USING PROPORTIONS TO DO MOLE

CONVERSIONS

The first step to doing conversions with proportions is to understand what an equality is.

In most cases, an equality compares two things where one of the numbers is 1.

Examples: ◦ 1 foot = 12 inches◦ 1 meter = 100 cm◦ 1 year = 365 days

Note that these equalities are reversible◦ 1 foot = 12 inches = 1 foot

EQUALITIES

Given value (unit) Unknown value (unit) Standard value (unit) Standard value (unit) How many inches are in 4.9 feet?

◦ The given value is 4.9 and the unit is feet◦ The unknown value is unknown and the units are

inches.◦ The equality we use for the standard value is

1 ft = 12 inches

USING AN EQUALITY IN A PROPORTION

=

Given value (unit) Unknown value (unit) Standard value (unit) Standard value (unit) 4.9 feet ? Inches 1 foot 12 inches

4.9 feet ? Inches 1 foot 12 inches

To solve: Cross multiply and divide.

SETTING UP A PROPORTION

=

=

=

4.9 feet ? Inches 1 foot 12 inches 4.9 = ? 1 12 12 X 4.9 = 58.8

SOLVING WITH ALGEBRA

12 X X 12

=

1 MOLE = MOLAR MASS◦ Molar mass = the atomic mass of an element

from the Periodic Table or the mass of all the elements in a compound.

◦ 1 mole Ne = 20.18g◦ 1 mole NaCl = 58.44g

1 MOLE = 6.02 X 1023

EQUALITIES USING MOLES, GRAMS AND PARTICLES

RELATIONSHIPS BETWEEN MOLES, GRAMS AND PARTICLES

8 moles CO2 ______g CO2 ______p CO2

______mole H2 2.02g H2 ______ p H2

___mole C3H8 _____ g C3H8 3.01x1023pC3H8

3 mole NaCl _____ g NaCl _____ p NaCl

_____mole H2SO4 392.36 g H2SO4 _____ p H2SO4

RELATIONSHIPS BETWEEN MOLES, GRAMS AND PARTICLES

8 moles CO2 352.08g CO2 4.8 x 1023p CO2

1 mole H2 2.02g H2 6.02 x 1023 p H2

0.5 mole C3H8 22.06 g C3H8 3.01x1023pC3H8

3 mole NaCl 175.32 g NaCl 1.8 x 1023 p NaCl

4 mole H2SO4 392.36 g H2SO4 2.4 x 1023p H2SO4

How many grams of CH4 are in 8.16 moles?

8.16 mole CH4 ? grams CH4

1 mole CH4 16.05 g CH4

8.16mole x 16.05g = 131g CH4

How many particles are in 8.16 moles CH4? 8.16 mole CH4 ? particles CH4

1 mole CH4 6.02 x 1023p CH4

8.16mole x 6.02 x 1023p = 4.91 x 1024p

USING PROPORTIONS FOR MOLE/GRAM/PARTICLE CONVERSIONS

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=

THE SIGNIFICANCE OF A BALANCED EQUATION 2 1/4 cups all-purpose flour 1 teaspoon baking soda 1 teaspoon salt 1 cup (2 sticks) butter, softened 3/4 cup granulated sugar 3/4 cup packed brown sugar 1 teaspoon vanilla extract 2 large eggs 2 cups (12-oz. pkg.)

NESTLÉ® TOLL HOUSE® Semi-Sweet Chocolate Morsels 1 cup chopped nuts Yields 60 What amounts would I use if I wanted to make 120?

THE SIGNIFICANCE OF A BALANCED EQUATION In a recipe, the amounts called for can be

doubled or halved. As long as all the amounts are treated the same way, the recipe should turn out.

Balanced equations can be treated the same way.

THE SIGNIFICANCE OF A BALANCED EQUATION When an equation is balanced, the

coefficient numbers represent numbers of atoms, molecules or formula units.

As long as the coefficient numbers remain proportional, the equation remains balanced.

THE SIGNIFICANCE OF A BALANCED EQUATION 2H2 + O2 2H2O Two molecules H2 + one molecule O2 forms

two molecules H2O. What if I changed the 2 in front of H2 to a

4? What would happen to the other coefficient numbers?

What if I changed the 2 in front of the H2O to 12?

THE SIGNIFICANCE OF A BALANCED EQUATION What if I changed the coefficient in front of

O2 to 6.02x 1023? As long as the proportions within the

equation remain the same, the equation remains balanced.

THE SIGNIFICANCE OF A BALANCED EQUATION Therefore, balanced equations can be used

to determine how much of one substance would be needed or produced from another substance in the equation.

This is called stoichiometry.

N2 + 3H2 2NH3# of atoms 2 atoms N 6 atoms H 2 atoms N

6 atoms H

# of moles 1 moles N2 3 mole H2 2 moles NH3

# of particles

6.02x1023 1.8x1024 1.2x1024

Mass 28.02 6.06g 34.08g

Total mass reactants34.08g

Total mass products34.08g

MOLE TO MOLE RATIOS 2C2H6 + 7O2 4CO2 + 6H2O

If 4 moles of C2H6 are reacted, how many moles of O2 are needed?◦ Since C2H6 doubled, O2 will also double.

If 8.94 mole H2O are produced, how many mole, O2 reacted?◦ 8.94 mole H2O = ? Mole O2

6 mole H2O 7 mole O2

= 10.4 mole O2

GRAM GRAM CONVERSIONS

Using Proportions Given amount (g A) = Unknown amount (g B)

Total molar mass A Total molar mass B (molar mass A x # moles) (molar mass B x # moles)

2C2H2 + 5O2 4CO2 + 2H2O

EXAMPLE: How many grams of O2 are needed to produce 135 g H2O

135g H2O = ? g O2 = 599 g O2

36.04g H2O 160g O2

Total molar mass Total molar mass (2 x 18.02g) (5 x 32.00g)