Stoichiometry-Classnotes

Stoichiometry & the Mole

And No, Not this Mole!

 The mole is the standard method in chemistry for communicating how much of a substance is present.  This is the fundamental definition of what one mole is. One mole contains as many entities as there are in 12 grams of carbon-12 (or 0.012 kilogram).  In one mole, there are 6.022 x 1023 atoms. (Avogadro's Number)  Here is another way: there are 6.022 x 1023 atoms of carbon in 12 grams of carbon-12.  One mole of ANYTHING contains 6.022 x 1023 atoms or molecules.

Molecular Weight The molecular weight of a substance is the weight in atomic mass units of all the atoms in a given formula.

What does this mean??????? Lets break it down. How to calculate the molecular weight of a substance Example #1 - Al2(SO4)3 There are: • two atoms of aluminum and the atomic weight of Al is 26.98 amu. • three atoms of sulfur and the atomic weight of S is 32.06 amu. • twelve atoms of oxygen and the atomic weight of O is 16.00 amu. First multiply: 2 x 26.98 = 53.96 total weight of all Al in formula 3 x 32.06 = 96.18 total weight of all S in formula 12 x 16.00 = 192.00 total weight of all O in formula Then add: 53.96 + 96.18 + 192.00 = 342.14 amu. This answer, 342.14 amu, represents the molecular weight of Al2(SO4)3 You might be asking why I used oxygen at 16.00 and not 15.9994. Actually, we will be using the complete molecular weight in our calculations. This will promote consistency in our final answers.

Mole Conversions

Converting Grams to Moles 1. Example 1: 22g of Cu. Make a T-chart

Finding Molecules of Cu:

When in doubt:

1.0mol of any compound or element = the MW = 6.023X1023atoms

Percent Mass Composition  Percent composition is the percent by mass of each element present in a compound. Example Glucose, C6H12O6 1. figure out the molar mass from the formula. 2. figure out the grams each atom contributes by multiplying the atomic weight by the subscript. 3. divide the answer for each atom by the molar mass and multiply by 100 to get a percentage.

Elemental Part Weight

X 100% = %

Comp. Total Molecular Weight

Remember, you may figure out the last percentage by subtracting the total percent from 100, as will be done in a moment. Step One: mass of one mole = 180.16 g Step Two: Carbon = 6 x 12.011 g = 72.066 g Hydrogen = 12 x 1.008 = 12.096 g The oxygen percentage will be arrived at by subtraction. Step Three: Carbon's percentage: (72.066 g / 180.16 g) x 100 = 40.00 % Hydrogen's percentage: (12.096 g / 180.16 g) x 100 = 6.71 % Oxygen's percentage: 100 - (40.00 + 6.71) = 53.29 %

Determining the Molecular & Empirical Formulas Consider the data obtained by analyzing a compound. A chemist determines that the substance contains potassium, chlorine and oxygen. Here is some sample data: element mass, g potassium 0.479 chlorine 0.434 oxygen 0.588

What is the chemical formula of this compound? It might be tempting to say K0.479Cl0.434O0.588 but of course that is not correct.

The numbers in chemical formulas are always integers since they represent actual whole atoms which combine. The fact that they represent the numbers of atoms and not their masses should suggest a direction to follow. If we knew the moles of K, Cl and O, we would be that much closer to knowing the formula.

element potassium determine the chlorine oxygen element.

moles 0.0123 0.0122 0.0368

Calculating the moles of each individual element to Chemical formula will set up a direct ratio between each

A closer look at those decimals will show that there is a nearly integer ratio hidden in the data. One way to get it out is to divide all of the values by the smallest value. That makes the smallest number 1 and hopefully all the other values larger integers.

Therefore, the formula for this compound is KClO3. Such a formula is often called the empirical formula. It gives the smallest integer ratio, which represents the proportions in which the atoms combine in the compound. It may also represent the actual or molecular formula.

 Important: If the number of moles is 0.5, such as 0.5 or 3.5, you must round up and that integer must be multiplied by each molar quantity by that factor. 

