Properties of matter fall into two categories A Chemical properties those properties that describe how a material reacts with other materials in reactions that change its identity. We will deal with chemical properties in a later unit. B Physical properties those readily observable properties which can be noted without changing the material, such as color, state (solid, liquid or gas) hardness, texture, odor, solubility, density and specific heat.
As we make our observations about size and quantity, always using the metric system, we have two types of measurements (1) fundamental quantities those which can be directly measured such as length, mass, temperature and time, and (2) derived quantities those which must be arrived at by using combinations of the fundamental measurements. Area, volume, density and heat are derived quantities.
By this time, the students will be familiar with the concepts of area and volume. In this unit we will concentrate on the concepts of density and heat content, important measurements in chemistry, which will be unfamiliar to the student..
Density is a measure of the relative heaviness of a substance. In order to compare “heaviness”, it is necessary to compare the masses of equal volumes of that material. The volume unit usually used is cm3 and the mass used is grams. Density is then expressed in grams per cubic centimeter g/cm3. Density is an identifying characteristic of any substance. One cubic centimeter of silver will always have a mass of 10.5 grams. One cubic centimeter of lead will always have a mass of 11.35 grams. Thus we can say the density of lead is 11,35 g/cm3’ and the density of silver is 10.5 g/cm3. These densities, along with the densities of all other elements and many other materials can be found in a table of physical properties in any chemistry text or handbook of physics and chemistry.
If we know the mass and volume of any object, we can compute its density. Density = mass/volume (D M/V). Once again, water becomes the standard. We know that 1 cm of water has a mass of 1 gram. It follows that the density of water will then be 1 gram/1 cm
3
= 1 g/cm
3
. Density expressed in grams per cubic centimeter is also called specific gravity.
The following 5 examples of density problems have been completed using our problem solving method.
Example 1 Find the density of sulfuric acid if 15 cm
3
has a mass of 27.6 g.
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Step 1
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Unknown density of sulfuric acid data V = 15 cm
3
M = 27.6 g
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Step 2
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To find density, we must divide the mass by the volume. We have the necessary data for this.
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Step 3
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Mass = 27.6 g Volume = 15 cm
3
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D = M/V = 27.6g / 15 cm
3
= 1.84 g/cm
3
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Step 4
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In our formula correct? Did we substitute correct numbers? Is our arithmetic correct? Have we handled units correctly? If we can answer yes to all these questions, then our problem is complete.
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Example 2 Find the density of a block of metal which is 16 mm long, 12 mm wide and 5 cm high. The mass of this block is 108960 mg.
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Step 1
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unknown density of block
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data mass = 108960 mg
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length = 16mm
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width = 12 mm
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height 5 cm
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Step 2
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We know that to find our density we want the mass in grams and the volume in cubic centimeters.
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We have been given the mass, but the unit is milligrams.
We will need to convert these to grams . There are 1000 mg in each gram, so we must divide the number of mg by 1000.
We have not been given the volume. We do, however have the necessary data to compute the volume.
If we look at the units given, we will note that some are mm and some are cm. We will need to convert all these measurements to cm and compute the volume using the formula A = L, x W x H
The final step will be to compute density using D=M/V.
Step 3
(figure available in print form)
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Step 4
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Check all calculations as in example 1.
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Example 3 Find the mass in grams of a bar of silver 6 cm long, 10 cm wide and 4 cm high. Silver has a density of 10.5 g/cm
3
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Step 1
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unknown the mass of silver bar
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data D = 10.5 g/cm
3
L = 6 cm W = 10 cm H= 4 cm
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Step 2
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Density tells us how much each cm
3
of a substance weighs. Therefore, since we know the density, we have only to find out how many cm
3
we have (volume) and to multiply the volume by the density. M=D x V We have not been given the volume, but we have the necessary measurements to compute it.
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Step3
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Volume = L x W x H = 6 cm x 10 cm x 4 cm = 240 cm
3
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Mass= D x V = (10.5 g/cm
3
) (240 cm3) = 2520 g.
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Step 4
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Check all calculations.
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In all density problems, only three quantities are needed, mass, volume and density. I feel it is very important to spend some time with the students in discovering how these quantities are interrelated, in the hope that they will be able to see other similar relationships in other types of problems as the course progresses. This should prove much more valuable than simply providing formulas for them to use.
In problems 1 and 2 we were given mass and volume and asked to find density. We find we can use the formula D= M/V in all problems of this type. In problem 3 we were given the density and volume, and asked to find the mass. We find the formula M= V x D will solve problems of this type. The only way a problem involving density could be presented would be to give us the density and the mass, and ask us to find the volume. Let us demonstrate a problem of this type.
Example 4 Find the volume of a container necessary to hold 200 g of mercury. The density of mercury is 13.55 g/cm
3
.
