity series table than Br2, which tells us that F forms a negative ion more easily than Br; therefore, if F is reacted with Br-1, the following two half-reactions will take place to cause the following reaction: 2 (Br-1 → Br2 + e-) F2 + 2e- → 2 F- ↓ F2 + 2 Br-1 → 2 F-1 + Br2

2.4: EMPIRICAL FORMULA

If the formula of a compound is known, the % composition of the compound can be determined. This process can be done in reverse: The formula of the compound can be determined if the % of each element present in the compound is known. The formula calculated from % composition is known as the empirical formula (or the simplest formula). The actual molecular formula is some multiple of this simplest formula, which is determined by knowing the molecular weight.

4. How many moles of NH4NO2 would be required to form 25 grams of N2 in the following reaction? NH4NO2 → N2 + 2 H2O

NH4NO2 → N2 + 2 H2O (25 grams N2 / 28.02) = 0.8922 moles N2 0.8922 moles N2 x (1 / 1) = 0.892 mol of NH4NO2

10. Show the calculation of the volume of 0.238 M solution which can be prepared using 13.4 grams of Ca3(PO4)2.

mlsolution = (gsolute x 1000) / (MW x M) = (13.4 x 1000) / (310.18 x 0.238) = 182 ml

Show the calculation of the volume of 0.987 M solution which can be prepared using 24.6 grams of NaNO3.

mlsolution = (gsolute x 1000) / (MW x M) = (24.6 x 1000) / (85.0 x 0.987) = 293 ml

To determine the empirical formula:

(1) Divide each element % by its exact atomic weight to give a set of numbers. (2) Divide the smallest of this set of decimal numbers into each of the numbers (including itself) to yield a second set of numbers. (3a) Round off each of the second set of decimal numbers to a whole number. OR (3b) If the numbers derived from the division in step 2 are recognized as exact decimal equivalents of fractions (such as n.25 = 1/4, n.333 = 1/3, n.5 = 1/2, n.666 = 2/3, n.75 = 3/4), multiply all of the numbers derived by division by the denominator of the recognized fraction to give whole numbers. (4) Each whole number is the number of atoms of that element in the empirical formula.

TYPES OF REACTIONS

1. C5H12 + 8 O2 → 5 CO2 + 6 H2O Combustion 2. Mg + 2 HCl → MgCl2 + H2 Single Replacement 3. H2 + Cl2 → 2 HCl Combination 4. 2 KClO3 → 2 KCl + 3 O2 Decomposition 5. C6H12O6 + 6 O2 → 6 CO2 + 6 H2O Combustion 6. FeS + 2 HCl → H2S + FeCl2 Double Replacement 7. HCl + NaOH → NaCl + H2O Double Replacement 8. Al + Fe(NO3)3 → Fe + Al(NO3)3 Single Replacement 9. 2 S + 3 O2 → 2 SO3 Combination 10. H2SO4 → SO3 + H2O Decomposition

Balancing redox equations should be done in this three-step process:

1. Determine the charge of each atom (including those that are part of polyatomic groups) and determine which two atoms' charges are changing and what the changes are. 2. Equate the changes. Multiply the changes by numbers to produce two products (results of each multiplications) that are equal. The numbers by which the changes were multiplied become the coefficients of the compounds containing the changing atoms (on both sides of the equation). 3. Balance the remainder of the atoms in the equation by making use of the coefficients determined in step #2.

stoichiometric calculations

1. Obtain a balanced equation. 2. Set up the form around the balanced equation: a. Molecular weights above each formula b. Moles and grams written successively below each formula 3. Put the information given in the problem in its correct place (under substance it belongs with; this is what we call the known substance since we have information given about it). 4. Do the calculations using the following equations.

