BellworkNovember 5th

• Dig around in your brain and fill in the following…–1 mole = ___________________ atoms

–1 mole = ___________________ liters

–1 mole = ___________________ g H2O

–3 mole = ________________ molecules

CHEMISTRY-1 CHAPTER 9

Stoichiometry

Just What is a Chemical Ratio?• A mole ratio is like making a s’more.

• The “chemical reaction for a s’more would be…– 3 chocolate squares + 1 marshmallow + 2 graham crackers 1 s’more

• If you want to make 3 s’mores, how many marshmallows would you need?

• 3 s’mores = 3 marshmallows

• If you have 27 chocolate squares, how many smores can you make?

• 3 chocolate squares = 1 s’more so 27 chocolate squares = 9 s’mores

9.1 Stoichiometry – study of calculations of quantities in chemical reactions using

balanced chemical equations.

2Mg + O2 2MgO

2 moles Mg + 1mole O2 2 moles MgO

• The mole ratios can be obtained from the coefficients in the balanced chemical equation.

• What are the mole ratios in this problem?

• Mole ratios can be used as conversion factors to predict the amount of any reactant or product involved in a reaction if the amount of another reactant and/or product is known.

2Mg + O2 2MgO

Mole Ratio! What’s that mean?

Well, a stoichiometry problem gives you an amount of one chemical and asks you to solve for a different chemical.

To get from one type of chemical to another, a MOLE RATIO must be found between the two chemicals. You get the MOLE RATIO from the

BALANCED CHEMICAL EQUATION!

Moles of substance

given

Mass of substance

given

R.P. of substance

given

Volume of substance

given

Atoms, formula units, molecules

grams

L, mL

mol

MM1mol

22.4

L1m

ol

6.02 x 10 23

1mol

R.P. of substance wanted

Atoms, formula units, molecules

Mass of substance wanted

grams

Volume of substance wanted

L, mL

Moles of substance wanted

mol

22.4 L

1mol

MM 1mol

6.02

x 1

023

1mol

Coefficients in a balanced equation!

Always follow the road map…

Do not make up new paths!

A balanced chemical equation tells the quantity of reactants and products as well as what they are.

2 mol 1 mol 2 mol

*the coefficients are*

The MOLE RATIOMOLE RATIO is your mechanism of transition between the chemical that is your

starting given and the chemical you are solving for.

The MOLE RATIO is the bridge between the two

different chemicals

given

moles moles

? want(given) (want)

MoleRatio

9.2

• Lets get back to those s’mores. If you have the following reaction…

• 3 chocolate squares + 1 marshmallow + 2 graham crackers 1 s’more

• You have 325 chocolate squares and you want to make the maximum number of s’mores possible. Show a calculation for this reaction.

• 325 c.s. 1 s’more = 108.3333 or 108 s’mores• 3 c.s.

Mole - Mole Calculations – Mole ratios are used to calculate the number of moles of product from the given number of moles of reactant, or vice versa.

EXAMPLE

• Calculate the number of moles of Na2O that will be produced when 5.00 moles of Na completely react with oxygen gas.

4Na + O2 2Na2O

5.00 mol Na

4 mol Na

2 mol Na2O = 2.50 mol Na2O

mol-mol ratio

Mole-Mass Calculations:

• We use mole ratios from balanced chemical equations and the molar conversions from Ch. 7 to calculate amounts (grams) of substances needed or produced in chemical reactions.

EXAMPLE

• How many grams of KClO3 must decompose to produce KCl and 1.45 moles O2?

2KClO3 → 2KCl + 3O2

1.45 moles O2

3 mol O2

2 mol KClO3

1 mol KClO3

122.548 g KClO3=

118 g KClO3

MM

mol-mol ratio

Mass-Mass Calculations:

• Same as mole-mass calculations, with an additional step of converting mass to moles.

EXAMPLE• When 24.0 g of Na are mixed with Cl2, are

52.0 g of NaCl produced? Explain.