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Step 1
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unknown volume of mercury
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data m = 200 g D =13.55 g/cm
3
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Step 2
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Density tells us that each cm3 weighs 13.55 g.
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We have 200 g. We need to find out how many 13.55 g units are contained in our 200 g. V = M/D
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Step 3
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D = 13.55 g/cm3
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(figure available in print form)
We now have reasoned out three formulas which we can use to solve any conceivable density problem, once we understand our problem. D = M/V V = M/D M = D x V
The learners should now be led to see that once we have derived any one of the formulas used to solve density problems, it is then only a matter of simple manipulation to get the other two. If D= M/V than M = D x V and V = M/D. This type of relationship will be encountered many times throughout the duration of the chemistry course.
Our young problem solver should now be ready to use his knowledge of density to solve a problem of slightly more complexity, involving an equation with one unknown quantity. Simple equations with one unknown should be reviewed here as they too will be used again and again during the year.
Example 5 We have 500 cubic centimeters of a liquid which has a mass of 2000 grams. How much water must be added to this liquid if we want to obtain a mixture with a density of 2 grams per cubic centimeter.
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Step 1
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unknown volume of water to be added
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data V of liquid = 500 cm3 Mass of liquid = 2000 g
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Step 2
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The density of water is 1 gram per cubic centimeter. This tells us that for every cubic cm of water we add, the mass will increase by one gram. We know that our final density will be 2 g/cm3. If we call our unknown quantity of water “x”, then our new mass will be 2000 + x grams and our new volume will be 500 + x cm
3
. D = M/V, therefore we can say
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(figure available in print form)
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Step 3
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v of liquid = 500 cm
3
M of liquid = 2000 g
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Density of mixture = 2 g/cm
3
Density water = 1 g/cm
3
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Let quantity of water to be added x
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(figure available in print form)
____
Our new mass will now be 2000 + 1000 = 3000 g
____
Our new volume will now be 500 + 1000 = 1500 cm
3
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Step 4
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In this case, we can check our answer by computing the density of our new quantities.
|
(figure available in print form)
____
It would appear from this that our problem has been solved correctly.
____
We needed to add 1000 grams of water ( 1 liter) to the liquid to bring the density to 2 g/cm
3
The following list of problems of increasing difficulty are suggested as types for the student to solve. More than one of each type should be assigned. As much attention should be paid to the students method of solving the problem as to his answer.
Suggested problems (answers in parentheses)
-
1. Find the density of ice if 10 cubic centimeters weighs nine grams. (.9 g/cm
3
)
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2, 72 cubic centimeters of alcohol weighs 57.6 grams. Find its density. ( O.8 g/cm
3
)
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3. What is the mass of a 8 centimeter cube which has a density of 0.67 g/cm
3
? ( 343.03 g)
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4. Find the mass of a cylinder of aluminum that has a radius of 3 centimeters and is 10 centimeters high. The density of aluminum is 2.7 g/cm
3
. (763.02 g)
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5. Find the volume of 130 grams of nickel. The specific gravity of nickel is 8.9. (14.61 cm
3
)
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6. A 22.4 liter container full of sea water has a mass of 25.96 kilograms. The empty container weighs 3 kilograms. Find the density of sea water. (1.024 g/cm
3
)
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7. A tube which has a diameter of 20 centimeters and is 30 centimeters high, is half full of a liquid. If the empty tube weighs 34 grams and the density of the liquid is 2.2 g/cm
3
, what is the total mass of the tube and its contents? (10,396 g)
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8. We have 27 kilograms of a liquid whose density is 1.21 g/cm
3
. How many cubic cm of another liquid with a density of .87 g/cm
3
would be have to be added to bring the total weight to 35 kilograms? (9195 cm
3
)
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9. What would be the density of the resulting solution if 1 liter of water was added to 1000 grams of a liquid which has a density of 3 g/cm
3
? (2 g/cm
3
)
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10. Liquid “A” has a density of 3 g/cm
3
and liquid “B” has a density of 2 g/cm
3
. Given 1500 grams of liquid “A”, how many grams of liquid “B” would have to be added to it to obtain a mixture of a density 2.4 g/cm
3
? (1500 grams)
We should now familiarize the students with the use of lab equipment and lab safety procedures. He should be taught to read the triple beam balance with accuracy, and to read the scale on a graduated cylinder, noting the meniscus. This can best be accomplished by weighing and measuring a rarity of materials.
It is also necessary to review percentages with the class. While the following presentation may seem very elementary, I have found that a little time spent now saves a great many problems later on.
The word percent (%) means “out of 100”. To find what percent any part is of its whole, we must first make a fraction and then multiply it by 100.
