MathConnections
3.5 Calculations Using Moles and Molar Masses
(c) Hg(ONC)2: 1 mole Hg: 1 × 200.59 g/mol = 200.59 g/mol 2 moles O: 2 × 15.9994 g/mol = 31.9988 g/mol 2 moles N: 2 × 14.0067 g/mol = 28.0134 g/mol 2 moles C: 2 × 12.011 g/mol = 24.022 g/mol Molar mass 284.62 g/mol Analyze Your Answer Looking at these three compounds, we see that nitroglyc- erin has the smallest molar mass even though it clearly contains the most atoms per molecule. Does this make sense? The key factor is that both lead azide and mercury fulminate contain elements with large molar masses, so it is not surprising that nitro- glycerin is the least massive of these three molecules.
Discussion The number of signifi cant fi gures to which the molar masses of the elements are known varies from one element to another, as you can see if you examine the periodic table inside the back cover of this book. So the appropriate number of signifi cant fi gures in the molar masses here varies according to the elements in each molecule.
Check Your Understanding What are the molar masses of the following com- pounds, which are used in the preparation of various explosives? (a) H2SO4, (b) HNO3, (c) (NH2)2HNO3
3.5 Calculations Using Moles
Now we can use the molar mass to convert from mass to moles:
650.5 g TNT × ——————— 1 mol TNT
227.133 g TNT = 2.864 mol TNT Finally, we can convert from moles to molecules using Avogadro’s number:
2.864 mol TNT × 6.022 ————————————— × 1023 molecules TNT
1 mol TNT = 1.725 × 1024 molecules TNT Analyze Your Answer Admittedly, it is diffi cult to have an intuitive feeling for the number of molecules in any sample. In assessing our answer, we can begin by noting that it is a very large number. This makes sense because we are calculating the number of molecules in a macroscopic amount of TNT. We could also check a little more closely by looking at the intermediate step in which we found the num- ber of moles. Because the molar mass of TNT is slightly larger than 200 g/mol, we should have roughly 3 moles in the 650-g sample. This is consistent with the value we found.
Discussion The mass of this sample corresponds to a little less than 1.5 pounds, so the sample would probably be roughly the size of a brick. Note the enormous number of molecules in this fairly small amount of material.
Check Your Understanding Liquid nitromethane was once widely used as an explosive but is now much more common as a fuel for high performance engines in drag racing. How many moles are present in 1.00 gallon of pure nitromethane (CH3NO2), which has a density of 1.137 g/mL?
In some cases, we might calculate the number of moles of a substance needed in a particular reaction. Then, if we want to carry out that reaction, we would need to know what mass of sample to prepare. We can easily invert the process above to convert from the number of moles to mass.
E X A M P L E P RO B L E M 3 . 7
A demolition engineer is planning to use the explosive ethylenedinitramine (C2H6N4O4, also known as halite) to bring down an abandoned building. Calculations show that 315 moles of the compound will provide the necessary explosive force. How many pounds of C2H6N4O4 should be used?
Strategy We are asked to convert from a number of moles to a mass. We know that the link between these two quantities is the molar mass of the compound, so we can start by calculating that. Once we know the molar mass, we can use it to fi nd the required mass of explosive. Finally, we can convert that result from grams to pounds.
Solution We will need to calculate the molar mass of the compound.
C2H6N4O4: 2 moles C: 2 × 12.011 g/mol = 24.022 g/mol 6 moles H: 6 × 1.0079 g/mol = 6.0474 g/mol 4 moles N: 4 × 14.0067 g/mol = 56.0268 g/mol 4 moles O: 4 × 15.9994 g/mol = 63.9976 g/mol
Molar mass = 150.094 g/mol
Now we can use the molar mass and the required number of moles to fi nd the appropriate mass.
315 moles C2H6N4O4× 150.094 g C2H6N4O4
——————————
1 mole C2H6N4O4 = 4.73 × 104 g C2H6N4O4
A common student error in this type of calculation is to divide by Avogadro’s number rather than multiply. This leads to a very small numerical answer.
A number of molecules less than one is not possible, of course, and is a sure sign of an error in the calculation.
A common student error in this type of calculation is to divide by Avogadro’s number rather than multiply. This leads to a very small numerical answer.
A number of molecules less than one is not possible, of course, and is a sure sign of an error in the calculation.
