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# Mass Relationships in Chemical Reactions - PowerPoint PPT Presentation

Mass Relationships in Chemical Reactions. Micro World atoms & molecules. Macro World grams. Atomic mass is the mass of an atom in atomic mass units (amu). By definition: 1 atom 12 C “weighs” 12 amu. On this scale 1 H = 1.008 amu 16 O = 16.00 amu. 3.1.

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### Mass Relationships in Chemical Reactions

atoms & molecules

Macro World

grams

Atomic mass is the mass of an atom in atomic mass units (amu)

By definition:

1 atom 12C “weighs” 12 amu

On this scale

1H = 1.008 amu

16O = 16.00 amu

3.1

100

Natural lithium is:

7.42% 6Li (6.015 amu)

92.58% 7Li (7.016 amu)

Average atomic mass of lithium:

= 6.941 amu

3.1

Pair = 2

The mole (mol) is the amount of a substance that

contains as many elementary entities as there

are atoms in exactly 12.00 grams of 12C

1 mol = NA = 6.0221367 x 1023

3.2

shoes

Molar mass is the mass of 1 mole of in grams

marbles

atoms

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)

3.2

S

C

Hg

Cu

Fe

3.2

Molecular mass (or molecular weight) is the

sum of the atomic masses (in amu) in a molecule.

1 X 32.07 g

1S

+ 2 x 16.00 g

2O

SO2

64.07 g/mole

For all practical purposes think of amu as grams

What is the molecular weight ofSO2

1 molecule SO2 = 64.07 amu

1 mole SO2 = 64.07 g SO2

3.3

and formula weights (ionic compounds) are

found exactly the same way

Try a few

Find the formula weight for NaCl

Find the molecular weight for H2SO4

A. Relationship between balanced chemical

equations andmass or volume

1. Mass - mass problems

a. The first thing to do is balanced the

chemical equations

i. this will allow you to find the mole

ratio between the reactants andproducts

2H2 + O2 --> 2H2O

2 moles 1 mole 2 moles

4.04 g 32.0 g 36.0 g

when 200.0 g of KClO3 decomposes?

Oxygen is the other product

1st write the balanced equation

2moles 2moles 3moles

2nd find mole ratio

2KClO3 ---> 2KCl + 3 O2

122.6g/m 74.55g/m 32.00g/m

200.0 g x

4th find mwt or fwt

b. change grams of given to moles of given

using themolecular or formula weight

c. change the moles of given to moles of

d. change the moles of asked to grams of

asked using themolecular or formula weight

(1 mole KClO3)

(200.0 g KClO3)

(2 moles KCl)

( 74.55 g KCl)

( 122.6 g KClO3 )

(2 mole KClO3)

(1 mole KCl)

Given

mole ratio

M.W

243.2 g KCl

How many grams of NaOH is required to completely

neutralize 155 grams of HCl? The product of this

reaction is HOH and NaCl.

1 mole

1 mole

1 mole

NaOH + HCl ===> HOH + NaCl

X g

155 g HCl

Na = 1 x 23.0 g = 23.0 g

H = 1 x 1.01 g = 1.01 g

Cl =1 x 35.5 g = 35.5 g

O = 1 x 16.0 g = 16.0 g

H = 1.01 g = 1.01 g

170. g NaOH

40.0 g/mole

36.5 g/mole

1 mole NaOH

40.0 g NaOH

(

(

)

1 mole HCl

)

(

)

(

)

155 g HCl

1 mole HCl

1 mole NaOH

36.5 g HCl

How many grams of Potassium is required to produce

200. gof KOH when it is involved in a single

replacement reaction with water?

Hydrogen is the other product

2 moles

1 moles

2 moles

2 moles

2

K + HOH ===> KOH + H2

2

2

200. g KOH

X g

39.1 g/mole

56.1 g/mole

39.1g K

)

)

(

2 moles K

)

(

1 mole KOH

)

(

(

200.g KOH

1 mole K

56.1 g KOH

2 moles KOH

grams CH3OH

moles CH3OH

moles H2O

grams H2O

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?

molar mass

H2O

molar mass

CH3OH

coefficients

chemical equation

18.0 g H2O

(

)

(

4 mol H2O

)

)

(

1 mol CH3OH

(

)