24.0 g Na22.990 g Na

MM

1 mol Na

2 mol Na

Na + Cl2 → NaCl2 2

mol-mol ratio

2 mol NaCl

1 mol NaCl

MM

58.443 g NaCl

61.0 g NaCl

=

Are 52.0 g produced? Actually more than 52.0 g were produced.

• CaCO3, limestone, is heated to produce calcium oxide, CaO, and CO2. What mass of limestone is required to produce 156.0 g of CaO?

156.0 g CaO

56.077 g CaO

MM

1 mol CaO

1 mol CaO

CaCO3 (s) CaO (s) + CO2 (g)

mol-mol ratio

1 mol CaCO3

1 mol CaCO3

MM

100.086 g CaCO3

278.4 g CaCo3=

BellworkNovember 6th

• Magnesium metal reacts with oxygen in a synthesis reaction. Write and balance the equation, then fill in the chart concerning the particles and mass.

Total Particles of Reactants (look at coefficient)

Total Particles of Products

Total Moles of Reactants

Total Moles of Products

Total Mass of Reactants

Total mass of Products

BellworkNovember 7th

• A chemist combines 6.32g of C2H2 and 12.2g of oxygen. How many grams of carbon dioxide are produced?

A quick reminder…

• Let’s quickly remind ourselves of just how awesome a mole is!

• 1 mole = ________________particles

• 1 mole = ________________ liters

• 1 moles = _______________ grams

6.02 x 1023

22.4

The mass from the periodic table!

• At STP, how many grams of oxygen are needed to produce 19.8 liters SO3 according to the balanced equation below?

19.8 L SO3

22.4 L SO3

Volume @ STP

1 mol SO3

2 mol SO3

2SO2 (g) + O2 (g) 2SO3 (g)

mol-mol ratio

1 mol O2

1 mol O2

MM

31.998 g O2

14.1 g O2=

Combination Calculations:

BellworkTuesday, November 10th • A student drops a 3.40g piece of zinc

into hydrochloric acid. How many liters of hydrogen gas are produced if the reaction occurs at STP?

BellworkTuesday, November 10th • How many grams of aluminum sulfate

are produced if 23.33 g Al reacts with 74.44 g CuSO4?

9.3Reactants in Excess and Limiting Reactants

• Limiting Reactant - the reactant that is used up in a reaction. When you run out of a reactant, the reaction stops and no more product is formed.

• Excess Reactant - the reactant that is not used up in a chemical reaction.

• The limiting reactant determines the amount of product formed.

Important

Back to S’mores

• The s’mores can teach us one more lesson; the lesson of limiting reagents. Lets look at our balanced equation one more time.

• 3 chocolate squares + 1 marshmallow + 2 graham crackers 1 s’more

• If your friend is throwing a slumber party and has 137 chocolate squares and 47 marshmallows, how many total smores can be made.

• You will need to do the calculation with BOTH starting reactants.

• 137 choc.squares 1 s’more = 45.6 s’mores

• 3 choc.squares

• 47 marshmallows 1 s’more = 47 s’mores

• 1 marshmallow

• So, in this case, what LIMITS our production of s’mores?

• Chocolate squares! So chocolate squares are the limiting reagent!

Steps to Determining the Limiting Reactant

1. Find out how much could be produced from each amount of reactant.

2. Which ever produces less is limiting the amount produced. It is therefore the Limiting Reactant.

EXAMPLE• If 6.70 mol of Na reacts with 3.20 mol Cl2,

what is the limiting reagent? How many moles of NaCl will be produced?

Na + Cl2 → NaCl2 2

6.70 mol Na

3.20 mol Cl2

2 mol Na

2 mol NaCl

1 mol Cl2

2 mol NaCl

=

=

6.70 mol NaCl

6.40 mol NaCl

Which amount produced is less?

This is the amount produced, and determines which is the limiting reactant

Cl2 is the limiting reactant

• To determine how much excess reactant is left after a reaction, subtract how much of the excess reactant reacted from how much excess reactant you started with.