Molecular Formula = Molar Mass (given) / Empirical Formula

Hydrates Some substances incorporate water into their structure when they crystallize. While this water is not exactly chemically bonded to the compound (it can be "boiled off"), it usually is combined in a definite ratio (ie. BaCl2 * 2 H2O , CuSO4*5H2O). Determining the formula of a hydrate is very similar to an empirical formula problem.

Example: 4.94 g of a hydrate of NiSO3 are heated until all the water is driven away. The mass of the remaining anhydrous substance is 2.78 g. What is the formula of the hydrate?

Step 1: Determine the mass of water and anhydrous material. anhydrous NiSO3 = 2.78 g 2.78 g = 2.16 g

water of hydration = hydrate - anhydrous = 4.94 g -

Step 2: Determine the moles of water and anhydrous material 2.78 g NiSO3 | 1 mole NiSO3 | 139 g

= 0.02 moles NiSO3

2.16 g water | 1 mole water = 0.12 moles water | 18 g water Step 3: Determine the simplest whole number ratio of moles water/ moles anhydrous substance. 0.12 moles water / 0.02 moles NiSO3 = 6 / 1 Step 4 Write the correct formula NiSO3 *6H2O

Types of Chemical Reactions Synthesis: A synthesis reaction is when two or more simple compounds combine to form a more complicated one. These reactions come in the general form of: A + B ---> AB  That means that two pieces join together to produce one, more complex compound. These pieces can be elements or simpler compounds. Complex simply means that the product compound has more atoms than the reactant molecules. One example of a synthesis reaction is the combination of iron and sulfur to form iron (II) sulfide: 8 Fe + S8 ---> 8 FeS 2Mg + O2 ---> 2MgO 2H2 + O2 ---> 2H2O CaO + CO2 ---> CaCO3 Na2O + CO2 ---> Na2CO3 Single displacement: This is when one element trades places with another element in a compound. These reactions come in the general form of: A + BC ---> AC + B AX + Y ---> YX + A  One cation replaces another.  Element Y has replaced A (in the compound AX) to form a new compound YX and the free element A. Remember that A and Y are both cations (positivelycharged ions) in this example.

One example of a single displacement reaction is when magnesium replaces hydrogen in water to make magnesium hydroxide and hydrogen gas: Mg + 2 H2O ---> Mg(OH)2 + H2 Cu + AgNO3 ---> Ag + Cu(NO3)2 Fe + Cu(NO3)2 ---> Fe(NO3)2 + Cu

Double displacement: This is when the anions and cations of two different molecules switch places, forming two entirely different compounds. These reactions are in the general form: AB + CD ---> AD + CB AB + XY ---> AY + XB  During double replacement, the cations and anions of two different compounds switch places.  A and X are the cations (positively-charged ions) in this example, with B and Y being the anions (negatively-charged ions). One example of a double displacement reaction is the reaction of lead (II) nitrate with potassium iodide to form lead (II) iodide and potassium nitrate: Pb(NO3)2 + 2 KI ---> PbI2 + 2 KNO3 2KOH + H2SO4 ---> K2SO4 + 2H2O Decomposition: A decomposition reaction is the opposite of a synthesis reaction - a complex molecule breaks down to make simpler ones. These reactions come in the general form: AB ---> A + B  During decomposition, one compound splits apart into two (or more pieces). These pieces can be elements or simpler compounds. One example of a decomposition reaction is the electrolysis of water to make oxygen and hydrogen gas: 2 H2O ---> 2 H2 + O2 2HgO ---> 2Hg + O2 CaCO3 ---> CaO + CO2 Na2CO3 ---> Na2O + CO2 Combustion: A combustion reaction is when oxygen combines with another compound to form water and carbon dioxide. These reactions are exothermic, meaning they produce heat. An example of this kind of reaction is the burning of naphthalene: C10H8 + 12 O2 ---> 10 CO2 + 4 H2O

Acid-base: This is a special kind of double displacement reaction that takes place when an acid and base react with each other. The H+ ion in the acid reacts with the OH- ion in the base, causing the formation of water. Generally, the product of this reaction is some ionic salt and water: HA + BOH ---> H2O + BA One example of an acid-base reaction is the reaction of hydrobromic acid (HBr) with sodium hydroxide: HBr + NaOH ---> NaBr + H2O

