# Chemical Equations - PowerPoint PPT Presentation

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Chemical Equations. Balancing equations and applications. Objectives. Write and balance chemical equations Perform calculations using moles and molar masses Determine percent composition from formula Determine empirical formula from percent composition Determine molecular formula.

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Chemical Equations

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## Chemical Equations

Balancing equations and applications

### Objectives

• Write and balance chemical equations

• Perform calculations using moles and molar masses

• Determine percent composition from formula

• Determine empirical formula from percent composition

• Determine molecular formula

### The chemical equation

aA + bB = cC + dD

The Law of Conservation of Matter states that matter is neither created nor destroyed

All the atoms on the left must be the same as those on the right

ELEMENT or

COMPOUND

coefficient

Reactant

side

Product

side

### Chemical book-keeping

• The key to writing correct equations is to ask the question, “Have I gained or lost any atoms?”

• Another thing is to put down the correct formula for each reactant or product

• Formulas cannot be changed in order to balance the equation

In the reaction of hydrogen with oxygen to produce water, the reactants are the elements H2 and O2, and the product is H2O

Count the atoms: 4 H and 2 O 4 H and 2 O

The big number multiplies every atom after it

The subscript only multiplies the atom before it

### Balance the equations

• CH4 +O2 = CO2 + H2O

• CH4 + 2O2 = CO2 + 2H2O

• C3H8 + O2 = CO2 + H2O

• C3H8 + 5O2 = 3CO2 + 4H2O

• N2 + H2 = NH3

• N2 + 3H2 = 2NH3

• Do balancing equation exercises

### Molecules or moles

• The numbers (coefficients) in chemical equation can refer to molecules

• But for practical applications, we need a more useful number: we cannot count molecules

### The Mole

• The mole is a unit of quantity used in chemistry to measure the number of atoms or molecules

• DEFINITION:

• The number of atoms in exactly 12 g of 12C

• A mole of anything always has the same number of particles: atoms, molecules or potatoes – 6.02 x 1023 – Avogadro’s number

### Atomic and molecular weights

• Two scales:

• Atomic mass unit scale

• The mass of an individual atom or molecule in atomic mass units (amu)

• Molar mass scale

• The mass of a mole of atoms or molecules in grams

• Confusing?...

### The Good News

• The weight of an atom in amu has the same numerical value as its molar mass in grams

• The atomic mass of carbon is 12 amu

• The molar mass of carbon is 12 g

• The formula mass of H2O is 18 amu

• The molar mass of H2O is 18 g

### Examples

• How many atoms are in 6.94 g of lithium if the atomic mass of Li is 6.94 amu?

• 6.02 x 1023

• What is the molar mass of H2O if the atomic mass of H = 1 amu and O = 16 amu

• 18 g

### How many H atoms in 1 mol of CH4?

• 1 mol = 6.02 x 1023 particles

### How many moles of O2 in 64 g of oxygen?

• Atomic mass of O = 16 AMU

### Calculate the formula mass of Na2SO4 in AMU.

Use atomic mass Na = 23 AMU, O = 16 AMU, S = 32 AMU

### Significance of formula unit

• Ionic compounds do not contain molecules. Simplest formula is the formula unit

• Covalent compounds, the molecular formula is the formula unit

### Percent composition and empirical formula

• Chemical analysis gives the mass % of each element in the compound

• Molar masses give the number of moles

• Obtain mole ratios

• Determine empirical formula

### Determining percent composition

• Percent composition is obtained from the actual masses.

Example:

Sample contained 0.4205 g of C and 0.0795 g of H.

Total mass = 0.5000 g (0.4205 + 0.0795)

Therefore: in 100 g there are:

(84.10 %)

(15.90 %)

Percent composition: 84.10 % C, 15.90 % H

### Percent composition from formula

• What is percent composition of C5H10O2?

