Problem Solving

•

Chemistry



college



about solving problems life is all

• •

Chemistry:

it makes sense!

Develop a logical plan (series of steps) from your known to your unknowns

http://www.geneseo.edu/~mcknight/

Problem Solving and Dimensional Analysis

• Many problems in chemistry involve using relationships to convert one to another

unit

of measurement • Conversion factors are relationships between two units – May be exact or measured • Conversion factors generated from equivalence statements – e.g., 1 inch = 2.54 cm can give 2 .

54 cm or 1 in 1 in 2 .

54 cm

Problem Solving and Dimensional Analysis

• •

Arrange conversion factors so given unit cancels

– Arrange conversion factor so given unit is on the bottom of the conversion factor

May “string” conversion factors

– So we do not need to know direct relationships, as long as we can find steps that leads to the desired units (known  unknown) given unit  desired unit given unit  desired unit

“Must have a Plan”

• a conceptual plan is a visual outline that shows the strategic route required to solve a problem • for unit conversion , the plan focuses on units and how to convert one to another • for problems that require equations , the conceptual plan focuses on solving the equation to find an unknown value

Concept Plans and Conversion Factors

• Convert inches into centimeters 1) Find relationship equivalence: 1 in = 2.54 cm 2) Write concept plan in cm 3) Change equivalence into conversion factors with starting units on the bottom 2.54

cm 1 in

A Systematic Approach

• Sort the information from the problem – identify the given quantity and unit, the quantity and unit you want to find, any relationships implied in the problem • Design a strategy to solve the problem (

roadmap

) – Concept plan • sometimes may want to work backwards • each step involves a conversion factor or equation • Apply the steps in the concept plan – check that units cancel properly – multiply terms across the top and divide by each bottom term • Check the answer – double check the set-up to ensure the unit at the end is the one you wished to find – check to see that the size of the number is reasonable • since centimeters are smaller than inches, converting inches to centimeters should result in a larger number

• • • • •

Example: Convert 1.76 yd. to centimeters

Sort information Strategize Follow the concept plan to

solve

the problem Sig. figs. and round Check

Given: Find: Concept Plan:

yd 1.76 yd length, cm m cm

Relationship

1.094 yd = 1 m 1 m = 100 cm

Solution:

1 .79

yd  1 m 1.094

yd  1 00 cm 1 m  160 .

8775 cm

Round: Check:

160.8775 cm = 161 cm Units & magnitude are correct

Practice – Convert 30.0 mL to quarts

(1 L = 1.057 qt)

(Hint: 1000 mL makes 1 L)

• • • • •

Convert 30.0 mL to quarts

Sort information Strategize Follow the plan to

solve

the problem Sig. figs. and round Check

Given: Find: Concept Plan:

mL 30.0 mL volume, qts L qt

Relationship:

1 L = 1.057 qt 1 L = 1000 mL

Solution:

30 .

0 mL  1 L 1000 mL  1 .057

qt 1 L  0 .

03171 qt

Round:

0.03171 qt = 0.0317 qt

Check:

Units & magnitude are correct

Problem Solving with Equations

• When solving a problem using an equation, you are usually given all the variables except the one you want to find • Solve the equation for the variable you wish to find, then substitute and compute

Using Density in Calculations

Density



Mass Volume Volume



Mass Density

Mass



Density



Volume

m, V Concept Plans: D m, D V, D V m

Density Calculations

•

We can use density as a conversion factor between mass and volume!!

– density of H 2 O = 1.0 g/mL \ H 2 O – density of Pb = 11.3 g/cm 3 \ 1.0 g H 2 O = 1 mL 11.3 g Pb = 1 cm 3 Pb How much does 4.0 cm 3 of lead weigh?

4.0 cm 3 Pb x

11.3 g Pb 1 cm 3 Pb

= 45 g Pb

Question:

The mass of fuel in a jet must be calculated before each flight to ensure that the jet is not too heavy. A 747 jet is fueled with 173,231 L of jet fuel. If the density of the fuel is 0.738 g/mL, what is the mass of the fuel in kilograms?

• • • • • Example: What is the mass in kg of 173,231 L of jet fuel whose density is 0.738 g/mL?

Sort information

Given: Find:

173,231 L density = 0.738 g/mL mass, kg Strategize

Concept Plan:

L mL g kg

Relationship

1 mL = 0.738 g, 1 mL = 10 -3 1 kg = 1000 g L Follow the concept plan to

solve

the problem Sig. figs. and round

Solution:

1 73,231 L  1 mL 10 3 L  0.738

g 1 mL  1 kg 1000 g  1 .

33  10 5 kg

Round:

1.3 x 10 5 kg Check

Check:

Units & magnitude are correct

• •

Counting Atoms by Moles

If we can find the mass of a particular number of atoms, we can use this information to convert the mass of an element sample into the number of atoms in the sample.

The number of atoms we use is

6.022 x 10 23

and we call this a

mole

– 1 mole = 6.022 x 10 23 • entities Like 1 dozen = 12 entities

Avogadro’s Number

Example: Calculate the number of atoms in 2.45 mol of copper

Given: Find: Concept Plan: Solution: Check:

2.45 mol Cu atoms Cu mol Cu 6 .

022 atoms Cu  10 23 atoms 1 mol 1 mol = 6.022 x 10 23 atoms 2 .

45 mol Cu  6 .

022  10 23 atoms 1 mol  1.48

 10 24 atoms Cu since the number is slightly greater than twice Avogadro’s number, it make sense

• • • •

Relationship Between Moles and Mass

The mass of one mole of atoms is called the

molar mass

The molar mass of an element, in grams, is numerically equal to the element’s atomic mass, in amu The lighter the atom, the less a mole weighs The lighter the atom, the more atoms there are in 1 g

Mole and Mass Relationships

Substance

hydrogen

Weight of Pieces in 1 atom 1 mole

1.008 amu 6.022 x 10 23 atoms

Weight of 1 mole

1.008 g carbon 12.01 amu 6.022 x 10 23 atoms 12.01 g oxygen 16.00 amu 6.022 x 10 23 atoms 16.00 g sulfur calcium chlorine copper 32.06 amu 6.022 x 10 23 atoms 40.08 amu 6.022 x 10 23 atoms 35.45 amu 6.022 x 10 23 atoms 63.55 amu 6.022 x 10 23 atoms 32.06 g 40.08 g 35.45 g 63.55 g 1 mole sulfur 32.06 g 1 mole carbon 12.01 g

Example: Calculate the # moles of carbon in 0.0265 g of pencil “lead” Given: Find: Concept Plan: Solution: Check:

0.0265 g C mol C g C mol C 1 mol 12.01

g 1 mol C = 12.01 g 0 .

0265 g C  1 mol 12.01

g  2.21

 10 3 mol C since the given amount is much less than 1 mol C, the number makes sense

Example: How many copper atoms are in a copper penny weighing 3.10 g?

Given: Find:

3.10 g Cu atoms Cu

Concept Plan: Solution: Check:

g Cu 1 mol 63.55

g 1 mol Cu = 63.55 g, 1 mol = 6.022 x 10 23 mol Cu atoms Cu 6.022

 10 23 atoms 1 mol 3 .10

g Cu  1 mol Cu 63.55

g Cu  6 .

022  10 23 atoms 1 mol  2.94

 10 22 atoms Cu since the given amount is much less than 1 mol Cu, the number makes sense