Balancing chemical reactions. This is an example of an unbalanced chemical reaction. There are two oxygen atoms on the left but only one on the right.

balancing chemical reactions

This is an example of an unbalanced chemical reaction. There are two oxygen atoms on the left but only one on the right.

In next slide we show procedure to balance chemicalreactions.

H2O H2 O2+

Procedure for balancing chemical reactions:

1. Choose the molecule with the greatest number of elements. Writein pencil a 1 before this molecule.

2. Find an element in this compound which appears only once more in the reaction. Balance for this element. Write coefficient in pencil.

3. Find another element which appears once more and balance forthis element. Reiterate until all molecules have a coefficient.

4. Multiply all coefficients by divisors's least common multiple.

1 H2O H2 O2+

1 H2O H2 1/2 O2+

1 H2O 1H2 1/2 O2+

2 H2O2H2 1 O2+

NH3 + O2 → N2 + H2O

Please balance the following equation

atomic and molecular mass

Concept 1:There are 90 or so naturally occuring atoms.They are the elements.

Concept 2: The different elements/atoms all have more or less a circular shape. A good analogy is eggs.

Egg tradition is to countthe eggs by the dozen

A dozen hummingbird eggs weigh 12 grams

A dozen quail eggs weigh 140 grams.

A dozen chicken eggs weigh 600 grams

A dozen turkey eggs weigh 1000 grams

A dozen ostrich eggs weigh 10,000 grams

1

Hu 12 g/dz

6

Qu 140 g/dz

20

Ch 600 g/dz

30

Tu 1000 g/dz

100

Os10000 g/dz

1

Hu 12 g/dz

6

Qu 140 g/dz

20

Ch 600 g/dz

30

Tu 1000 g/dz

How much does a dozen chicken eggs weigh?

How much do 3 dozen quail eggs weigh?

If I have 2400 grams of chicken eggs how manydozens of chicken eggs do I have?

If I have 2500 grams of chicken eggs how manydozens of chicken eggs do I have?

B is for boron. Boron is the fifth smallest element.

We count atoms not by the dozenbut by the mole.

There are 10.81 grams per moleof boron atoms.

Atoms are very small. A dozen atoms would not weigh any appreciable amount.

A million atoms is too small a number to be weighedby any ordinary machine.

Nor could a trillion or a quadrillion atoms be weighedan ordinary machine either.

The problem with counting atoms

We can weigh 100,000,000,000,000,000,000,000 atoms.

A mole is 602,214,129,000,000,000,000,000.

couple = 2

dozen = 12

gross = 144

ream = 500

mole = 602,214,129,000,000,000,000,000.

Words used in English to count groups of people or objects

How much does one moleof hydrogen atoms weigh?

How much do two moles of bismuth (Bi) atoms weigh?

If I have 10 grams of carbonatoms, how many moles doI have?

How much does two moleof H atoms together withone mole of oxygen atomsweigh?

How much does one moleof hydrogen atoms weigh?

How much do two moles of bismuth (Bi) atoms weigh?

If I have 10 grams of carbonatoms, how many moles doI have?

How much does two moleof H atoms together withone mole of oxygen atomsweigh?

The molecular weight orMW is the weight of amole of the molecule.

What is the molecular weight of NH3?

How much does two molesof NH3 weigh?

What is the MW of molecular oxygen?

How much does 2.5 molesof oxygen molecules weigh?

If I have 10 grams of molecularoxygen, how many moles do Ihave?

combustion analysis

The molecular formula of ethane is C2H6.

ethane

molecular and empirical formula

the empirical formula is themolecular formula coefficientsdivided by these coefficient’sgreatest common divisor.

The empirical formula of ethane is CH3.

Two types of chemical formula

We determine the empirical formula by combustion analysis.

an unknown organic compound containingC, H and O (for simplicity ignore N or other elementsfor now)

CnHmOp

the empirical formula of an unknown organic compound

Goal: determine values of n, m, and p.

The unbalanced chemical reaction for burning is:

CnHmOp + O2 CO2 + H2O

We find n, m, and p by burning the compound

For this course, burning and combustion mean the same thing .

unbalanced reaction

CnHmOp + O2 CO2 + H2O

1 CnHmOp + O2 n CO2 + m/2 H2Ocomplicatednumber of

balanced reaction

Strategy for combustion analysis

strategy (part 1)

1. Weigh the compound before burning and weigh the carbon dioxide and water after burning.

2. Use number of grams of CO2 and H2O to figure out the number ofmoles of CO2 and H2O. These values will get us values n and m.

3. Use initial weight of compound and n and m to get p.

CnHmOp + O2 n CO2 + m/2 H2Ocomplicatednumber of

Strategy for combustion analysis

strategy (part 2)

1. Weigh the compound before burning and weigh the carbon dioxide and water after burning.

2. Use number of grams of CO2 and H2O to figure out the number ofmoles of CO2 and H2O. These values will get us values n and m.

3. Use initial weight of compound and n and m to get p.

CnHmOp + O2 n CO2 + m/2 H2Ocomplicatednumber of

1. Weigh the original CnHmOp sample.

2. Burn sample compound and weigh in grams the H2O and CO2 produced by burn.

3. Convert grams of H2O and CO2 to number of moles of H2O and CO2.

4. Number of moles of CO2 is n. Number of moles of H2O is m/2.

5. Calculate the weight in grams of m H and and n C.

6. Compare weight from (5) to weight determined in (1). Difference is weight of oxygen atoms.

7. Convert weight of oxygen atoms in (6) to number of moles of oxygen atoms. This number is p.

8. Use n, m. and p and knowledge of fractions to deduce the empirical chemical formula.

Procedure for combustion analysis problems

CnHmOp CnHmOp + O2 n CO2 + m/2 H2Ocomplicatednumber of

A 30.5 g sample of a pure compound containing only C, H, and O is burnt in a combustion analysis apparatus. 66 gof carbon dioxide and 40.5 g of water are accumulated.What is the empirical formula of this compound?

1. Weigh the original CnHmOp sample.

2. Burn sample compound and weigh in grams the H2O and CO2 produced by burn.

3. Convert grams of H2O and CO2 to number of moles of H2O and CO2.

4. Number of moles of CO2 is n. Number of moles of H2O is m/2.

5. Calculate the weight in grams of m H and and n C.

6. Compare weight from (5) to weight determined in (1). Difference is weight of oxygen atoms.

7. Convert weight of oxygen atoms in (6) to number of moles of oxygen atoms. This number is p.

8. Use n, m. and p and knowledge of fractions to deduce the empirical chemical formula.