Stoichiometry (part II) Stoichiometry (part II). 1 mole of anything = 6.022 x 10 23 units of that...

21
Stoichiomet Stoichiomet ry ry (part II) (part II)

Transcript of Stoichiometry (part II) Stoichiometry (part II). 1 mole of anything = 6.022 x 10 23 units of that...

Page 1: Stoichiometry (part II) Stoichiometry (part II). 1 mole of anything = 6.022 x 10 23 units of that thing (Avogadro’s number) = molar mass of that thing.

StoichiometryStoichiometry

(part II)(part II)

Page 2: Stoichiometry (part II) Stoichiometry (part II). 1 mole of anything = 6.022 x 10 23 units of that thing (Avogadro’s number) = molar mass of that thing.

1 mole of anything

= 6.022 x 1023 units of that thing (Avogadro’s number)

= molar mass of that thing

Thus, the molar mass of an element is numerically equal to its atomic mass (reflected on the periodic table)

molar mass = atomic mass

Review Slide:

Page 3: Stoichiometry (part II) Stoichiometry (part II). 1 mole of anything = 6.022 x 10 23 units of that thing (Avogadro’s number) = molar mass of that thing.

Mass percent of an element:

• Calculate the mass % of Fe in iron(III) oxide, (Fe2O3):

Percent Composition of Compounds

Review Slide:

Page 4: Stoichiometry (part II) Stoichiometry (part II). 1 mole of anything = 6.022 x 10 23 units of that thing (Avogadro’s number) = molar mass of that thing.

Review Problem:

In 1987 the first substance to act as a superconductor at a temperature above that of liquid nitrogen (77K) was discovered. The approximate formula of this substance is YBa2Cu3O7.

Calculate the percent composition by mass of this material

Page 5: Stoichiometry (part II) Stoichiometry (part II). 1 mole of anything = 6.022 x 10 23 units of that thing (Avogadro’s number) = molar mass of that thing.

Empirical formula - Simplest whole-number ratio

> Base the calculation on 100g of compound.

> Determine the number of moles of each element present in 100g of compound using the atomic masses of the elements present.

> Divide each value of the number of moles by the smallest of the values.

* If each resulting number is a whole number (after appropriate rounding), these numbers represent the subscripts of the elements in the empirical formula.

* If the numbers obtained in the previous step are not whole numbers, multiply each number by an integer so that the results are all whole numbers

Formulas

Page 6: Stoichiometry (part II) Stoichiometry (part II). 1 mole of anything = 6.022 x 10 23 units of that thing (Avogadro’s number) = molar mass of that thing.

Molecular formula Actual formula of the compound

= (empirical formula)n

where:

n => integer = molar mass

empirical formula mass

Formulas

Page 7: Stoichiometry (part II) Stoichiometry (part II). 1 mole of anything = 6.022 x 10 23 units of that thing (Avogadro’s number) = molar mass of that thing.

•7

e.g. #1

The composition of adipic acid is 49.3% C, 6.9% H, and 43.8% O (by mass). The molar mass of the compound is about 146 g/mol.

What is the empirical formula? What is the molecular formula?

Page 8: Stoichiometry (part II) Stoichiometry (part II). 1 mole of anything = 6.022 x 10 23 units of that thing (Avogadro’s number) = molar mass of that thing.

Chemical Stoichiometry

Stoichiometry – The study of quantities of materials consumed and produced in chemical reactions.

Stoichiometric Calculations

• Chemical equations can be used to relate the masses of reacting chemicals.

Review Slide:

Page 9: Stoichiometry (part II) Stoichiometry (part II). 1 mole of anything = 6.022 x 10 23 units of that thing (Avogadro’s number) = molar mass of that thing.

• The number of atoms of each type of element must be the same on both sides of a balanced equation.

• Subscripts must not be changed to balance an equation.

• A balanced equation tells us the ratio of the number of molecules which react and are produced in a chemical reaction.

• Coefficients can be fractions, although they are usually given as lowest integer multiples.