2.8: BALANCING REDOX EQUATIONS; Rules for oxidation numbers

1. The sum of all the total charges in a compound is equal to zero (0). 2. The charge of an element is zero (0) (an element is an atom that is not combined with any other element). 3. The normal charge of H is +1. 4. The normal charge of O is -2. 5. The charge of a group I, II, or III metal atom is positive (+) and equal to its periodic table group number. (K = +1, Ca = +2, Al = +3) 6. The charge of any other (not in group I, II, or III) metal atom (in the center of the table) cannot be determined from its periodic table position since it has no group number. We can determine the charge of this atom by knowing the total charges of the other atoms (or polyatomic groups) with which it is combined and using rule #1. (Mn = +7 in KMnO4) since K = +1, 4 O = (4 x -2) so Mn = + 7 for sum of charges = 0 [Cr = +3 in Cr(NO3)3] since 3 NO3 = (3 x -1) so Cr = +3 for sum = 0 7. The charge of a nonmetal (which is combined only with a metal) is negative and equal to (its Group Number - 8). (S = -2 in Na2S) 8. The charge of any other nonmetal is determined by knowing the total charges of the other atoms with which it is combined and using rule #1 (N = + 5 in KNO3) sinceK = +1, 3 O = (3 x -2) so N = +5 for sum of charges = 0

Will the following reaction occur?

2 K + Pb+2 → 2 K+ + Pb K is higher on the metal activity series table than Pb, which tells us that K forms a positive ion more easily than Pb; therefore, if K reacts with Pb+2, the following two half-reactions will take place to cause the following reaction: 2 (K → K+ + e-) Pb+2 + 2e- → Pb 2 K + Pb+2 → 2 K+ + Pb

% Composition of a compound is:

32.37% Na; 32.37% Na ÷ 22.99 = 1.408 22.58%S; 22.58% S ÷ 32.07 = 0.704 (smallest number of the set) 45.05% O; 45.05% O ÷ 16.00 = 2.816 0.704 is the smallest of this set of numbers, so it is divided into each of the set of numbers. Na = 1.408 ÷ 0.704 = 2 Na S = 0.704 ÷ 0.704 = 1 S O = 2.816 ÷ 0.704 = 4 O Na2SO4

3. How many grams of SO2 would be formed from 60 grams of FeS2 in the following reaction? 4 FeS2 + 11 O2 → 2 Fe2O3 + 8 SO2

4 FeS2 + 11 O2 → 2 Fe2O3 + 8 SO2 (60 grams FeS2 / 119.99) = 0.5000 moles FeS2 0.5000 moles FeS2 x (8 / 4) = 1.00 mole SO2 1.00 mole SO2 x (64.07) = 64.1 grams of SO2

2.1: MOLECULAR WEIGHT

A compound is made up of two or more elements combined in a definite ratio that is represented by a molecular formula. Each of these elements has a certain atomic weight, which can be found in the periodic table. The sum of the atomic weights of the atoms in the molecular formula is called the formula weight or molecular weight or formula mass.

5) Single Replacement Reaction

A reaction in which a (1) metallic element reacts with an ionic compound to replace the metal ion from the ionic compound as a metal element or a reaction in which a (2) nonmetal element reacts with an ionic compound to replace the nonmetal ion as a nonmetal element. One of the materials gains electrons (undergoing reduction) as one of the other materials loses electrons (undergoing oxidation). 1) Zn + Cu+2, SO4-2 → Zn+2, So4-2 + Cu 2) 2K+1, 2Br-1 + Cl2 → 2K+1, 2Cl-1 + Br2

2) Decomposition Reactions

A reaction in which a single reactant is converted to multiple products CaCO3 → CaO + CO2 Mg(OH)2 → MgO + H2O

3) Combustion Reactions

A reaction in which hydrocarbon (compound containing only carbon and hydrogen) or C,H,O compound is reacted with elemental oxygen (O2) to form carbon dioxide and water 2 C4H10 + 13 O2 → 8 CO2 + 10 H2O C6H12O6 + 6 O2 → 6 CO2 + 6 H2O

