AP Chem U2-1 Chemical Formulas Lab
Skill set
Gravimetric determination of components in a reaction
Determination of empirical and molecular ratios
Structure and use of hot water baths to speed processes
Use of crucibles in high temp reaction systems
Balance crucible cover
250 mL beaker clay triangle
Bunsen burner 6 M HCI
Evaporating dish granular zinc, about 2 g
50 mL graduated cylinder powered sulfur, about 3 g
ring stand and ring corborundum boiling chips, or glass beads
stirring rod Wire gauze
Just as a secretary uses shorthand in order to take dictation at a greater speed, chemists use a shorthand notation of their own to indicate the exact chemical composition of compounds (chemical formulas). We then use these chemical formulas to indicate how new compounds are formed by chemical combinations of other compounds (chemical reactions). However, before we can learn how chemical formulas are written, we must first acquaint ourselves with the symbol used to denote the elements from which these compounds are formed and then work toward determining what ratio these atoms like to combine.
We generally use two letters to symbolize a chemical element. These symbols are derived, as a rule, from the first two letters of first syllable of the elements name.
We can
combine the symbols to denote the formulas for compounds in the following
fashion. The formula CO (read “carbon monoxide”) means that one atom of carbon,
C, combines with one oxygen, O, to form the compound carbon monoxide.
Similarly, in the compound CHCI3, one atom of carbon, C, is combined
with one atom of hydrogen, H, and three atoms of chlorine. The subscripts, 3 on
the chlorine and 1 (understood) on both the carbon and hydrogen, imply exactly
this. Thus, our chemical shorthand indicates the
precise combining ratios of the chemical elements that form the molecule. Each
and every molecule of carbon monoxide contains one atom of carbon and one of
oxygen, just as each and every molecule of chloroform contains exactly one atom
of each carbon and hydrogen and three of chlorine. Every chemical compound has
this property; that is, every compound is composed of definite numbers of
whole atoms in fixed proportions.
It is important to know something about masses of atoms and molecules. We can measure the masses of individual atoms with a high degree of accuracy. We know, for example, the hydrogen-1 atom has a mass of 1.674 x 10-24g. Because it is cumbersome to express such small masses in grams we use a unit called the atomic mass unit, or amu. An amu equals 1.66054 x 10-24 g. Most elements occur as mixtures of isotopes. The average atomic mass of each element expressed in listed both in the table of elements and in the periodic table in Appendices C and D, respectively, are in amu.
The formula weight of a substance is merely the sum of atomic weights of each atom in its chemical formula. For example, HNO3, nitric acid has a formula weight of 63 amu.
FW = (AW of H) + (AW of N) + 3(AW of O)
= 1.0 amu + 14.0 amu + 3(16.0 amu)
= 63.0 amu
If the chemical formula of a substance is its molecular formula, then the formula weight is also called the molecular weight. For example, the molecular formula for formaldehyde is CH2O. The molecular weight of formaldehyde is therefore
MW = 12 amu + 2(1.0 amu) + 16.0 amu
= 30.0 amu
For ionic substances such as NaCl that exist as three dimensional arrays of ions, it is not appropriate to speak for molecules. Similarly, the terms molecular weight and molecular formula are inappropriate for these ionic substances. It is correct to speak of their formula weight, however. Thus, the formula weight of NaCi is:
FW = 23.0 amu + 35.5 amu
= 58.5 amu
New compounds are made in laboratories every day, and the formulas of thses compounds must be determined. The compounds are often analyzed for their percentage composition (i.e., the percentage by mass of each element present in the compound). The percentage composition is useful information in establishing the formula for the substance. If the formula of a compound is known by calculating the percentage composition is a straight-forward matter. In general, The percentage of an element in a compound is given by:
(Number
of atoms of element)(AW) X 100
FW of compound
If we wanted to know the percentage of formaldehyde, CH2O, whose formula weight is 30.0 amu we proceed as follows:
%C = 12.0 amu x 100 = 4.0.%
30.0 amu
%H = 2(1.0 amu) x 100 = 6.7%
30.0 amu
%O = 16.0 amu x 100 = 53.3%
30.0 amu
Even the smallest samples we use in the laboratory contain an enormous number of atoms. A drop of water contains about 2 x 1021 water molecules! The unit that the chemist uses for dealing with such a large number of atoms, ions or molecules is the mole, abbreviated mol. Jus as the unit dozen refers to 12 objects, the mole refers to a collection of 6.02 x 1023 H2O molecules and a mol of sodium contains 6.02 x 1023Na atoms. The mass (in grams) of 1 mol of a substance is called its molar mass. The molar mass (in grams) of any substance is numerically equal to its formula weight.
