Chapter 3: Stoichiometry

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

• In order to understand stoichiometry, we must first understand atomic masses

Atomic Masses

• Elements occur in nature as mixtures of isotopes

• Carbon = 98.89% 12C 1.11% 13C <0.01% 14C

Carbon atomic mass = 12.01 amu

Try this…

• Find the average atomic mass of lithium.

Naturally occurring lithium is:

7.42% 6Li (6.015 amu)

92.58% 7Li (7.016 amu)

7.42(6.015) 92.58 7.016

100

6.941amu

How about this one?Fill in the blanks.

Isotope Mass (amu) Abundance27.98

4.70%29.97 3.09%

28Si29Si32Si

The mole

• A unit to count the number of molecules

• 1 mol = 6.022 x 1023

Dozen = 12

Pair = 2

Samples of 1 mole ofCu, Al, Fe, S, I, and Hg

Molar mass is the mass of 1 mole of in gramseggsshoes

marblesatoms

1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g

1 12C atom = 12.00 amu

1 mole 12C atoms = 12.00 g 12C

1 mole lithium atoms = 6.941 g of Li

For any element atomic mass (amu) = molar mass (grams)

Try this!

Which of the following 100.0 g samples contains the greatest number of atoms? a) Magnesiumb) Zincc) Silver

Rank the following according to number of atoms (greatest to least):

a) 100.0 g of silverb) 62.0 g of zincc) 21.0 g of magnesium

Molar mass

• My expectation is that you do NOT need a refresher in molar mass! So try this!!!

• What is the molar mass of isopentyl acetate (C7H14O2)?

If 1 g of C7H14O2 is released in a bee sting, how many molecules are released?

• HINT: – g = 1x10-6g– g mol molecules

• How many atoms of carbon are present?

How many C atoms in bee sting?

• HINT: each molecule of isopentyl acetate has 7 atoms of carbon

Percent Composition or Mass %

• There are two ways to describe the composition of a compound.– The number of constituents

atoms – The percent of each of its

elements (by mass)

isopentyl acetate (C7H14O2)

• Let’s look at mass percent of an element

• Find the mass % of C

molarmassof elementmass% 100%

molarmassof compoun

n

d

12.01mass% 100%

13 .18

7

0

mass%of C 64.58%

Try this one.

Consider separate 100.0 gram samples of each of the following: H2O, N2O, C3H6O2, CO2

• Rank them from highest to lowest percent oxygen by mass.

Formulas

• molecular formula = (empirical formula)n

[n = integer]

• molecular formula = C6H6 = (CH)6

• empirical formula = CH– The lowest ratio of atoms in a

molecule (mole ratio)

Working backwards

• From the percent composition, you can determine the empirical formula.

Determine the empirical formula of a compound that has the following percent composition by

mass: K 24.75%, Mn 34.77% , O 40.51%Assume 100 g sample of the compound

K

1mol Kn 24.75g 0.6330molK

39.10g K

Mn

1mol Mnn 34.77g 0.6329molMn

54.94g Mn

O

1mol On 40.51g 2.532molO

16.00g O

nK = 0.6330, nMn = 0.6329, nO = 2.532

KMnO4

0.6330K 1.0

0.6329

0.6329Mn 1.0

0.6329

2.532O 4.0

0.6329

This one has a twist!

• A compound is made of only sulfur and nitrogen. It is 69.6% S by mass. Its molar mass is 184 g/mol. What is its empirical and molecular formula?

How to “Read” Chemical Equations

2 Mg + O2 2 MgO

2 atoms Mg + 1 molecule O2 makes 2 formula units MgO

2 moles Mg + 1 mole O2 makes 2 moles MgO

48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO

NOT

2 grams Mg + 1 gram O2 makes 2 g MgO

Things to know about equations

• They are sentences• They describe what happens in

a chemical reaction• Reactants Products• They should be balanced• They have the same number of

each kind of atoms on both sides

Try thisChromium compounds exhibit a variety of bright

colors. When solid ammonium dichromate (NH4)2Cr2O7, a vivid orange compound is ignited a spectacular reaction occurs. Assume that the products are chromium (III) oxide, nitrogen gas, and water vapor. Write and balance the equation.