4) Double Replacement Reactions

A reaction in which two ionic compounds exchange their ions to form a non-ionic (molecular) compound product as (1) a solid insoluble precipitate, (2) a gas, or (3) a molecular compound, the formation of which drives the reaction to completion. (1) Ba+2(NO3-1)2 + (K+)2SO4 → BaSO4 ↓ + 2 (K+),(NO3-1) (precipitate) (2) Ca+2 CO3-2 + 2 H+Cl- → Ca+2(Cl-)2 + CO2 ↑ + H2O (gas) (3) Ca+2(OH-1)2 + 2 H+Cl- → Ca+2(Cl-)2 + H2O (molecular)

2.7: TYPES OF REACTIONS II

A single replacement reaction occurs when: 1. a metallic element reacts with an ionic compound to replace the metal ion from the ionic compound as a metal element, or 2. a nonmetal element reacts with an ionic compound to replace the nonmetal ion as a nonmetal element. This is another reaction in which a molecular material (the metal or nonmetal element) is formed as a product as in the double replacement reactions. This single replacement reaction is one example of a very prevalent type of reaction know as an oxidation-reduction (or redox) reaction in which electrons are transferred to cause oxidation number changes. One of the materials will gain electrons (undergoing reduction) as one of the other materials loses electrons (undergoing oxidation). (1) Zn + Cu+2, SO4-2 → Zn+2, SO4-2 + Cu (2) 2 K+1,2 Br-1 + Cl2 → 2 K+1, 2 Cl-1 + Br2

Now let's balance another equation: Original equation: Al4C3 + H2O → Al(OH)3 + CH4

Al4C3 + H2O → Al(OH)3 + CH4 (original equation) Al4C3 + H2O → 4 Al(OH)3 + CH4 (to balance Al) Al4C3 + H2O → 4 Al(OH)3 + 3 CH4 (to balance C) Al4C3 + 12 H2O → 4 Al(OH)3 + 3 CH4 (to balance H and O)

Will the following reaction occur?

Au + Mg+2 → Au+2 + Mg Mg is higher on the metal activity series table than Au, which tells us that Mg forms a positive ion more easily than Au; therefore, if Au reacts with Mg+2, the reaction proposed above will not take place because Mg+2 is already present as the ion and will remain in that form.

Bi(OH)3 + Na2SnO2 → Bi + Na2SnO3 + H2O

Bi(OH)3: each OH is -1 (total is -3), so Bi is +3 Bi (on right side) is uncombined so Bi = 0 Na2SnO2: Na is metal in group I = +1 (total is +2), each O is -2 (total is -4), so Sn is +2 Na2SnO3: Na is metal in group I = +1 (total is +2), each O is -2 (total is -6), so Sn is +4 Since Bi (on left side) is +3 and Bi (on right side) is 0: Bi changes by 3 Since Sn (on left side) is +2 and Sn (on right side) is +4: N changes by 2 Multiply Bi compounds by 2 and Sn compounds by 3 and after balancing other atoms = 2 Bi(OH)3 + 3 Na2SnO2 → 2 Bi + 3 Na2SnO3 + 3 H2O

SINGLE DISPLACEMENT REACTIONS

Br2 + 2 I-1 → I2 + 2 Br-1 Will occur since Br2 is more active than I2 Cu + Sn+2 → Will not occur since Sn is more active than Cu Cl2 + 2 F-1 → Will not occur since F2 is more active than Cl2 (???) Ba + Fe+2 → Fe + Ba+2 Will occur since Ba is more active than Fe

Calculate the number of moles in 10.0 grams of each of the following compounds:

Ca3(PO4)2 10.0 g ÷ 310.18 = 0.0322 mol (3 sig fig because of 10.0 g) C3H5O2Cl 10.0 g ÷ 108.52 = 0.0921 mol Al2(SO4)3 10.0 g ÷ 342.17 = 0.0292 mol Ca3(PO4)2 0.0500 mol x 310.18 = 15.5 g C3H5O2Cl 0.0500 mol x 108.52 = 5.43 g Al2(SO4)3 0.0500 mol x 342.17 = 17.1 g

2.2: MOLES

Chemical compounds react with one another in amounts that are based on their molecular weights; this chemically reactive amount of compound is called a mole.

