a chemical reaction shows the process in which a substance ... · pdf fileto roast 10.0 mol of...

A chemical reaction shows the process in

which a substance (or substances) is

changed into one or more new substances

Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.

A chemical equation uses chemical symbols to

show what happens during a chemical reaction

reactants product

(g) (g) (l)

“Two molecules of hydrogen react with one molecule of

oxygen to yield two moles of water”

Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.

The Law of Conservation of Mass states that

matter is neither created nor destroyed

Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.

To conform with the Law of Conservation of Mass, there

must be the same number of each type of atom on both

sides of the arrow. Hence, we balance the equation by

adding coefficients before each chemical symbol

Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.

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Translate the statement

Balance the atoms

Specify states of matter

Adjust the coefficients

Check the atom balance

A quick note on balancing chemical equations

Calculating the amounts of

reactant and product

Stoichiometry of a double

cheeseburger

2 bun slices + 2 cheese slices

+ 2 burger patties =

In a balanced equation, the number of moles of

one substance is equivalent to the number of

moles of any of the other substances

2CO(g) + O2(g) 2CO2(g)

2 mol CO = 1 mol O2

Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.

Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.

2 mol CO

1 mol O2

= 11 mol O2

2 mol CO= 1

In a balanced equation, the number of moles of

one substance is equivalent to the number of

moles of any of the other substances

2CO(g) + O2(g) 2CO2(g)

2 mol CO = 2 mol CO2

Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.

Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.

2 mol CO

2 mol CO2

= 12 mol CO2

2 mol CO= 1

In a balanced equation, the number of moles of

one substance is equivalent to the number of

moles of any of the other substances

2CO(g) + O2(g) 2CO2(g)

1 mol O2 = 2 mol CO2

Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.

Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.

1 mol O2

2 mol CO2

= 12 mol CO2

1 mol O2

= 1

The amount of one substance in a reaction

is related to that of any other

Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.

All alkali metals react with water to produce

hydrogen gas and the corresponding

alkali metal hydroxide

2Li(s) + 2H2O(l) 2LiOH(aq) + H2(g)

How many moles of H2 will be formed by the

complete reaction of 6.23 moles of Li with water?

nH2 = x6.23 mol Li 1 mol H2

2 mol Li

= 3.12 mol H2

2Li(s) + 2H2O(l) 2LiOH(aq) + H2(g)

How many grams of H2 will be formed by the

complete reaction of 80.57 g of Li with water?

2Li(s) + 2H2O(l) 2LiOH(aq) + H2(g)

mH2 = x x80.57 g Li 1 mol Li 1 mol H2

6.941 g Li 2 mol Li

2.016 g H2x

1 mol H2

= 11.70 g H2

In a lifetime, the average American uses about

794 kg of copper in coins, plumbing, and wiring.

Copper is obtained from sulfide ores (such as

Cu2S) by a multistep process. After an initial

grinding, the first step is to “roast” the ore (heat it

strongly with O2) to form Cu2O and SO2

2Cu2S(s) + 3O2(g) 2Cu2O(s) + 2SO2(g)

How many moles of oxygen are required

to roast 10.0 mol of Cu2S?

2Cu2S(s) + 3O2(g) 2Cu2O(s) + 2SO2(g)

nO2 = x10.0 mol Cu2S 3 mol O2

2 mol Cu2S

= 15.0 mol O2

How many grams of SO2 are formed when

10.0 mol of Cu2S is roasted?

2Cu2S(s) + 3O2(g) 2Cu2O(s) + 2SO2(g)

mSO2 = x x10.0 mol Cu2S 2 mol SO2 64.07 g SO2

2 mol Cu2S 1 mol SO2

= 641 g SO2

How many grams of O2 are required to form

2.86 kg of Cu2O?

2Cu2S(s) + 3O2(g) 2Cu2O(s) + 2SO2(g)

mO2 = x x2.86 kg Cu2O 1000 g Cu2O 1 mol Cu2O

1 kg Cu2O 143.10 g Cu2O

3 mol O2 32.00 g O2x x

2 mol Cu2O 1 mol O2

= 960 g O2

Within the cylinders of a car’s engine, the

hydrocarbon octane (C8H18), one of many

components of gasoline, mixes with oxygen from

the air and burns to form carbon dioxide and water

vapor.

a. How much carbon dioxide ( in kg) is produced

when 2.00 kg of octane is burned ?

b. How much oxygen(in kg) is required to burn the

same amount of octane?