=

209 g CH3OH

1 mol H2O

2 mol CH3OH

32.0 g CH3OH

235 g H2O

3.8

2 mole

1mole

2 mole

2

2

x

120.g

O = 2 x 16.0 g = 32.0 g/mole

Hg = 1 x201 = 201g

O = 1 x 16.0 = 16.0 g

217 g/mole

1 mole HgO

)

32.0 g O2

(

)

(

(

1 mole O2

)

(

)

120. gHgO

2 moles HgO

1 mole O2

217 g HgO

= 8.85 g O2

2 moles

2 moles

1 moles

2

2

70.0 g

x

34.0 g/mole H2O2

32.0 g/mole O2

)

(

)

1 mole H2O2

)

(

1 mole O2

)

(

32.0 g O2

(

70.0 g H2O2

2 moles H2O2

1mole O2

34.0 g H2O2

= 32.9 g O2

2 moles Al

6 moles HCl

2 moles AlCl3

3 moles H2

2

6

2

3

x g

40.00 g Al

2.016 g/mole

26.98 g /mole Al

1 mole Al

)

)

(

3 moles H2

)

(

2.016 g H2

(

)

(

40.00 g Al

1 mole H2

26.98 g Al

2 moles Al

1 mole 1 mole 1 mol 1mole

84.0 g CuO

X g H2

79.4 g/mole

2.00 g/mole H2

2.00 g H2

(

)

(

)

)

)

(

1 mole CuO

(

1 mole H2

84.0 g CuO

1 mole H2

79.4 g CuO

1 mole CuO

5. How many grams of potassium chlorate are required in the preparation of 90.0 grams of oxygen?

KClO3 ===> KCl + O2

Mass Changes in Chemical Reactions of hydrogen are produced?

• Write balanced chemical equation

• Convert quantities of known substances into moles

• Use coefficients in balanced equation to calculate the number of moles of the sought quantity

• Convert moles of sought quantity into desired units

3.8

6 green used up of hydrogen are produced?

6 red left over

Limiting Reagents

3.9

Method 1 of hydrogen are produced?

• Pick A Product

• Try ALL the reactants

• The reactant that gives the lowest answer will be the limiting reactant

Limiting Reactant: Method 1 of hydrogen are produced?

• 10.0g of aluminum reacts with 35.0 grams of chlorine gas to produce aluminum chloride. Which reactant is limiting, which is in excess, and how much product is produced?

2 Al + 3 Cl2 2 AlCl3

• Now Cl2:

10.0 g Al 1 mol Al 2 mol AlCl3 133.5 g AlCl3

27.0 g Al 2 mol Al 1 mol AlCl3

= 49.4g AlCl3

35.0g Cl2 1 mol Cl2 2 mol AlCl3 133.5 g AlCl3

71.0 g Cl2 3 mol Cl2 1 mol AlCl3

= 43.9g AlCl3

Method 2 of hydrogen are produced?

• Convert one of the reactants to the other REACTANT

• See if there is enough reactant “A” to use up the other reactants

• If there is less than the GIVEN amount, it is the limiting reactant

• Then, you can find the desired species

5.0 g of H of hydrogen are produced? 2 reacts with 10.0 g of O2. How many grams of H2O

is produced?

2 moles

1moles

2 moles

2

H2 + O2 H2O

2

5.0 g

10.0 g

x

2.0 g/mole

32.0g/mole

18 .0g/mole

1 mole H2

1 mole O2

(

(

)

(

32 g O2

)

)

(

)

5.0 g

= 40. g O2

2.0g H2

2mole H2

1 mole O2

2 mole H2O

)

(

1 mole O2

)

(

(

18.0 g H2O

)

= 11.2 g H2O

(

10.0 g

)

32.0g O2

1mole O2

1 mole H2

Do You Understand Limiting Reagents? of hydrogen are produced?

In one process, 124 g of Al are reacted with 601 g of Fe2O3

2Al + Fe2O3 Al2O3 + 2Fe

g Al of hydrogen are produced?

mol Al

mol Fe2O3 needed

g Fe2O3 needed

OR

g Fe2O3

mol Fe2O3

mol Al needed

g Al needed

Calculate the mass of Al2O3 formed.

