# Mass Relationships in Chemical Reactions - PowerPoint PPT Presentation

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Mass Relationships in Chemical Reactions. Chapter 3. 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

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## Mass Relationships in Chemical Reactions

Chapter 3

Micro World

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

(7.42% x 6.015) + (92.58% x 7.016)

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

Dozen = 12

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

eggs

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

One Mole of:

S

C

Hg

Cu

Fe

3.2

1 g = 6.022 x 1023 amu

1 amu = 1.66 x 10-24 g

= molar mass in g/mol

M

3.2

NMSI Renee Mc Cormick

Do You Understand Molar Mass?

1 mol K

6.022 x 1023 atoms K

x

x

=

1 mol K

39.10 g K

How many atoms are in 0.551 g of potassium (K) ?

1 mol K = 39.10 g K

1 mol K = 6.022 x 1023 atoms K

0.551 g K

8.49 x 1021 atoms K

3.2

1S

32.07 amu

2O

+ 2 x 16.00 amu

SO2

SO2

64.07 amu

Molecular mass (or molecular weight) is the sum of

the atomic masses (in amu) in a molecule.

For any molecule

molecular mass (amu) = molar mass (grams)

1 molecule SO2 = 64.07 amu

1 mole SO2 = 64.07 g SO2

3.3

Do You Understand Molecular Mass?

8 mol H atoms

6.022 x 1023 H atoms

1 mol C3H8O

x

x

x

=

1 mol C3H8O

1 mol H atoms

60 g C3H8O

How many H atoms are in 72.5 g of C3H8O ?

1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O

1 mol C3H8O molecules = 8 mol H atoms

1 mol H = 6.022 x 1023 atoms H

72.5 g C3H8O

5.82 x 1024 atoms H

3.3

Mole fractions:

Xa= na_____ Where n is the number of moles

na+nb....

Use 1 mole to determine the ratio amounts:

What is the mole fraction of Na in Na2O.

Given the answer can you work backwards and find the moles of O?

• mass spectrometer—a device for measuring the mass of atoms or molecules

• atoms or molecules are passed into a beam of high-speed electrons

• this knocks electrons OFF the atoms or molecules transforming them into cations o apply an electric field

• this accelerates the cations since they are repelled from the (+) pole and attracted toward the (−) polesend the accelerated cations into a magnetic field

• an accelerated cation creates it’s OWN magnetic field which perturbs the original magnetic fieldthis perturbation changes the path of the cation

• the amount of deflection is proportional to the mass; heavy cations deflect little

• ions hit a detector plate where measurements can be obtained.

Heavy

Heavy

Light

Light

KE = 1/2 x m x v2

v = (2 x KE/m)1/2

F = q x v x B

3.4

2 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

### Types of Formulas

• 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.

### To obtain an 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

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.

require mole ratios so convert grams to moles

moles of N = 2.34g of N = 0.167 moles of N

14.01 g/mole

moles of O = 5.34 g = 0.334 moles of O

16.00 g/mole

Formula:

### Calculation of the Molecular Formula

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

N2O4

### Empirical Formula from % Composition

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 grams 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

• Find the empirical formula using a equation.

g CO2

g H2O

mol CO2

mol H2O

mol C

mol H

g H

g C

Combust 11.5 g ethanol

Collect 22.0 g CO2 and 13.5 g H2O

6.0 g C = 0.5 mol C

1.5 g H = 1.5 mol H

g of O = g of sample – (g of C + g of H)

4.0 g O = 0.25 mol O

Empirical formula C0.5H1.5O0.25

Divide by smallest subscript (0.25)

Empirical formula C2H6O

3.6

Mass Changes in Chemical Reactions

• 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

### Other units

• Molarity

• Moles solute / L solution

• Gases

• 22.4 L = 1 mole of ANY GAS at STP

2CH3OH + 3O2 2CO2 + 4H2O

grams CH3OH

moles CH3OH

moles H2O

grams H2O

4 mol H2O

18.0 g H2O

1 mol CH3OH

x

=

x

x

2 mol CH3OH

32.0 g CH3OH

1 mol H2O

Methanol burns in air according to the equation

If 209 g of methanol are used up in the combustion,

what mass of water is produced?

molar mass

CH3OH

molar mass

H2O

coefficients

chemical equation

209 g CH3OH

235 g H2O

3.8

6 green used up

6 red left over

Limiting Reagents

3.9

### Method 1

• Pick A Product

• Try ALL the reactants

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

### This is the method I use. Pick your favorite and use it.

• Limiting reagent

• To determine the limiting reagent requires that you do two stoichiometry problems.

• Figure out how much product each reactant makes.

• The one that makes the least is the limiting reagent.

LimitingReactant

### Limiting Reactant: Method 1

• 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

• 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

• Limiting Reagents: shortcut • Limiting reagent problems can be solved another way (without using a chart)...

• • Do two separate calculations using both given quantities. The smaller answer is correct.

• Q-How many g NO are produced if 20gNH3 is burned in 30g O2 ?

Do You Understand Limiting Reagents?

2Al + Fe2O3 Al2O3 + 2Fe

g Al

mol Al

mol Fe2O3 needed

g Fe2O3 needed

OR

g Fe2O3

mol Fe2O3

mol Al needed

g Al needed

1 mol Fe2O3

160. g Fe2O3

1 mol Al

=

x

x

x

27.0 g Al

2 mol Al

1 mol Fe2O3

need 367 g Fe2O3

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

Calculate the mass of Al2O3 formed.

367 g Fe2O3

124 g Al

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

3.9

2Al + Fe2O3 Al2O3 + 2Fe

g Al

mol Al

mol Al2O3

g Al2O3

1 mol Al

x

27.0 g Al

1 mol Al2O3

102. g Al2O3

=

x

x

2 mol Al

1 mol Al2O3

Use limiting reagent (Al) to calculate amount of product that

can be formed.

234 g Al2O3

124 g Al

3.9

### Finding Excess Practice

• 10.0g of aluminum reacts with 35.0 grams of chlorine gas 2 Al + 3 Cl2 2 AlCl3

• We found that chlorine is the limiting reactant, and 43.8 g of aluminum chloride are produced.

35.0 g Cl2 1 mol Cl2 2 mol Al 27.0 g Al

71 g Cl2 3 mol Cl2 1 mol Al

= 8.8 g Al USED!

10.0 g Al – 8.8 g Al = 1.2 g Al EXCESS

Given amount of excess reactant

Amount of excess reactant actually used

Note that we started with the limiting reactant! Once you determine the LR, you should only start with it!

% Yield =

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.

% Yield = what you got

what you could have got

x 100

3.10

3H2 (g) + N2 (g) 2NH3 (g)

NH3 (aq) + HNO3 (aq) NH4NO3 (aq)

2Ca5(PO4)3F (s) + 7H2SO4 (aq)

fluorapatite

3Ca(H2PO4)2 (aq) + 7CaSO4 (aq) + 2HF (g)

Chemistry In Action: Chemical Fertilizers

Plants need: N, P, K, Ca, S, & Mg