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INTRODUCTION TO CHEMISTRY 30. COMPONENTS OF AN ATOM-  The modern atom as viewed by scientists today consists of three main particles located in two regions.

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Presentation on theme: "INTRODUCTION TO CHEMISTRY 30. COMPONENTS OF AN ATOM-  The modern atom as viewed by scientists today consists of three main particles located in two regions."— Presentation transcript:

1 INTRODUCTION TO CHEMISTRY 30

2 COMPONENTS OF AN ATOM-  The modern atom as viewed by scientists today consists of three main particles located in two regions.  The first of these two regions is the nucleus, or central core of the atom which is composed of positively charged protons and neutrons with a neutral charge.

3 ATOM REVIEW  It is believed that the neutrons are needed to hold the positively charged protons together in the nucleus. The force that holds these particles together is termed the nuclear binding force and it is believed to be one of the strongest forces that exists in nature.  The nucleus takes up a very small portion of the atom. If the atom was the size of a football field the nucleus would be the size of a household fly on the 55 yard line.  The second region surrounds the nucleus and is termed an electron cloud. The cloud holds the third particle which is a negatively charged electron.

4 CHARACTERISTICS OF THE ATOM  The number of protons in the nucleus of an atom is termed its atomic number. This number is distinctive (characteristics) for the atoms of each element (it can be found on the periodic table).  Atomic Number = # of protons

5  The number of neutrons in the nucleus of an atom is not distinctive and may vary. This creates varieties of atoms we call isotopes. Isotopes are atoms of the same element with different numbers of neutrons.

6  Another characteristic of an atom of an element is it's atomic mass. Protons and Neutrons have approximately the same mass, while electrons have very little mass in comparison to either Protons or Neutrons. The atomic mass is therefore determined from the number of protons and neutrons. It is often called the mass number (#).  Mass # (Atomic Mass) = # of Protons + # of Neutrons.  Since atoms of an element always have the same number of protons, isotopes are atoms of the same element with different mass numbers.

7  The number of electrons in a neutral atom (no charge) is equal to the number of protons (atomic #). Atoms may either gain or lose electrons during chemical interactions with other atoms. If they gain electrons they become negatively charged, if they lose electrons they become positively charged. We term these charged atoms ions.

8 ASSIGNMENT  Components of the atom

9 NAMING COMPOUNDS  Covalent compounds-  (contain nonmetal atoms combining with other nonmetal atoms)  Electrons are shared between atoms  To Name:  The first atom is named in full  The second atoms name is shortened and -ide added.  The prefixes below are added to the first and second name to indicate the number of atoms present in the compounds.  The prefix mono is typically not placed on the first atoms name.

10 HYDROCARBONS (STRAIGHT CHAINED)  Hydrocarbons are compounds that contain only carbon and hydrogen. (C n H 2n )  To Name:  The first part of the name is a prefix that indicates the number of carbon atoms. The prefixes are the same as those used by Binary Molecular compounds above except for the first four. These four are: # of Carbon Atoms Prefixes 1meth - 2eth - 3prop - 4but -

11  The second part of the name is a suffix that describes if the compound contains single bonds, a double or a triple bond between the carbon atoms. The following table illustrates the naming and gives examples. Group Bonds between carbons Suffix Ratio of Carbon to Hydrogen Alkaneall single- aneC n H 2n+2 Alkeneone double- eneC n H 2n Alkyneone triple- yneC n H 2n-2

12  Assignment: naming covalent compounds

13  Ammonium (NH 4 +) compounds  Ammonium compounds contain the complex cation NH 4 +1.  To Name: (using a table of ions)  The first part of the name is Ammonium  Second part of name is the name of the anion. FormulaName (NH 4 ) 2 SO 4 Ammonium sulfate (NH 4 ) 3 PO 4 Ammonium phosphate

14 MONOVALENT IONIC CATIONS  Ionic compounds containing monovalent cations (cation with one charge)  To Name (using a table of ions):(using a table of ions):  Locate the name of the metal cation (positive ion). Make sure it has only one charge.  Locate the name of the anion.  Place names together. FormulaName AgClSilver chloride Mg(NO 3 ) 2 Magnesium nitrate

15 DIVALENT CATIONS  Ionic compounds containing divalent cations (metal ions with two possible charges)  To Name using given the formula:  Locate the name of the anion and cation in a table of ions or periodic table.  Using the concept that the total charge of the anion and cation must be equal we can determine what the charge on the cation must be (anion charges are always fixed.) See example below Formula Charge on Anion Total charge on anion Total charge on Cation Number of Cations Charge on Each Cation Fe 2 O 3 O 2- 2

16 EXAMPLES  Find the charges on the cations:  FeO- NiF 2

17  To determine name then we:  Name the cation  Determine the charge on the cation using the method above.  Apply the correct Roman Numerals  Finally name of the anion Formula Name of Cation Total Charge on Anion Total charge on Cation Charge on Cation Name Cu 2 OCopper-2+2+1Copper (I) oxide CrCl 3 Chromiu m -3+3 Chromium (III) chloride

18 EXAMPLES  TiC-PbO 2 -  Assignment: naming ionic compounds

19 MOLE THEORY  Definition: The mole is defined as the atomic or molecular weight of a substance expressed in grams. It is a metric unit for amount of substance and has the abbreviation mol  One mole of a substance then is the atomic weight of the element expressed in grams.  The number of particles in this amount is termed Avogadro's Number and has been estimated at 6.023* 10 23 particles.  Whenever we have Avogadro's number of particles in a sample we have one mole.