2Al + Fe2O3 Al2O3 + 2Fe

(

)

(

)

)

(

)

1 mol Al

(

1 mol Fe2O3

160. g Fe2O3

=

367 g Fe2O3

124 g Al

1 mol Fe2O3

27.0 g Al

2 mol Al

3.9

need 367 g Fe of hydrogen are produced? 2O3

Have more Fe2O3 (601 g) so Al is limiting reagent

2Al + Fe of hydrogen are produced? 2O3 Al2O3 + 2Fe

g Al

mol Al

mol Al2O3

g Al2O3

Use limiting reagent (Al) to calculate amount

of product that can be formed.

(

)

(

)

(

)

1 mol Al

1 mol Al2O3

102. g Al2O3

(

)

=

234 g Al2O3

124 g Al

27.0 g Al

2 mol Al

1 mol Al2O3

3.9

% Yield of hydrogen are produced? =

x 100

Actual Yield

Theoretical Yield

Theoretical Yield is the amount of product that would

result if all the limiting reagent reacted.

Actual Yield is the amount of product actually obtained

from a reaction.

3.10

2 of hydrogen are produced? x (12.01 g)

6 x (1.008 g)

1 x (16.00 g)

n x molar mass of element

%C =

%H =

%O =

x 100% = 34.73%

x 100% = 13.13%

x 100% = 52.14%

x 100%

46.07 g

46.07 g

46.07 g

molar mass of compound

C2H6O

Percent composition of an element in a compound =

n is the number of moles of the element in 1 mole of the compound

52.14% + 13.13% + 34.73% = 100.0%

3.5

You try two of hydrogen are produced?

Find percent composition of

H2O

H3PO4

To obtain an of hydrogen are produced? Empirical Formula

1. Determine the mass in grams of each element present, if necessary.

2. Calculate the number of moles of each element.

3. Divide each by the smallest number of moles to obtain the simplest whole number ratio.

• If whole numbers are not obtained* in step 3), multiply through by the smallest number that will give all whole numbers

*Be careful! Do not round off numbers prematurely

e.g. problem of hydrogen are produced?

Find empirical formula for a compound found to

contain

60.8 % Na

28.60 % B

10.60 % H

2.644 moles of hydrogen are produced?

60.80 % Na = 60.80 g Na

(

)

(

1 mole Na

)

1

=

=

23.00 g

)

(

2.644 moles

28.60 % B = 28.60g B

(

)

(

1 mole B

)

2.646 moles

=

=

1

(

10.81 g B

)

2.644 moles

10.60 % H =

(

10.60g H

)

(

1 mole H

)

10.52 moles

=

4

=

(

1.008 g H

)

2.644 moles

Na1

B1

H4

Stop both 6th and 8th of hydrogen are produced?

Empirical Formula from % Composition of hydrogen are produced?

A substance has the following composition by

mass: 60.80 % Na ; 28.60 % B ; 10.60 % H

What is the empirical formula of the substance?

Consider a sample size of 100 grams

This will contain 60.80 g of Na , 28.60 grams

of B and 10.60 grams H

Determine the number of moles of each

Determine the simplest whole number ratio

You try one of hydrogen are produced?

A compound was found to contain 11 % Hydrogen

and 89 % O. What is the empirical formula?

Types of Formulas of hydrogen are produced?

• Empirical Formula

The formula of a compound that expresses the smallest whole number ratio of the atoms present.

Ionic formula are always empirical formula

• Molecular Formula

The formula that states the actual number of each kind of atom found in one molecule of the compound.

require of hydrogen are produced? mole ratios so convert grams to moles

A sample of a brown gas, a major air pollutant, is found to contain 2.34 g N and 5.34g O. Determine a formula for this substance.

(

)

(

)

1 mole N2

0.167 moles of N

moles of N =

2.34 g N

=

14.01g N2

0.167 moles of N

)

(

)

(

1 mole O2

0.334moles of O

moles of O =

5.34g O

=

16.0 g O2

0.167 moles of N

NO2

Calculation of the Molecular Formula of hydrogen are produced?

From the previous problem, it was found a compound has an empirical formula of NO2. The colourless liquid, used in rocket engines has a molar mass of 92.0 g/mole. What is the molecular formula of this substance?

empirical formula mass: 14.01+2 (16.00) = 46.01 g/mol

n = molar mass = 92.0 g/mol emp. f. mass 46.01 g/mol

n = 2 So the molecular formula is twice the size of the empirical formula

NO2 = N2O4

8 of hydrogen are produced? th period start here

6 of hydrogen are produced? th period start here