2C8H18(l) + 25O2 (g) 16CO2 (g) + 18H2O (g)

How many double cheeseburgers can you

make from 8 bun slices + 2 cheese slices + 6

burger patties ?

What is the limiting factor?

8 bun slices

2 cheese slices

6 burger patties

# Cheeseburgers=

2 cheese slices [1 cheeseburger/

2 cheese slices]

Limiting Reactants

The reactant used up first in a chemical

reaction is called the limiting reactant. Excess

reactants are present in quantities greater than

necessary to react with the quantity of the

limiting reactant.

A + B C + D

Given the amounts of A and B, which is the

limiting reactant? How much C and D are

produced?

Urea is prepared by reacting ammonia with

carbon dioxide:

2NH3(g) + CO2(g) (NH2)2CO(aq) + H2O(l)

In one process, 637.2 g of NH3 are allowed to

react with 1142 g of CO2.

(a) Which is the limiting reactant?

(b) How much urea (in grams) is produced?

(c) How much of the excess reactant (in grams)

is left at the end of the reaction?

Strategy

• Check if the equation is balanced.

• Convert mass of each reactant to moles.

• Calculate the amount of product formed

from the each of the reactants.

• The reactant the produces the less

amount is the limiting reactant.

1. The reaction between aluminum and iron (III) oxide can generate temperatures around 3000⁰C and is

used in welding metals:

Al + Fe2O3 -- Al2O3 + 2Fe

In one process, 124 g of Al are reacted with 601 g of

ferric oxide.

(a) Which is the limiting reactant?

(b) How much Al2O3 (in grams) is produced?

(c) How much of the excess reactant (in grams) is left at

the end of the reaction?

2. Titanium is a strong & light metal used in rockets

& aircrafts. It is prepared by the reaction between

titanium (IV) chloride with molten magnesium at around 1000⁰C:

TiCl4 + Mg Ti + 2MgCl2

In a certain industrial operation, 3.54 x 107g of TiCl4are reacted with 1.13 x 107 g of magnesium.

(a)Which is the limiting reactant?

(b)How much Ti (in grams) is produced?

(c)How much of the excess reactant (in grams) is left

at the end of the reaction?

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Mass % of element X =

atoms of X in formula x atomic mass of X (amu)

molecular (or formula) mass of compound (amu)

x 100

Mass % of element X =

moles of X in formula x molar mass of X (g/mol)

mass (g) of 1 mol of compound

x 100

Mass Percent from the Chemical Formula

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Sample Problem 3.3 Calculating Mass Percents and Masses of

Elements in a Sample of a Compound

PLAN:

SOLUTION:

PROBLEM: In mammals, lactose (milk sugar) is metabolized to glucose

(C6H12O6), the key nutrient for generating chemical potential

energy.

(a) What is the mass percent of each element in glucose?

(b) How many grams of carbon are in 16.55 g of glucose?

We have to find the total mass of

glucose and the masses of the

constituent elements in order to

relate them.

Per mole glucose there are 6 moles of

C, 12 moles of H, 6 moles of O(a)

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Sample Problem 3.3 Calculating the Mass Percents and Masses of

Elements in a Sample of a Compoundcontinued

6 mol C x 12.01 g C

mol C = 72.06 g C 12 mol H x

1.008 g H

mol H = 12.096 g H

6 mol O x 16.00 g O

mol O= 96.00 g O

M = 180.16 g/mol

(b)mass percent of C =

72.06 g C

180.16 g glucose= 0.4000 x 100 = 40.00 mass % C

mass percent of H =12.096 g H

180.16 g glucose= 0.06714 x 100 = 6.714 mass % H

mass percent of O =96.00 g O

180.16 g glucose= 0.5329 x 100 = 53.29 mass % O

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Empirical and Molecular Formulas

Empirical Formula -

Molecular Formula -

The simplest formula for a compound that agrees with

the elemental analysis and gives rise to the smallest set

of whole numbers of atoms.

The formula of the compound as it exists; it may be a

multiple of the empirical formula.