Nonmetal Activity Series

F2 (most active gain electron - ion) Cl2 Br2 I2 (least active)

2. How many moles of Fe would be formed from 50 grams of Fe3O4 in the following equation? Fe3O4 + 4 H2 → 3 Fe + 4 H2O

Fe3O4 + 4 H2 → 3 Fe + 4 H2O (50 grams Fe3O4 / 231.55) = 0.2159 moles Fe3O4 0.2159 moles Fe3O4 x (3 / 1) = 0.648 mol of Fe

To balance a reaction equation, numbers (called coefficients) may be placed in front of each formula, but the subscript numbers indicating how many atoms of each element are in each formula may not be changed

Original equation: Al(OH)3 + H2SO4 → Al2(SO4)3 + H2O NOTICE: The Al, O, H, and S atoms are not equal on each side of the equation. The equation is unbalanced (1 Al on left, 2 Al on right, etc). Balanced equation: 2 Al(OH)3 + 3 H2SO4 → Al2(SO4)3 + 6 H2O NOTICE: There are 2 Al, 18 O, 12 H, and 3 S atoms on each side of the equation. The equation is balanced.

In some redox equations, one of the changing atoms is polyatomic, such as Cl2, Br2, C2O4-2, Cr2O7-2. In these cases, as shown below, the change of that atom must be doubled (or tripled, etc.) before equating the changes.

FeCl2 + K2Cr2O7 + HCl → FeCl3 + CrCl3 + KCl + H2O FeCl2: Each Cl is nonmetal in group VII = -1 (total = -2), so Fe = +2. FeCl3: Each Cl is nonmetal in group VII = -1 (total = -3), so Fe = +3. K2Cr2O7: K group I metal = +1 (total = +2), each O is -2 (total = -14), so Cr total is +12; each Cr is +6. CrCl3: Each Cl is nonmetal in group VII = -1 (total = -3), so Cr = +3. Since Fe (on left side) is +2 and Fe (on right side) is +3, Fe changes by 1. Since each Cr (on left side) is +6 and Cr (on right side) is +3, Cr changes by 3 x 2 (since Cr2O7-2) = 6. Multiply Fe compounds by 6 and Cr2O7-2 compound by 1, and, after balancing other atoms (which requires placing a 2 in front of CrCl3) = 6 FeCl2 + 1 K2Cr2O7 + 14 HCl → 6 FeCl3 + 2 CrCl3 + 2 KCl + 7 H2O This is an oxidation-reduction (redox) reaction: 6 FeII - 6 e- → 6 FeIII (oxidation) 2 CrVI + 6 e- → 2 CrIII (reduction) FeCl2 is a reducing agent, K2Cr2O7 is an oxidizing agent.

Type #4 - Predicting Precipitation Reactions by Using the Solubility Rules

How do we know when a precipitation reaction will occur to form an insoluble precipitate rather than the reaction just forming a mixture of the four ions? This is based on a set of rules known as solubility rules, which allows you to predict when a reaction will occur between two ions to form an insoluble molecular precipitate. This set of solubility rules is as follows: Compounds that are mostly insoluble in water are: Compounds containing the S-2 (sulfide) ion: exceptions are NH4+ (ammonium) and group I and II sulfides Compounds containing the OH-1 (hydroxide) ion: exceptions are NH4+ and group I hydroxides Compounds containing the CO3-2 (carbonate) ion: exceptions are NH4+ and group I carbonates Compounds containing the PO4-3 (phosphate) ion: exceptions are NH4+ and group I phosphates Compounds that are mostly soluble in water are: Compounds containing the NH4+ and group I and II ions Compounds containing the NO3-1 (nitrate) and C2H3O2-1 (acetate) ions Compounds containing the Cl-1 (chloride), Br-1 (bromide) or I-1 (iodide) ions: exceptions are Ag+, Pb+2, and Hg2+2 compounds Compounds containing the SO4-2 ion: exceptions are Sr+2, Pb+2, Ba+2, and Hg2+2 compounds

