Mass Relationships in Chemical...

Mass Relationships in Chemical Reactions

Chapter 4

Some images copyright © The McGraw-Hill Companies, Inc.

ed. Brad Collins

Sunday, August 18, 13

Pair = 2

Dozen = 12

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

Avogadro’s number (NA) 4.1


Molar mass is the mass of 1 mole of in grams

eggsshoes

marblesatoms

1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g1 12C atom = 12.00 amu

1 mole 12C atoms = 12.00 g 12C1 mole lithium atoms = 6.941 g of Li

For any element atomic mass (amu) = molar mass (grams)

(M)molecules

4.1


One Mole of:

C S

Cu Fe

Hg

4.1


1 amu = 1.66 x 10-24 g or 1 g = 6.022 x 1023 amu

1 12C atom12.00 amu

x 12.00 g6.022 x 1023 12C atoms

= 1.66 x 10-24 g1 amu

M = molar mass in g/molNA = Avogadro’s number

4.1


Molar Mass Conversion FactorsWe will use two equalities in molar mass calculations

For mass to moles and moles to mass: 1 mol X = molar mass X (g)

For # atoms to moles and moles to # atoms: 1 mol X = 6.022 x 1023 X atoms

4.1


Do You Understand Molar Mass?

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 1 mol K39.10 g K

x x 6.022 x 1023 atoms K1 mol K

=

8.49 x 1021 atoms K

4.1


Practice• How many sulfur atoms are in 16.3 grams of sulfur?

4.1


Percent composition of an element in a compound =

n x molar mass of elementmolar mass of compound x 100%

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

C2H6O

%C = 2 x (12.01 g)46.07 g x 100% = 52.14%

%H = 6 x (1.008 g)46.07 g x 100% = 13.13%

%O = 1 x (16.00 g)46.07 g x 100% = 34.73%

52.14% + 13.13% + 34.73% = 100.0%

4.2


Chemical composition of a lead bullet.4.2


Elemental Analysis• Determines the percent composition to determine the empirical formula

• Example: Formaldehyde

• 40.00 % carbon

• 6.714% hydrogen

• 53.29% oxygen

• Assume percentages are grams in a 100-gram sample

• Convert grams to moles:

• 40.00 g C x 1 mol C/12.01 g C = 3.33 mol C

• 6.714 g H x 1 mol H/1.008 g H = 6.66 mol H

• 53.29 g O x 1 mol O/16.00 g O = 3.33 mol O

• Write formula with mol of each element as subscripts: C3.33H6.66O3.33

• Convert to whole numbers by dividing each subscript by the smallest subscript: CH2O4.2


Practice• Determine the empirical formula of vitamin C

(ascorbic acid) given the following elemental analysis data:

• Carbon, 40.92%

• Hydrogen, 4.58%

• Oxygen, 54.50%

4.2


Formulas determined from percent composition data are always empirical formulasTo determine a molecular formula you must know its molar mass.

Example Ascorbic Acid: molar mass = 176.12

First: Calculate Empirical Formula (see previous example)

Second: Calculate empirical molar mass

Third: Determine whole number multiple to calculate the accurate molar mass and determine the molecular formula.

Molecular Formulas from Percent Composition Data

4.2


3 ways of representing the reaction of H2 with O2 to form H2O

A process in which one or more substances is changed into one or more new substances is a chemical reaction

A chemical equation uses chemical symbols to show what happens during a chemical reaction

reactants products4.3


How to “Read” Chemical Equations2 Mg + O2 2 MgO

IS:2 atoms Mg + 1 molecule O2 makes 2 formula units

MgO2 moles Mg + 1 mole O2 makes 2 moles MgO

48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO

IS NOT:2 grams Mg + 1 gram O2 makes 2 g MgO

4.3


Balancing Chemical Equations

1. Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation.

Ethane reacts with oxygen to form carbon dioxide and waterC2H6 + O2 CO2 + H2O

2. Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts.

2C2H6 NOT C4H124.3



3. Start by balancing those elements that appear in only one reactant and one product.

C2H6 + O2 CO2 + H2O start with C or H but not O

2 carbonon left

1 carbonon right multiply CO2 by 2

C2H6 + O2 2CO2 + H2O

6 hydrogenon left

2 hydrogenon right multiply H2O by 3

C2H6 + O2 2CO2 + 3H2O 4.3



4. Balance those elements that appear in two or more reactants or products.

2 oxygenon left

4 oxygen(2x2)

C2H6 + O2 2CO2 + 3H2O

+ 3 oxygen(3x1)

multiply O2 by 72

= 7 oxygenon right

C2H6 + O2 2CO2 + 3H2O72

remove fractionmultiply both sides by 2

2C2H6 + 7O2 4CO2 + 6H2O

4.3



5. Check to make sure that you have the same number of each type of atom on both sides of the equation.

2C2H6 + 7O2 4CO2 + 6H2O

Reactants Products

4 C12 H14 O

4 C12 H14 O

4 C (2 x 2) 4 C12 H (2 x 6) 12 H (6 x 2)14 O (7 x 2) 14 O (4 x 2 + 6)

4.3


Practice: Balancing EquationsFe (s) + O2 (g) Fe2O3 (s)

C2H4 (g) + O2 (g) CO2 (g) + H2O (g)

Al (s) + HCl (aq) AlCl3 (aq) + H2 (g)

4.3


StoichiometryDefinition: The quantitative study of reactants and products in a chemical reaction.Uses the coefficients in the chemical equation to stand for the number of moles of reactants and products.

