Mass Relationships

Mass RelationshipsLab Data

All masses are relative to one another.Cu:O = 4:1 O:H = 16:1 Cu:H= 64:1Cu:O:H 64:16:1Must choose a standard, but which one?C = 12 O = 16 F = 19 H = 1

H hard to work with, variable resultsF gives best results, very hard to duplicateO,C used most often Eventually settled on C

Chlorine ProblemTwo naturally occuring isotopes

Cl mass number 35 Actual relative mass = 34.97 amuCl mass number 37 Actual relative mass = 36.97 amu

Cl 35 - 75.53% abundance in natureCl 37 - 24.47% abundance in nature

34.97 x .7553 = 26.4136.97 x .2447 = 9.04 35.45 amu average atomic massThis is the number found on the periodic tables

PracticeSr relative mass 83.9 - 0.50 % abundanceSr relative mass 85.9 - 9.9 % abundanceSr relative mass 86.9 - 7.0 % abundanceSr relative mass 87.9 - 82.6 % abundance

Average atomic mass = ?

PracticeSr relative mass 83.9 0.50 % abundanceSr relative mass 85.9 9.9 % abundanceSr relative mass 86.9 7.0 % abundanceSr relative mass 87.9 82.6 % abundance

Average atomic mass = 87.61amu

Discovery Activity

PracticePractice: molar amounts of elements

1 mole K atoms = ? grams 2 moles Mg atoms = ? grams .3 moles Al atoms = ? grams

Chemical Formulastells: * atom ratio within the molecule * mole ratio within the substance as well

formula weight(molar mass) sum total of masses of all atoms in the formula

ex. CaCO3 (Ca) 40.08 x 1 = 40.08 (C) 12.01 x 1 = 12.01 (O) 16.00 x 3 = 48.00 100.09

Formula Weight Practice

NaCl = ?

K2I = ?

Ca(OH)2 = ?

% Composition

% Compositionin a sample of a compound,

total mass of one element x 100 = % total mass of compound sample

ex. compound containing K, Cr, O

25.0 gram sample contains 6.64 g K, 8.84 g Cr, 9.52 g O 6.64 g K x 100 = 26.7 % K 8.84 g Cr x 100 = 35.4 % Cr25.0 g 25.0 g 9.52 g O x 100 = 38.1 % O25.0 g

Empirical Formulasimplest ratio of elements in a compoundmay not be actual, just simplest

total mass of one element = element quotient formula weight

ratio of element quotients = empirical formula 6.64 g K = 0.170 8.84 g Cr = 0.170 9.52 g O = 0.59539.10 52.00 16.00 if ratio is not obvious, divide all quotients by the smallest0.170 = 1 0.595 = 3.50.170 0.170No half atoms, so actual ratio must be 2:2:7empirical formula = K2Cr2O7

Empirical vs Actualsimplest ratio of elements in a compound CH2O

empirical F.W. = 30 actual F.W. = 180

actual formula must be 6 times the empirical or C6H12O6

Reaction TypesFirst 5 types - most common

A) Composition

A + B → AB

B) Decomposition

AB → A + B

C) Single Replacement

X + AB → A + XB

D) Double Replacement

XA + ZB → ZA + XB

E) Combustion

CxHz + O2 → CO2 + H2O

Terms

XA + ZB → ZA +

XB

reactants reaction symbol

products

reaction = equationspecifically: reactions happens in the lab equations symbolize reactions on paper.

Writing Double Replacement Reactions

1. Write the names of the reactants.

2. Switch the first names of the reactants. 3. Write these names on the right side of the arrow.

4. Equation needs to be written all on one line.

5. Starting on the far right, write the formulas.

6. Moving to the left, continue to write formulas.

Examplecalcium + sodium →chloride carbonate

calcium + sodium →sodium + calciumchloride carbonate chloride carbonate

CaCl2 + Na2CO3 → NaCl + CaCO3




2. Switch the first names of the reactants. 3. Write these names on the right side of the arrow.

4. Starting on the far right, write the formulas.

5. Moving to the left, continue to write formulas.

Balancing Equations CaCl2 + Na2CO3 → NaCl + CaCO3

Must be the same number and kind of atoms on each side of the arrow.


1 Ca 2 Na 1 Na 1 Ca2 Cl 1 CO3 1 Cl 1 CO3

Use coefficients to multiply the number of molecules to achieve balance.