Predicting Single Replacement Reactions by Using the Activity Series

How do we know whether a single replacement reaction will occur to form a metal or nonmetal element? This is based on set reactivity rules that allow you to predict that a reaction will occur between a metal and a metal ion or a nonmetal and a nonmetal ion. This set of predictors is known as the metal and nonmetal activity series and is shown below. Any element (Metal 1) higher on the metal activity series is more likely to react with an ion of a metal (Metal 2) lower on the activity series to form an ion of Metal 1 and the element of Metal 2. Any element (Nonmetal 1) higher on the nonmetal activity series is more likely to react with an ion of a nonmetal (Nonmetal 2) lower on the activity series to form an ion of Nonmetal 1 and the element of Nonmetal 2.

KMnO4 + KNO2 + HCl → MnCl2 + KCl + KNO3 + H2O

KMnO4 + KNO2 + HCl → MnCl2 + KCl + KNO3 + H2O KMnO4: K is metal in group I = +1, each O is -2 (total is -8), so Mn is +7 MnCl2: each Cl is nonmetal in group VII = -1 (total = -2), so Mn = +2 KNO2: K is metal in group I = +1, each O is -2 (total is -4), so N is +3 KNO3: K is metal in group I = +1, each O is -2 (total is -6), so N is +5 Since Mn (on left side) is +7 and Mn (on right side) is +2: Mn changes by 5 Since N (on left side) is +3 and N (on right side) is +5: N changes by 2 Multiply Mn compounds by 2 and N compounds by 5 and after balancing other atoms = 2 KMnO4 + 5 KNO2 + 6 HCl → 2 MnCl2 + 2 KCl + 5 KNO3 + 3 H2O

KMnO4 + H2C2O4 + H2SO4 → MnSO4 + K2SO4 + CO2 + H2O

KMnO4: K group I metal = +1, each O is -2 (total = -8), so Mn = +7 MnSO4: each SO4 is -2, so Mn is +2 H2C2O4: each H = +1 (total = +2), each O is -2 (total = -8), so C total is +6, each C is +3 CO2: each O is -2 (total = -4), so C is +4 Since Mn (on left side) is +7 and Mn (on right side) is +2: Mn changes by 5 Since each C (on left side) is +3 and C (on right side) is +4: C changes by 1 x 2 (since H2C2O4) = 2 Multiply Mn compounds by 2 and H2C2O4 compound by 5 and after balancing other atoms (which requires placing a 10 in front of CO2) = 2 KMnO4 + 5 H2C2O4 + 3 H2SO4 → 2 MnSO4 + K2SO4 + 10 CO2 + 8 H2O This is an oxidation-reduction (redox) reaction: 2 MnVII + 10 e- → 2 MnII (reduction) 10 CIII - 10 e- → 10 CIV (oxidation) KMnO4 is an oxidizing agent, H2C2O4 is a reducing agent.

Metal Activity Series - (Reaction #5)

Li-Most Acitve; Likely to lose electron+ion K Ba Ca Na Mg Al Mn Zn Cr Fe Co Ni Sn Pb H2 Cu Ag Hg Pt Au (Least active)

DOUBLE DISPLACEMENT PRECIPITATION REACTIONS

Mg+2, SO4-2 + 2 Na+1, 2 OH-1 → Mg(OH)2 ↓ + 2 Na+1, SO4-2 Ni+2, Cl-1 + Na+1, Br-1 → Will not occur since neither set of ions forms a precipitate Ag+1, NO3-1 + Na+1, I-1 → AgI ↓ + Na+1, NO3-1 Ca+2, Cl2-1 + Al+3, 3 NO3-1 → Will not occur since neither set of ions forms a precipitate Pb+2, 2 NO3-1 + 2 Na+1, 2 Br-1 → PbBr2 ↓ + 2 Na+1, 2 NO3-1

