Moles. Balanced chemical equations Molar ratios Mass Composition Empirical and Molecular Mass Predicting Quantities Equations

Moles

• Balanced chemical equations

• Molar ratios

• Mass Composition

• Empirical and Molecular Mass

• Predicting Quantities

• Equations

By definition:

1 atom

C “weighs” 12 amu On this scale

1

H = 1.008 amu

16

O = 16.00 amu

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

Micro World

atoms & molecules

Macro World grams

3.1

3.2

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

C 1 mol = N

= 6.0221367 x 10

Avogadro’s number (N

)

Molar mass is the mass of 1 mole of in grams Na atoms

Pb atoms Kr atoms

Li atoms

1 mole

C atoms = 6.022 x 10

atoms = 12.00 g 1 mole

C atoms = 12.00 g

C

1 mole lithium atoms = 6.941 g of Li

For any element

atomic mass (amu) = molar mass (grams)

3.2

Species

Quantity

Number of H atoms

H

1 mole

6.022 x 10 ²³

Species

Quantity

Number of H ₂ molecules

H ₂

1 mole

6.022 x 10 ²³

Species

Quantity

Number of Na atoms

Na

1 mole

6.022 x 10 ²³

Species

Quantity

Number of C ₆ H ₆ molecules

C ₆ H ₆

1 mole

6.022 x 10 ²³

1 mol of atoms = 6.022 x 10 ²³ atoms

6.022 x 10 ²³ molecules 6.022 x 10 ²³ ions

1 mol of molecules =

1 mol of ions =

Converting Units

Unit Factor Label Method

I. Write Unit of Answer

II. Write Starting Quantity from Problem

III. Add appropriate unit factor(s) to cancel out

starting quantity and put in unit of answer.

Understanding 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 10

atoms K

0.551 g K 1 mol K 39.10 g K

x x 6.022 x 10

atoms K 1 mol K = 8.49 x 10

atoms K

3.2

Atomic mass iron = 55.85

How many moles of iron does 25.0 g of iron represent?

Conversion sequence: grams Fe → moles Fe

1 mol Fe (grams Fe)

55.85 g Fe

 

 

 

1 mol Fe (25.0 g Fe)

55.85 g Fe

  =

 

  0.448 mol Fe

Set up the calculation using a conversion factor

between moles and grams.

Molar mass Na = 22.99 g

Conversion sequence: atoms Na → grams Na

23

22.99 g Na (atoms Na)

6.022 x 10 atoms Na

 

 

 

What is the mass of 3.01 x 10 ²³ atoms of sodium (Na)?

23

23

22.99 g Na (3.01 x 10 atoms Na)

6.022 x 10 atoms Na

  =

 

  11.5 g Na

Set up the calculation using a conversion factor

between grams and atoms.

Molecular mass (or molecular weight) is the sum of the atomic masses (in amu) in a molecule.

SO

1S 32.07 amu

2O + 2 x 16.00 amu SO

64.07 amu

For any molecule

molecular mass (amu) = molar mass (grams)

1 molecule SO

= 64.07 amu 1 mole SO

= 64.07 g SO

3.3

2 C = 2(12.01 g) = 24.02 g 6 H = 6(1.01 g) = 6.06 g 1 O = 1(16.00 g) = 16.00 g 46.08 g

Calculate the molar mass of C ₂ H ₆ O.

1 Li = 1(6.94 g) = 6.94 g 1 Cl = 1(35.45 g) = 35.45 g 4 O = 4(16.00 g) = 64.00 g 106.39 g

Calculate the molar mass of LiClO ₄ .

Calculate the molar mass of (NH ₄ ) ₃ PO ₄ .

3 N = 3(14.01 g) = 42.03 g

12 H = 12(1.01 g) = 12.12 g

1 P = 1(30.97 g) = 30.97 g

4 O = 4(16.00 g) = 64.00 g

149.12 g

In dealing with diatomic elements (H ₂ , O ₂ ,

N ₂ , F ₂ , Cl ₂ , Br ₂ , and I ₂ ), distinguish between

one mole of atoms and one mole of

molecules.

Calculate the molar mass of 1 mole of H atoms.

1 H = 1(1.01 g) = 1.01 g

Calculate the molar mass of 1 mole of H ₂ molecules.

