Stoichiometry - Loudoun County Public Schools

Stoichiometry - Loudoun County Public Schools

STOICHIOMETRY Molar mass, Percent composition, Moles, Conversions, Empirical formulas, Molecular formulas 1 THE MOLE SI unit for amount A counting unit for measuring large quantities of small items Atoms molecules Particles 1 mole = 6.022 x 10 23 atoms = molar mass 2 MOLAR MASS The amount of mass (grams) in one mole of a substance For elements

Average atomic mass periodic table For compounds Must add up each elements mass Subscripts are factored in by multiplication Units are g/mol May be called formula mass or molecular mass 3 STEPS TO SOLVE 1. Find elements on periodic table 2. Write down the mass 3. Multiply the mass by the subscript 4. Add together 4 SAMPLE: DETERMINE MOLAR MASS 1. Ca 1.40.08 g/mol

2. Fe 2. 55.847 g/mol 3. NaCl Calculatio ns are FUN! Na: 22.989 g/ mol Cl: 35.453 g/mol 58.443 g/mol 5 MOLAR MASS (CONT.) 4. Fe2O3

Fe: 55.847 x (2) = 111.694 g/mol O: 15.999x (3)= 47.997 g/mol 159.691 g/mol 5. Na2O Na: 22.989 x(2) = 45.978 g/mol O: 15.999= 15.999 g/mol 61.977 g/mol 6 PERCENT COMPOSITION Steps to solve 1. Find the total mass of the COMPOUND 2. Take the mass of each element and divide it by the total mass 3. Multiply by 100 7 SAMPLE: CALCULATE THE PERCENT COMPOSITION C: CO2

12.011 43.991 O: 31.998 C: 12.011 g/mol .2730 100 27.3 % .727 100 72.7% 43.991 O: (2)15.999 g/mol = 31.998 g/mol 44.009 g/mol

8 SAMPLE 2 H2SO4 H: 2.016 98.076 .0206 100 2.06 % S: 32.064 98.076 O: (4)15.999 g/mol = 63.996 g/mol .327 100 32.7 % .653

100 65.3% H: (2)1.008 g/mol = 2.016 g/mol S: 32.064 g/mol 98.076 g/mol O: 63.996 98.076 9 COMPOSITION OF HYDRATES Hydrate: crystal contains water within Water can fit into the salt crystal Happens in fixed ratios Each salt that forms a hydrate has a DIFFERENT water ratio Each salt crystal is unique

Anhydrous: water has been removed Steps to solve 1. Calculate the total mass of the compound INCLUDING THE water 2. Divide the water mass by total mass 3. Multiply by 100 Achieved through drying 1 That was nothing. Time for Coco! SAMPLE PROBLEM: HYDRATE What percentage of water is found in CuSO4 5H2O Cu: 63.54

S: 32.064 O: (4)15.999= 63.996 90.05 249.65 .361 100 H2O: (5)18.01= 90.05 249.65 36.1% H2O 11 THE MOLE CONVERSION Particles b ym ma ola r

ss Xb Xb ma y Mo lar ss y x 2 2 6.0 Mole by 23 10

10 x 22 0 . 6 Ions Atoms Molecules Formula units 23 by 22.4 L x by 22.4 L Mass

(grams) Volume (Liters) Gases at STP 1 STOICHIOMETRIC CONVERSIONS: MASS TO MASS You can relate any unit to another by this method Relate one unit to another conversion factors Steps to convert 1.Identify what you start with 2.Determine what units you end in A. Might have to do

side calculations 3.Set up conversion factor 1 EXAMPLE How many moles are in 28 grams of CO2? 28 grams CO2 1 mole CO 2 44.009 g CO 2 .636 moles CO2 _______ C: 12.011g/mol O: (2)15.999 g/mol = 31.998 g/mol 1 EMPIRICAL FORMULA When the elements are in the smallest whole

number ratio within compound Steps to solve When given a compound Divide out by the common multiple Steps to solve When given percentages 1. Change percent sign to grams (NO MATH) 2. Convert masses to moles A. Using conversions 3. Re-divide ALL moles amounts by the smallest mole amount A. Multiply if not whole numbers 4. Write compound with subscripts 1 EXAMPLE: WHATS MY EMPIRICAL FORMULA

1. C6H6 1. CH 2. C8H18 2. C4H9 3. C2H6O2 3. CH3O 4. X39Y13 4. X2Y Piece of Cake! 1 EXAMPLE: WITH PERCENT A compound is found to contain 36.48% Sodium, 25.41% Sulfur, and 38.11% Oxygen. Find its empirical formula. Eleme Percen

nt t Mass Convert to moles Re-divide Subscri pts Na 36.48 % 36.48 g 36.48 g x = 1.587 moles = 2.01

2 S 25.41 % 25.41 g 25.41 g x = 0.792 moles 1.00 1 O 38.11 % 38.11 g 38.11 g

x = 2.382 moles = 3.01 3 Na2SO3 1 MOLECULAR FORMULA Multiple of an empirical formula So many compounds with the same Empirical formula 1

STEPS TO SOLVE 1. Complete an empirical formula process if needed 2. Find the mass of the EMPIRICAL FORMULA 3. Divide the Empirical formula mass by the molecular mass A. Molecular mass is normally given in the problem B. Answer is the common multiple 4. Multiple the SUBSCRIPTS by the common multiple 1 EXAMPLE All done! Time to go sledding! Yippee!! A compound with an empirical formula of C4H4O and a molecular mass of 136 grams. What is the molecular formula of this compound? C: 4 (12.011 ) = 48.044 136 g

=1.997 68.079 g H:4 (1.009 ) = 4.036 O: 15.999 68.079 C4H4O x 2 = C8H8O2 2

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