Experiment 4: Chemical and Physical Changes
Page 32: Physical and chemical properties
| Substance (Formula) | Color | Form | Odor | H2O Solubility | Behavior upon heating |
| Water (H2O) | |||||
| Iodine (I2) | |||||
| Salt (NaCl) | |||||
| Sugar (C12H22O11) | |||||
| Rubbing alcohol (C3H8O) | |||||
| Napthalene(C10H8) | |||||
| Copper(II) bromide (CuBr2) | |||||
| Mineral oil (hydrocarbons) |
This is the table in the lab manual. The physical properties are "Color, Form, and Odor" By "Form" they mean solid or liquid. The chemical properties are "H2O solubility" and "Behavior when heating"
a. Solutions: Comparison between physical change and chemical change.
| Observation | Physical or chemical change? | |
| NaHCO3 in water | ||
| NaHCO3 in 1M HCl solution |
NaHCO3 + HCl -> NaCl + H2O + CO2
This is a chemical change because the starting chemicals changed into different chemicals.
| Observation after mixing with 0.1 M AgNO3 solution | Physical or chemical change? | |
| Salt solution | ||
| Sugar solution | ||
| Tap water | ||
| Distilled water |
Here we are looking for signs of a chemical change. When you mix the salt solution with the silver nitrate solution, you will see the solution turn white. After sitting for awhile, the white will settle to the bottom. Therefore, the mixing created small white particles that settles to the bottom. They call this a precipitate. A precipitate is a good indicator that some chemical change (chemical reaction) occurred. You shouldn't expect anything to happen when adding distilled water because the 0.1M AgNO3 solution already has water in it. Tap water is not pure, so there may be something in it that will produce a precipitate or at least a cloudiness (small amount of precipitate). Here's the reaction you are looking for:
NaCl(aq) + AgNO3(aq) -> AgCl(s) + NaNO3(aq)
The "(s)" means that AgCl is a solid. This is the precipitate that you are looking for.
Some physical properties are color, form (gas/liquid/solid), odor, melting point, boiling point, malleable (easy to bend or reshape), crystalline, amorphous (no crystal structure), density, electrical conductivity (good conductor or insulator), and more. These are properties of the substance itself. Physical properties do not mention how the element or compound behaves with another substance. Without using equipment, you can identify the physical properties of copper regarding color, form, and malleability (it bends easily without breaking).
Freezing a saltwater solution:
In general freezing or melting are physical changes. They don't usually cause a chemical change. Also, if you can imagine that after freezing you let it thaw and it will return to the same condition before freezing. That is just a physical change.
As you know boiling an egg causes a permanent change in the egg because when the egg cools down it doesn't look the same as it did before you heated it up. The heating of the egg causes the protein in the egg to change shape by breaking bonds and forming some new ones. An analogy would be a basket. Toss it gently back and forth and there's no change to the basket (warming); however, if you throw it too fast (high temperature), the woven fibers in the basket will break and rearrange themselves. That is a like a chemical change. The basket is not going to return to its original shape after such vigorous handling. That's like a chemical change.
This is easy. It simply changes from liquid to gas, but it is still water.
This is easy. When the combustion products of gasoline go out the tail pipe (water and carbon dioxide), you no longer have gasoline anymore. There was definitely a chemical change.
The question to determine chemical or physical change is will sugar remain sugar in water or does it undergo a chemical change? Also, does it cause water to undergo a chemical change? Our experience is that sugar in water is still sweet, so it still appears to be unchanged. Also, if sugary water spills and the water evaporates, sugar remains. So this is a physical change. Be aware that there are some substances that are added to water that will undergo a chemical change. For example, when magnesium sulfate is added to water, it bonds to the water and forms a new compound called magnesium sulfate heptahydrate.
This is like burning gasoline. Combustion is always a chemical change. Heat begins to break the fuel down and oxygen comes in to combine with the carbon and hydrogen atoms to make carbon dioxide and water.
color_______ density________ flammability____________
boiling point__________ reactivity of a substance___________
Chemical properties is finding out information about how a substance reacts with other substances. "Flammability" is information about how a substance will react with oxygen when heated. Density, for example, can be determined by weighing a certain volume of a substance. It doesn't require any other substances.
This is easy. You can do this without any help.