CaCl2 + Na2CO3 → 2 NaCl + CaCO3

1 Ca 2 Na 2 Na 1 Ca2 Cl 1 CO3 2 Cl 1 CO3

Stoichiometry stoy key om e tree

nn

The numerical relationship between the molecules in a reaction.

CaCl2 + Na2CO3 → 2 NaCl + CaCO3basic 1 molecule 1 molecule 2 molecules 1 moleculeratio

367 molecules 367 molecule 734 molecules 367 molecule

1 mole 1 mole 2 moles 1 mole 7 mole 7 mole 14 moles 7 mole .09 moles .09 moles .18 moles .09 moles

Stoichiometry 2

nn

CaCl2 + Na2CO3 → 2 NaCl + CaCO3

Since moles and coefficients both are related to the number of molecules, the coefficients in the equation, and the theoretical/actual moles in the reaction MUST have exactly the same ratio.

coefficient CaCl2 = moles CaCl2 = 1 coefficient Na2CO3 moles Na2CO3 1

coefficient CaCl2 = moles CaCl2 = 1 coefficient NaCl moles NaCl 2

coefficient CaCl2 = moles CaCl2 = 1 coefficient CaCO3 moles CaCO3 1

Stoichiometry 3

nn

2 AlCl3 + 3 CaCO3 → 3 CaCl2 + Al2(CO3)3

Since moles and coefficients both are related to the number of molecules, the coefficients in the equation, and the theoretical/actual moles in the reaction MUST have exactly the same ratio.

coefficient AlCl3 = moles AlCl3 = 2 coefficient CaCO3 moles CaCO3 3

coefficient AlCl3 = moles AlCl3 = 2 coefficient CaCl2 moles CaCl2 3

coefficient AlCl3 = moles AlCl3 = 2 coefficient Al2(CO3)3 moles Al2(CO3)3 1

Practice

nn

2 AlCl3 + 3 CaCO3 → 3 CaCl2 + Al2(CO3)3

If 1.6 moles of AlCl3 is used to react with CaCO3, how many moles of CaCl2 are produced, how many moles of Al2(CO3)3 are produced, and how many moles of CaCO3 are needed?

coefficient AlCl3 = moles AlCl3 = 2 = 1.6coefficient CaCO3 moles CaCO3 3 ?

coefficient AlCl3 = moles AlCl3 = 2 = 1.6coefficient CaCl2 moles CaCl2 3 ?

coefficient AlCl3 = moles AlCl3 = 2 = 1.6coefficient Al2(CO3)3 moles Al2(CO3)3 1 ?

Data

nn

2 HCl + Mg →H2 + MgCl2

Trial Mass HCl(g) Mass Mg(g) ml H2 produced1 .178 .020 21 1 .178 .040 42 1 .178 .080 82 1 .178 .120 122 1 .178 .160 122 1 .178 .200 122

Patterns??amount of H2 levels out

Data

nn

2 HCl + Mg →MgCl2 + H2

Trial Mass HCl(g) Mass Mg(g) ml H2 produced1 .178 .020 21 1 .178 .040 42 1 .178 .080 82 1 .178 SEP .120 SEP 122 1 .178 .160 122 1 .178 .200 122

Patterns??amount of H2 levels outcalled ‘ stoichiometric equivalence point’

Data

nn


Trial Mass HCl(g) Mass Mg(g) ml H2 produced1 .178 excess .020 21 1 .178 excess .040 42 1 .178 excess .080 82 1 .178SEP .120SEP 122 1 .178 .160 excess 122 1 .178 .200 excess 122


reactant that determines product amount called limiting

Data

nn


Trial Mass HCl(g) Mass Mg(g) ml H2 produced1 .178 excess .020 limiting 21 1 .178 excess .040 limiting 42 1 .178 excess .080 limiting 82 1 .178SEP .120SEP 122 1 .178 limiting .160 excess 122 1 .178 limiting .200 excess 122


reactant that determines product amount called limiting reactant that determines product amount called limiting

Practice

nn

2 Sb + 3 I2 →2 SbI3

a) if 1.20 moles of Sb react, how much SbI3 is produced?

b) if 2.40 moles of I2 react, how much SbI3 is produced?

c) if 1.20 moles of Sb react with 2.40 moles of I2, howmuch SbI3 is produced?

d) if 1.00 moles of SbI3 are produced, what per cent yield of final product is produced?

actual yield x 100 = % yield predicted yield

Mass Relationships

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