Thus:
One CH2O molecule weighs 30.0 amu; 1 mol CH2O weighs 30.0 g
And contains 6.02 x 1023 CH2O molecules.
One Na atom weighs 23.0 amu; 1 mol Na atom weighs 23.0 g
And contains 6.02 x 1023 Na atoms.
It is simple matter to calculate the number of moles of any substance whose weight we can obtain. For example, suppose we have one quart of rubbing alcohol (generally isopropyl alcohol) and know its density to be 0.785 g/mL and we want to know how many moles this is. First we need to convert the volume to our system units. Since 1 qt is 0.946 L and 1 L contains 1000 mL, our quart of isopropoyl alcohol is
(1qt) 0.946 L 1000 mL = 946 mL
qt L
we can now calculate the weight:
946 mL x 0.785 g/mL = 743 g
We next need the chemical formula for isopropyl alcohol. This is C3H7OH. The molecular weight is, therefore,
Weight Carbon 3 x 12.0 = 36.0 amu
Weight hydrogen 8 x 1.0 = 8.0 amu
Weight oxygen 1 x 16.0 = 16.0 amu
Molecular weight C3H7OH = 60.0 amu
Hence,
Moles of C3H7OH = (743 g C3H7OH) 1 mol C3H7OH = 12.4 mol
60.0 g C3H7OH
Thus our quart of rubbing alcohol contains 12.4 mol of isopropyl alcohol. It should now be apparent to you how much information is contained in a chemical formula. After you have some feeling for the nature of different chemical substances, you should be able to read the labels of any household items and know what the chemical contents are and why they are there. We have often heard that phrase “What’s in a name?” The answer to this question when the name is that of a chemical compound takes on considerable meaning as we gain familiarity with that compound. For example the simple name carbon dioxide tells us precisely
1. The elements present, carbon and oxygen;
2. The combining ratios of these elements, one atom of carbon for every two atoms of oxygen in each molecule of carbon dioxide;
3. The molecular weight;
4. And, After we learn some chemistry, that we are speaking of a colorless gas, heavier than air, which will not support combustion and which is lethal simply because it will not allow us to take in oxygen when there is a lot of it around. We will also learn that it can be solidified be compression to produce the familiar dry ice.
All of this dialogue is simply to illustrate that chemical formulas are tremendously important, and it behooves us in the very beginning to have a firm understanding of just exactly what they are. It should also be evident that we must have some way of determining them precisely if they do contain so much information.
Suppose you were working in a hospital laboratory today, and suppose further that this morning the emergency ward were admitted a patient complaining of severe stomach cramps and labored respiration and that this patient died within minuets of being admitted. Relatives of the patient later told you that he may have ingested some rat poison. You therefore had his stomach pumped to verify this or simply to determine the cause of death. One of the more logical things to do would be to attempt to isolate the agent that caused death and perform chemical analyses on it. Let’s suppose that this was done, and that the analyses showed that the isolated chemical compound contained, by weight, 60.0 percent potassium. 18.5 percent carbon, and 21.5 percent nitrogen. What is the chemical formula for this compound? One simple and direct way of making the necessary calculations is as follows.
Assume you had 100 g of the compound. This 100 g therefore would contain
(100 g)(0.600) = 60.0 g potassium
(100 g)(0.185) = 18.5 g carbon
(100 g)(0.215) = 21.5 nitrogen
Chemical formulas tell what elements are present and the ratio of the number of atoms of the constituent elements. Hence, the next step is to determine the number of moles of each element present.
Moles potassium = 60.0 g K = 1.54 mol K
39.0 g K/mol K
Moles carbon = 18.5 g C = 1.54 mol C
12.0 g N/ mol N
Moles nitrogen = 21.5 g N = 1.54 mol N
14.0 g N / mol N
Hence the chemical formula is K1.54/1.54C1.54/1.54N1.54/1.54. But molecules are not formed from partial atoms; therefore, the above number must be changed to whole numbers. This is accomplished be division of all subscripts by the smallest subscript. In this case, they are all the same.