This is like the earlier experiments. The percentages of KBr and water are just fractions of the whole weight. KBr is this fraction of the solution: 1.79g/5.00g. If we divide 5.00g into 1.79g, we get the decimal fraction of 0.358 (notice grams cancelled). Percentage is how many hundreds so we re-read 0.358 as 35.8 hundredths (%). Or we can do this math
0.358 x 100 = 35.8 = 35.8%
100 100
If KBr weighed 1.79 grams then water has to be the remaining weight of 3.21g. (5.00g-1.79g). Water's percentage is done the same way. Start with the common fraction 3.21g/5.00g, go to the decimal fraction and then to percent.
This is pretty easy. Density is defined as weight (mass) per a specific volume. Normally in chemistry we use the mass as grams and the volume as milliliters (mL) or the equivalent cubic centimeters. We are given the mass as 44.867g. The volume is calculated of course by subtracting the 30.0 mL from the 33.5 mL to find the volume change. That will be 3.5 mL. So our density is 44.867grams per 3.5 milliliters. The standard density is given as weight per one milliliter.
44.867g = ??? g
3.5 mL 1 mL
The above shows the calculations. Note that 3.5mL has only two significant figures. In other words, the accuracy of the volume was only measured to two significant digits (figures). That means we are not justified in reporting a density that has more than 2 significant figures. We have to round it to 2 significant digits.
You've done problems like this before. This is a refresher. You use factor-label method (also called dimensional analysis). By looking at the labels (dimensions) you set up factors (fractions) that will achieve the final label. In this case the final label or dimension is mass (grams).
19.3 g x 25.0 cm3 = ? g
cm3
Notice that with this setup the cubic centimeters cancel leaving just the grams. Remember to round to 3 significant digits because the measurements here are to 3 significant digit accuracy. Note that sometimes you may have to invert the density in order to get the correct final label.
8. How many grams of NaCl are in 50.0 mL of a 22.9% by weight NaCl solution that has a density of 1.32 g/mL?
This problem is a tricky because there is extra information given that might confuse you. It's a good problem because it forces you to read the details. Notice is says "22.9% by weight". Remember, this way of measuring concentration is saying that NaCl is 22.9% of any weight (grams) of the solution. In other words, if you had 50.0 grams of the salt solution, 22.9% of that 50.0 grams would be NaCl. However, notice that they did not say 50.0 grams; they said 50.0 mL of the NaCl solution. So knowing the concentration is 22.9% by weight doesn't help. Fortunately, they also gave the density as 1.32g/mL. Density tells us grams in a certain volume. So that will work because the mL will cancel giving us just grams:
50.0mL x 1.32g =
mL
When you find the answer remember to round it to 3 significant digits because our measurements only had 3 significant digits.
Calculate the weight in grams and pounds (1 lb = 454 g) of a gold brick 35.00 cm long with a square cross section 9.50 cm by 9.50 cm. The density of gold is 19.3 g/cm3.
This problem is pretty easy. Find the volume then use that to multiply by density, which will cancel out volumes (cm3)
Volume is L x W x H. Here they give 35.00 cm x 9.50 cm x 9.50 cm, which gives us 3158.75 cm3. Now you think you should round this to 3 significant figures because 9.50 only has 3; however, since there's more calculations to follow, it is best not to round off these answers. Wait until you get the final answer.
Looking at 3158.75 cm3 and 19.3g/cm3, you should now begin to see how the units of cm3 will cancel when multiplying.
3158.75 cm3 x 19.3 g =
cm3
You see that the cm3 will cancel giving you the final units of grams. To get lbs use the conversion numbers.
grams gold x 1 lb = ? lb
454 g
Remember to round to 3 significant digits because our least accurate measurements were only 3 significant digits.
Here they ask for % water, but they are assuming you know that means % by weight because only weights were given. This problem is pretty easy. Just subtract the crucible's weight from the 65.082 g to get the weight of the solution (water and salt). Subtract the crucible's weight from 63.664 g to get just the weight of the salt. The weight of the solution turns out to be 65.082g-61.084g=0.3998g. Just like in question 5, you find the percent of salt by making a common fraction
grams salt = decimal fraction = percent salt.
0.3998g solution
Also like question 5, to find the water's weight just subtract the salt's weight from the total solution weight (0.3998). Then make the water's weight a fraction of the solution's weight and convert it to a percentage like before.