2 H = 2(1.01 g) = 2.02 g

How many grams of (NH ₄ ) ₃ PO ₄ are contained in 2.52 moles of (NH ₄ ) ₃ PO ₄ ?

Conversion sequence: moles (NH ₄ ) ₃ PO ₄

→ grams (NH ₄ ) ₃ PO ₄

4 3 4

4 3 4

149.12 grams (NH ) PO Use the conversion factor:

1 mole (NH ) PO

4 3 4

(2.52 mol (NH ) PO ))

4 3 4

149.12 g (NH ) PO 1 mol (NH ) PO

 

 

 

4 3 4

= 376g (NH ) PO

The molar mass of (NH ₄ ) ₃ PO ₄ is 149.12 g.

(56.04 g N )

1 mol N 28.02 g N

 

 

 

23

2 2

6.022 x 10 molecules N 1 mol N

 

 

 

56.04 g of N ₂ contains how many N ₂ molecules?

The molar mass of N ₂ is 28.02 g.

Conversion sequence: g N ₂ → moles N ₂ → molecules N ₂

Use the conversion factors

2 2

1 mol N 28.02 g N

23

2 2

6.022 x 10 molecules N 1 mol N

24

= 1.204 x 10 molecules N

2 2

1 mol N 28.02 g N

 

 

 

(56.04 g N )

23

2 2

6.022 x 10 molecules N 1 mol N

 

 

 

56.04 g of N ₂ contains how many N ₂ atoms?

The molar mass of N ₂ is 28.02 g.

Conversion sequence: g N ₂ → moles N ₂ → molecules N ₂ → atoms N

Use the conversion factors

2 2

1 mol N 28.02 g N

23

2 2

6.022 x 10 molecules N

1 mol N

2 atoms N 1 molecule N

24

= 1.204 x 10 molecules N

2 atoms N 1 molecule N

 

 

 

If the formula of a compound is known, a two-step process is needed to calculate the percent composition.

Step 1 Calculate the molar mass of the formula.

Step 2 Divide the total mass of each

element in the formula by the

molar mass and multiply by

100.

total mass of the element

x 100 = percent of the element

molar mass

Step 1 Calculate the molar mass of H ₂ S.

2 H = 2 x 1.01g = 2.02 g 1 S = 1 x 32.07 g = 32.07 g 34.09 g

Calculate the percent composition of hydrosulfuric

acid H ₂ S.

Calculate the percent composition of hydrosulfuric acid H ₂ S.

Step 2 Divide the mass of each element by the molar mass and multiply by 100.

H 5.93%

S 94.07%

32.07 g S

S: (100) 94.07%

34.09 g

 

  =

 

2.02 g H

H: (100) = 5.93%

34.09 g

 

 

 

Percent Composition

From Experimental Data

Percent composition can be calculated from experimental data without knowing the composition of the compound.

Step 1 Calculate the mass of the compound formed.

Step 2 Divide the mass of each element by the total mass of the

compound and multiply by 100.

Step 1 Calculate the total mass of the compound 1.52 g N

3.47 g O 4.99 g

A compound containing nitrogen and oxygen is found to contain 1.52 g of nitrogen and 3.47 g of oxygen.

Determine its percent composition.

= total mass of product

Calculate the percent composition of hydrosulfuric acid H ₂ S.

Step 2 Divide the mass of each element by the total mass of the

compound formed.

3.47 g O

(100) = 69.5%

4.99 g

 

 

 

1.52 g N

(100) = 30.5%

4.99 g

 

 

 

30.5%

O 69.5%

Percent composition of an element in a compound =

n x molar mass of element

molar mass of compound x 100%

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

C

H

O

%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%

3.5

Empirical Formula versus

Molecular Formula

• The empirical formula or simplest formula gives the smallest whole-number ratio of the atoms present in a compound.

• The empirical formula gives the

relative number of atoms of each

element present in the compound.

• The molecular formula is the true formula of a compound.

• The molecular formula represents the

total number of atoms of each element

present in one molecule of a compound.

C ₂ H ₄ Molecular Formula

CH ₂ Empirical Formula

C:H 1:2 Smallest Whole

Number Ratio

C ₆ H ₆ Molecular Formula

Empirical Formula CH

C:H 1:1 Smallest Whole

Number Ratio

H ₂ O ₂ Molecular Formula

HO Empirical Formula

H:O 1:1 Smallest Whole

Number Ratio

Two compounds can have identical

empirical formulas and different molecular

formulas.

