Stoichiometry Review

Spring Final Exam

PbO2

HNO3

Pb: 1 (207.2 g) = 207.2 g

O: 2 (16.0 g) = 32.0 g 239.3 g

H: 1 (1.0 g) = 1.0 g

N: 1 (14.0 g) = 14.0 g 63.0 g

O: 3 (16.0 g) = 48.0 g

Molar Massthe mass of one mole of a substance

percentage composition: the mass % of each element in a compound

Find % composition.

PbO2

(NH4)3PO4

% of element =g element

molar mass of compoundx 100

207.2 g Pb 239.2 g : = 86.6% Pb

32.0 g O 239.2 g = 13.4% O

31.2 g P 149.0 g = 20.8% P

64.0 g O 149.0 g = 43.0% O

42.0 g N 149.0 g = 28.2% N

12.0 g H 149.0 g = 8.1% H

:

::

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(see calcs above)

Finding an Empirical Formula from Experimental Data

1. Find # of g of each element.

2. Convert each g to mol.

3. Divide each “# of mol” by the smallest “# of mol.”4. Use ratio to find formula.

A compound is 45.5% yttrium and 54.5% chlorine.Find its empirical formula.

YCl3

Yg 45.5

Yg 88.9 Ymol 1

Ymol 0.512 0.512 1

Cl g 54.5

Cl g 35.5Cl mol 1

Cl mol 1.535 0.512 3

(How many empiricals “fit into” the molecular?)

To find molecular formula… A. Find empirical formula.

B. Find molar mass of empirical formula.C. Find x = MM molecular MM empirical D. Multiply all parts of empirical formula by x.

A carbon/hydrogen compound is 7.7% H and has amolar mass of 78 g. Find its molecular formula.

emp. form. CH

mmemp = 13 g 78 g 13 g

= 6 C6H6

H g 7.7

H g 1.0H mol 1

H mol 7.7 7.69 1

C g 92.3

C g 12.0C mol 1

C mol 7.69 7.69 1

A compound has 26.33 g nitrogen, 60.20 g oxygen,and molar mass 92 g. Find molecular formula.

mmemp = 46 g = 2 N2O4 92 g 46 g

NO2

N g 26.33

N g 14.0N mol 1

N mol 1.881 1.881 1

O g 60.20

O g 16.0O mol 1

O mol 3.763 1.881 2

Island Diagram:

a. Diagram has four islands.

b. “Mass Island” for elements or compounds

c. “Particle Island” for atoms or molecules

d. “Volume Island”: for gases only

1 mol @ STP = 22.4 L = 22.4 dm3

1 mol = 6.02 x 1023 particles

MOLE(mol)

Mass(g)

Particle(at. or m’c)

1 mol = molar mass (in g)

Volume(L or dm3)

1 mol = 22.4 L

1 mol = 22.4 dm3

1.29 mol

What mass is 1.29 mol ?

How many molecules is 415 L at STP?sulfur dioxide

Fe2+ NO31– Fe(NO3)2

( )1 mol179.8 g

= 232 g

sulfur dioxideSO2

415 L ( )1 mol22.4 L ( )1 mol

6.02 x 1023 m’c

= 1.12 x 1025 m’c

Iron (II) nitrate

22.4 L

( )1 mol

87.3 L

What mass is 6.29 x 1024 m’cules ? Al3+ SO4

2– Al2(SO4)3

aluminum sulfate aluminum sulfate

= 3580 g

( )1 mol

342.3 g( )1 mol6.02 x 1023 m’c

6.29 x 1024 m’c

At STP, how many g is 87.3 L of nitrogen gas?N2

( )1 mol28.0 g = 109 g

•During a chem. rxn.; atoms are rearranged (NOT created or destroyed!)•Chemical equations must be balanced to show the relative amounts of all substances.•Balanced means: each side of the equations has the same # of atoms of each element.CH4 + O2 —> H2O + CO2

CH4 + 2O2 —> 2H2O + CO2

Chemical Reactions--Chemical Equations--

Unbalanced

Balanced

Cellulose reacts with oxygen gas to form carbon dioxide gas & liquid water.

C6H10O5 + 6O2 6CO2 + 5H2O

6 C 6 10 H 10 17 O 17 Balanced!!!

Balanced?

Nitroglycerin decomposes to form nitrogen gas, oxygen gas, carbon dioxide gas & water vapor

2 C3H5(NO3)3 3N2 + O2 + 6CO2 + 5H2O

6 C 6 10 H 10 6 N 6 18 O 19

Not Balanced!

Balanced?

• When balancing equations, you may change coefficients as much as you need

to, but you may never change subscripts because you can’t change what substances are involved.

Balancing Chemical Equat.

H2 (g) + O2 (g) H2O (l)

1. Atom Inventory:

Reactants Products

2 H 2

2 O 12 Add coefficients to balance

3. Double check to make sure it is all BALANCED!

2 2

244

Example:

• Balancing Chemical Equations practice

Balancing Chemical Equat.