Notes on Balancing Chemical EquationsReview Slide:

Page 10: Stoichiometry (part II) Stoichiometry (part II). 1 mole of anything = 6.022 x 10 23 units of that thing (Avogadro’s number) = molar mass of that thing.

Limiting reactant – the reactant that is consumed first and therefore limits the amounts of products that can be formed.

• Determine which reactant is limiting to calculate correctly the amounts of products that will be formed.

Limiting ReactantsReview Slide:

Page 11: Stoichiometry (part II) Stoichiometry (part II). 1 mole of anything = 6.022 x 10 23 units of that thing (Avogadro’s number) = molar mass of that thing.

• An important indicator of the efficiency of a particular laboratory or industrial reaction.

Reaction Yield

% yield

= (actual yield / theoretical yield) x 100%

Review Slide:

Page 12: Stoichiometry (part II) Stoichiometry (part II). 1 mole of anything = 6.022 x 10 23 units of that thing (Avogadro’s number) = molar mass of that thing.

• For 1 mole of an ideal gas at 0°C and 1 atm, the volume of the gas is 22.42 L.

• STP = standard temperature and pressure 0°C and 1 atm Therefore, the molar volume is 22.42 L

at STP.

Molar Volume of an Ideal Gas

V = nRT =

P

(1 mol) 0.08206 L atm (273.2K) = 22.42L mol K

1 atm

Page 13: Stoichiometry (part II) Stoichiometry (part II). 1 mole of anything = 6.022 x 10 23 units of that thing (Avogadro’s number) = molar mass of that thing.

Calculate the volume that 0.881 moles of oxygen gas at STP will occupy.

e.g. #2

Page 14: Stoichiometry (part II) Stoichiometry (part II). 1 mole of anything = 6.022 x 10 23 units of that thing (Avogadro’s number) = molar mass of that thing.
Page 15: Stoichiometry (part II) Stoichiometry (part II). 1 mole of anything = 6.022 x 10 23 units of that thing (Avogadro’s number) = molar mass of that thing.

D = density of gas T = temperature in Kelvin P = pressure of gas R = universal gas constant

Molar Mass of a Gas

MM = Molar Mass = DRT

P

Page 16: Stoichiometry (part II) Stoichiometry (part II). 1 mole of anything = 6.022 x 10 23 units of that thing (Avogadro’s number) = molar mass of that thing.

What is the density of F2 at STP (in g/L)?

e.g. #3

Page 17: Stoichiometry (part II) Stoichiometry (part II). 1 mole of anything = 6.022 x 10 23 units of that thing (Avogadro’s number) = molar mass of that thing.

- concentration of solutions

- expressed as molar concentrations in M (mol/L)

Molarity (M)

Page 18: Stoichiometry (part II) Stoichiometry (part II). 1 mole of anything = 6.022 x 10 23 units of that thing (Avogadro’s number) = molar mass of that thing.

Oxalic Acid (H2C2O4) is used for the removal of rust (Fe2O3). Calculate the number of grams of rust that can be removed with 1 L 0f 0.10M solution of oxalic acid. The equation is shown as:Fe2O3 + H2C2O4 => Fe(C2O4)3 + H2O + H

e.g. #4a

Page 19: Stoichiometry (part II) Stoichiometry (part II). 1 mole of anything = 6.022 x 10 23 units of that thing (Avogadro’s number) = molar mass of that thing.

Which is the limiting reactant if Fe(C2O4)3 is our concerned product in e.g. #4a?

e.g. #4b

Page 20: Stoichiometry (part II) Stoichiometry (part II). 1 mole of anything = 6.022 x 10 23 units of that thing (Avogadro’s number) = molar mass of that thing.

e.g. #4c

What mass of product will be produced in e.g. #4a?

Page 21: Stoichiometry (part II) Stoichiometry (part II). 1 mole of anything = 6.022 x 10 23 units of that thing (Avogadro’s number) = molar mass of that thing.

e.g. #4d

What would the % yield of the reaction on e.g.#4a if only 2.5g of Fe(C2O4)3 are actually isolated at the end?