Mn(NO3)2 + NaBiO3 + HNO3 → HMnO4 + Bi(NO3)3 + NaNO3 + H2O

Mn(NO3)2: each NO3 is -1 (total is -2), so Mn is +2 HMnO4: H = +1, each O is -2 (total is -8), so Mn is +7 NaBiO3: Na is metal in group I = +1, each O is -2 (total is -6), so Bi is +5 Bi(NO3)3: each NO3 is -1 (total is -3), so Bi is +3 Since Mn (on left side) is +2 and Mn (on right side) is +7: Mn changes by 5 Since Bi (on left side) is +5 and Bi (on right side) is +3: N changes by 2 Multiply Mn compounds by 2 and Bi compounds by 5 and after balancing other atoms = 2 Mn(NO3)2 + 5 NaBiO3 + 16 HNO3 → 2 HMnO4 + 5 Bi(NO3)3 + 5 NaNO3 + 7 H2O

2.6: TYPES OF REACTIONS - 5 types

Molecular and Ionic Equations For a reaction involving ionic materials, we can write the reaction equation in different ways depending on what information we want to express. We can write the equation as a molecular equation, an ionic equation, or a net ionic equation: In a molecular equation the substances are written as molecular substances even though they may actually exist in solution as ions. A molecular equation is shown below: CaCO3 + 2 HCl → CaCl2 + CO2 + H2O You could also write this as an ionic equation, which tells what is actually happening at the ionic level in a solution. An ionic equation is shown below: Ca+2, CO3-2 + 2 H+, 2 Cl- → Ca+2, 2 Cl- + CO2 + H2O You could also write this as a net ionic equation, which shows the actual reaction that occurs at the ionic level. In this net ionic equation, spectator ions (ions that appear on both sides of the equation in the same form) are canceled out to yield the net ionic equation. CO3-2 + 2 H+ → CO2 + H2O

1. How many grams of NaCl would be formed from 100 grams of Na2CO3 in the following reaction? Na2CO3 + 2 HCl → 2 NaCl + CO2 + H2O

Na2CO3 + 2 HCl → 2 NaCl + CO2 + H2O (100 grams Na2CO3 / 105.99) = 0.9435 moles Na2CO3 0.9435 moles Na2CO3 x (2 / 1) = 1.887 moles NaCl 1.887 moles NaCl x 58.44 = 110.3 grams of NaCl

NaCrO2 + NaClO + NaOH → Na2CrO4 + NaCl + H2O

NaCrO2: Na is metal in group I = +1, each O is -2 (total is -4), so Cr is +3 Na2CrO4: Na is metal in group I = +1 (total is +2), each O is -2 (total is -8), so Cr is +6 NaClO: Na is metal in group I = +1, each O is -2, so Cl is +1 NaCl: Na is metal in group I = +1, so Cl is -1 Since Cr (on left side) is +3 and Cr (on right side) is +6: Cr changes by 3 Since Cl (on left side) is +1 and Cl (on right side) is -1: Cl changes by 2 Multiply Cr compounds by 2 and Cl compounds by 3 and after balancing other atoms = 2 NaCrO2 + 3 NaClO + 2 NaOH → 2 Na2CrO4 + 3 NaCl + H2O

2.9: EQUATION CALCULATIONS

Once an equation is balanced, it can be used to calculate how many grams or moles of one material involved in the equation will react with another material or how many grams or moles of one material will be formed from another material. These calculations are called stoichiometric calculations.

2.10: SOLUTION CONCENTRATION

Solutions are one of the types of mixtures known as a homogenous mixture, in which two or more materials are combined to form a single phase that has the same composition, properties, and appearance throughout. Salt or sugar dissolved in water is an example of a solution. The minor component of the mixture is the material being dissolved, also called the solute (salt or sugar in the examples given). The major component is the material doing the dissolving, also called the solvent (water in the examples given). Solutes can be of two types: (1) electrolytes, which are ionic, or (2) very polar compounds, which form ions when they dissolve. Solutions of electrolytes (like salt) conduct an electric current. Solutions of solutes that form molecules when they dissolve (like sugar) do not conduct an electric current and are called nonelectrolytes.

concentration

The amount of solute present in a certain amount of solvent is referred to as the concentration of the resulting solution. Solutions are often described as concentrated or dilute, which refers to the relative amount of solute present. "Concentrated" means there is a relatively large amount of solute, and "dilute" means there is a relatively small amount of solute. There are three common concentration terms used in chemistry: mass percent, molarity, and molality.

