matter and energy chapter 1 cem131. matter and its classification

Matter and Energy

Chapter 1CEM131

Matter and Its Classification

MatterCan be subdivided into…

Pure Substances

Mixtures

Solutions

Heterogeneous Mixtures

Pure Substances

Mixtures

Compounds

Elements

Ionic Compounds

Molecular Compounds

Compounds

Elements

Elements can be dividedinto three Groups

Metals

Metalloids

Nonmetals

Chemical Formulas

Oxygen

Carbon

Hydrogen

Oxygen = 1

Carbon = 2

Hydrogen = 6

Oxygen = O

Carbon = C

Hydrogen = H

Oxygen = 1 OCarbon = 2 CHydrogen = 6 H

Oxygen = OCarbon = C2

Hydrogen = H6

Oxygen = OCarbon = C2

Hydrogen = H6Bonds are the holes in this model… little hard to see

So we start with C2

Then we add the rest

C2H5OH

Wait… why not C2H6O or C2OH6?

CH3OH

Physical and Chemical Properties

What is a Physical Property?

Observations can be either

Qualitative or Quantitative

Qualitative is Subjective

Observations can be either

Qualitative or Quantitative

Quantitative is Objective

What types of Quantitative measures can we do?

Mass

Volume

Density

Dimensional Analysis

Ooo, pretty fractions

Try this one.𝑋=( 12 )

2

×12+ 14

Answer = 3/8

1. Square 1/2 = 1/4 2. Multiple that by 1/2 = 1/83. Add that to 1/4, with the least common

denominator as 8 = 3/8

How about this one?

Solve the following for R.

Answer:

1. divide nT from both side of the equation.

Now a little more… try to figure out these conversions.

Hint, check your books for the conversion factors.

Complete the following conversions

A. 1.54 kg = __________gB. 3.46 cm = __________ µmC. 12.4 ml = __________ lD. 2.3 × 103 in = ___________ m

15403460000.0124

58.42

Try these, they are density equations.

A liquid that has a mass of 125 g per 120 ml, what is the density in g/ml?

1.04 g/ml

Now to ramp it up a bit…Calculate the density of the following materials in g/ml.

A. 49 lbs./fl. oz.B. 3.79 × 10-4 kg/l

A. 751.65 g/mlB. 3.79 × 10-4 g/ml

How to use Dimensional Analysis

Let us look at this one again…Calculate the density of the following materials in g/ml.A. 49 lbs./fl. oz.

What are you trying to determine?

Let us look at this one again…Calculate the density of the following materials in g/ml.A. 49 lbs./fl. oz.

How many grams/ml?

Right, but let’s rewrite that…

Let us look at this one again…Calculate the density of the following materials in g/ml.A. 49 lbs./fl. oz.

?𝒈𝒎𝒍

=¿

Now put in what we know already…

Let us look at this one again…Calculate the density of the following materials in g/ml.A. 49 lbs./fl. oz.

?𝒈𝒎𝒍

=𝟒𝟗 𝒍𝒃𝒔𝒇𝒍 .𝒐𝒛 .

Now we need to go from one set of units to the other, do them one at a time. Let’s start with the pounds…

Let us look at this one again…Calculate the density of the following materials in g/ml.A. 49 lbs./fl. oz.

?𝒈𝒎𝒍

=𝟒𝟗 𝒍𝒃𝒔𝒇𝒍 .𝒐𝒛 .

×𝟒𝟓𝟑 .𝟔𝒈𝟏.𝟎 𝒍𝒃𝒔 .

Notice how pounds are above and below the fraction line. These cancel out each other. If we solve it here, we would be in g/fl. oz.

Let us look at this one again…Calculate the density of the following materials in g/ml.A. 49 lbs./fl. oz.

?𝒈𝒎𝒍

=𝟒𝟗 𝒍𝒃𝒔𝒇𝒍 .𝒐𝒛 .

×𝟒𝟓𝟑 .𝟔𝒈𝟏.𝟎 𝒍𝒃𝒔 .

Now do the same with the fluid ounces to milliliters.