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Sample Problem 3.4 Determining an Empirical Formula from Masses

of Elements

PROBLEM:

PLAN:

Elemental analysis of a sample of an ionic compound showed

2.82 g of Na, 4.35 g of Cl, and 7.83 g of O. What is the empirical

formula and name of the compound?

Once we find the relative number of moles of each element,

we can divide by the lowest mol amount to find the relative

mol ratios (empirical formula).

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Sample Problem 3.4 Determining an Empirical Formula from Masses

of Elements

SOLUTION:2.82 g Na x

mol Na

22.99 g Na= 0.123 mol Na

4.35 g Cl xmol Cl

35.45 g Cl= 0.123 mol Cl

7.83 g O xmol O

16.00 g O= 0.489 mol O

Na1.00 Cl1.00 O3.98 NaClO4

NaClO4 is sodium perchlorate.

continued

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Sample Problem 3.5 Determining a Molecular Formula from Elemental

Analysis and Molar Mass

PROBLEM:

PLAN:

During physical activity, lactic acid (M = 90.08 g/mol) forms in

muscle tissue and is responsible for muscle soreness.

Elemental analysis shows that this compound contains 40.0

mass % C, 6.71 mass % H, and 53.3 mass % O.

(a) Determine the empirical formula of lactic acid.

(b) Determine the molecular formula.

a) Assume 100 g of lactic acid and find the mass of each

element. Convert mass of each to moles, get a ratio and

convert to integer subscripts.

b) Divide molar mass by empirical mass to get the multiplier

then write the molecular formula accordingly.

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Sample Problem 3.5 Determining a Molecular Formula from Elemental

Analysis and Molar Mass

continued

SOLUTION: In 100.0 g of lactic acid, there are:

40.0 g C 6.71 g H 53.3 g O

40.0 g C x

6.71 g H x

53.3 g O x

mol C

12.01 g C

mol H

1.008 g H

mol O

16.00 g O

= 3.33 mol C

= 6.66 mol H

= 3.33 mol O

We convert the grams to moles and get a ratio for the empirical formula.

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Sample Problem 3.5 Determining a Molecular Formula from Elemental

Analysis and Molar Mass

continued

SOLUTION:

3.33 mol C 6.66 mol H 3.33 mol O

C3.33 H6.66 O3.33

3.33 3.33 3.33

CH2O empirical formula

mass of CH2O

molar mass of lactate 90.08 g

30.03 g3 C3H6O3 is the

molecular formula

We now divide by the smallest number and get a ratio for the empirical

formula.

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Combustion apparatus for determining formulas

of organic compounds.Figure 3.5

CnHm + (n + ) O2 = n CO(g) + H2O(g)m

2m

2

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Sample Problem 3.6 Determining a Molecular Formula from

Combustion Analysis

PLAN:

PROBLEM: Vitamin C (M = 176.12 g/mol) is a compound of C,H, and O

found in many natural sources, especially citrus fruits. When a

1.000-g sample of vitamin C is placed in a combustion chamber

and burned, the following data are obtained:

mass of CO2 absorber after combustion = 85.35 g

mass of CO2 absorber before combustion = 83.85 g

mass of H2O absorber after combustion = 37.96 g

mass of H2O absorber before combustion = 37.55 g

What is the molecular formula of vitamin C?

The difference in absorber mass before and after combustion

is the mass of oxidation product of the element. Find the mass

of each element from its combustion product, convert each to

moles and determine the formula. Using the molar mass and

empirical mass, determine the molecular formula.

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SOLUTION:

CO2 85.35 g - 83.85 g = 1.50 g H2O 37.96 g - 37.55 g = 0.41 g

There are 12.01 g C per mol CO2. 1.50 g CO2 x12.01 g C

44.01 g CO2

= 0.409 g C

0.41 g H2O x 2.016 g H

18.02 g H2O= 0.046 g H

There are 2.016 g H per mol H2O.

O must be the difference: 1.000 g - (0.409 + 0.046) = 0.545 g O

0.409 g C

12.01 g C

0.046 g H

1.008 g H

0.545 g O

16.00 g O

= 0.0341 mol C = 0.0456 mol H = 0.0341 mol O

C1.00H1.3O1.00 C3H4O3 176.12 g/mol

88.06 g= 2.000 = 2 C6H8O6

Sample Problem 3.6 Determining a Molecular Formula from Combustion

Analysiscontinued

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