List what type the following reactions are: 1) NaOH + KNO3 --> NaNO3 + KOH 2) CH4 + 2 O2 --> CO2 + 2 H2O 3) 2 Fe + 6 NaBr --> 2 FeBr3 + 6 Na 4) CaSO4 + Mg(OH)2 --> Ca(OH)2 + MgSO4 5) NH4OH + HBr --> H2O + NH4Br 6) Pb + O2 --> PbO2 7) Na2CO3 --> Na2O + CO2 1) 2) 3) 4) 5) 6) 7)

double displacement combustion single displacement double displacement acid-base synthesis decomposition

Predict the following reaction type by looking at the following reactions. 1. RbClO3 ---> RbCl + O2 2. Cr2(SO3)3 + H2SO4 ---> Cr2(SO4)3 + SO2 + H2O 3. NaCl + O2 ---> NaClO3 4. Ra + Cl2 ---> RaCl2 5. RaCl2 ---> Ra + Cl2 6. C2H5OC2H5 + O2 ---> CO2 + H2O 7. Cl2 + MgI2 ---> I2 + MgCl2 8. C4H9OH + O2 ---> CO2 + H2O

Answers: 1. Decomposition 2. Double Replacement 3. Synthesis 4. Synthesis 5. Decomposition 6. Combustion 7. Single Replacement 8. Combustion 9. Single Replacement 10. Double replacement

9. Fe + CuSO4 ---> Cu + FeSO4 10. AgC2H3O2 + K2CrO4 ---> Ag2CrO4 + KC2H3O2

STEPS TO TAKE IN WRITING CORRECTLY BALANCED NET IONIC EQUATIONS I. Examine each reactant. Ask the following series of questions: 1. Is the reactant an acid? If not, go to Question 2 below. Is the acid a strong acid? (Memorize the 8 strong acids, given on another sheet.) − If the reactant is a strong acid, write it as separate ions, e.g. H+(aq) + NO3 (aq). If the reactant is a weak acid, write it as an undissociated molecule, e.g. HNO2(aq). 2. Is the reactant a salt (including hydroxides)? If not, go to Question 3 below. Is the salt soluble in water? (Memorize the solubility rules, given on another sheet.) − If the reactant is a soluble salt, write it as separate ions, e.g. Na+(aq) + Br (aq). If the reactant is an insoluble salt, write it as a solid, e.g. AgBr(s). 3. If a reactant is neither an acid nor a salt, is it a weak base, either ammonia or an amine? (Check for nitrogen (N) in the chemical formula.) Also check to see if it is a basic anion. If the reactant is a weak base, write it as an undissociated molecule, e.g. C3H7NH2(aq). II. Pair all possible cations and anions to check for all possible products. For each possible product, follow the procedures in Part I above, and make the tests below. III. Check for possible acid-base reactions. Both molecules and ions can be acids or bases. An acid reacting with a base will transfer protons (H+) to the base. − If the base is a hydroxide (OH ), proton transfer will form water, H2O. In strongly acidic solution, anions of weak acids become fully protonated. − For example, the anion S 2 will react with 2 H+ to form H2S(g). IV. Check for unstable products that decompose to form gases. For example, any H2CO3 made as a product decomposes to form CO2(g) + H2O. V. Eliminate any spectator ions (bystander ions, non-participating ions), that is, ions present on both sides of the chemical reaction equation, which do not take part in the reaction at all. VI. Now balance the chemical equation so that the total numbers of atoms of each element, in whatever chemical form or combination, are the same on each side of the equation. Check that total net charge is the same on each side of the equation. SUMMARY − GENERAL RULES USED IN WRITING CORRECTLY BALANCED NET IONIC EQUATIONS 1. Write all soluble, strong electrolytes as separate ions: M n+(aq), X q-(aq). Ions in solution must always be written with the proper charge superscript. 2. Write all gases, insoluble solids, nonelectrolytes, and weak electrolytes in molecular form. For example, CO2(g); Fe(OH)3(s); CH3OH(aq); CH3COOH(aq). Do not write charges with any solids or with molecules in solution. 3. In the final net reaction equation, do not include any substance or ion that is present but that does not take part in the reaction. 4. While not strictly necessary, it is useful and good practice to specify in parentheses, as shown below, the state of any substance or ion taking part in a chemical reaction. Ions or non-electrolyte or weak electrolyte molecules dissolved in water: M n+(aq), Q(aq). Insoluble solids, or solid precipitates formed: R(s) or R↓ . Gases as reactants or gases evolved as products: G(g) or G. Important Note: Students often confuse the property of being soluble in water or insoluble in water with the property of being a strong electrolyte or a weak electrolyte. There is simply no connection whatever