Mass Percent

The concentration term mass percent refers to the mass of solute present per 100 grams of solution (solute + solvent) and can be determined by the following formula: A solution prepared by dissolving 10.5 grams of C2H5OH in 230 grams of water would have what mass percent? Mass percent = 10.5 g / (10.5 g + 230 g) x 100% = 4.37% (by mass)

Molality (m)

The concentration term molality (m) refers to the moles of solute present per kilogram of solvent and can be determined by the following formula: Note here that the solute quantity is expressed in moles (grams / molecular weight) and quantity of solvent expressed in kilograms, since this is usually measured in grams (kg = grams / 1000). The symbol m is used for molality. A solution prepared by dissolving 10.5 grams of C2H5OH in 300 grams of water would have what molarity? molality = (10.5 g / 46.068) / (300 g / 1000 g/kg) = 0.760 m C2H5OH

Molarity (M)

The concentration term molarity (M) refers to the moles of solute present per liter of solution and can be determined by the following formula: Note here that the solute quantity is expressed in moles (grams / molecular weight) and volume is of the total solution (not solvent added), since volume is usually measured in ml (liters = ml / 1000). The symbol M is used for molarity. A solution prepared by dissolving 10.5 grams of C2H5OH in enough water to make 230 ml of solution would have what molarity? Molarity = (10.5 g / 46.068) / (230 ml / 1000 ml/L) = 0.991 M C2H5OH

2.8: BALANCING REDOX EQUATIONS

There is a special type of chemical reaction equation known as a redox reaction. In this type of equation, two of the atoms in the compounds undergo changes in their charges: The charge of one of the atoms increases, and the charge of the other decreases. An increase in charge occurs when that element loses one or more electrons, a process called oxidation. A decrease in charge occurs when that element gains one or more electrons, a process called reduction. In this reaction, all electrons lost by the one element must be gained by the other atom.

2.5: BALANCING CHEMICAL EQUATIONS

When certain chemical materials are added to one another, they undergo a chemical reaction in which the atoms of the materials separate from one another and recombine in a new way to form new materials. This chemical reaction can be described by a chemical reaction equation in which the reactants (starting materials) are written on the left side of the equation and the products (final materials) are written on the right side of the equation, separated by an arrow which points from left to right (some sites use an equals sign here but this is incorrect because the only thing equal about this equation is the number of atoms, once balanced), an example of which is shown below: NaOH + HCl → NaCl + H2O The reaction equation is read as follows: Sodium hydroxide + Hydrochloric acid YIELDS Sodium chloride + Water

Zn + KNO3 + HCl → ZnCl2 + KCl + NH4Cl + H2O

Zn + KNO3 + HCl → ZnCl2 + KCl + NH4Cl + H2O Zn (on left side) is uncombined so Zn = 0 ZnCl2: each Cl is nonmetal in group VII = -1 (total = -2), so Zn = +2 KNO3: K is metal in group I = +1, each O is -2 (total is -6), so N is +5 NH4Cl: Cl is nonmetal in group VII = -1, each H is +1 (total is +4), so N is -3 Since Zn (on left side) is 0 and Zn (on right side) is +2: Zn changes by 2 Since N (on left side) is +5 and N (on right side) is -3: N changes by 8 Multiply Zn compounds by 4 and N compounds by 1 and after balancing other atoms = 4 Zn + 1 KNO3 + 10 HCl → 4 ZnCl2 + 1 KCl + 1 NH4Cl + 3 H2O

Ca3(PO4)2

calcium phosphate Molecular Weight = 3 Ca= 3 x 40.08=120.24 2 P=2 x 30.97=61.94 8 O=8 x 16.00=128.00 Total 310.18

2.3: PERCENT COMPOSITION

he molecular formula represents the definite ratio of elements in a compound. The weight of each element present in the compound represents a certain percentage of the total weight of the compound. The percentage of each element present in a compound is called the % composition of the compound.