Course Outline

Prep Chem

04/25/09

�Before You Begin:

To master this material you will need to be familiar with inorganic nomenclature and formulas. You can review these topics in Compounds.

Chemical Changes

Chemistry is the study of matter and the changes in matter. A chemical reaction is the process by which a substance changes into a different substance. During these transformations matter is not created nor does it vanish. A fundamental law about the nature of matter is the Law of Conservation of Mass. This law states that during any chemical change, mass is neither lost nor gained; matter is merely transformed from one substance into another.

We can understand why the Law of Conservation of Mass works if we think of Dalton’s model of the atom. Atoms are not created nor destroyed during chemical changes; they are just re-grouped into different compounds.

Chemical Reaction Equations

We use chemical reaction equations to describe the details of a chemical change. This type of shorthand gives a surprising amount of information in as few symbols as possible.

Reactants and Products

The reactants are the substances that are present before the chemical change takes place. They are the things that are present at the starting point. By convention, the chemical symbols for the reactants are written on the left hand side of the chemical reaction equation.

The products are the substances that are formed during the chemical change. They are the things that are present at the end. By convention, the chemical symbols for the products are written on the right hand side of the chemical reaction equation.

A reaction arrow is the symbol that signifies the actual change. A reaction arrow separates the reactants from the products. While a typical chemical reaction equation will include much more information, they all have the substances present before and after the chemical change with a reaction arrow to separate them. It has the basic form: HCl + NH₃ à NH₄Cl

The chemical information in the reaction equation can be thought of as number and kinds of atoms needed and formed. Note that the number and kinds of atoms on both sides of the reaction arrow are the same. In this case, one nitrogen atom, one chlorine atom, and four hydrogen atoms take part in this particular chemical change. Because a mole is a collection of atoms or molecules, it is more convenient to consider reaction equations as information about the number and kinds of moles of substances taking part in the chemical change. In the above example, one mole of hydrogen chloride (or hydrochloric acid) and one mole of ammonia react to form one mole of ammonium chloride. Remember to consider all of the information associated with the chemical formulas. We know that hydrogen chloride (or hydrochloric acid) is an acid, ammonia is a molecular compound, and ammonium chloride is an ionic compound.

Coefficients:

In most chemical changes, the relative amounts of each of the reactants and products are not one to one. Information about the relative number of moles of each type of substance is indicated by the coefficients. These are numbers written in front of the chemical formulae. In the previous example, there weren’t any numbers written in front of the chemical formulae, so the relative mole ratios were one to one to one.

For example, in this reaction equation, the relative number of moles is two moles of hydrogen, one mole of oxygen, and two moles of water: 2H₂ + O₂ à 2H₂O. As before, the number and kinds of atoms on each side of the reaction arrow is the same (four hydrogen atoms and two oxygen atoms).

State and Solution Information:

Information about the states of matter for the reactants and products may be included in a reaction equation. If so, the abbreviation for the state is written in parenthesis following the chemical formula.

If the two previous reaction equations were to include state information they would look like this:

HCl_(g) + NH_3(g) à NH₄Cl_(s)

2H_2(g) + O_2(g) à 2H₂O_(l)

Translating Chemical Reaction Equations:

A great deal of information is imbedded in any reaction equation. We can write a single line of chemical symbols to represent a paragraph of words. For example, this reaction equation is translated in the next paragraph.

CaCO_3(s) + 2HNO_3(aq) à  Ca(NO₃)_2(aq) + CO_2(g) + H₂O_(l)

“Calcium carbonate and nitric acid react to form calcium nitrate, carbon dioxide, and water. The relative proportions are one mole of calcium carbonate to two moles of nitric acid reactants form one mole each of the products calcium nitrate, carbon dioxide, and water. Calcium carbonate is an ionic compound; it consists of one calcium two positive ion for every one carbonate two negative ion (the carbonate ion has one carbon atom and three oxygen atoms). Nitric acid is an acid; it consists of one hydrogen one positive ion for every one nitrate one negative ion (the nitrate ion has one nitrogen atoms and three oxygen atoms). Calcium nitrate is an ionic compound; it consists of one calcium two positive ion for every two nitrate one negative ions. Carbon dioxide is a molecular compound; it consists of molecules of two oxygen atoms bonded to one carbon atom. Water is a molecular compound; it consists of two hydrogen atoms bonded to one oxygen atom. In this particular reaction, solid calcium carbonate is reacted with aqueous nitric acid; the calcium nitrate that forms is an aqueous solution; the carbon dioxide is a gas; and the water is a liquid.”

The information that this reaction equation conveys is

1. The identities of the substances involved.
2. Which of the substances are the reactants and which of the substances are the products.
3. The types of substances involved in the reaction (element versus compounds, and ionic, molecular, etc) as well as the number and kinds of atoms/ions which make up these substances.
4. The states of matter for the substances in this particular reaction.

The single line of symbols in a chemical reaction equation is a lot shorter!

4Concept Check: Translate this reaction equation into words. Include information about the products and reactants as well as their names.