K1.54/1.54C1.54/1.54N1.54/1.54=KCN
The smallest whole-number mole ratio is 1:1:1. Since KCN is a common rat poison, we may justifiably conclude that the relatives’ suggestion of rat poison ingestion as the probably cause of death is correct.
The above calculation has given us what is known as the empirical formula. There is another type of chemical formula the molecular formula. The distinction between these two is simple that the empirical formula represents the smallest whole number ratio and of the combining atoms in a chemical compound, whereas the molecular formula gives the actual number of atom in a molecule. Recall, however, as we stated earlier, that not all compounds exist as discrete molecules. This is true for most ionic compounds, whereas covalent compounds do exist as discrete molecules. The distinction between empirical and molecular formulas may be clarified by the following example.
A chemical compound was found by elemental analyses to contain 92.3 percent carbon and 7.7 percent hydrogen by weight and to have a molecular weight of 78. The empirical formula may be obtained just as in the previous example-that is, in 100 g of the compound there are 92.3 g C and 7.7 g H. Hence,
Moles C = 92.3 g C = 7.7 mol C
12 g C/ mol C
Moles H = 7.7 g H = 7.7 mol H
1.0 g H/ mol H
The empirical formula is then C7.7H7.7, or CH, whose formula weight is 12 + 1 = 13. but the molecular weight of the compound is 78. Therefore, there are 78/13 = 6 empirical-formula weight. The molecular formula is them C6H6.
In this experiment you will determine the empirical formulas of two chemical compounds. One is copper sulfide, which you will prepare according to the following chemical reaction:
xZn(s) + yS8(s) à CuxSy(s)
The other is zinc chloride, which you will prepare according to the chemical reaction:
XZn(s) + yHCl(aq) -à ZnxCly(s) à ZnxCly(s) + y H2(g)
2
The objective is to determine the combining ratios of the elements (that is, to determine x and y) and to balance the chemical equations given above.
CAUTION: Zinc chloride is caustic and must be handled carefully in order to avoid any contact with your skin. Should you come in contact with it, immediately wash the area with copious amounts of water. Clean and dry your evaporating dish and place it on the wire gauze resting on the iron ring. Heat the dish with your Bunsen burner, gently at first, and then more strongly, until all of the condensed moisture has been driven off. This should require heating for about 5 min. Allow the dish to cool to room temperature on a heat-resistant pad (do not place the hot dish on the counter top) and weight it. Record the weight of the empty evaporating dish.
Obtain a sample of granular zinc from your laboratory instructor and ass about 0.5 g of it to the weighted evaporating dish. Weight the evaporating dish containing the zinc and record the total weight. Calculate the weight of the zinc.
Slowly, and with constant stirring, add 15 mL of 6 M HCl to the evaporating dish containing the zinc. A vigorous reaction will ensue, and hydrogen gas will be produced. NO FLAMES ARE PERMITTED IN THE LABORATORY WHILE THIS REACTION IS TAKING PLACE, SONCE WET HYDROGEN GAS IS VERY EXPLOSIVE. If any undisovled zinc remains after the reaction ceases, add an additional 5 mL of acid. Continue to add 5 mL portions of acid as needed until all the zinc has dissolved.
Set up a steam bath as illustrated in Figure 5.1 using a 250-mL beaker, and place the evaporating dish on the steam bath. Heat the evaporating dish very carefully on the steam bath until most of the liquid has disappeared. Then remove the steam bath and heat the dish on the wire gauze. During this last stage of heating, the flame must be carefully controlled or there will be spattering, and some loss of product will occur. DO NOT HEAT TO THE POINT THET THE COMPOUND MELTS, OR SOME WILL BE LOST DUE TO SUBLIMATION. Leave the compound looking somewhat pasty while hot.
Allow the dish to cool to room temp. and weigh it. Record the weight. After this first weighing, heat the dish again very gently; cool it and reweigh it. If these weighings do not agree within 0.02 g, repeat the heating and weighing until two successive weighings agree. This is known as drying to constant weight and is the only way to be certain that all moisture is driven off. Zinc chloride is very deliquescent and so should be weighed quickly.
Calculate the weight of zinc chloride. The different in weight of the zinc and the zinc chloride is the weight of the chlorine. Calculate the weight of chlorine in zinc chloride. From this information you can readily calculate the empirical formula for zinc chloride and balance the chemical equation for its formation. Perform these operations on the report sheet.
