Chapter 6 Chemical Calculations

Chapter 6 Chemical Calculations

Submicroscopic

Chapter Outline

1. Formula Masses (Ch 6.1)

2. Percent Composition (supplemental material)

3. The Mole & Avogadro’s Number (Ch 6.2)

4. Molar Mass (Ch 6.3)

5. Chemical Formula & Mole Calculations

(Ch 6.4 & 6.5)

6. Empirical & Molecular Formula

(supplemental material & Lab Exp 8)

7. Chemical Equations & Stoichiometric

Calculations (Ch 6.6, 6.7, 6.8)

8. Heat of Reaction (supplemental material)

9. Percent Yield (supplemental material)

10. Limiting Reagent (supplemental material)

1. Formula Masses

(Ch 6.1)

Molecular mass/weight – sum of atomic masses of

the atoms in a molecule.

Example:

________

M.W. = 46.0 amu

C: 2 x 12.0 amu = 24.0 amu

H: 6 x 1.0 amu = 6.0 amu

O: 1 x 16.0 amu = 16.0 amu

46.0 amu

Formula mass/weight – sum of atomic masses of

atoms in an ionic substance (formula unit, NOT a

molecule).

Example: Ammonium sulfide _____________

F.W. = 68.1 amu

N: 2 x 14.0 amu = 28.0 amu

H: 8 x 1.0 amu = 8.0 amu

S: 1 x 32.1 amu = 32.1 amu

68.1 amu

Due to its offensive smell, ammonium sulfide it is the active ingredient in a variety of foul pranks, including the common stink bomb.

2. Percent Composition

(supplemental material)

% Composition = # of g of each element

in 100 g of a compound

= mass of element x 100

total mass

Example: Ammonium sulfide (NH

₄

)

₂

S

F.W. = 68.1 amu

% N = 2 x 14.0 amu N x 100 = 41.1

11

% N

68.1 amu (NH

₄

)

₂

S

% H = 8 x 1.0 amu H x 100 = 11.7

47

% H

68.1 amu (NH

4

)

2

S

% S = 32.1 amu S x 100 = 47.1

36

% S

68.1 amu (NH

₄

)

₂

S

3. The Mole & Avogadro’s Number

(Ch 6.2)

1 dozen = 12 objects

1 ream = 500 sheets

1 mole = _________

objects

6.02

21415

x 10

23

Avogadro’s Number

Sample problem:

How many He atoms are in 2.55 moles of He?

information x _________ = information

given factor sought

2.55 mole He x 6.02 x 1023 He atoms = __________ He atoms

4. Molar Mass

(Ch 6.3)

1 mole

12

C = exactly 12 g

12

C

demo sample

Molar mass of

12

C = __________

We now have three conversion factors to relate moles, number of atoms, and mass of Carbon:

6.02 x 10

23

C atoms 12 g C

1 mole 1 mole

6.02 x 10

23

C atoms

Figure 6.5

1 mole of S, Zn, C, Mg, Pb, Si, Cu, Hg

(start counterclockwise from yellow S and Hg in center)

12_{C is our standard and when we compare it to other}

elements, we find that there are Avogadro’s number of atoms of any element in a sample whose mass in grams is numerically equal to its atomic weight.

Mg = 24.31 amu

1 mole Mg = 24.31 g Mg = 6.02 x 1023 atoms Mg Pb = ________ amu

Now we can do the same for ionic compounds as well as for molecules because the molar mass is the mass (in grams) of a substance that is numerically equal to the substance’s formula mass.

Ammonium sulfide (NH4)2S formula mass = 68.1 amu

68.1 g (NH4)2S = 1 mole

(molar mass or formula weight, F.W.: ____________)

68.1 g (NH4)2S = 6.02 x 1023 formula units of (NH4)2S

Carbon dioxide CO2 formula mass = 44.01 amu

44.01 g CO2 = 1 mole

(molar mass or molecular weight, M.W.: _____________)

Sample Problem:

If 7.50 moles of ammonia, NH3, are required for a certain

experiment, what mass of ammonia is needed? Formula mass = 3 x 1.0 (H) + 14.0 (N) = 17.0 amu 1 mole NH3 (molar mass) = 17.0 g NH3

