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Dr. Ron Rusay

Chemical Reactions

Stoichiometry

- The Mole
- % Composition: Determining the Formula of an Unknown Compound
- Writing and Balancing Chemical Equations
- Calculating the amounts of Reactant and Product
- Limiting Reactant

The Mole

- The numberof carbon atoms in exactly 12 grams of pure 12C. The numberequals
6.02 10 23

- 1 mole of anything = 6.02 10 23 units
- 6.02 10 23“units” of anything: atoms, people, stars, $s, etc., etc. = 1 mole

12 red marbles @ 7g each = 84g

12 yellow marbles @4g each=48g

55.85g Fe = 6.022 x 1023 atoms Fe

32.07g S = 6.022 x 1023 atoms S

Atomic and Molecular WeightsMass Measurements

- 1H weighs 1.6735 x 10-24 g and 16O 2.6560 x 10-23 g.
- DEFINITION: mass of 12C = exactly 12 amu.
- Using atomic mass units:

- 1 amu = 1.66054 x 10-24 g
- 1 g = 6.02214 x 1023 amu

1.06

10

g

A mole of copper atoms has a mass of 63.55 g.

The mass of 1 copper atom is

63.55 g/(6.022

10

) = 1.06

10

g.

ANSWER–

22

´

23

–

22

´

´

Atomic and Molecular Weights

- Formula Weight a.k.a. Molecular Weight
- Formula weights (FW): sum of Atomic Weights (AW) for atoms in formula.
- FW (H2SO4) = 2AW(H) + AW(S) + 4AW(O)
- = 2(1.0 amu) + (32.0 amu) + 4(16.0)
- = 98.0 amu

Atomic and Molecular Weights

- Molecular weight (MW) is the weight of the molecular formula in amu.
- MW of sugar (C6H12O6 ) = ?
- MW = 6(12.0 amu) + 12(1.0 amu) + 6(16.0 amu)
- = 180 amu

Molar Mass

- A substance’s molar mass (equal to the formula weight: atomic or molecular weight in grams) is the mass in grams of one mole of the element or compound.
- C = 12.01 grams per mole (g/mol)
- CO2 = ??
- 44.01 grams per mole (g/mol) 12.01 + 2(16.00) = 44.01

46.07

The molar mass is the sum of masses of all the

atoms in the molecule.

2

12.01 + 6

1.008 + 1

16.00 = 46.07

ANSWER´

´

´

CH

Cl

The molar mass has units of g/mol.

(12.8 g/0.256 mol) = 50.0 g/mol. The molecule

with the closest molar mass is CH

ANSWER3

Cl.

3

278 g

The molar mass of sodium hydroxide, NaOH, is

22.99 g/mol + 16.00 g/mol + 1.008 g/mol = 40.00

g/mol. Convert to grams: 6.94 mol

(40.00 g/mol)

= 278 g.

ANSWER´

Percent Composition

- Mass percent of an element:
- For iron in (Fe2O3), iron (III) oxide = ?

A

ll of these

The ratio of C to H in C

H

is 1:1. This is the

same ratio found for each of the compounds, so

all have the same percent composition by mass.

ANSWER8

8

QUESTION

Morphine, derived from opium plants, has the potential for use and abuse. It’s formula is C17H19NO3. What percent, by mass, is the carbon in this compound?

42.5%

27.9%

71.6%

This cannot be solved until the mass of the sample is given.

ANSWER

Morphine, derived from opium plants, has the potential for use and abuse. It’s formula is C17H19NO3. What percent, by mass, is the carbon in this compound?

42.5%

27.9%

71.6%

This cannot be solved until the mass of the sample is given.

(17 12) / ((17 12) + (19 1) + (1 14) + 3 16)) = 0.716

0.716 100 = 71.6 %

4.04 g

The molar mass of K

CrO

is 2

39.10 + 52.00 +

7

16.00 = 242.2. The mass fraction of potassium

is (2

39.10)/242.2 = 0.3229.

0.3229

12.5 g = 4.04 g.

ANSWER´

2

7

´

´

´

Formulas: Dalton’s Law

- Dalton’s law of multiple proportions:
When two elements form different compounds, the mass ratio of the elements in one compound is related to the mass ratio in the other by a small whole number.