between these two properties. The rules and the exceptions regarding each of these properties must be learned independently. 1. Substances said to be soluble in water, whether they are solids, liquids, or gases, will dissolve completely in water (up to a maximum concentration, which may be quite large; certain liquid substances are in fact miscible with water in all proportions). Examples of soluble substances: NaCl; KNO3; HCl; NH3; CH3COOH; CH3OH. a. Substances that are soluble in water may be strong electrolytes: all of the dissolved substance dissociates entirely into ions, and none of it exists as undissociated molecules dissolved in the solution. Examples: HCl; HNO3; NaCl; KBr; Na2SO4; NaOH. b. Substances that are soluble in water may be weak electrolytes: most of the dissolved substance exists as undissociated molecules in the solution, and only a small fraction is dissociated into ions. Examples: HF; HNO2; CH3COOH; NH3; C5H5N. c. Substances that are soluble in water may be non-electrolytes: all of the dissolved substance exists as undissociated molecules in solution, and no ions at all are formed. Examples: methanol, CH3OH; acetone, CH3COCH3; glucose, C6H12O6. 2. Substances said to be insoluble in water actually do dissolve, but only to a very small extent (the maximum dissolved concentration may be extremely small). Examples: AgCl; Ni(OH)2. What does dissolve, however little that is, may be a strong electrolyte, or may be a weak electrolyte, or may be a nonelectrolyte. a. Substances that are insoluble in water may be strong electrolytes: all of the dissolved substance, however little that is, dissociates entirely into ions, and none of it exists as undissociated molecules dissolved in the solution. Example: AgCl. b. Substances that are insoluble in water may be weak electrolytes: most of the dissolved substance, however little that is, exists as undissociated molecules in solution, and only a small fraction is dissociated into ions. Example: C8H17COOH. c. Substances that are insoluble in water may be non-electrolytes: all of the dissolved substance, however little that is, exists as undissociated molecules in solution, and no ions are formed at all. Examples: H2; P4; S8; cholesterol, C27H46O. Example: When a salt solution is added to another salt solution, a precipitate (an insoluble compound) may form. For example, when aqueous silver nitrate, AgNO3, is added to aqueous sodium chloride, NaCl, a white solid forms and settles out of solution. The balanced equation for this reaction is AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq) To focus on the formation of the precipitate AgCl, it is useful to write the equation to show the ions separately. This is how they exist in solution: Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq) → AgCl(s) + Na+(aq) + NO3-(aq) Now we can see that only the Ag+ ion and the Cl- ion are involved in the reaction. The sodium and nitrate ions are called spectators because they are unchanged in the reaction. Cancel out the spectator ions, and what is left is called the net ionic equation. Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq) → AgCl(s) + Na+(aq) + NO3-(aq) or Ag+(aq) + Cl-(aq) → AgCl(s) Net ionic equations can be written for any ion exchange reaction in solution. To write net ionic equations, follow these simple rules:

1. Write a balanced equation showing the reactant(s) and product(s) before dissociation. 2. Repeat the equation with reactant(s) and product(s) dissociated where appropriate. 3. Cancel all spectator ions and rewrite the remaining net ionic equation.

