Stoichiometric Calculations

Stoichiometry Stoichiometry

A. Proportional A. Proportional RelationshipsRelationshipsA. Proportional A. Proportional RelationshipsRelationships

I have 5 eggs. How many cookies can I make?

3/4 c. brown sugar1 tsp vanilla extract2 eggs2 c. chocolate chipsMakes 5 dozen cookies.

2 1/4 c. flour1 tsp. baking soda1 tsp. salt1 c. butter3/4 c. sugar

5 eggs 5 doz.

2 eggs= 12.5 dozen cookies

Ratio of eggs to cookies

A. Proportional A. Proportional RelationshipsRelationshipsA. Proportional A. Proportional RelationshipsRelationships

StoichiometryStoichiometry• mass relationships between

substances in a chemical reaction• based on the mole ratio

Mole RatioMole Ratio• indicated by coefficients in a

balanced equation

2 Mg + O2 Mg + O22 2 MgO 2 MgO

What can we do with Stoichiometry?What can we do with Stoichiometry?What can we do with Stoichiometry?What can we do with Stoichiometry?

For the generic equation:

RA +RB → P1 + P2

Given the… …one can find the…

Amount of RA (or RB) Amount of RA (or RB) that is needed to react with it

Amount of RA or RB Amount of P1 or P2 that will be produced

Amount of P1 or P2 you need to produce

Amount of RA &/or RB

you must use

Given the equation…Given the equation…Given the equation…Given the equation…

2TiO2 + 4Cl2 + 3C → 2TiCl4 + CO2 + 2CO How many mol chlorine will react with 4.55

mol carbon? X mol Cl2 = 4.55 mol C (4mol Cl2/ 3mol C)

6.07 mol Cl2 will react with 4.55 mol carbon

Given the same Given the same equation…equation…Given the same Given the same equation…equation…

2TiO2 + 4Cl2 + 3C → 2TiCl4 + CO2 + 2CO What mass titanium (IV) oxide will react

with 4.55 mol carbon? X g TiO2 =

4.55 mol C(2 mol TiO2/3 mol C)(79.9 g TiO2/1 molTiO2)

242 g TiO2 will react with 4.55 mol carbon

And with the same And with the same equation…equation…And with the same And with the same equation…equation…

2TiO2 + 4Cl2 + 3C → 2TiCl4 + CO2 + 2CO How many molecules titanium (IV)

chloride can be made from 115 g titanium (IV) oxide?

X m’c TiCl4 = 115g TiO2 (1mol TiO2/79.9g TiO2)(2 mol TiCl4/2 mol TiO2)

(6.02 x 1023 m’c TiCl4/1 mol TiCl4)

8.7 x 1023 m’cules TiCl4 can be made from 115 g TiO2

B. Stoichiometry StepsB. Stoichiometry StepsB. Stoichiometry StepsB. Stoichiometry Steps

1. Write a balanced equation.2. Identify known & unknown.3. Line up conversion factors.

• Mole ratio - moles moles• Molar mass - moles grams• Molarity - moles liters soln• Molar volume - moles liters gas

Core step in all stoichiometry problems!!

• Mole ratio - moles moles

4. Check answer.

1 mol of a gas=22.4 Lat STP

C. Molar Volume at STPC. Molar Volume at STPC. Molar Volume at STPC. Molar Volume at STP

Standard Temperature & Pressure0°C and 1 atm

C. Molar Volume at STPC. Molar Volume at STPC. Molar Volume at STPC. Molar Volume at STP

Molar Mass(g/mol)

6.02 1023

particles/mol

MASSIN

GRAMSMOLES

NUMBEROF

PARTICLES

LITERSOF

SOLUTION

Molar Volume (22.4 L/mol)

LITERSOF GASAT STP

Molarity (mol/L)

Stoichiometry “Pop” QuizStoichiometry “Pop” QuizStoichiometry “Pop” QuizStoichiometry “Pop” Quiz Iron (III) reacts with water…

• Write the complete balanced equation• How many liters of hydrogen (@ STP) will be produced from

8 moles of iron? • How many grams of water are needed to produce 68 grams

of Iron (III) oxide?• 5.2 x 1025 particles of iron will produce how many moles of

hydrogen? For the complete combustion of butane (C4H10)…

• Write the complete balanced equation• In a room at 0°C and 1 atm pressure, how many m’cules

oxygen react with 632 L butane?• If 316 L liters of carbon dioxide gas are produced, how much

butane (in moles) was combusted? Silver nitrate reacts with sodium chloride…

• Write the complete balanced equation (including states of matter)

• How much precipitate (in g) is formed when 1.5 L of 0.10M silver nitrate is used?