Calculating

Empirical Formulas

Step 1 Assume a definite starting quantity (usually 100.0 g) of the compound, if not given, and express the mass of each element in grams.

Step 2 Convert the grams of each element into moles of each

element using each element’s

molar mass.

Step 3 Divide the moles of atoms of each element by the moles of atoms of the element that had the smallest value

– If the numbers obtained are whole numbers, use them as subscripts and write the empirical formula.

– If the numbers obtained are not

whole numbers, go on to step 4.

Step 4 Multiply the values obtained in step 3 by the smallest numbers that will

convert them to whole numbers

Use these whole numbers as the subscripts in the empirical

formula.

FeO _1.5

Fe _{1 x 2} O _{1.5 x 2} Fe ₂ O ₃

• The results of calculations may differ from a whole number.

– If they differ ±0.1 round off to the next nearest whole number.

2.9 → 3

– Deviations greater than 0.1 unit from a

whole number usually mean that the

calculated ratios have to be multiplied by

a whole number.

A gaseous compound containing carbon and hydrogen was analyzed and found to consist of 83.65% carbon by mass. Determine the empirical formula of the

compound.

1.Obtain the mass of each element present (in grams).

Assume you have 100 g of the compound.

83.65% C = 83.65 g C (100.00 – 83.65)

16.35% H = 16.35 g H

2. Determine the number of moles of each type of atom

present.

3. Divide the number of moles of each element by the smallest number of moles to convert the smallest number to 1. If all of the numbers so obtained are

integers (whole numbers), these are the subscripts in

the empirical formula. If one or more of these numbers

are not integers, go on to step 4.

4. Multiply the numbers you derived in step 3 by the smallest integer that will convert all of them to whole numbers.

This set of whole numbers represents the

subscripts in the empirical formula.

The balanced chemical equation represents the ratio Of one species to any others in the equation:

NaOH (aq) + HCl (aq)  NaCl (aq) + H

O (l) 1mole of NaOH produces 1mole of NaCl

1 mol NaCl 1mol NaOH

2H

O

(l)  2 H

O (l) + O

(g)

2 moles of H

O

(l) produces 1 mole of O

gas 1 mol O

2 mol H

O

(l)

Mole composition gives ratio of each element In a compound

0.345 moles of Al

(CO

)

has the following moles of Al in the compound:

Mol ratio of Al in Al

(CO

)

:

0.345 mol Al

(CO

)

X 2 mol Al = 0.690 mol Al 1 mol Al

(CO

)

Moles of compound X mol ratio (element/molecule)

= mol of element

Mass of Aluminum is obtained from the molar mass:

0.690 mol Al X 26.98 g = 18.62 g Al mol Al

Mol X grams/mol = mass

The mass of any substance can be determined if

The molar mass is known. The molar mass comes

From the periodic table.

Al

(CO

)

continued:

C: 0.345 mol X 3 mol C = 1.04 mol C X 12.011g/mol 1 mol Al

(CO

)

= 12.43 g C

O: 0.345 mol X 9 mol O = 3.11 mol O X 15.999g/mol 1 mol Al

(CO

)

= 49.677 g O

Formula Weight converted to mass from moles:

2 mol X 26.98g/mol (Al) =53.96g 3mol X 12.011g/mol (C) =36.033g 9 mol X 15.999g/mol (O) = 143.991 g

total molar mass 233.984 g in 1 mol Al

(CO

)

233.984 g/mol X 0.345 mol = 80.724 g Al

(CO

)

The empirical formula for a compound is the Smallest whole number ratio between the Elements that make up the compound.

This is not necessarily the actual formula, especially For non-ionic compounds such as organic compounds.

To determine the empirical formula, we must know

The percent composition of each element in a compound The percent composition is a MASS PERCENT and is

Determined for 100 grams of the substance. This has

The effect of converting the percent of each element

Into a RATIO!

Determination of Empirical Formula from Mass %

1. Assume 100 grams of material

2. Determine moles of each element in compound

3. Divide all subscripts in empirical formula

by lowest number because the number of

atoms must be an integer

A compound is discovered to be 63.6% nitrogen

And 36.4% oxygen. What is the empirical formula?

