(A) Unit Conversionsand

(B) Chemical Problem Solving

Chemistry 142 BJames B. Callis, Instructor

Winter Quarter, 2006

Lecture #2

Friedrich von Hardenberg (Novalis), 1772-1801

All Measured Quantities Consist All Measured Quantities Consist of a Number and a Unitof a Number and a Unit

Example calculations using units

Length : A car is 12 feet long, not “12 long”. A person is 6 feet tall, not “6 tall”.

P 2-1 Area : A carpet measuring 3.0 feet(ft) by 4.0 ft has an area of: Area =

P 2-2 Speed and Distance : A car traveling 350 miles(mi) in 7.0 hours(hr) has a speed of:

Speed =

SI system of unitsSI system of units

Derived SI UnitsQuantity Definition of Quantity SI unit

Area Length squared m2

Volume Length cubed m3

Density Mass per unit volume kg/m3

Speed Distance traveled per unit time m/s

Acceleration Change in speed per unit time m/s2

Force Mass times acceleration of object kg m/s2 ( = newton, N)Pressure Force per unit area kg/(m s2) ( = pascal, Pa)Energy Force times distance traveled kg m2/s2

( = joule, J)

Definitions - Mass & Weight

Mass - The quantity of matter an object contains

kilogram - ( kg ) - the SI base unit of mass, is a platinum - iridium cylinder kept in Paris as a standard!

Weight - depends upon an object’s mass and the strength of the gravitational field pulling on it, i.e. w = f = ma.

Conversion Factors : I

Equivalence statements can be turned into conversion factors by dividing one side into the other.

1 mile = 5280 ft

1 in = 2.54 cm

In converting one set of units for another, the one desired is on top in the conversion factor, and the ‘old’ one is on the bottom so the old units are canceled out.

P 2-3: Convert 29,141 ft into miles.

1 mi 5280 ft15280 ft 1 mi

1 in 2.54 cm12.54 cm 1 in

Conversion Factors - II

1.61 km = 1 mi

P 2-4: Convert 5.519 miles into kilometers:

Conversions in the metric system are easy, as 1 km = 1000 m and 1 meter (m) = 100 centimeters (cm) and 1 cm = 10 millimeters (mm).

P 2-5: Convert 8.89 km into cm

Conversion Factors - III

Conversion factors can be combined.P 2-6: Convert 3.56 lbs/hr into units of

milligrams/sec:

LiterLiter = = LL = dm= dm33

cmcm33 = = mLmLmm33

Conversion Factors - IVmetric volume to liters

P 2-7: The volume of the world’s oceans is 1.35 x 109 km3. How many liters of water are in the oceans?

conversion factors: 1 km = 1000 m1 L = 1 dm3 = 10-3 m3 or 1000 L = 1 m3

How to Solve Chemistry Problems(1) Pose the Problem: Employ the Find, Given, Using

Approach Find the unknown quantity, given known quantities and conversion factors, using known or derived formulas (equations). Very often, the problem is stated only in terms of the knowns and the unknown. It is up to you to discover the necessary conversion factors and formulae.

(2) Solve the Problem Symbolically

Rearrange and combine the formulae so that the unknown is on the left side of the equation.

How to Solve Chemistry Problems (cont.)

(3) Perform a units analysis to determine what conversion factors are needed.

(4) Place all numerical factors and their factor labels into the expression (formula) to be evaluated.

(5) Perform the Calculations.

How to Solve Chemistry Problems (cont.)

(6) Check the result:

– Do the numbers make sense?

– Do the units in the factor labels cancel properly to give the expected units for the answer?

Calculate the mass in grams of 1.00 ft3 of Lead (density=11.3 g/ cm3)?

Step 1: Pose the problem.

Step 2: Solve the formula for mass symbolically.

Problem 2-8: Calculation Involving Density (1)

Step 3: Perform units analysis to determine what conversion factors are needed.

Step 4: Place all numerical factors and factor labels into the expression (formula) to be evaluated.

Problem 2-8: Calculation Involving Density (2)

Step 5: Perform the Calculations.

Step 6: Check the Result:Do the numbers make sense?

Do the units in the factor labels cancel properly to give the expected units for the answer?

Problem 2-8: Calculation Involving Density (3)

Buoyancy Method of Density Determination

Based on Archimedes principle:

A body immersed in a fluid is buoyed up by a A body immersed in a fluid is buoyed up by a force equal to the weight of the displaced fluid.force equal to the weight of the displaced fluid.

If a solid weighs x g in air and y g in a fluid of density, z g/cm3, then the density of the solid is:

fluidsolidfluid

solidair

solidair

solid

solidairsolid

mmm

Vm

/1

Find the density of a sample of brass in g/cm3, given its mass in air (75.14 g) and water (66.21 g) and pwater = 0.998 g/cm3), using the buoyancy method.

Step 1: Pose the problem.

Step 2: Solve the formula for density symbolically.

Problem 2-9: Calculation Involving Density (1)

fluidsolidfluid

solidair

solidair

solid

solidairsolid

mmm

Vm

/1

Problem is correctly posed above.

Step 3: Perform units analysis to determine what conversion factors are needed.

Step 4: Place all numerical factors and factor labels into the expression (formula) to be evaluated.

Problem 2-9: Calculation Involving Density (2)

Step 5: Perform the Calculations.

Step 6: Check the Result:Do the numbers make sense?Ans =

Do the units in the factor labels cancel properly to give the expected units for the answer?

Ans =

Problem 2-9: Calculation Involving Density (3)

gcmggsolid

998.0/121.66g14.7514.75

3

Answers to Problems in Lecture 2

1. 12 ft2

2. 50 mi/hr3. 5.5191 mi4. 8.886 km5. 8.89 x 105 cm6. 448 mg/s7. 1.35 x 1021 L8. 3.20 x 105 g = 3.20 x 102 kg9. 8.39 g/cm3