For example, suppose we have a pie cut in 4 equal pieces. If two are eaten, what percent is gone? 2 out of a possible 4 are gone, or 2/4. The percent would be 2/4 x 100 = 50%. Similarly, if we had eaten 3 pieces, we would have had 3 out of 4 gone, or 3/4 x 100 = 75%.
In chemistry labs it is usually necessary to check our results against some standard. We do this by computing percentage of error. This means, how many part per hundred are we off from the correct or accepted result.
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Example:
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We are given a chain of pure aluminum. We are asked to find its density experimentally, using only a balance, a graduated cylinder and water. We are then asked to cheek our results against the accepted density of aluminum, 2.7 g/cm
3
.
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First, we must solve our problem finding the density in the
let
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Step 1
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Unknown density of chain
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Step 2
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In order to find density, we know we must have mass and volume because D= M/V. We can easily find the mass by weighing our chain on the balance. We can find the volume of the chain by displacement of water in a graduated cylinder. Because 1 ml of water is the same as 1 cm3, if we drop a chain in a cylinder partly filled with water, the volume of the chain will be the difference between the beginning volume in the cylinder and the final reading. Once we have these facts we can easily compute our density.
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Step 3
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Mass of chain = 85.7 g
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Volume of chain =
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initial level of water = 25 m1
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Final level of water chain= 58.2 m1
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chain 58.2 25 = 33.2 ml (33.2 cm
3
)
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(figure available in print form)
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Step 4
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Is our method logical? Are our measurements correct? Have we handled units correctly? Can we verify our calculations’
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Knowing that the correct or accepted density of aluminum is 2.7 g/cm
3
, we can now calculate our percentage of error. First, we must know what our error is. This will be the difference between our accepted value and our experimental value. In this case, the accepted density was 2.7 g/cm3 and our experimental findings were 2.58 g/cm
3
. Our error will be as follows 2.7 g/cm
3
2.58 g/cm
3
= .12 g/cm
3
.
We were in error .12 parts of a possible 2.7, or .12/2.7. Now to make this fraction a percent, we multiply by 100. % error=.12 x 100 = 4.44%.
2.7
We can generalize this, and use the following formula for all percent of error calculations.
Error = accepted value experimental value
(figure available in print form)
Activity Density.
Students should be given several cylinders and blocks of known density. They should calculate the density using careful measurements and calculations and al80 by displacement of water for volume. They should then be asked to compute the percentage of error between their density and the accepted density.
Students will also find it interesting to try to identify objects of unknown metal by finding their density in the lab and using a density or specific gravity table. An abbreviated table will be found at the conclusion of this unit.
Temperature and Heat
Just as density is one property by which we can identify substances, specific heat is another. Temperature and heat are not the same thing. Most students will think the terms are interchangeable. Temperature is a measure of how hot or cold something is, not how much heat energy it contains. Heat tends to be transferred from one object to another until the object and its surroundings are the same temperature.
Scientists measure temperature on the Celsius scale. 0° C. is the freezing point of water. 100° C. is the boiling point of water. The interval between freezing and boiling is divided into 100 equal parts called Celsius degrees.
Heat is measured in units called calories. A calorie is the amount of heat required to raise the temperature of 1 cm3 of water 1° C.
The specific heat of a substance is the number of calories needed to raise the temperature of 1 gram of that substance 1° C. Therefore, the specific heat of water would be 1 calorie. Using this information we should now be able to solve problems involving calories and specific heat. Once again, two or three problems such as the examples which follow, should be worked through in class with the students before assigning a number of problems for them to solve alone.
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Example 1
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How many calories are required to raise the temperature of 10 grams of water from the freezing point to the boiling point.
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Step 1
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unknown calories required
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data substance is water, therefore freezing point (f.p.) is 0° C. and boiling point (b.p.) is 100° C.
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Mass of water 10 grams.
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Step 2
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To compute the number of calories needed to raise a given amount of any substance a specified number of degrees, we need to know how much it takes to raise 1 gram of the substance 1 degree, or the specific heat. We know that the specific heat of water is 1 calorie. We are given the mass of the water (10 grams). We can calculate the change in temperature. The Greek letter delta is used to symbolize “a change in”, and t denotes temperature.
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(figure available in print form)
____
We can reason that if 1 calorie (c) will raise the temperature of 1 gram by 1 degree, then it would take 100 calories to raise the temperature of one gram by 100 degrees, In our problem however, we do not have just one gram, we have 10. It will then be necessary to multiply the number of calories it takes to raise one gram the specified number of degrees by the number of grams that are involved.
____
The following formula should do these things calories = t x specific heat x mass in grams
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Step 3
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Substance water
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Specific heat 1 c/g mass 10 grams initial t = 0° C final t =100° C.
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t = final t initial t = 100 ° C 0° C = 100 0 C.