2Fe(NO₃)_3(aq) + 3Na₂S_(aq) à Fe₂S_3(ppt) + 6NaNO_3(aq)

Answer: Iron (III) nitrate reacts with sodium sulfide to form iron (III) sulfide and sodium nitrate. All of these substances are ionic compounds. The iron (III) nitrate has one iron +3 charge ion for every three nitrate -1 ions. The sodium sulfide has two sodium +1 ions for every one sulfide -2 ions. The iron (III) sulfide has two iron +3 ions for every three sulfide -2 ions. The sodium nitrate has one sodium +1 ion for every one nitrate -1 ion. The iron (III) ion, sodium ion, and the sulfide ion are all monatomic ions. Nitrate is a polyatomic ion. It has one nitrogen atom and three oxygen atoms. The relative amounts of products and reactants for this reaction are two moles of iron (II) nitrate and three moles of sodium sulfide reactants form one mole iron (III) sulfide and six moles sodium nitrate products. The iron (III) nitrate, sodium sulfide and sodium nitrate are aqueous solutions. The iron (III) sulfide is a precipitate.

Types of reactions

There are many different methods of categorizing chemical reactions. One is to classify reactions based on changes in the number of reactants and products. By this method, most of the chemical reactions studied in beginning chemistry fall into one of four categories: combination, decomposition, single replacement or metathesis.

Combination:

A combination reaction, sometimes called addition reaction, is one in which two reactants form a single product. Some more complicated combination reactions have more reactants and/or products, but the general trend is for the number of different chemical substances to decrease as the reaction progresses. An example of a combination reaction is HCl_(g) + NH_3(g) à NH₄Cl_(s).

Decomposition:

A decomposition reaction is one in which a single reactant breaks apart into more than one product. Decomposition reactions are the reverse of combination reactions. An example of a decomposition reaction is the decomposition of hydrogen peroxide: 2H₂O_2(aq) à H₂O_(l) + O_2(g).

Single Displacement:

During a single replacement reaction, a reactant element takes the place of a chemically similar element that is part of a compound. An element and a compound react to form a different element and compound. An example is

Zn_(s) + Cu(NO₃)_2(aq) à Cu_(s) + Zn(NO₃)_2(aq). Note that the zinc metal takes the place of the copper ion in the compound. Zinc is more similar to copper than to nitrogen or oxygen.

Double Displacement:

During a double replacement reaction, two compounds exchange elements to form two different compounds. One type of double replacement reaction is metathesis. In a metathesis reaction, a pair of ionic compounds exchanges ions in an aqueous solution. An example of a metathesis reaction is NaCl_(aq) + AgNO_3(aq) à AgCl_(ppt) + NaNO_3(aq).

4Concept Check: What types of reactions are these?

CH₃Br_(g) + HCl_(g) à CH₃Cl_(g) + HBr_(g)

Mg_(s) + O_2(g) à 2MgO_(s)

2NaHCO_3(s) à Na₂CO_3(s) + H₂O_(g) + CO2_(g)

2Al_(s) + 3Pb(NO₃)_2(aq) à 2Al(NO₃)_3(aq) + 3Pb_(s

Answer:

CH₃Br_(g) + HCl_(g) à CH₃Cl_(g) + HBr_(g) is a double displacement (not metathesis because it doesn’t take place in an aqueous solution) reaction.

Mg_(s) + O_2(g) à 2MgO_(s) is a combination reaction.

2NaHCO_3(s) à Na₂CO_3(s) + H₂O_(g) + CO_2(g) is a decomposition reaction.

2Al_(s) + 3Pb(NO₃)_2(aq) à 2Al(NO₃)_3(aq) + 3Pb_(s) is a single displacement reaction.

Stoichiometry

Balancing Reaction Equations

The Law of Conservation of Mass requires that the number and type of atoms be the same on both sides of a reaction arrow. Students are expected to deduce the reaction coefficients if they are not given. This is called balancing the reaction. There are lots of tricks to balancing reaction equations, but they all rely on ‘trial and error.’ You must fill in coefficients and count atoms to see if the atom counts on both sides of the equation match.

!Warning! When you balance chemical reaction equations, you can adopt a systematic method that helps reduced the time involved in the trial and error process, or you can leap right in. In either case, watch for these common problems:

1. Do not change subscripts! You balance a reaction equation by adjusting coefficients only. If you are given a verbal description, double check to make sure that your chemical formulae are correct before you start to balance the equation. A tiny mistake with a subscript can result in a reaction that is impossible to balance no matter how hard you try.
2. Do not forget that hydrogen, nitrogen, oxygen, and the halogens are all diatomic elements.
3. When counting atom totals, remember that subscripts after a parenthesis indicate a multiple of everything inside the parenthesis. Ca₃(PO₄)₂ has two phosphorus and eight oxygen atoms.
4. When counting atom totals, remember that coefficients indicate multiples of the entire compound. 2Ca₃(PO₄)₂  has six calcium, four phosphorus, and 16 oxygen atoms.
5. When an element appears in more than one reactant or product, you have to include ALL of the atoms in the count. Ca₃(PO₄)₂ + NaNO₃ has 11 oxygen atoms if both appear on the same side of the reaction arrow.