Convert to mass: 63.6% N = 63.6 g N

100 g Compound

36.4% O= 36.4 g O

100 g compound

II. Convert to moles: 63.6 g N X 1 mol N = 4.54 mol N

14.0 g N

36.4 g O X 1 mol O = 2.28 mol O

15.999 g O

The ratio of N to O atoms will be the same as the Ratio of N to O moles:

N: 4.54 = 1.99 O: 2.28 = 1.00 2.28 2.28

III Divide all mols by the SMALLEST number of mols

Empirical Formula= N

O

For organic compounds and some non-organic Compounds, the molecular formula is the actual

Ratio of elements. This is a multiple of the empirical Formula.

We can discover the molecular formula from the

Empirical one if we determine the molar mass from

An experiment. We simply compare the experimental

Molar mass with the empirical one.

Determination of Molecular

Formula from Empirical Formula

olarMass EmpiricalM

larMass MoleculeMo

Multiplier =

Only Empirical Formula can be

Determined by Mass Percent

A phosphorous compound has an empirical formula of P

O

. It has a molar mass of 284 g/mol,

determined in the laboratory.

The empirical formula weight is:

(2 x 30.974 g + 5 x 15.999 g)/mol= 141.943 g/mol Compare the masses in the molar ratio:

284 g/mol = 2; therefore:(P

O

)

is the molecular 141.943g/mol

Formula: P

O

Molar Mass from Experiment = multiplier for E.F.

Empirical formula weight

The real usefulness of balanced equations can be Observed in the following series of predictions:

I Predict moles of any products of a reaction II Predict the amount of any and all reactants needed for a given amount of product

III Determine precisely how much of all reactants are used, and which are present in excess.

The basis of chemical calculations for reactions

Is the balanced reaction equation.

So far, we have seen mole ratio’s from an equation That tells us how much to expect in the reaction:

N

(g) + H

(g)  NH

(g)

N

(g) + 3H

(g)  2NH

(g) (Balanced equation) Think: mols reactant x moles product = mols product

mols reactant

What we are asked for goes on top of the ratio. What we Are given goes on bottom.

How many moles of NH

can be produced from 5.00 mol H

5.00 mol H

(given) x 2 mol NH

= 3.33 mol NH

formed

3 mol H

In the laboratory, we must measure the mols by Mass. This requires an initial conversion to mols From the given mass of substance:

How many moles of NH

are formed from 33.6 g Of N

:

33.6 g x 1 mol N

x 2 mol NH

= 2.40 mol NH

28.0 g N

1 mol N

Mass reactant x 1 x mol product = mol product

F.W react. mol reactant

Finally, we must determine the mass of a product In the laboratory, since we cannot evaluate moles:

What mass of H

is needed to produce 119 g of NH

What is asked for: mass of H

What is given: mass of NH

Mass reactant x 1 x mol product x FW prod =mass F.W react. mol reactant product 119 g NH

x1 mol NH

x 3 mol H

x 2.02 g H

= 21.2 g H

17.0 g NH

2 mol NH

mol H

Memorize these steps:

Mass react x 1 mol react x mol product x prod. g = mass

grams react mol react mol prod.

Mass react x 1 mol react x mol product x prod. g = mass grams react mol react mol prod.

SiO

is etched with HF (aq):

SiO

(s) + 4 HF (aq)  SiF

(g) + 2 H

O (l)

If there is 4.86 moles HF and 60 g SiO

, will there be Any glass left over?

4.86 mol HF x 1 mol SiF

x 104 g SiF

= 126 g SiF

; 4 mol HF mol SiF

126 g SiF

x 1 mol SiF

x SiO

x 60g SiO

= 72.7 g SiO

104 g SiF

SiF

mol SiO

Since I have only 60g (1 mol) of SiO

, I will run out of

Glass before I use up the acid! SiO

is LIMITING

4 Ag (s) + 2 H

S (g) + O

(g)  2 Ag

S (s) + 2 H

O (l) IF there is 0.145 mol Ag, 0.0872 mol H

S, and excess O

, how much Ag

S is produced and what mass of

Reactant is in excess?