And this one

• At 1000oC, ammonia gas, (NH3) reacts with oxygen gas to form gaseous nitric oxide, and water vapor. This reaction is the first step in the commercial production of nitric acid by the Ostwald process. Write and balance the equation for this reaction.

StoichiometryHow much do you remember?

• Methanol burns in air according to the equation

2CH3OH + 3O2 2CO2 + 4H2O

If 209 g of methanol are used up in the combustion, what mass of water is produced?

A good FRQ for class.• Methane (CH4) reacts with the oxygen in

the air to produce carbon dioxide and water.

• Ammonia (NH3) reacts with the oxygen in the air to produce nitrogen monoxide and water.

• What mass of ammonia would produce the same amount of water as 1.00 g of methane reacting with excess oxygen?

1.4g ammonia

Limiting Reactant

• Reactant that determines the amount of product formed.

• The one you run out of first.• Makes the least product.• Three methods to solve:

– Ratio method shown in text– Have versus need – Do 2 stoichiometry problems to find the

amount of product. The reactant that produces the LEAST amount is the limiting reagent.

Let’s try together!

• Nitrogen gas can be prepared by passing gaseous ammonia over solid copper(II) oxide at high temperatures. The other products of the reaction are solid copper and water vapor. If a sample containing 18.1g of NH3 is reacted with 90.4g of CuO, which is the limiting reactant? How many grams of nitrogen gas will be formed?

• Have versus need method:g NH3mol NH3 mol CuO neededg CuO needed

2NH3 + 3CuO N2 +3Cu + 3H2O

• The problem says the we HAVE 90.4g of CuO. • Using ALL of the NH3 given we NEED 127g of

CuO; therefore, the CuO is the limiting reactant since we don’t have enough of CuO

• You can do this starting with the CuO as well. Let’s double check our work.

33

3 3

1molNH 3molCuO 79.55gCuO18.1gNH 127gCuOneeded

17.04gNH 2molNH 1molCuO

2NH3 + 3CuO N2 +3CuO + 3H2O

• The problem says the we HAVE 18.1g of NH3. • Using ALL of the CuO given we NEED 12.9g of

NH3; therefore, the CuO is the limiting reactant since we have an excess of NH3.

• There is an excess of 5.2g of NH3

3 33

3

2molNH 17.04gNH1molCuO90.4gCuO 12.9gNH needed

79.55gCuO 3molCuO 1molNH

2NH3 + 3CuO N2 +3CuO + 3H2O

• Do the stoichiometry problem twice. The reactant that gives you the LEAST amount of product is the limiting reactant.

• The CuO is the limiting reactant. This confirms our results from the have versus need method.

2 22

2

1molCuO 1molN 28.02gN90.4gCuO 10.6gN made

79.55CuO 3molCuO 1molN

3 2 23 2

3 3 2

1molNH 1molN 28.02gN18.1gNH 14.9gN made

17.04gNH 2molNH 1molN

Try one on your own.

• When 124 grams of aluminum are reacted with 601 grams of iron(III) oxide, aluminum oxide and solid iron are produced. Calculate the mass of aluminum oxide formed.

Excess Reactant (reagent)

• The reactant you don’t run out of.• The amount of stuff you make is the

yield.• The theoretical yield is the amount you

would make if everything went perfect.• The actual yield is what you make in the

lab.

Percent Yield

actual%yield 100%

theoreticalOR

what yougot%yield 100%

what youcouldhave

Try this

Methanol is the simplest alcohol. It is used as a fuel in race cars and is a potential replacement for gasoline. Suppose 68.5kg CO(g) is reacted with 8.6kg H2(g). Calculate the theoretical yield of methanol.

If 3.57x104g CH3OH is actually produced, what is the percent yield?