Support a clean, dry porcelain crucible and cover on a clay triangle and dry by heating to a dull red in a Bunsen flame. Allow the crucible and cover to cool to room temperature and weight them. Record the weight.
Place a 1.5-2.0 g of tightly wound copper wire or copper turnings in the crucible and weight the copper, crucible and lid. Calculate the weight of copper. Record your results.
In the hood, add sufficient sulfur to cover the copper, place the crucible with cover in place on the triangle, and heat the crucible gently until sulfur ceases to burn (blue flame) at the end of the cover. Do not remove the cover while the crucible is hot. Finally, heat the crucible to dull redness for about 5 min.
Allow the crucible to cool to room temperature. This will take about 10 min. Then weigh with the cover in place. Record the weight. Again cover the contents of the crucible with sulfur and repeat the heating procedure. Allow the crucible to cool and reweigh it. Record the weight. If the last two weighings do not agree to within 0.02 g, the chemical reaction between the copper and sulfur is incomplete. If this is found to be the case, add more sulfur and repeat the heating and weighing until a constant weight is obtained.
Calculate the weight of copper sulfide obtained. The difference in weight between the copper sulfide and copper is the weight of sulfur in copper sulfide. Calculate this weight. From this information the empirical formula for copper sulfide can be obtained and the chemical equation for its production can be balanced. Perform these operations on your report sheet.
5. A substance was found by analysis to contain 65.95 percent barium and 34.05 percent chlorine. What is the empirical formula for the substance?
6. What is the law of definite proportions?
7. How do empirical and molecular formulas differ?
8. What is the weight in grams of one copper atom?
9. Soda-lime glass is prepared by fusing sodium carbonate, Na2CO3; lime-stone, CaCO3; and sand, SiO2. The composition of the glass varies, but the commonly accepted reaction for its formula is
Na2CO3(s) + CaCO3(s) + 6SiO2(s) à Na2CaSi6O14(s) + 2CO2(g)
Using this equation, how many kilograms of sand would be required to produce enough glass to make five thousand 400-g wine bottles?
10. Caffeine, a stimulant found in coffee and tea contains 49.5 percent C, 5.15 percent H, 28.9 percent N, and 16.5 percent O by mass. What is the empirical formula of caffeine? If its molar mass is about 195 g, what is its molecular formula?
11. An analysis of an oxide of nitrogen with a molecular weight of 92.02 amu gave 69.57 percent oxygen, and 30.43 percent nitrogen. What are the empirical and molecular formulas for this nitrogen oxide? Complete and balance the equation for its formation from the elements nitrogen and oxygen.
12. How many lithium atoms are present in 0.01456 g of lithium?
REPORT SHEET
A. Zinc Chloride
1. Weight of evaporating dish and zinc
2. Weight of evaporating dish
3. Weight of zinc
4. Weight of evaporating dish and zinc chloride
First weighing
Second weighing
Third weighing
5. Weight of zinc chlorine
6. Weight of chlorine in zinc chloride
7. Empirical formula for zinc chloride (show calculations)
8. Balanced chemical equation for the formation of zinc chloride from zinc and HCl
B. Copper Sulfide
1. Weight of crucible, cover, and copper
2. Weight of crucible and cover
3. Weight of copper
$. Weight of crucible, cover, and copper sulfide
First weighing
Second weighing
Third weighing
5. Weight of copper sulfide
6. Weight of sulfur in copper sulfide
7. Empirical formula for copper sulfide (show calculations)
8. Balanced chemical equation for the formation of copper sulfide from copper and sulfur
Analysis Questions part 2
1. Can you determine the molecular formula of a substance from its percent composition?
2. Given tat zinc chloride has a formula weight of 136.28, what is its formula?
3. Can you determine the atomic weights of zinc or copper by the methods used in this experiment? How? What additional information is necessary in order to do this?
4. How many grams of zinc chloride could be formed from the reaction of 9.76 g of zinc with excess HCl?
5. How many kilograms of copper sulfide could be formed from the reaction of 2.70 mol of copper with excess sulfur?
6. If copper (I) sulfide is partially roasted in air (reaction with O2) sulfite is first formed. Subsequently, upon heating, the copper sulfite thermally decomposes to copper (I) oxide and sulfur dioxide. Write balanced chemical equations for these two reactions.