Ag: 0.145 mol x 2 mol Ag

S = 0.0725 mol Ag

S 4 mol Ag

H

S: 0.0872 mol H

S x 2 mol Ag

S = 0.0872 mol Ag

S 2 mol H

S

Ag is limiting and H

S is present in excess. We must use the limiting reagent quantity to predict actual products:

0.0725 mol Ag

S x 248 g Ag

S = 18.0 g Ag

S formed

mol Ag

S

A reaction in the laboratory results in a product mass That is not quite what the balanced equation calculation Predicts.

The difference between the theoretical amount and the Actual one is called Percent Yield:

Actual Yield (mass) x 100% = percent yield

Theoretical yield (mass)

2S (s) + 3O ₂ (g) ->2 SO ₃ (g)

Reactant Mixture Product Mixture

S

O

1. Write balanced chemical equation

2. Convert quantities of known substances into moles 3. Use coefficients in balanced equation to calculate the

number of moles of the sought quantity

4. Convert moles of sought quantity into desired units

Mass Changes in Chemical Reactions

Memorize these steps:

Mass react x 1 mol react x mol product x product g = mass grams react mol react mol prod.

molar mass of reactant

molar mass of product mol ratio from

balanced equation

Volume reactant

Grams reactant

Moles reactant Molarity of

reactant

Atoms or molecules Of reactant

* *

Molarity of product

Atoms or molecules Of products moles product

moles reactant a,b,c or d From balanced

equation

Density

(g/mL) (mol/g)

1/N_a Volume

(L)

Volume^-1 (L^-1)

(g/mol) 1/density (mL/g)

N_a

Stoichiometric Relationships

a Reactant A + b Reactant B  c Product C + d Product D a Product A + b Product B  c Reactant C + d Reactant D

or

© 2004 Randal Hallford

Methanol burns in air according to the equation 2CH

OH + 3O

2CO

+ 4H

O

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

grams CH

OH moles CH

OH moles H

O grams H

O molar mass

CH

OH

coefficients

chemical equation

molar mass H

O

209 g CH

OH 1 mol CH

OH 32.0 g CH

OH

x 4 mol H

O

2 mol CH

OH

x 18.0 g H

O

1 mol H

O

x =

235 g H

O

2S (s) + 3O ₂ (g) ->2 SO ₃ (g)

Reactant Mixture Product Mixture

S O

Excess Reagent

Limiting Reagent

Limiting Reagent Problems

1. Balance chemical equation 2. Determine limiting reagent

• Do two separate calculations for the amount of product each reactant would produce if they were the limiting reagent

• The reactant that gives the lower number is the limiting reagent.

3. The amount of product produced is the

number calculated by limiting reagent

Limiting Reagent Problems

4. Determination of the amount of excess reagent left over

• Calculate the amount of excess reagent (ER) used in chemical reaction

• Subtract the ER used from original amount of

ER.

Calculating Amounts of Reactant and Product in Reactions Involving a Limiting Reactant

PROBLEM: A fuel mixture used in the early days of rocketry is composed of two liquids, hydrazine(N₂H₄) and dinitrogen tetraoxide(N₂O₄), which ignite on contact to form nitrogen gas and water vapor.

How many grams of nitrogen gas form when 1.00x10²g of N₂H₄ and 2.00x10²g of N₂O₄ are mixed?

PLAN: We always start with a balanced chemical equation and find the number of mols of reactants and products which have been given.

In this case one of the reactants is in molar excess and the other will limit the extent of the reaction.

mol of N₂ mol of N₂

divide by M

molar ratio mass of N₂H₄

mol of N₂H₄

mass of N₂O₄

mol of N₂O₄

limiting mol N₂

g N₂

multiply by M

Calculating Amounts of Reactant and Product in Reactions Involving a Limiting Reactant

SOLUTION: N₂H₄(l) + N₂O₄(l) N₂(g) + H₂O(l) 1.00x10²g N₂H₄ = 3.12mol N₂H₄

mol N₂H₄ 32.05g N₂H₄

3.12mol N₂H₄ 3 mol N₂ = 4.68mol N₂ 2mol N₂H₄

2.00x10²g N₂O₄ = 2.17mol N₂O₄ mol

N₂O₄

92.02g N₂O₄

2.17mol N₂O₄ 3 mol N₂ = 6.51mol N₂ mol N₂O₄

N₂H₄ is the limiting reactant because it produces less product, N₂, than does N₂O₄.

