the factor label method u a way to solve math problems in chemistry u used to convert km to miles, m...

The factor label methodThe factor label method A way to solve math problems in chemistry Used to convert

km to miles, m to km, mol to g, g to mol, etc. To use this we need: 1) desired unit, 2) given

unit, 3) conversion factors Conversion factors are relationships

expressed as a fractionE.g. for 1 km=0.6 miles the conversion factor is

km 1

miles 0.6 or

miles 0.6

km 1

The steps to followThe steps to followNow we are ready to solve problems using the

factor label method. The steps are:

1. Write down the starting unit

2. Write down the desired units

3. Determine what conversion factors you can use

4. Make sure the conversion factor has the starting unit on the bottom and the next unit on top

5. Multiply and divide to eliminate units you don’t want and leave units you do want

Factor label exampleFactor label exampleQ - How many kilometers are in 47 miles?

(note: 1 km = 0.621 miles)

First write down the desired quantity

# km



Next, equate desired quantity to the given quantity

# km = 47 mi



Now we have to choose a

conversion factor

# km = 47 mi



What conversion factors are possible?

# km = 47 mi 1 km 0.621 mi

0.621 mi 1 km



Pick the one that will allow you to cancel

out miles

# km = 47 mi 1 km 0.621 mi

0.621 mi 1 km

Pick the one that will allow you to cancel

out miles



# km = 47 mi 1 km 0.621 mi

0.621 mi 1 km



Multiply given quantity by chosen conversion factor

# km = 47 mi x 1 km 0.621 mi



Cross out common factors

# km = 47 mi x 1 km 0.621 mi



Are the units now correct?

# km = 47 x 1 km 0.621



Yes. Both sides have km as units.

= 47 x 1 km 0.621

# km



Now finish the math.

# km = 47 x 1 km 0.621

= 75.7 km



The final answer is 75.7 km

# km = 47 x 1 km 0.621

= 75.7 km

Moles in chemical equations The coefficients tell us how

many moles of each substance

2 Al2O3 Al + 3 O2

2 Na + 2 H2O 2 NaOH + H2

A balanced equation is a Molar Ratio

Mole to Mole conversions 2 Al2O3 Al + 3 O2

• each time we use 2 moles of Al2O3 we make 3 moles of O2

2 moles Al2O3

3 mole O2

or2 moles Al2O3

3 mole O2

Mole to Mole conversions How many moles of O2 are

produced when 3.34 moles of Al2O3 decompose?

2Al2O3 Al + 3O2

3.34 mol Al2O3 2 mol Al2O3

3 mol O2 = 5.01 mol O2

If you know the amount of ANY chemical in the reaction,

you can find the amount of ALL the other chemicals!

Practice: 2C2H2 + 5 O2 4CO2 + 2 H2O

• If 3.84 moles of C2H2 are burned, how many moles of O2 are needed?(9.6 mol)

•How many moles of C2H2 are needed to produce 8.95 mole of H2O? (8.95 mol)

•If 2.47 moles of C2H2 are burned, how many moles of CO2 are formed? (4.94 mol)

End Day 1

Calculations with Molar MassCalculations with Molar Mass

molar mass

Grams Moles

Molar mass is a conversion factor for grams to moles

Converting Moles and GramsConverting Moles and Grams

Aluminum is often used for the structure of light-weight bicycle frames. How many grams of Al are in 3.00 moles of Al?

3.00 moles Al ? g Al

1. Molar mass of Al = 27 g/mol

1 mole Al = 27.0 g Al

2. Conversion factors for Al

27.0g Al or 1 mol Al

1 mol Al 27.0 g Al

3. Setup 3.00 moles Al x 27.0 g Al

1 mole Al

Answer = 81.0 g Al

Practice How many grams are there in 2 moles of H20?

2 mol H20 (18.0 g H20 /1 mol H20) =

36.0 g H20 How many moles of NaOH are in 160 grams?

160 g NaOH (1 mol NaOH / 40.0 g NaOH)=

4.0 mol NaOH How many grams are there in 3.5 moles of NaCl?

3.5 mol NaCl (48.5 g NaCl/1 mol NaCl)=

169.75 g NaCl How many moles of AgNO3 are in 87 grams?

87 g AgNO3 (1 mol AgNO3/ 169.9 g AgNO3)=

.51 mol AgNO3

End Day 2

Mass to mass conversions Must go through moles!!!! Use factor label method and make sure

of your units. Remember that the top of one

conversion must cancel with the bottom of the next.

Mass-Mass Problem:Mass-Mass Problem:

6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed?

4Al + 3O2 2Al2O3

=6.50 g Al

? g Al2O3

1 mol Al

26.98 g Al 4 mol Al

2 mol Al2O3

1 mol Al2O3

101.96 g Al2O3

(6.50 x 1 x 2 x 101.96) ÷ (26.98 x 4 x 1) = 12.3 g Al2O3

are formed

Another example:

4 Al + 3 O2 2 Al2O3

How many grams of Al is needed to react with 100 grams O2 ?

100 g O2 X 1 mol O2 X 4 mol Al X 27.0 g Al =

32.0 g O2 3 mol O2 1 mol Al

(100 X 1 X 4 X 27.0) / (32.0 X 3 X 1) =

112.5 g Al

Another example: If 10.1 g of Fe are added to a

solution of CuSO4, how many grams of Cu would form?

2Fe + 3CuSO4 Fe2(SO4)3 + 3Cu

Answer = 17.2 g Cu

the factor label method u a way to solve math problems in chemistry u used to convert km to miles, m...

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