7.50 moles NH3 x 17.0 g NH3 = _______ g NH3

5. Chemical Formula & Mole Calculations

(Ch 6.4 & 6.5)

The subscripts in a chemical formula give the number of moles of atoms present in 1 mole of the substance:

Example: Ammonium sulfide (NH4)2S

For N: 2 moles of N atoms or 1 mole (NH4)2S

1 mole (NH4)2S formula units ___ moles N

For H: 8 moles of H atoms or 1 mole (NH4)2S

1 mole (NH4)2S formula units ___ moles H

For S: 1 mole of S atoms or 1 mole (NH4)2S

Sample calculation:

How many H atoms are in 35.6 g of (NH4)2S?

(NH4)2S formula mass = 68.1 amu

1 mole (NH4)2S = 68.1 g

1 mole (NH4)2S = ___ moles H atoms

1 mole H atoms = 6.02 x 1023 H atoms

Strategy:

mass (NH4)2S → moles (NH4)2S → moles H → atoms H

35.6 g (NH4)2S x 1 mol (NH4)2S x 8 mol H x 6.02 x 1023 H

68.1 g (NH4)2S 1 mol (NH4)2S 1 mol H

Fig 6.7 “Transitions” allowed in solving chemical-formula bases problems:

Drill problem:

How many g of (NH4)2S are required to obtain ___ moles of NH4+?

(NH4)2S = 68.1 g/mol (molar mass)

moles NH4+ → mole (NH4)2S → g (NH4)2S

0.50 moles NH4+ x 1 mole (NH4)2S x 68.1 g (NH4)2S

2 moles NH4+ 1 mol (NH4)2S

6. Empirical & Molecular Formula

(supplemental material & Lab Exp 8)

_______ C

₆

H

₁₂

O

₆

Molecular Formula

CH

₂

O Empirical Formula

The Empirical Formula (E.F.) is the simplest ratio of atoms in a compound.

_____ acid C

₂

H

₄

O

₂

Molecular Formula

CH

₂

O Empirical Formula

Both compounds have the same composition:

40.0 % C, 6.7 % H, 53.3 % O

If we are given the experimentally determined composition of a compound, we can calculate the E.F.

Sample calculation. Composition: 40.0 % C, 6.7 % H, 53.3 % O

Step 1: assume a 100 g sample and convert to ________ For C: 40.0 g C x 1 mole C = 3.33 moles C

12.0 g C

For H: 6.7 g H x 1 mole H = 6.7 moles H

1.0 g H

For O: 53.3 g O x 1 mole O = 3.33 moles O

16.0 g O

Step 2: divide each number of moles by the smallest of the numbers to obtain mole ratios = E.F.

For C: 3.33/3.33 = 1.0

For H: 6.7/3.33 = 2.0

For O: 3.33/3.33 = 1.0

Drill problem:

Composition of Borazole = 40.28%B, 52.20%N, 7.52%H Molar mass = 80.5 amu

Calculated the molecular formula

B: 40.28g x 1 mole = 3.726 moles = 1.0 10.81 g 3.726 N: 52.20g x 1 mole = 3.726 moles = 1.0 14.01 3.726 H: 7.52g x 1 mole = 7.45 moles = 2.0 1.01 g 3.726

E.F.= _____ E.F. mass: 10.81 + 14.01 + (2x1.01) = 26.84

Molar mass = ______ amu = 3

E.F. mass ______ amu

Mole ratios must be within 0.1 of a whole number. If they are not, each result must be multiplied by the same

multiplication factor until every value is + 0.1 of a whole number.

Example for a hypothetical set of mole ratios obtained from % composition:

2.247 for C 1.98 for H 1.000 for O

The result for C is not + 0.1 of a whole number; therefore, each result must be multiplied by an integer until all of the values are. For the above example, the multiplication

factor that works is 4: 2.247 x 4 = 8.988 for C 1.98 x 4 = 7.92 for H 1.000 x 4 = 4.000 for O

7

.