Formulas & Multiple ProportionsComponents of acid rain, SO2(g) and SO3(g)

- Compound A contains:
1.000 g Sulfur & 1.500 g Oxygen

- Compound B contains:
1.000 g Sulfur & 1.000 g Oxygen

- Mass ratio A: 2 to 3; Mass ratio B: 1 to 1
- Adjusting for atomic mass differences: AW sulfur is 2x the AW oxygen; the atom ratios therefore are S1O3 and S1O2 respectively

Formulas &Molecular Representations

- molecular formula = C6H6Benzene
- empirical formula = CH = C6/6H6/6
- molecular formula = (empirical formula)n
[n = integer] (CH)6

- Other representations: Lewis Dot formulas, structural formulas, 2-D, 3-D

Formulas &Molecular Representations

Empirical Formula Determination

- 1. Use percent analysis.
Let 100 % = 100 grams of compound.

- 2. Determine the moles of each element.
(Element % = grams of element.)

- 3. Divide each value of moles by the smallest of the mole values.
- 4. Multiply each number by an integer to obtain all whole numbers.

QUESTION

The dye indigo is a compound with tremendous economic importance (blue jeans wouldn’t be blue without it.) Indigo’s percent composition is: 73.27% C; 3.84% H; 10.68%N and 12.21% O. What is the empirical formula of indigo?

C6H4NO

C8H3NO

C8H5NO

I know this should be whole numbers for each atom, but I do not

know how to accomplish that.

ANSWER

The dye indigo is a compound with tremendous economic importance (blue jeans wouldn’t be blue without it.) Indigo’s percent composition is: 73.27% C; 3.84% H; 10.68%N and 12.21% O. What is the empirical formula of indigo?

C6H4NO

C8H3NO

C8H5NO

I know this should be whole numbers for each atom, but I do not

know how to accomplish that.

Empirical & Molecular Formula Determination

The Molecular Formula is the important objective.

The Molar Mass (molecular weight) must be determined in order to reach this objective since the Molar Mass may not be equal to the Empirical formula’s Mass.

The experimental process involves different processes as does the calculations.

Using mass percent data and molar mass is the most straightforward.

Combustion analysis is more involved.

COMPARISON of CALCULATIONS

Empirical & Molecular Formula Determination Empirical Formula: C10 H12 N1 O1 Molecular Formula = C20 H24 N2 O2

Quinine:

C 74.05%, H 7.46%, N 8.63%, O 9.86%

- 74.05/12.01, 7.46/1.008, 8.63/14.01, 9.86/16.00
C6.166 H7.40 N0.616 O0.616

Empirical Formula Weight = ?

- Molecular Weight = 324.42
Molecular Formula = 2x empirical formula

Records a mass spectrum

A mass spectrum records only positively charged

fragments

m/z = mass to charge ratio of the fragment

http://chemconnections.org/pdb/Quinine.html

From the structures, determine the

molecular formula of quinine.

A Carbon atom is at each angle.

Each C has 4 bonds (lines + Hs).

Hs are not always drawn in &

must be added.

A)C18H24NO2 B)C20H20NO3C)C20H24N2O2D)C20H26N2O2

http://chemconnections.org/pdb/Quinine.html

From the structures, determine the

molecular formula of quinine.

C = 20

H = 24

N = 2

O =2

A Carbon atom is at each angle.

Each C has 4 bonds (lines + Hs).

Hs are not always drawn in &

must be added.

C20H24N2O2

A)C18H24NO2 B)C20H20NO3C)C20H24N2O2D)C20H26N2O2

ANSWER

B)

C

H

8

8

The mass of CH is 13.018. 13.018 divides into

104.1 about 8 times. Therefore there are 8 CH

groups in this compound. C

H

is the molecular

8

8

formula.

QUESTION

Examine the condensed structural formulas shown below:

I) Acetic acid (main ingredient in vinegar), CH3COOH

II) Formaldehyde (used to preserve biological specimens), HCHO

III) Ethanol (alcohol in beer and wine), CH3CH2OH

For which molecule(s) are the empirical formula(s) and the molecular formula(s) the same?

A) II B) I and II C) II and III D) I, II, and III

ANSWER

Examine the condensed structural formulas shown below:

I) Acetic acid (main ingredient in vinegar), CH3COOH

II) Formaldehyde (used to preserve biological specimens), HCHO

III) Ethanol (alcohol in beer and wine), CH3CH2OH

For which molecule(s) are the empirical formula(s) and the molecular formula(s) the same?

A) II B) I and II C) II and III D) I, II, and III

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