Stoichiometry

The Mole The quantity of a substance is measured in metric units, called moles. A mole of a substance is the quantity of that substance that contains the same number of particles as Avagadro’s number.

Avagadro’s number



6.023 · 1023 = atoms

1 mol.

-

particles

-

molecules

-

formula units

-

ions

6.023 · 1023 is equal to 6.02300000000000000000.

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Molar Mass Molar Mass is the mass of one mol of a substance. You determine molar mass using the atomic mass values in the periodic table.

Example: What is the molar mass of Water – H2O? Hydrogen  2 · 1.01 = 2.02 Oxygen  1 · 16.00 = 16.00 = 18.02 Therefore, water (H2O) has a molar mass of 18.02 g.

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Percent Composition of Molar Mass To find the percent composition of molar mass, simply take the total of each element.

Example: take 117.3g of Potassium, divide it by the overall total of molar mass (212.27g), and then multiply it by 100. This is the percent composition of molar mass for whichever element you are solving for.

K  3 · 39.00 = 117.30 P  1 · 30.97 = 30.97 O  4 · 16.00 = 64.00 Total: 212.27g.

K  117.30 ÷ 212.27 · 100 = 55.26% P  30.97 ÷ 212.27 · 100 = 14.59% O  64.00 ÷ 212.27 · 100 = 30.15%

Ensure that all percent totals for each element equal a combined number of 100, or a number close to 100, and remember to round two places after the decimal.

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The Empirical Formula The empirical formula shows the lowest whole number ratio of the elements in a compound (the simplest formula of a compound).

1. When you are given the percent composition, you assume that the percentage is grams (g). So 14.9% = 14.9 grams.

2. Convert the grams to mols.

14.9 g of H

=

1 mol 1.01 g

=

14.8 mols of H

3. Find the ratio.

Example: H14.8 O2.0

4. Divide both sides by the lowest number that is visible. So 2.0 is the lowest number visible.

Example: H14.8 ÷ 2.0

O2.0 ÷ 2.0 Total: H7.4 and O1 H7O is the final answer

Things To Remember:

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You may have to multiply your “total” to obtain a final answer that does not have decimals.

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The Molecular Formula of a Compound The molecular formula, also known as the actual formula, describes the number of atoms of each element that make up a molecule.

Molecular formula subscripts = n x Empirical formula subscripts. (n = 1, 2, 3...)

Example: The empirical formula of ribose (sugar) is CH2O. In a separate experiment, using a mass spectrometer, the molar mass of ribose was determined to be 150 g/mol. What is the molecular formula of ribose?

1. Find the molar mass.

CH2O C -> 1 x 12.0 = 12.0 H -> 2 x 1.01 = 2.02 O -> 1 x 16.0 = 16.00 Total: 30.02g.

2. Divide the molar mass by the molar mass that you've found.

150 g/mol ÷ 30.02 g/mol = 5 (rounded)

3. Multiply the answer you obtain from Step 2, with the compound formula given.

Molecular Formula = (CH20)x5 = C5H10O5

C5H10O5 is the final answer, and is also known as the molecular formula of Ribose (sugar).

Things to Remember: •

If a percent is given in the question, simply use the Empirical Formula before going into the steps given above for the Molecular Formula of a Compound.

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Stoichiometric Calculations Stoichiometry is the solution of problems involving specific quantities of substances. When substances react, a chemical equation is used to describe the change. The symbols and formulas indicate what substances are involved in the change. The coefficients in the equation provide a mole ration. The mole ratio tells us how many moles of one particular substance are used up in the reaction.

Example: How many grams of oxygen would be needed to react with 60g of carbon to produce carbon dioxide?

1. Write your equation.

O2 + C  CO2

2. Balance your equation.

O2 + C  CO2

3. Decide the known and unknown quantities.

O2 + C  CO2 Xg

60g

You know that you are looking for the amount of grams of Oxygen that would be needed to react with 60 g of Carbon, so you know that Oxygen is the unknown quantity, while Carbon is the known quantity.

4. Convert your grams to mols (if needed).

5. Do the mole-mole ratio.

Answer from Step 4 = Coefficient of unknown quantity ÷ Coefficient of known quantity

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1 mol of O2 5 mols of C

=

1 mol of C

=

5 mols of O2

6. Convert mols to grams.

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Limiting Reactants (Reagents) In most chemical reactions, only one of the reactants is completely consumed. Any reactant which is not completely used up is present in excess.

Example: Suppose that 14g of Fe and 12g of S are available for reaction. How many grams of Iron (II) Sulfide are produced?

1. Write out the chemical equation and balance it. To do this, look at what the problem is telling you.

In this formula, it is saying that you have 14g of Fe and 12g of S, so:

Fe + S

It is also telling us that Iron (II) Sulfide is produced, so:

Fe + S  FeS

When a question says that it has “produced” something, it simply means that that chemical formula is on the right side of the yield sign (). Similarly, to know what to write on the left side of the , which is Fe + S in this problem, you usually use the chemical formulas in which the problem tells you that one chemical is added to another, or that, one or more chemical is available, and together they produce another chemical.