• How many molecules of sodium chloride do you need to react with that amount of silver nitrate?

The Limiting Reactant The Limiting Reactant The Limiting Reactant The Limiting Reactant

A balanced equation for making a Big Mac® might be:

3 B + 2 M + EE B3M2EE

With… …and……one can

make…

30 Mexcess B and

excess EE15 B3M2EE

30 Bexcess M and

excess EE10 B3M2EE

30 M30 B and excess

EE10 B3M2EE

A balanced equation for making a tricycle might be:

3 W + 2 P + S + H + F W3P2SHF

With… …and……one can

make…

50 Pexcess of all other reactants

25 W3P2SHF

50 Sexcess of all other reactants

50 W3P2SHF

50 P50 S and excess

of all other reactants

25 W3P2SHF

Solid aluminum reacts w/chlorine gas to yield solid aluminum chloride.

2 Al(s) + 3 Cl2(g) 2 AlCl3(s)

If 125 g aluminum react w/excess chlorine, how many g aluminum chloride are made?

= 618 g AlCl3

3

333 AlClmol 1

AlClg 133.5

Almol 2 AlClmol 2

Alg 27 Almol 1

Alg 125 AlClgX

If 125 g chlorine react If 125 g chlorine react w/excess aluminum, how w/excess aluminum, how many g aluminum chloride many g aluminum chloride are made?are made?

If 125 g chlorine react If 125 g chlorine react w/excess aluminum, how w/excess aluminum, how many g aluminum chloride many g aluminum chloride are made?are made?

= 157 g AlCl3 If 125 g aluminum react w/125 g

chlorine, how many g aluminum chloride are made?

157 g AlCl3 We’re out of Cl2.

3

3

2

3

2

223 AlClmol 1

AlClg 133.5

Cl mol 3 AlClmol 2

Cl g 71Cl mol 1

Cl g 125 AlClgX

limiting reactant (LR): the reactant that runs out first

• Amount of product is based on LR.

Any reactant you don’t run out of is an excess reactant (ER).

ExampleLimiting Reactant

Excess Reactant(s)

Big Macs buns meat

tricycles pedals W, S, H, F

Al / Cl2 /

AlCl3

Cl2 Al

How to Find the Limiting How to Find the Limiting ReactantReactantHow to Find the Limiting How to Find the Limiting ReactantReactant

For the generic reactionRA + RB P

Assume that the amounts of RA and RB are given.

Should you use RA or RB in your calculations?

How to find LR steps…How to find LR steps…How to find LR steps…How to find LR steps…

1. Calc. # of mol of RA and RB you have.

2. Use the mole ratio from the balanced equation to calculate the required amount of the other reactant.

3. If resulting amount is less than what is given, that reactant is the LR.

% Yield% Yield% Yield% Yield

molten + solid molten + solidsodium aluminum aluminum sodium oxide oxide Al3+ O2– Na1+ O2–___Na(l) + ___Al2O3(s) ___Al(l) +___Na2O(s)

6 Na(l)+ 1 Al2O3(s) 2 Al(l) + 3 Na2O(s)

Find mass of aluminum produced if you start w/575 g sodium and 357 g aluminum oxide.

= 189 g Al

% Yield% Yield% Yield% Yield

This amount of product is the theoretical yield.• amt. we get if reaction is perfect• found by calculation

Now suppose that we perform this reaction in the lab and get only 172 grams of aluminum. Why?• couldn’t collect all Al• not all Na and Al2O3 reacted• some reactant or product spilled and

was lost

% Yield Calculation:% Yield Calculation:% Yield Calculation:% Yield Calculation:

% yield can never be > 100%

100 x yieldltheoretica

yieldactual yield%

Find % Yield for the Previous Find % Yield for the Previous ProblemProblemFind % Yield for the Previous Find % Yield for the Previous ProblemProblem

91.0% 100 x Alg 189 Alg 172

100 x yld.theo.

yld.act. yield%