EXAMPLE• When 0.500 mole of aluminum reacts with

0.720 mole of iodine, to form aluminum iodide, AlI3(s), what mass of AlI3 is produced? How much of the excess reactant will remain?

I will separate this problem into two different steps.

1.Find the limiting reactant and amount produced (as in previous example)

2.Find amount in excess (left over)

1. When 0.500 mole of aluminum reacts with 0.720 mole of iodine, to form aluminum iodide, AlI3(s), what mass of AlI3 is produced?

Al + I2 → AlI32 23

0.500 mol Al

0.720 mol I2

2 mol Al

3 mol l2

2 mol AlI3

2 mol AlI3

1 mol AlI3

407.7 g AlI3 = 204 g AlI3

1 mol AlI3

407.7 g AlI3 = 196 g AlI3

amount produced

Limiting Reactant? I2

2. How much of the excess reactant will remain?

First…take the Limiting Reactant and the amount with which you began……then convert to find the amount that reacted with it…

0.720 mol I2

3 mol l2

2 mol Al

…then subtract the amount used from the amount given (starting amount)

= 0.480 mol Al

0.500 mol Al 0.480 mol Al- = 0.020 mol Al

this is the excess (left over)

Al + I2 → AlI32 23

given in the problem

BellworkWednesday, November 11th

• 5.9 L of carbon dioxide is combined with 8.4 g MgO in a synthesis reaction to form magnesium carbonate. How many grams of magnesium carbonate is created in this reaction?

Percent Yield

• Theoretical Yield – The maximum amount of product that could be formed from given amounts of reactants. (What you should get) (You calculate this using dimensional analysis)

• Actual Yield - amount of product actually obtained in the reaction (What you really get) (You either get this in a lab, or are given it in a problem)

• Percent Yield describes how much product was actually made in the lab versus the amount that theoretically could be made.

Percent yield tells you how close you were to the 100% mark.

• Reactions do not always work perfectly. Experimental error (spills, contamination) often means that the amount of product made in the lab does not match the ideal amount that could have been made.

% yield = Actual Yield x 100

Theoretical Yield

Amount you actually obtain in an experiment

Amount you should obtain in an experiment (this we will calculate)

EXAMPLE

12.5 g of copper are reacted with an excess of chlorine gas, then 25.4 g of copper(II) chloride are obtained. Calculate the theoretical yield and the percent yield.

First write the balanced equation, then find the

theoretical yield (amount you “should get”)

12.5 g Cu

63.546gCu

1 mol Cu

First write the balanced equation, then find the theoretical yield (amount you “should get”)

Cu + Cl2 → CuCl2

1 mol Cu

1 mol CuCl21 mol CuCl2

134.452 g CuCl2

MM mol-mol ratio MM= 26.4 g CuCl2

this is the theoretical yield

Now do the % yield calculation

% yield = Actual Yield x 100

Theoretical Yield

% yield = 25.4 g CuCl2

26.4 g CuCl2=X 100

from calculation

given in problem

96.2 %

When I was a sophomore in college, we had a lab where we were supposed to isolate caffeine from tea leaves. Most of my caffeine was washed down the drain in a freak accident. Although I should have had 5.0 g of caffeine, I only ended up with 0.040 g of caffeine and a bad grade on the lab. What was my percent yield?

0.040 g

5.0 gX 100 = 0.80 %

What is the theoretical yield if 5.50 grams of hydrogen react with nitrogen to form ammonia?

H2 + N2 NH33 2

5.50 g H2

2.016 g H2

1 mole H2

3 moles H2

2 moles NH317.031 g NH3

1 mol NH3

= 31.2 grams NH3 = Theoretical Yield!!!!

Only 20.4 grams of ammonia is actually produced in the lab. What is the percent yield?

20.4 g NH3

31.2 g NH3

X 100 = 65.4 % yield