• 1 mol C5H10O2 contains 5 mol C, 10 mol H and 2 mol O atoms

• Mass of each element

• Total mass = 102.13 g

### Convert masses into percents

• Percent composition:

58.80 % C + 9.870 % H + 31.33 % O = 100.00%

### Empirical formula from percent composition: 84.1 % C, 15.9 % H

• Convert percents into moles

84.10 g of C ≡ 7.00 mol C

15.9 g of H ≡ 15.8 mol H

• Determine mole ratio

Mole ratio H:C =

• Simplest formula (decimal form): C1H2.26

• Make smallest integers by multiplying

C4H9

May require rounding. Errors in real data cause problems

• Do percent composition and empirical formula exercises

### What is percent C content of C2H6?

• Molar mass C = 12 g/mol; molar mass H = 1 g/mol

### Empirical formula with more than two elements

• Percent composition of vitamin C is:

• 40.9 % C, 4.58 % H, 54.5 % O

• Convert into moles

• Determine mole ratios

• Find lowest whole numbers

### Inaccuracy can lead to ambiguous or incorrect formulas

• What if H:C is 2.20 rather than 2.26? An error of only 3 %

• Formula becomes C5H11 rather than C4H9

• What if H:C is 2.30 rather than 2.26? An error of only 2 %

• Formula becomes C3H7

• Sometimes chemical intuition is required: we know there is FeO, Fe3O4 and Fe2O3; so a formula FeO3 would indicate an error

### Rounding or not: the role of chemical intuition

• Formulae are always written with integers

• Experimental ratios are always fractions

• Two choices:

• Round to nearest whole number

• Multiply top and bottom to find ratio of whole numbers with same value

• Choice depends on the type of substance

### Hydrocarbons: A case for not rounding

• There are millions of different hydrocarbons

• What if H:C is 2.20 rather than 2.26? An error of only 3 %

• Formula becomes C5H11 rather than C4H9

• What if H:C is 2.30 rather than 2.26? An error of only 2 %

• Formula becomes C3H7

• All formulae are reasonable

• So how do I know what the composition is?

### Inorganic compounds: Rounding makes sense

• Inorganic compounds tend to have few compositions

• Iron forms three oxides: FeO, Fe3O4 and Fe2O3

• Experimental formula FeO1.75 would indicate Fe2O3 not FeO2

### Practice empirical formula problem

• A compound contains 62.1 % C, 5.21 % H, 12.1 % N and 20.7 % O. What is the empirical formula?

### Empirical and molecular formula

• Percent composition gives the empirical (simplest) formula. It says nothing about the molecular formula.

• Molecular formula describes number of atoms in the molecule

• May be much larger than the empirical formula in the case of molecular covalent compounds

• For ionic compounds empirical formula = “molecular” formula

### Determination of molecular formula

• Require:

• Empirical formula from percent composition analysis

• Molar mass from some other source

• Number of empirical formula units in molecule:

• There are n (AaBbCc) in molecule:

• Molecular formula is AnaBnbCnc

### Molecular formula of vitamin C

• Empirical formula of vitamin C is C3H4O3

• Molar mass vitamin C is 176.12 g/mol

• Mass of empirical formula = 88.06 g/mol

• (3 x 12.01 + 4 x 1.008 + 3 x 16.00)

• Number of formula units per molecule =

• Molecular formula = 2(C3H4O3) = C6H8O6

### Ionic solids do not have molecular formulas

• Infinite lattices like ionic solids and covalently bonded lattices are not molecular.

• The formula used for an ionic compound is the same as the empirical formula – with one or two exceptions

• Hg2Cl2 rather than HgCl

• The molecular formula indicates the number of atoms in the molecule

• The structural formula indicates how those atoms are arranged

• C2H6O is the molecular formula for ethanol and dimethyl ether

• Structural formula for ethanol is CH3CH2OH

• Structural formula for ether is CH3OCH3

• In a recipe we would need to use the structural formula to identify the correct reagent

### In some cases the mole contents of the compound are obtained indirectly

• Analysis of the hydrocarbon is performed by combustion.

• Mole ratios of the elements are derived from the mole ratios of the combustion products CO2 + H2O

(1 mol H2O ≡ 2 mol H)

(1 mol CO2 ≡ 1 mol C)

### The molecular or empirical formula can be used to determine the percent composition

• Formula of aspirin is C9H8O4

• Molar mass is 180 g

• Mass of C = 108 g

• Mass of H = 8 g

• Mass of O = 64 g

• % C = 108/180 x 100 %

• % H = 8/180 x 100 %

• % O = 64/180 x 100 %