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calories =
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t x specific heat x mass
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= 100 x 1 c/g x 10 g
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= 1000 c
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Step 4
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Check our work
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Example 2 If adding 1000 calories to 500 grams of a given substance raises its temperature from 28 C. to 32 C., what is the specific heat of that substance?
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Step 1
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unknown specific heat
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data mass = 500 grams
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calories = 1000
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initial t = 28° C
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final t = 32° C.
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Step 2
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We know that specific heat is the number of calories required to raise the temperature of 1 gram of the substance by 1 degree. If we know how many grams we have and how many calories were added, we can compute how many calories were added per gram by dividing the number of calories by the number of grams. In this case it would be 1000 c / 500 g = 2 c/g. Adding these 2 calories per gram changed the temperature from 28° to 32°, or 4°. If 2 calories will change the temperature by 4 degrees, then 1/4th of 2 calories should change the temperature by 1 degree. We now have a formula to solve our problem, or any other like it.
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(figure available in print form)
Step 3
(figure available in print form)
Problems (answers in parentheses)
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1. How many calories would be needed to raise the temperature of 750 grams of water from 4° C. to 81° C? (57750 c)
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2. How many calories would change the temperature of 300 grams of a metal with a specific heat of .63 calories from 2¼ C to 43¼C (8505 c)
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3. Find the specific heat of a cylinder whose mass is 740 grams, if the addition of 600 calories raises the temperature from 10° C to 11.5¼ C. (.54 calories)
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4. What is the mass of a piece of metal if its specific heat is .88 calories and the addition of 1000 calories raises the temperature from 10° C to 25° C. (75.76 g)
This section should be completed by asking the students to experimentally determine the specific heat of samples of aluminum, copper, iron and lead. An outline for this lab procedure is attached, along with an abbreviated table of the physical properties of some common metals. Samples of these metals should be readily available in any high school physics or chemistry lab.
As we complete this unit, it is hoped that both teacher and student will have established a pattern that can be used as each new concept is introduced during the year. We should have now reviewed all the mathematics necessary for a high school chemistry course, and the repeated need to use this math should make further review, as such, unnecessary.
Experiment The specific heat of metals.
Purpose:
To measure and compare the specific heats of several metals and to compute the experimental percent of error. The following materials will be necessary to complete the experiment: samples of A1, Cu, Fe, and Pb; balance: beakers (2 250 ml); bunsen burner; graduated cylinder (100 ml); paper towels; ring stand and ring; thermometer; utility clamp; wire gauze.
Procedure
: The following method is to be used to measure the specific heat of each of the four samples given.
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1, The mass of the metal sample should be determined to the nearest mg using the balance.
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2. The metal should be suspended from a strong piece of thread hanging from the clamp above a 250 ml beaker half full of water. The clamp should then be lowered until the metal is completed submerged in the water, but does not touch the bottom. The clamp is also used to support the thermometer. The apparatus is shown in diagram 1 following this experiment.
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3. Allow the water to come to a boil and keep it boiling until the metal is removed. Record the temperature of the boiling water.
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4. Carefully measure out 100 ml of water in your graduated cylinder. Pour the water into the second 250 ml beaker, which should be carefully placed on a folded paper towel to insulate the beaker from the lab table. Since we know that 1 ml of water has a mass of 1 gram, we can safely assume that the mass of our water is 100 grams.
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5. Carefully record the temperature of the 100 grams of water, carefully reading the thermometer to the closest .1 degree.
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6. Quickly transfer the metal from the boiling water to the water in the second beaker.
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7. Being very careful not to break the thermometer, stir the water around the metal and record the temperature to the nearest O.1 degree, as soon as the temperature stops rising.
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8. Repeat this procedure with each of the 3 other samples.
The following table for the recording of data should be prepared before the experiment is begun. The data will need to be recorded four times, once for each type of metal.
Data Table
1. type of metal _________
2. mass of metal ____
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3. t of boiling water and metal _________
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4, mass of cool water ________
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5. initial t of cool water ______
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6. final t of cool water and metal ___
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7. t. gain of cool water _________(final t of cool water and meta1 minus the initial t of cool water)
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8. t loss of metal ___ (t of boiling water and metal minus final t of cool water and metal)
Calculations
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1. We know the specific heat of water is 1 c/g. This allows us to calculate the total heat gained or lost by any mass of water undergoing heat change as calories = m x t
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2. All heat lost by the metal should be gained by the water if we are careful to insulate the beaker so that other heat transfer will not occur. Since we have insulated our beaker with a paper towel, which does not conduct heat, we can safely use the formula specific heat =
-
(figure available in print form)
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3. Compare the experiential specific heat with the accepted specific heat. ( found in the following table of properties of common metals). Compute the percent of error in each case.
(figure available in print form)