Chemical Equations & Stoichiometric Calculations

(Ch 6.6, 6.7, 6.8)

Summary of submicroscopic and macroscopic levels of a chemical equation:

2 Na(s) + 2 H2O → 2 NaOH(aq) + H2(g)

2 atoms + 2 molecules → 2 formula units + 1 molecule

2 moles + 2 moles → 2 moles + 1 mole 2x23=46g + 2x18=36g → 2x40=80g + 1x2=2g ______ reactants → ______ products

The coefficients in a balanced equation give the numerical relationships among formula units consumed or produced in a chemical reaction.

Keys to the calculations = _______________

N

2

+ 3 H

2

→ 2 NH

3 1 mole N2 3 moles H2 2 moles NH3

3 moles H2 1 mole N2 1 mole N2

1 mole N2 3 moles H2 2 moles NH3

Figure 6.9 In solving stoichiometric calculations, only the following “transitions” are allowed:

Sample calculation:

How many g O2 are needed to convert 45 g glucose

(C6H12O6) into CO2 and H2O?

First write a balanced equation and calculated the molecular masses of glucose and oxygen:

C6H12O6 + 6 O2 → 6 CO2 + 6 H2O

M.W. 180 amu + 32 amu

Then set up the calculations according to Figure 6.9

45 g glu x 1 mol glu x ___mol O2 x 32 g O2

180 g glu 1 mol glu 1 mol O2

Drill Problem: Figure 6.10

The chemical equation for the deployment of airbags is

2 NaN3(s) → 2 Na(s) + 3 N2(g)

How many g NaN3 would have to

decompose on order to generate

253 million molecules of N2?

F.W. NaN3 = 65.0 amu

Avogadro’s # = 6.02 x 1023 molecules N2 → moles N2 → moles NaN3 → g NaN3

2.53x108 molecules N2 x 1 mole N2 = 4.203 x 10-16 moles N2

6.02 x 1023molecules N2

4.203 x 10-16 moles N₂ x ___ moles NaN₃ x 65.0 g NaN₃

___ moles N2 ___mol NaN3

8.

Heat of Reaction (Ch 9.5 & supplemental material)

When the chemical energy stored in reactants is greater than that stored in the products, energy is released by the reaction, and it is termed exothermic.

The change in energy is called the enthalpy change and is represented by

∆H

; the value is negative for an exothermic reaction.

Example: combustion of propane is exothermic.

The heat released can be represented with energy on the product side:

C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) + 530 kcal

However, the reaction is also represented by placing the enthalpy change to the right of the equation:

An endothermic reaction is a chemical reaction in which a continuous input of energy is needed for the reaction to occur. Energy is a reactant.

Example: photosynthesis is

endothermic

∆H = positive

6 CO2(g) + 6 H2O(g) + 678 kcal → C6H12O6(aq) + 6 O2(g)

6 CO2(g) + 6 H2O(g) → C6H12O6(aq) + 6 O2(g)

Sample Problem:

How much energy is produced when 0.50 g of butane, C4H10, is burned in

a butane lighter? C4H10 = 58.1 g/mole

2 C4H10(g) + 13 O2(g) → 8 CO2(g) + 10 H2O(l) + 1365 kcal

The equation shows that 1365 kcal of heat are produced when 2 moles of butane undergo combustion.

0.50g C4H10 x 1 mol C4H10 x 1365 kcal = _____ kcal

58.1 g C4H10 ___ moles C4H10

Is this reaction exothermic or endothermic? What is the sign for ∆H?

9. Percent Yield

(supplemental material)

The theoretical yield in a reaction is the amount of product that could be obtained if a given reactant reacted completely. In most cases the actual yield is smaller that the theoretical yield because of side reactions, incomplete reaction, or other experimental limitations. The discrepancy between the theoretical yield and the actual yield is reported as the percent yield, which is calculated as shown:

% yield

=

actual yield

x 100

theoretical yield

The theoretical yield is calculated from the given amount of the specified reactant. The actual yield is identified in the problem.

Pure iron can be produced from iron oxide in a blast furnace.

Sample Problem. When 884 g of Fe2O3 was reduced with excess

carbon, 507 g of Fe were obtained. What was the percent yield?