20  To give you a sense of it's size, imagine that each particle in a mole was a piece of paper. If we were to stack this paper one sheet on top of another, a mole of paper would stretch from the surface of the earth to the planet Pluto

21 GRAM MOLECULAR WEIGHT (MOLAR MASS)  To calculate the gram molecular weight of a substance made up of more than one element (compound), we add up the atomic weights of the elements that comprised the compound. Atomic weights are listed in the periodic table.  Example: NaCl  Element Number of Atoms Atomic Weight (g/mol) Total weight (g/mol) Na122.99 Cl135.45 Gram Molecular weight ( Molar Mass ) ; 1 mole = 58.44 g/mol

22 Element Number of Atoms Atomic Weight (g/mol) Total weight (g/mol) C112.01 H41.014.04 Molar Mass =16.05 g/mol Example: CH 4

23  Example Ca(NO 3 ) 2  Assignment: Molar Mass sheet Element Number of Atoms Atomic WeightTotal weight Ca140.08 N214.0128.02 O616.0096.00 Gram Molecular weight ( Molar Mass ) ; 1 mole = 166.10 g

24 CALCULATIONS WITH THE MOLE  The mole has three values that can be used to do calculations with the mole. They are:  mass  volume (one mole in the gas state at 0 C and 100 kPa.(called standard temperature and pressure STP = 24.5 L.)  # of particles

25 STEPS:   Write down the given information.  Write down what you are trying to find.  Use the mole triangle to determine what operation you will perform.  Complete all required calculations and write down the final answer- **be sure to include units**

26 EXAMPLES  1. Calculate the mass in grams of 35 moles of CaCO 3.  Given information is 35 moles of CaCO 3.  Asked to calculate the mass in grams.  The mole triangle indicates that you should multiply the 35 moles by the molar mass of CaCO 3 Step 1; Calculation of molar mass 1 * 40.08 g/mole = 40.08 g 1 * 12.01 g/mole = 12.01 g 3 * 16.00 g/mole = 48.00 g 1 mole = 100.09 g/mole Step 2; Calculation of Answer 35 moles * 100.09 g/mol = 3503.15 g

27  Calculate the # of molecules in 820 L of SO 2 (g) given off by a chemical plant at STP.  The given information is 820 L of SO 2 (g) at STP.  Information you are asked to find is the # of particles.  The mole triangle indicates that you should divide 820 L by the molar volume to calculate moles  then multiply the moles by 6.023 * 10 23 p/mol. Step 1;Calculation of moles 820 L / 25.4 L/mol. = 32.28 mol. Step 2; Calculation of Answer 32.28 mol. * 6.023 *10 23 p/mol. = 1.9 *10 24 particles

28  Assignment: Mole Calculations

29 WORD AND FORMULA EQUATIONS  1. Two types of equations are written by chemists:  word equations: describe the substances that react in a chemical reaction (termed reactants), and the products that are formed, along with their states  formula equations are a shorthand method used to describe the same reactions. These are of two types:  Skeleton equations: (unbalanced) which lists the correct formula of each reacting substance and product substances, and their states.  Balanced equations: which list the correct formulas, states and balances the equation for the number of atoms present. That is it takes into account the Law of Conservation of Mass, and makes sure there is the same number and type of atom in the reactant and product. In Chemistry 30 we should only use balanced equations.

30  We balance equations by changing the coefficients or numbers in front of the substance.  WE NEVER CHANGE THE FORMULAS OF SUBSTANCES IN ORDER TO BALANCE.  Counting the atoms correctly is therefore critical. The balance (coefficient) we use is always multiplied by the subscripts used in each formula, to indicate how many atoms are represented. If atoms are in two different reactant or product compounds, they are added together to determine how many are present in total.

31 DETERMINE THE # OF ATOMS OF EACH TYPE PRESENT IN THE FOLLOWING REACTANTS. Reactants# of each atom PbS + 2 PbO Pb = S = O = Ca(NO 3 ) 2 + 2 KOH Ca = N = O = K = H = 2 NH 4 NO 3 + H 2 S N = H = O = S = Fe(NO 3 ) 3 + 3 LiOH Fe = N = O = Li = H = Ca 3 (PO 4 ) 2 + 3 H 2 SO 4 Ca = P = O = H = S =