4.68mol N₂

mol N₂ 28.02g N₂

= 131g N₂

2 3 4

Limiting Reagents

124 g of Al are reacts with 601 g of Fe

O

: 2Al + Fe

O

Al

O

+ 2Fe

Calculate the mass of Al

O

formed.

g Al mol Al mol Fe

O

needed g Fe

O

needed OR

g Fe

O

mol Fe

O

mol Al needed g Al needed

124 g Al 1 mol Al 27.0 g Al

x 1 mol Fe

O

2 mol Al

x 160. g Fe

O

1 mol Fe

O

x = 367 g Fe

O

Start with 124 g Al need 367 g Fe

O

Have more Fe

O

(601 g); Al is limiting reagent

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

g Al mol Al mol Al

O

g Al

O

124 g Al 1 mol Al 27.0 g Al

x 1 mol Al

O

2 mol Al

x 102. g Al

O

1 mol Al

O

x = 234 g Al

O

2Al + Fe

O

Al

O

+ 2Fe

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 = Actual Yield Theoretical Yield

x 100

Calculating Percent Yield

PROBLEM: Silicon carbide (SiC) is an important ceramic material that is made by allowing sand(silicon dioxide, SiO2) to react with

powdered carbon at high temperature. Carbon monoxide is also formed. When 100.0kg of sand are processed, 51.4kg of SiC are recovered. What is the percent yield of SiC in this process?

PLAN:

write balanced equation

find mol reactant & product

find g product predicted

percent yield

actual yield/theoretical yield x 100 SOLUTION:

SiO₂(s) + 3C(s) SiC(s) + 2CO(g)

100.0kg SiO₂ mol SiO₂ 60.09g SiO₂ 10³g SiO₂

kg SiO₂

= 1664 mol SiO₂

mol SiO₂ = mol SiC = 1664 1664mol SiC 40.10g SiC

mol SiC

kg 10³g

= 66.73kg

x100 =77.0%

51.4kg 66.73kg

If a sample of acetylene is burned, and releases 550 KJ of Energy, how many grams of CO

can we expect to form?

2 C

H

(g) + 5O

(g)  4CO

(g) + 2H

O (l)  H=-2602 KJ We can relate the moles of CO

to the amount of heat, since It is part of the balanced equation:

550 KJ x 4 mol CO

x 44.0 g CO

= 37.2 g CO

2602 KJ mol CO

3 ways of representing the reaction of H

with O

to form H

O

Chemical Equations

reactants products

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 water C

H

+ O

CO

+ H

O

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.

2C

H

NOT ^C

^H

Balancing Chemical Equations

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

C

H

+ O

CO

+ H

O start with C or H but not O

2 carbon on left

1 carbon

on right multiply CO

by 2 C

H

+ O

2CO

+ H

O

6 hydrogen on left

2 hydrogen

on right multiply H

O by 3

C

H

+ O

2CO

+ 3H

O

Balancing Chemical Equations

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

2 oxygen on left

4 oxygen (2x2)

C

H

+ O

2CO

+ 3H

O

+ 3 oxygen (3x1)

multiply O

by 7 2

= 7 oxygen on right C

H

+ O 7

2CO

+ 3H

O

2

remove fraction

multiply both sides by 2

2C

H

+ 7O

4CO

+ 6H

O

Balancing Chemical Equations

5. Check to make sure that you have the same

number of each type of atom on both sides of the equation.

2C

H

+ 7O

4CO

+ 6H

O

Reactants Products 4 C

12 H 14 O

4 C 12 H 14 O

4 C (2 x 2) 4 C

12 H (2 x 6) 12 H (6 x 2)

14 O (7 x 2) 14 O (4 x 2 + 6)

I. Formulas show chemistry at a standstill. Equations show chemistry in action.

A. Equations show:

1. the reactants which enter into a reaction.

2. the products which are formed by the reaction.

3. the amounts of each substance used and each substance produced.

B. Two important principles to remember:

1. Every chemical compound has a formula which cannot be altered.

2. A chemical reaction must account for every atom that is used. This is an

application of the Law of Conservation of Matter which states that in a chemical reaction atoms are neither created nor destroyed.