2 CO (g) + O2 (g) 2 CO2 (g)2 molecules 1 molecule 2 molecules

2 * 6.022E23 molecules 6.022E23 molecules 2 * 6.022E23 molecules

2 mol 1 mol 2 mol

Mole ratio: The ratio of moles of a reactant or product to any other reactant or product.

• Mole ratio of CO and CO2 is 2:2

• Mole ratio of CO and O2 is 2:1

Stoichiometric calculations use mole ratios to calculate the amount of a product produced or reactant consumed.

4.4


4.4


1. Write balanced chemical equation2. Convert quantities of known substances into moles3. Use coefficients in balanced equation to calculate the

number of moles of the sought quantity4. Convert moles of sought quantity into desired units

Mass Changes in Chemical Reactions

4.4


Methanol burns in air according to the equation2CH3OH + 3O2 2CO2 + 4H2O

If 209 g of methanol are used up in the combustion, what mass of water is produced?

grams CH3OH moles CH3OH moles H2O grams H2O

molar massCH3OH

coefficientschemical equation

molar massH2O

209 g CH3OH1 mol CH3OH32.0 g CH3OH

x4 mol H2O

2 mol CH3OHx

18.0 g H2O1 mol H2O

x =

235 g H2O

4.4


Yields in Chemical Reactions• Yield is the amount of product that is produced in a

chemical reaction.

• Several definitions:

• Theoretical Yield is the calculated amount of product that would result from the amount of reactants present.

• Actual Yield is the amount of product that is actually obtained from a reaction

• % Yield = Actual Yield ÷ Theoretical Yield x 100

4.4


Actual chemical reactions rarely involve exact stoichiometric amounts.

If an large amount of one reagent is added, some of it may be left over after the reaction is complete.

Reactant used up first is the Limiting Reagent

Reactant left over is the Excess Reagent

Limiting Reagents

4.5


6 green used 6 red left

Limiting Reagents

4.5


Solving Limiting Reagent Problems

Use the reaction coefficients to calculate the amount of each reactant needed to react completely with the other reactant.

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

Example:If we have 125.0 g of CO and 62.5 g of O2, which one is the excess reagent? Which is the limiting reagent?

Once you know the limiting reagent, use that amount to calculate the amount of product that can be produced.

COO2

62.5 g O2 * 1 mol O2/32.00 g O2 * 2 mol CO2/1 mol O2 = 3.91 mol CO2

4.5


Do You Understand Limiting Reagents?

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

2Al + Fe2O3 Al2O3 + 2FeCalculate the mass of Al2O3 formed.

g Al mol Al mol Fe2O3 needed g Fe2O3 needed

ORg Fe2O3 mol Fe2O3 mol Al needed g Al needed

124 g Al1 mol Al27.0 g Al

x 1 mol Fe2O3

2 mol Alx

160. g Fe2O3

1 mol Fe2O3x = 367 g Fe2O3

Start with 124 g Al need 367 g Fe2O3

Have more Fe2O3 (601 g) so Al is limiting reagent 4.5


Use limiting reagent (Al) to calculate amount of product thatcan be formed.

g Al mol Al mol Al2O3 g Al2O3

124 g Al1 mol Al

27.0 g Alx

1 mol Al2O3

2 mol Alx 102. g Al2O3

1 mol Al2O3x = 234 g Al2O3

2Al + Fe2O3 Al2O3 + 2Fe

4.5


Combustion Reactions• Reactions can be classified into different types depending on the

reactants involved

• Combustion reactions are any reaction between oxygen and another substance.

• Burning fuels (butane, ethanol, benzene)

• Cellular respiration

• Reactions of non-metals with oxygen (rusting)

• If reactant contains: carbon, hydrogen, another nonmetal, a metal

• Product is: CO2, H2O, a nonmetal oxide (P4O10), a metal oxide (Fe2O3)


3.6

g CO2 mol CO2 mol C g C

g H2O mol H2O mol H g H

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

Combust 11.5 g ethanolCollect 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

4.0 g O = 0.25 mol O

Empirical formula C0.5H1.5O0.25

Divide by smallest subscript (0.25)

Empirical formula C2H6O

Combustion Analysis


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