32 BALANCING  Balancing equations is basically a process of trial and error, called inspection, but a few hints can help.  1. Balance atoms that appear only once in reactant and product first, and atoms that appear more than once last.  Example #1: Reactant Product ___C 3 H 8 (g)+ ___O 2 (g)___CO 2 (g) + ___H 2 O (g) Step 1 Balance atoms that appear only once first (C and H ) _1_C 3 H 8 (g)+ ___O 2 (g) _3_CO 2 (g) + _4_H 2 O (g) Step 2 Balance atoms that appear more than once last (O) _1_C 3 H 8 (g)+ _5_O 2 (g) _3_CO 2 (g) + _4_H 2 O (g)

33  Balance polyatomic ions as a group, for example SO 4 2- ion. Caution: The ion must remain the same in reactant and product.  Example #2: Reactant Product ___Ca(NO 3 ) 2 (aq) + ___Na 3 PO 4 (aq)___Ca 3 (PO 4 ) 2 (s) + ___NaNO 3 (aq) Step 1 Balance atoms that appear only once ( Ca, Na ) _3__Ca(NO 3 ) 2 (aq) + ___Na 3 PO 4 (aq) ___Ca 3 (PO 4 ) 2 (s) + _3__NaNO 3 (aq) Step 2 Balance atoms in ions as groups (PO 4 ; NO 3 ) _3__Ca(NO 3 ) 2 (aq) + _2__Na 3 PO 4 (aq) _1__Ca 3 (PO 4 ) 2 (s) + _6__NaNO 3 (aq)

34  In some cases the # of atoms of an element in the reactants may be odd, while the # in the products will always be even (due to a even subscript). In this case you need to double the balance of all atoms already balanced and continue the balancing.  Example #3: Reactant Product ___CuFeS 2 (s) + ___O 2 (g)___Cu (s) + ___FeO (s) + ___SO 2 (g) Step 1 Balance atoms that appear only once first _1_CuFeS 2 (s) + ___O 2 (g) _1_Cu (s) + _1_FeO (s) + _2_SO 2 (g) Step 2 Balance oxygen now. Notice that there is five atoms in the products but the reactants will always be even. To balance, double all balances already made. Now continue by balancing the oxygen _2_CuFeS 2 (s) + _5_O 2 (g)_2_Cu (s) + _2_FeO (s) + _4_SO 2 (g)

35  Assignment: Balancing equations worksheet

36 STOICHIOMETRY  Chemical reactions are like the recipe for the cookies. The balances in the equations are mole recipes that tell us the # of moles that react and form.  Example:  Notice that all of the values in the table are in the same ratio as the balance in the equation. C 3 H 8 : O 2 is always 1 : 5, CO 2 : H 2 O is always 3: 4. We call these ratios, equation factors or mole ratio. The reactants and products in this equation always react and form in these Equation (balanced) _1_C 3 H 8 (g) + _5_O 2 (g) _3_CO 2 (g) + _4_H 2 O (l) If we had 2 moles of C3H8 (g) then 2 moles 10 moles 6 moles 8 moles

37 Moles of ------>C 3 H 8 (g)O 2 (g)CO 2 (g)H 2 O (l) Complete the rest of this table using the equation above 12 moles 2 moles 0.5 moles 5 moles

38 MOLE TO MOLE STOICHIOMETRY  The following technique can be used to predict the # of moles that will react or form in an equation.  Identify the given number of moles.  Identify the balance of the Given.  Identify the balance of the substance you wish to find  Multiply. Moles of Known (n)X Balance of Unknown Balance of Known =Moles of unknown (n)

39  Example:  Using the balanced equation below, predict  1. The number of moles of C 3 H 8 (g) that reacts to produce 8.4 moles of CO 2 (g). Equation (balanced)_1_C 3 H 8 (g)+_5_O 2 (g) _3_CO 2 (g) + _4_H 2 O (l) 8.4 moles of C 3 H 8 (g)X 3 CO 1 C 3 H 8 =25.2 moles CO 2 (g)

40  The number of moles of H 2 O (l) that form if 63.7 moles of O 2 reacts.  Answer: 63.7 moles O 2 (g)X 4 H 2 O (l) 5 O 2 (g) = 51.0 moles H 2 O (l)

41  Assignment: Mole to Mole stoichometry

42 MASS TO MASS  We can use a simple 3 step method to solve stoichiometric questions with balanced equations.  Identify the given and convert it to moles.  Identify the desired, and multiply the given number of moles by the mole ratio to produce moles of desired substance.  Convert moles of desired substance to the units asked for in the question.  Example:

43  What mass of methane gas in grams must burn to produce 365 grams of water, by the following chemical reactions ?  Given is 365 g of water. Calculate moles of water.  Identify desired; mass of Methane gas. Multiple by mole ratio _1_CH 4 (g)+_2_O 2 (g)_2_H 2 O (g)+_1_CO 2 (g) 365 g ÷18.02 g/mol=20.25 mol 20.25 mol H 2 O (l)X1 CH 4 (g)=10.13 mol CO 2 (g) 2 H 2 O (g)

44  Convert moles of desired to the units asked for in the question: Mass of Methane gas.  Assignment: Mass to Mass Stoichiometry 10.13 mol CO2X44.01 g/mol=446 g


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