Chapter 7

 Representative particle- refers to whether a substance commonly exists as atoms, ions, or molecules  Ex. Elements- Representative Particle is the atom

 7 elements exist as diatomic molecules  H 2 N 2 O 2 F 2 Cl 2 Br 2 I 2  Representative Particle of a molecular compound is a molecule Example CO SO 3

Formula unit (FU) -Representative Particle of ionic compounds Example NaCl AgNO 3 BaS Ca(C 2 H 3 O 2 ) 2

 Mole- Chemists quantity of a substance that represents 6.02x10

23 representative particles of that substance- called Avagadro’s number  ex. 1 dozen eggs  How many moles of Mg atoms in 3.01x10

22 atoms of Mg?

 # of moles in 1.20x10

25 atoms of P?

 # of atoms in .750 mol of Zn?

 # of molecules in 4 mol of glucose, C 6 H 12 O 6 ?

 # of moles in 1.20x10

24 of CO 2 ?

molecules

 To find the # of atoms in 1 mole of a compound, you must determine the # of atoms in a representative formula of that compound

 # of fluoride ions in 1.46 mol of Aluminum fluoride?

 # of C atoms in a mixture of 3 mol C 2 H 2 and .7 mol carbon monoxide?

 Atomic Mass – amu  Mass of single atom  Ex) C = 12 amu  Ex) H = 1 amu (Periodic Table)

 Gram Atomic Mass – gam  # of grams of an element that is numerically equal to the atomic mass in amu.

 Ex) Carbon – gam is 12 g  Ex) Oxygen – gam is 16 g

 GAM – mass of 1 mol of atoms of a mono-atomic element  Ex) Carbon – gam is 12 g/mol  Ex) Oxygen – gam is 16 g/mol

 GMM – mass of 1 mol of that compound.

 Ex) GMM of H 2 O 2

2 mol H x 1 g H = 2.0 g H 1 mol H 2 mol O x 16.0 g O = 32.0 g O 1 mol O

34 g

Examples

 Find the GMM of C 6 H 4 Cl 2

   C 6 x 12.01 = 72.06

H 4 x 1 = 4 Cl 2 x 35.45 = 70.9

146.96 g

 GFM – Mass of 1 mol of an ionic compound.

 Ex) GFM of Ammonium Carbonate

2 mol N x 14gN = 28 g 1 mol N 8 mol H x 1 g H = 8 g 1 mol H 1 mol C x 12 g C = 12 g 1 mol C 3 mol O x 16 g O = 48 g 1 mol O

96 g

Molar Mass 

Mass of a mole of any element or compound.



Ex) O

2

= 32 g/mol,



O = 16 g/mol

Mole Mass Conversion  Ex) # of grams in 7.20 mole of dinitrogen trioxide

2 mol N x 14 g N = 28 g N 1 mol N 3 mol O x 16 g O = 48 g O 1 mol O

76 g

Example Cont’d… 7.20 mol N 2 O 3 x 76 g N 2 O 3 1mol N 2 O 3

= 5.47 x 10 2 g N 2 O 3

Grams Moles  Ex) find # of moles 922g of iron(III) oxide.

922 g Fe 2 O 3 x 1 mol Fe 2 O 3 159.6 g Fe 2 O 3

= 5.78 mol Fe 2 O 3

 The Volume of a gas at Standard Temperature and Pressure is 22.4 L (STP).

 Std Temp = 0°C  Std Press = 1 atmosphere (atm)  22.4 = molar volume of a gas = 22.4 L 1 mol

 Ex) Determine the Volume in L of 0.600 mol of Sulfur Dioxide gas @ STP.

 Mol L  Known : 1 mol SO 2 = 22.4 L

.600 mol SO 2 x 22.4 L 1 mol SO 2

= 13.4 L SO 2

 Gas Density and Gram Molecular Mass   Density of gas – units g/L Ex. Density of carbon and oxygen is 1.969 g/L at STP. Determine gfm. Is compound CO or CO 2 ?

 Densities of A, B, and C are 1.25, 2.86, and .714 g/L at STP. Calculate gfm of each. Identify each substance as ammonia, sulfur dioxide, chlorine, nitrogen or methane.

Volume of gas at STP 1 mol/ 22.4 L 22.4 L/ 1 mol 1 mol/6.02x10

23 part Representative particles 6.02x10

23 part/ 1 mol Mole gfm/ 1 mol 1 mol/gfm Mass

   Ex. How many Carbon atoms are in a 50 carat diamond that is pure carbon?  50 carats= 10 g Mass in grams of an atom of nickel?