2. Find the known and unknown quantities.

In this equation, you know that both Fe and S are known, and the unknown is FeS. So:

Fe + S  FeS 14g 12g

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Limiting Reactants (Reagents) Continued… 3. Find the Limiting Reagent.

Take each known chemical, in this case Fe and S, and convert their known grams to mols.

14 g of Fe

=

12 g of S

=

1 mole 55.8 g

1 mole 32.1 g

=

=

0.25 moles of Fe.

0.37 moles of S.

Now that you have converted grams to moles you can find the Limiting Reagent.

Fe: (Limiting Reagent)

S:

0.25 1

0.37 1

To find the Limiting Reagent, take the answer for each chemical formula that you’ve obtained in Step 3 and divide it by each of the chemical formulas coefficient. For instance: Fe + S  FeS, 1 is the coefficient of Fe, as it is for S, so you divide by one. If the coefficient was 2, then you would divide by two, and so on.

Now that you’ve divided each one, the chemical that has the smallest number is our Limiting Reagent. So in this case Fe is our Limiting Reagent.

Things to Remember:

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Sometimes, you may be asked to find more than just the Limiting Reagent.

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Questions for Limiting Reactants (Reagents) 1. How many grams of sulfuric acid can be prepared from 50g of sulfur dioxide, 15g of oxygen, and an unlimited quantity of water?

2SO2 + O2 + 2H2O  2H2SO4

1 mole

50 g of SO2 =

15 g of O

1 mole

=

2

=

32.0 g

0.78

SO2:

=

64.1 g

2

O: 2

=

0.78 moles of SO2

0.47 moles of O

2

0.39 (Limiting Rea gent)

0.47 1

=

0.47

You’ve found the limiting reagent, now you must use Stoichiometry to complete the rest of the problem.

You must find the mole-to-mole ratio, as shown above, to find how many moles of H204 there are. You will always use the Limiting Reagent (SO2) in the mole-mole ratio. So what you are looking for goes on the top, and the limiting reagent goes on the bottom, as shown below.

0.78 moles of SO

2

=

2 Moles H204 2 Moles SO

=

0.78 moles of H 0

=

76.5g of H 0

2 4

2

Now, convert moles to grams.

0.78 moles of H 0 = 2 4

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98.12 g 1 mole

2 4

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The Yield of a Chemical Reaction The Theoretical Yield is the maximum amount of a product that can form in a chemical, and assumes that all of the Limiting Reagents has reached to form the product, whereas the Actual Yield is the amount of a product that is “actually” or experimentally obtained from a chemical reaction.

Actual Yield

Percent Yield =

x 100%

Theoretical Yield

Example: Suppose 7g of AgNO3 is added to a solution which contains an excess of dissolved KBr. If 7.32g of AgBr is obtained, what is the percent yield?

AgNO3 + KBr  AgBr + KNO3

7 g of AgNO

3

=

0.04 moles of AgNO3 =

1 mole 169.91 g

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3

=

1 mole of AgNO3

Percent Yield =

1:

0.04 moles of AgNO

1 mole AgBr

0.04 moles of AgBr

UNIT

=

=

0.04 moles of AgBr

187.8g

=

1 mole

(Theoretical Yield)

7.32 7.51

7.51 g of AgBr

x 100% = 97.5%

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Gas Stoichiometry Gas Stoichiometry involves volumes of gases, with one mol of gas taking up a certain volume depending on the temperature and pressure.

1. STP (Standard Temperature and Pressure: O°C and 101.3 kPa (1 mol of any gas occupies 22.4 L of space).

2. SATP (Standard Ambient Temperature and Pressure: 25°C and 100 kPA (1 mol of any gas occupies 24.8 L of space).

Gas Stoichiometry is essentially the same thing as basic Stoichiometric calculations, with the exception of possibly adding a step, depending on what the problem is asking.

Example: When 6.25 mols of butane burns with oxygen gas, what volume (in litres) of water vapour at SATP is produced?

Unbalanced: C4H10 + O2 ---> CO2 + H20 Balanced: C4H10 + 13O2 ---> 4CO2 + 5H20 Find known and unknown quantities: C4H10 + 13O2 ---> 4CO2 + 5H20 6.25 mols

XL

There is no need to convert anything because you are working with mols; however, if it were 6.25 grams, you would need to convert the grams to mols.

6.25 mols of C4H10 =

1 mol of H2O 1 mol of C4H10

= 31.25 mols H2O

This step, which in basic Stoichiometry is usually converting mols to grams, is different in Gas Stoichiometry; you are now converting mols to litres. The problem asks for SATP, so you know to use 24.8 L. If the question asked for STP the 24.8 L would be 22.4 L.

31.25 mols H2O =

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24.8 L 1 mol

= 775 L of H2O

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Things to Remember: •

If two known quantities are present, then you must find the Limiting Reagent.

•

You may need to convert litres to mols, and then mols to litres.

•

If two unknown quantities are present, then simply do a mole-to-mole ratio with each of the unknown quantities with the known quantity.

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