C. Some things to remember about writing equations:

1. The diatomic elements when they stand alone are always written H₂, N₂, O₂, F₂,

Cl₂, Br₂, I₂

2. The sign, --->, means "yields" and shows the direction of the action.

3. A small delta, (∆), above the arrow shows that heat has been added.

4. A double arrow, <--->, shows that the reaction is reversible and can go in

both directions.

5. Before beginning to balance an equation, check each formula to see that it is

correct. NEVER change a formula during the balancing of an equation.

6. Balancing is done by placing coefficients in front of the formulas to insure the

same number of atoms of each element on both sides of the arrow.

7. Always consult the Activity Series of metals and non-metals before attempting to write equations for replacement reactions.

8. If a reactant or product is solid, place (s) after the formula

9. If the reactant or product is a liquid, place (l) after the formula 10. If the reactant or product is a gas, place (g) after the formula 11. If the reactant or product is in water, place (aq) after the formula

12. A category of reaction will produce an unstable product which decomposes:

H₂CO₃ (aq)  H₂O (l) + CO₂ (g) carbonic acid

H₂SO₃ (aq)  H₂O (l) + SO₂ (g) sulfurous acid

NH₄OH (aq)  NH₃ (g) + H₂O (l) ammonium hydroxide

Rules for Writing Chemical Equations:

1. Write down the formula(s) for the reactants and add a “+” between them then put a “yields” arrow () at the end

2. Examine the formula(s) to determine which of four basic types of reactions will occur. On the basis of your decision, write down correct formula(s) for the products on the right side of the yield arrow.

3. Place a coefficient in front of each species in the reactants and the products to ensure that the conservation of matter is observed: “Balance” the reaction

II. Four Basic Types of Chemical reactions A. Synthesis

two or more elements or compounds combine to give a more complex product Examples of Synthesis reactions:

1. Metal + oxygen ---> metal oxide 2Mg_(s) + O_2(g) ----> 2MgO_(s)

2.

Nonmetal + oxygen ---> nonmetallic oxide

C_(s) + O_2(g) ----> CO_2(g) 3.

Metal oxide + water ---> metallic hydroxide

MgO_(s) + H₂O_(l) ----> Mg(OH)_2(s) 4.

Nonmetallic oxide + water ---> acid CO_2(g) + H₂O_(l) ----> ; H₂CO_3(aq)

5.

Metal + nonmetal ---> salt 2 Na_(s) + Cl_2(g) ----> 2NaCl_(s)

6. A few nonmetals combine with each other.

2P_(s) + 3Cl_2(g) ----> 2PCl_3(g)

B. Decomposition:

• A single compound breaks down into simpler compounds.

• Basic form: AX ---> A + X

Examples of decomposition reactions (when heated):

1. Metallic carbonates form metallic oxides and CO_2(g). CaCO_3(s) ----> CaO_(s) + CO_2(g)

2.

Most metallic hydroxides decompose into metal oxides and water. Ca(OH)_2(s) ----> CaO_(s) + H₂O_(g)

3.

Metallic chlorates decompose into metallic chlorides and oxygen.

2KClO_3(s) ----> 2KCl_(s) + 3O_2(g) 4.

Some acids decompose into nonmetallic oxides and water.

H₂SO₄ ----> H₂O_(l) + SO_3(g) 5.

Some oxides decompose.

2HgO_(s) ----> 2Hg_(l) + O_2(g) 6.

Some decomposition reactions are produced by electricity.

2H₂O_(l) ----> 2H_2(g) + O_2(g) 2NaCl_(l) ----> 2Na_(s) + Cl_2(g)

C. Replacement:

• a more active element takes the place of another element and frees the less active one.

• Basic form: A + BX ---> AX + B or AX + Y ---> AY + X

Examples of replacement reactions:

1. Replacement of a metal in a compound by a more active metal.

Fe_(s) + CuSO_4(aq) ----> FeSO_4(aq) + Cu_(s)

2. Replacement of hydrogen in water by an active metal.

2Na_(s) + 2H₂O_(l) ----> 2NaOH_(aq) + H_2(g) Mg_(s) + H₂O_(g) ----> MgO_(s) + H_2(g)

3. Replacement of hydrogen in acids by active metals.

Zn_(s) + 2HCl_(aq) ----> ZnCl_2(aq) + H_2(g)

4. Replacement of nonmetals by more active nonmetals.

Cl_2(g) + 2NaBr_(aq) ----> 2NaCl_(aq) + Br_2(l)

NOTE: Refer to the Activity Series for metals and nonmetals to predict products of replacement reactions. If the free element is above the

element to be replaced in the compound, then the reaction will occur.