How many molecules are in a 6 L balloon filled with carbon dioxide (@ STP)?

Percent Composition

 the percent by mass of each element in a compound.

Examples  Find the percent composition of K 2 CrO 4

% mass = grams of element x 100 grams of compound

 40.3 % K  26.8 % Cr  32.9 % O  They must add up to equal 100%

Example  An 8.20g piece of Mg combines completely with 5.40g of oxygen to form a compound. Calculate the % composition of the compound.

8.20g + 5.40g = 13.60g

% Mg = mass of Mg x 100 mass of compound 8.2 x 100 = 60.3% 13.6

Cont’d %O = mass of O x 100 mass of compound = 5.40 x 100 = 39.7% 13.6

Example  29g of silver combines with 4.3g of sulfur. Calculate % composition.

29 x 100 = 87.1% Ag 33.3 4.3 x 100 = 12.9%S 33.3

Example  222.6g of Sodium combines with 77.4g of Oxygen. Calculate % composition.

222.6 x 100 = 74.2 % Na 300 77.4 x 100 = 25.8% O 300

% Composition of a known compound % mass =

grams of element in 1 mol of cmpd x 100 gfm of compound

Examples  Calculate the % composition of ethane, C 2 H 6.

 C  H Cont’d 2 x 12 = 24g 6 x 1 = 6g 30 g

Cont’d 24 x 100 = 80 % C 30 6 x 100 = 20 % H 30

Examples  Calculate the % composition of:  a) C 3 H 8  b) Calcium Acetate  c) Hydrogen Cyanide

 a) 81.8%C, 18.2% H  b) 25.4% Ca, 30.4% C, 3.8% H, 40.5% O  c) 3.7% H, 44.4% C, 51.9% N

Examples



Calculate the mass of Carbon in 82 g of C

2

H

6

.

82g C 2 H 6 x 80g C = 66g C 100g C 2 H 6 *Based on previous example, Composition of C in C 2 H 6 is 80% C & 20% H

Examples



Calculate the amount of Hydrogen in the following compounds.

350 g C

3

H

8 350 g C 3 H 8 x 18.2g H = 100g C 3 H 8

63.7g H

124g Calcium Acetate

124 g Ca(OAc) 2 x 3.8g H = 100 g Ca(OAc) 2

4.71g H

378g Hydrogen Cyanide

378 g HCN x 3.7 g H = 100 g HCN

14 g H

Empirical Formulas

Empirical Formula -

 The lowest whole number ratio of the element in a compound.

 An empirical formula may not be the same as the molecular formula  Examples  CO 2 is both empirical and molecular   N 2 H 4 is molecular formula NH 2 is empirical formula because it is the simplest ratio of N : H

Examples

 Find the empirical formula of a compound that is 25.9% nitrogen and 74.1% oxygen.

25.9 g N x 1 mol N = 1.85 mol N 14 g N 74.1g O x 1 mol O = 4.63mol O 16 g O

1.85 mol = 1 mol N 1.85 mol 4.63 mol = 2.5 mol O 1.85 mol

N) 1 x 2 = 2 O) 2.5 x 2 = 5 empirical formula - N 2 O 5

Examples

 Find the empirical formulas for the following compounds.

79.8 %C , 20.2 %H



Empirical Formula CH

3

67.6% Mercury , 10.8% Sulfur, 21.6% Oxygen 

Empirical Formula HgSO

4

17.6% Sodium, 39.7% Chromium, 42.7% Oxygen 

Empirical Formula

Na 2 Cr 2 O 7

Calculating Molecular Formulas



You must know:



Empirical Formula



GFM

Examples

-gfm = 60g -Empirical formula = CH 4 N -efm = 30 gfm = 60 = 2 efm 30 2(CH 4 N) = C 2 H 8 N 2

 gfm = 181.5 g   Empirical formula = C 2 HCl efm = 60.5

gfm = 181.5 = 3 efm 60.5

3(C

2

HCl) = C

6

H

3

Cl

3

Methyl butanoate smells like apples. The Percent Composition is 58.8% C, 9.8%H, 31.4% O. The gfm is 102 g/mol - what is the molecular formula?

58.8 gC x 1 mol C = 4.9 mol C 12g C 9.8 gH x 1 mol H = 9.8 mol H 1g H 31.4 gO x 1 mol O = 1.96 mol O 16g O

4.9 = 2.5 mol C 1.96

9.8 = 5 mol H 1.96

1.96 = 1 mol O 1.96

C) 2.5 x 2 = 5 H) 5 x 2 = 10 0) 1 x 2 = 2 C 5 H 10 0 2