Fe2O3 + 3 C → 2 Fe + 3 CO

1. Calculate the theoretical yield:

g Fe2O3 → mol Fe2O3 → mol Fe → g Fe

884g Fe2O3 x 1 mol Fe2O3 x 2 moles Fe x 55.85g Fe = ___g Fe

159.7g Fe2O3 1 mol Fe2O3 1 mol Fe

2. Calculate the % yield:

% yield = actual yield x 100 = 507 g x 100 = _______ % theoretical yield ____ g

10.

Limiting Reagent

(supplemental material)

Two batteries are required for these flashlights to work.

So, if you have 10 flashlights and 17 batteries, how many working flashlights do you have?

The batteries are the LIMITING “REAGENT”. The flashlights are IN EXCESS.

Reaction A: Stoichiometric amounts of reactants are used.

NaOH + HCl → NaCl + H

₂

O

1 mol + 1 mol → 1 mol + 1 mol

Reaction B: One reactant is limiting.

NaOH + HCl → NaCl + H

₂

O

0.6 mol + 0.3 mol → ___ mol + ___ mol

Which is the limiting reagent? _______

Which reagent is in excess? _______ What are the amounts of each product?

How much of each compound is present at the end of the reaction?

NaOH =

HCl

=

NaCl =

H

₂

O

=

Please take note that you must calculate the theoretical yield of any reaction only from the limiting reagent!

Sample Problem:

If you have 25 moles of N2 reacting with 45 moles of H2

according to the following reaction, which is the limiting reagent?

N

₂

(g) + 3 H

₂

(g) → 2 NH

₃

(g)

moles of A x mole ratio = moles of B available needed

25 moles N

₂

x 3 moles H

₂

= ____ moles H

₂

1 mole N

₂

For 25 moles N2 we would need ____ moles H2 but only 45

moles H2 are available.

H

₂

= limiting reagent

You can do the same analysis starting with H2

45 moles H

₂

x 1 moles N

₂

= ____ moles N

₂

___ moles H

₂

Here is another way to finding the

limiting reagent

.

N₂

(g) + 3 H

₂

(g) → 2 NH

₃

(g)

25 moles 45 moles

The limiting reagent has the lowest mole-to-coefficient ratio:

25 moles N

₂

= 25 45 moles H

₂

= 15

1 mole N

₂

3 moles H

₂

Demo: Identify the limiting reagent

HC2H3O2(aq) + NaHCO3(s) → H2O(l) + CO2(g) + NaC2H3O2(aq) A B

0.19 mol + 0.048 mol in Rxn I

0.19 mol + 0.095 mol in Rxn II

0.19 mol + 0.19 mol in Rxn III

0.19 mol + 0.38 mol in Rxn IV

Reagent A and B are mixed and CO2 evolution is measured.

Which is the limiting reagent? Rxn I B is limiting

Rxn II B is limiting

Rxn III stoichiometric amounts (balanced)

The following drill problem summarizes the important chemical calculations you have learned in this chapter:

When aqueous solutions of CaCl2 and

AgNO3 are mixed, a white precipitate

of AgCl forms.

If 3.33 g CaCl2 are reacted with 8.50 g

AgNO3 and 5.63 g AgCl are obtained,

what is the % yield?

CaCl2(aq) + 2 AgNO3(aq) → 2 AgCl(s) + Ca(NO3)2(aq)

• Calculated formula masses • Convert g to moles

• Check if you have a limiting reagent • Calculate theoretical yield

CaCl2(aq) + 2 AgNO3(aq) → 2 AgCl(s) + Ca(NO3)2(aq)

3.33 g 8.50 g 5.63 g

111.0 amu 169.9 amu 143.3 amu 3.33 g CaCl2 x 1 mol = 0.0300 mol CaCl2

111.0 g

8.50 g AgNO3 x 1 mol = 0.0500 mol AgNO3

169.9 g

Mole-to-coefficient ratios:

0.0300 CaCl2 = 0.0300 0.0500 AgNO3 = 0.0250

______________

Theoretical yield:

0.0500 mol AgNO3 x 2 mol AgCl x 143.3 g AgCl = ___ g AgCl

2 mol AgNO3 1 mol AgCl

% Yield:

actual yield x 100 = 5.63 g x 100 = ______ %