4Concept Check: Balance these reaction equations:

K₂O_(s) + H₂O_(l) à KOH_(aq)

C₅H_10(l) + O_2(g) à CO_2(g) + H₂O_(l)

Barium nitrate aqueous solution reacts with sulphuric acid aqueous solution to form nitric acid aqueous solution and barium sulphate precipitate.

Answer:

K₂O_(s) + H₂O_(l) à 2KOH_(aq)

2C₅H_10(l) + 15O_2(g) à 10CO_2(g) + 10H₂O_(l)

Ba(NO₃)_2(aq) +H₂SO_4(aq) à 2HNO_3(aq) + BaSO_4(ppt)

Calculating the Mass of Products:

The coefficients of a balanced chemical reaction equation can be used as conversion factors to determine the amount of product that can be made from a given amount of reactant. Use the formula weight to find the moles of reactant. Then use the mole ratio to find moles of product.  The formula weight of the product can then be used to find grams of the product.

4Concept Check: When heated, sodium hydrogen carbonate decomposes to form sodium carbonate water and carbon dioxide. If we start with 5.5 grams of sodium hydrogen carbonate, how many grams of sodium carbonate will be produced?

Answer:

• Start with the balanced chemical reaction equation:

2NaHCO_3(s) à Na₂CO_3(s) + H₂O_(g) + CO_2(g)

• Find the formula weights of sodium hydrogen carbonate and sodium carbonate:

  NaHCO₃ is 22.99 + 1.01 + 12.01 + 3(16.00) = 84.01 g/mol

Na₂CO₃ is 2(22.99) + 12.01 + 3(16.00) = 105.99 g/mol

• Convert 5.5 grams of sodium hydrogen carbonate to moles:

5.5  g (1 mol/ 84.01 g) = 0.065 mol NaHCO₃

• Use the mole ratio to find the number of moles of product:

0.065 mol NaHCO₃ (1 mol Na₂CO₃/2mol NaHCO₃) = 0.033 mol Na₂CO₃

• Use the formula weight of Na₂CO₃ to find grams:

0.033 mol Na₂CO₃  (105.99 g/mol) = 3.5 grams Na₂CO₃

4Concept Check: Magnesium metal combines with oxygen gas to form magnesium oxide. If 1.775 grams of magnesium react with excess oxygen, how many grams of magnesium oxide form?

Answer: 2.943 grams of magnesium oxide are produced.

Limiting Reactant Problems:

In a reaction with more than one reactant, one of the reactants will run out first. The one with fewer moles adjusted for the mole ratio will be the one that runs out. This is called the limiting reactant because the amount of product is governed by it. The other reactant(s) is (are) called the excess reactant because some will be left over when the reaction is complete.

To calculate the mass of product, find which reactant is limiting. Use formula weight to find moles then mole ratio to find moles product. Do this for each reactant. The limiting reactant is the one that makes fewer moles of product (it can’t make more because it runs out too soon). Use this amount of moles of product to calculate grams of product.

4Concept Check: 2.55 grams of silver nitrate and 1.97 grams of potassium iodide are each dissolved in water and combined. The solutions react to form a precipitate of silver iodide and an aqueous solution of potassium nitrate. How many grams of the precipitate form?

Answer:

• Write the balance chemical reaction equation:

AgNO_3(aq) + KI_(aq) à AgI_(ppt) + KNO_3(aq)

• Find formula weights of the two reactants and the precipitate:

AgNO_3(aq)  is 107.87 + 14.01 + 3(16.00) = 169.01 g/mol

KI_(aq) is 39.10 + 126.90 = 166.00 g/mol

AgI_(ppt)   is 107.87 + 126.90 = 234.77 g/mol

• Use the formula weights and the reactant masses to find the moles of the two reactants:

2.55 g (1 mol/169.01 g) = 0.0151 mol silver nitrate

1.97 g (1 mol/166.00 g) = 0.0119 mol potassium iodide

• Use the mole ratios to find the moles of product that each would produce if it were limiting. In this case, both ratios are one to one, but that is not usually the case:

0.0151 mol silver nitrate à 0.0151 mol silver iodide product

0.0119 mol potassium iodide à 0.0119 mol silver iodide product

• Choose the smaller number. This is the amount of product that forms.

0.0119 mol silver iodide product forms and potassium iodide is the limiting reactant. Discard the other result.

• Use the formula weight and number of moles of product to find the grams of product: 0.0119 mol (234.77g/mol) = 2.79 g of silver iodide product.

4Concept Check: 2.89 grams of lead (II) nitrate and 1.05 grams of potassium iodide are each dissolved in water and combined. The solutions react to form a precipitate of lead (II) iodide and an aqueous solution of potassium nitrate. How many grams of precipitate form?

Answer: 2.01 grams of lead (II) iodide are produced. The lead (II) nitrate is the limiting reactant.

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