Balancing Equations Using Algebraic Methods Health hazard alert: The following technique is difficult and involves the creative use of algebra. It is recommended only for students who are attempting the balancing equation contest, who have a strong interest in algebra, and lots of time on their hands. It is well beyond the scope of normal balancing of equations and beyond the scope of chemistry. Read at your own risk. Your teacher will be happy to help you with these ideas to help you to apply them to any equations. A simple example: H2 + O2 = H2O (1) Insert lower case letters as coefficients of the balanced equation. a H2 + b O2 = c H2O (2) Write equations, balancing each atom, by multiplying subscripts and coefficients. for atom H: 2a = 2c, which reduces to "a=c"; for atom O: "2b = c" (3) Use algebra to relate all coefficients in a single line. "a = c = 2b" (4) Select a number, usually 1, for the coefficient with the largest multiplier (b). (5) Using simple algebra, if b = 1, then a = 2 and c = 2. (6) Write the balanced equation by substituting coefficients for a, b, and c. 2H2 + O2 = 2H2O A slightly more complicated example: Al(OH)3 + H2SO4 = Al2(SO4)3 + H2O (1) a Al(OH)3 + b H2SO4 = c Al2(SO4)3 + d H2O (2) for Al: "a = 2c"; for O: "3a + 4b = 12c + d"; for H: "3a + 2b = 2d"; for S: "b = 3c" (3) Combine the expressions for Al and S first, since they are the simplest. But "a = 2c" and "b = 3c" are not that easy to combine. The term in common is c. How can I get equal numbers of c? Aim for 6c. Thus: 3a = 6c and 2b = 6c. So "3a = 2b = 6c" The expression for H includes a new term, d. Let's solve for this in terms of any one of the other variables. for H: "2d = 3a + 2b". (You can also use the expression for oxygen, however this is a bit more complicated so it will take longer to get to the same place. You might wish to try it.) We already found that "2b = 3a = 6c", so we substitute using the relationship "2b = 3a". for H: 2d = 3a + 2b 2d = 3a + 3a 2d = 6a or "d = 3a" How can we relate our previous equation, "3a=2b=6c" with our new equation "d=3a"? Since "3a" is in the first equation, simply add "=d" "3a = 2b = 6c = d" (Note: This method is much rougher than it seems. Just what the coefficients are, is not important, so long as they come out in the correct ratios. You might have gotten "6a=4b=12c=2d" or "30a=20b=60c=10d" as examples. They will all reduce to the correctly balanced equation.) (4) Set c = 1 (5) Solve for a: 3a = 6, so a = 2. Solve for b: 2b = 6, so b = 3 Solve for d: d = 6 (6) 2 Al(OH)3 + 3 H2SO4 = 1 Al2(SO4)3 + 6 H2O

Limiting Reagents Most everyone understands the concept of a limiting reagent. It is the same idea as when you go to the grocery store and buy a package of hot dogs and a package of hot dog buns. When you get home, you realize that you bought 10 hot dogs but only 8 hot dog buns (you're two hot dig buns short!). If you want to put one hot dog in one bun then your limiting ingredient will be the hot dog buns. Likewise, your excess ingredient will be the hot dogs. The same thing happens in chemistry.  Limiting Reagent - Reagent that limits the amount of products that can be formed. For example, nitrogen gas is prepared by passing ammonia gas over solid copper(II) oxide at high temperatures. The other products are solid copper and water vapor. 2 NH3(g) + 3 CuO(s) --> N2(g) + 3 Cu(s) + 3 H2O(g) If 18.1 g of NH3 are reacted with 90.4 g of CuO, which is the limiting reagent? How many grams of N2 will be formed? First we compute the number of moles of NH3 (M.W. = 17.0 g/mole) and the number of moles of CuO (M.W. = 79.5 g/mole).

To determine which reagent is limiting we use the mole ratio from the chemical equation to convert moles NH3 to moles CuO.

So, only 1.14 moles of CuO is available, therefore CuO is the limiting reagent. That is, CuO will run out before the NH3 does. The mass of N2 produced will be

 % Yield = Actual Yield (grams) from the limiting Theoretical Yield (grams)

X 100%

The theoretical yield is determined

reagent and the calculated value is converted to grams.