If it is below, then no reaction occurs.

If the free element is above the element to be replaced in the compound,

Then the reaction will occur. If the free element is below it, no reaction occurs.

D. Ionic or Double Displacement:

• occurs between ions in aqueous solution. A reaction will occur when a pair of ions come together to produce at least one of the following:

1. a precipitate 2. a gas

3. water or some other non-ionized substance.

• Basic form: AX + BY ---> AY + BX Examples of ionic reactions:

1. Formation of precipitate.

NaCl _(aq) + AgNO_3(aq) ----> NaNO_3(aq) + AgCl_(s) BaCl_2(aq) + Na₂ SO_4(aq) ----> 2NaCl_(aq) + BaSO_4(s) 2.

Formation of a gas.

HCl_(aq) + FeS_(s) ----> FeCl_2(aq) + H₂S_(g) 3.

Formation of water. (If the reaction is between an acid and a base it is called a neutralization reaction.)

HCl_(aq) + NaOH_(aq) ----> NaCl_(aq) + H₂O_(l) 4.

Formation of a product which decomposes.

CaCO_3(s) + HCl_(aq) ----> CaCl_2(aq) + CO_2(g) + H₂O_(l)

Use the Solubility Rules to decide whether a product of an ionic reaction is insoluble in water and will thus form a precipitate. If a compound is soluble in water then it should be shown as being in aqueous solution, or left as separate ions. It is, in fact, often more desirable to show only those ions that are actually taking part in the actual reaction.

Equations of this type are called Net Ionic equations.

Combustion of Hydrocarbons:

Another important type of reaction, in addition to the four types above, is

that of the combustion of a hydrocarbon. When a hydrocarbon is burned with sufficient oxygen supply, the products are always carbon dioxide and water vapor. If the supply of oxygen is low or restricted, then carbon monoxide will be produced. This is why it is so dangerous to have an automobile engine running inside a closed garage or to use a charcoal grill indoors.

•

Hydrocarbon (C_xH_y) + O_2(g) ---> CO_2(g) + H₂O_(g)

•

CH_4(g) + 2O_2(g) ----> CO_2(g) + 2H₂O_(g)

•

2C₄H_10(g) + 13O_2(g) ----> 8CO_2(g) + 10H₂O_(g) NOTE:

• Complete combustion means the higher oxidation number is attained.

• Incomplete combustion means the lower oxidation number is attained.

• The phrase "To burn" means to add oxygen unless told otherwise.

Conservation of matter requires identical numbers of atoms on both sides Begin by inspecting the number of each type of atom on each side.

Nitrogen must separate before the reaction Hydrogen must separate before the reaction.

Re-combine in a different chemical species Same atoms, different ratio!

D. Ionic or Double Displacement:

• occurs between ions in aqueous solution. A reaction will occur when a pair of ions come together to produce at least one of the following:

1. a precipitate 2. a gas

3. water or some other non-ionized substance.

• Basic form: AX + BY ---> AY + BX Examples of ionic reactions:

1. Formation of precipitate.

NaCl _(aq) + AgNO_3(aq) ----> NaNO_3(aq) + AgCl_(s) BaCl_2(aq) + Na₂ SO_4(aq) ----> 2NaCl_(aq) + BaSO_4(s) 2.

Formation of a gas.

HCl_(aq) + FeS_(s) ----> FeCl_2(aq) + H₂S_(g) 3.

Formation of water. (If the reaction is between an acid and a base it is called a neutralization reaction.)

HCl_(aq) + NaOH_(aq) ----> NaCl_(aq) + H₂O_(l) 4.

Formation of a product which decomposes.

CaCO_3(s) + HCl_(aq) ----> CaCl_2(aq) + CO_2(g) + H₂O_(l)

Use the Solubility Rules to decide whether a product of an ionic reaction is insoluble in water and will thus form a precipitate. If a compound is soluble in water then it should be shown as being in aqueous solution, or left as separate ions. It is, in fact, often more desirable to show only those ions that are actually taking part in the actual reaction.

Equations of this type are called Net Ionic equations.