Let us look at this one again…Calculate the density of the following materials in g/ml.A. 49 lbs./fl. oz.

?𝒈𝒎𝒍

=𝟒𝟗 𝒍𝒃𝒔𝒇𝒍 .𝒐𝒛 .

×𝟒𝟓𝟑 .𝟔𝒈𝟏.𝟎 𝒍𝒃𝒔 .

×𝟏 𝒇𝒍 .𝒐𝒛 .𝟐𝟗 .𝟓𝟕𝒎𝒍

See how each step is another step towards what we were looking for, now the fl. oz. cancel.

Let us look at this one again…Calculate the density of the following materials in g/ml.A. 49 lbs./fl. oz.

?𝒈𝒎𝒍

=𝟒𝟗 𝒍𝒃𝒔𝒇𝒍 .𝒐𝒛 .

×𝟒𝟓𝟑 .𝟔𝒈𝟏.𝟎 𝒍𝒃𝒔 .

×𝟏 𝒇𝒍 .𝒐𝒛 .𝟐𝟗 .𝟓𝟕𝒎𝒍

Now when we plug and chug the only units left are g/ml… time to hit the calculator.

Let us look at this one again…Calculate the density of the following materials in g/ml.A. 49 lbs./fl. oz.

?𝒈𝒎𝒍

=𝟒𝟗 𝒍𝒃𝒔𝒇𝒍 .𝒐𝒛 .

×𝟒𝟓𝟑 .𝟔𝒈𝟏.𝟎 𝒍𝒃𝒔 .

×𝟏 𝒇𝒍 .𝒐𝒛 .𝟐𝟗 .𝟓𝟕𝒎𝒍

?𝒈𝒎𝒍

=𝟒𝟗×𝟒𝟓𝟑 .𝟔×𝟏𝟏×𝟏×𝟐𝟗 .𝟓𝟕

=𝟐𝟐𝟐𝟐𝟔 .𝟒𝟐𝟗 .𝟓𝟕

=𝟕𝟓𝟏 .𝟔𝟓𝒈𝒎𝒍

Significant Figures

You can only be as precise as your least precise measurement.

Not Accurate or Precise.

Preciseand Accurate

0

1

2

3

4

We account for precise through significant figures.

The question of zeros

Zeros that denote magnitude are not significant.

Zeros that are part of the measurement are significant

Examples

2305.0 = 5 significant digits

0.00456 = 3 significant digits

52,000,000 = 2-8 significant digits, these cases must be determined by context of the measurement.

Problems

1. 345.00 = ______ sig figs 2. 0.0030045 = ______ sig figs 3. 4500.001 = ______ sig figs 4. 4.6 × 106 = ______ sig figs

5

5

7

2

In math, do not consider exact numbers (such as conversion factors)

The answer must have the same number of significant figures as the least precise measured value in the equation.

Examples

2.5 X 3.25 = 8.1 (2 sig figs)

3 X 3.567 = 10 (1 sig fig)

45.3 / 67.9 = 0.667 (3 sig figs)

Problems

1. 3.45 × 3.45 = ____________ 2. 4.9/2 = ______________ 3. 3.45×104 × 9.5674 = _______________ 

11.9

2

3.30×105

Rounding Off AnswersThis is where a lot of students have problems, but the rules are simple.

If the first digit to be dropped is less than 5, leave the last digit kept unchanged.

If the first digit dropped is greater than 5 or is 5 followed by a digit other than zero, raise the last digit kept by one.

If the first digit dropped is 5 followed by only zeros or no other digits, then leave the kept digit unchanged if even and raise by one if odd.

Problems

Given a solution of density 3.87 lbs./gal. how many kilograms of the solution would you have with 1.5 liters of the solution?

0.70 kg

Problems

Nitroglycerin expands at a rate of 1200 times its original volume during detonation, in terms of metric unit volume, how much volume would be generated by a 1.000 pound mass that has a density of 6.870 lbs./ft3?

4947 liters

Problems

Given a solution of density 3.87 lbs./gal. how many kilograms of the solution would you have with 1.5 liters of the solution?

0.70 kg