Finally, we can convert this from grams to pounds:
4.73 × 104 g C2H6N4O4× 1 pound
————
453.59 g = 104 pounds C2H6N4O4 needed Analyze Your Answer We can assess the reasonableness of this answer by com- paring it to Example Problem 3.6. For these two problems, the molar masses of the compounds are roughly comparable. In this problem, we have about 300 moles, whereas in the previous one, we had about 3 moles, so it’s not surprising that our an- swer is close to 100 times larger than that in our last exercise.
Check Your Understanding Approximately 80 billion moles of ammonium nitrate (NH4NO3) were produced globally in 2008. How many tons of ammonium nitrate is this?
Elemental Analysis: Determining Empirical and Molecular Formulas
When a new molecule is synthesized, an elemental analysis is routinely performed to help verify its identity. This test, which measures the mass percentage of each ele- ment in the compound, is also frequently done as part of the process of identifying any substance whose composition is unknown. The mass percentages describe the compound’s composition, and so they must be related to its chemical formula. But the data obtained from elemental analysis describe the composition in terms of the mass of each element, whereas the formula describes the composition in terms of the num- ber of atoms of each element. So these are two different representations of very simi- lar information, and the molar masses of the elements provide a connection between them. The process of obtaining the empirical formula of a compound from its percent composition by mass is best illustrated by an example.
E X A M P L E P RO B L E M 3 . 8
The explosive known as RDX contains 16.22% carbon, 2.72% hydrogen, 37.84%
nitrogen, and 43.22% oxygen by mass. Determine the empirical formula of the compound.
Strategy The empirical formula is based on the mole ratios among the elements in the compound, and the data given are in terms of mass. Molar mass provides the link between mass and moles, as usual. We can begin by choosing a convenient mass of the compound—usually 100 g—and use the percentages given to fi nd the mass of each element in that sample. Then we will convert those masses into numbers of moles of each element. The ratios between those numbers of moles must be the same for any sample of the compound. Finally, we will need to convert those ratios into whole numbers to write the empirical formula.
Solution Consider a 100-g sample of RDX. From the percentages given, that sam- ple will contain 16.22g C, 2.72g H, 37.84g N, and 43.22g O. We will convert each of these masses into moles:
16.22g C × ————— 1 mol C
12.011 g C = 1.350 mol C in 100 g RDX
2.72g H × 1 mol H
—————
1.0079 g H = 2.70 mol H in 100 g RDX
37.84g N × —————— 1 mol N
14.0067 g N = 2.702 mol N in 100 g RDX
43.22g O × 1 mol O
——————
15.9994 g O = 2.701 mol O in 100 g RDX
These numbers tell us the ratios of C:H:N:O in the compound. We might be tempted to write a formula of C1.35H2.70N2.70O2.70. But we know that the correct empirical formula requires whole numbers of each element. In this case, it may be easy to see the correct ratios, but we need to approach such a problem systematically. The usual strategy is to divide all of the numbers of moles we obtained above by the smallest number of moles. This assures that the smallest resulting number will always be one.
In this case, we will divide all four numbers of moles by 1.350 because it is the smallest of the four.
1.350 mol C
——————
1.350 = 1 2.70 mol H
——————
1.350 = 2 2.702 mol N
——————
1.350 = 2.001 ≈ 2 2.701 mol O
——————
1.350 = 2.001 ≈ 2
The result is a small whole number ratio: 1 mole C:2 moles of H:2 moles N:2 moles O.
So the empirical formula is CH2N2O2. The mass percentage data allow us to determine only the empirical formula, not the molecular formula, because we do not have any information about how large or small a molecule of the compound might be.
Analyze Your Answer We know that the coeffi cients in a chemical formula must be whole numbers. The fact that all four of the coeffi cients we found are so close to integers means that our proposed formula is plausible.
Discussion In working with this type of problem, it is a good idea to use accurate molar masses for the elements and to carry as many significant figures as possible throughout the calculation. This ensures that no large rounding errors occur and helps make it easy to decide whether the fi nal coeffi cients obtained are whole numbers.
The steps we used above will produce whole numbers for all of the elements only if at least one of the subscripts in the empirical formula is a one. Otherwise, the result for one or more elements will still not be an integer. In nearly all such cases, though, such nonintegral results will be readily identifi ed as small rational fractions (e.g., 1.5 or 2.33). We can obtain integral values by multiplying all of the coeffi cients by an appropriate integer. This is demonstrated in the “Check Your Understanding” exercise that follows.
Check Your Understanding Nitroglycerin contains 15.87% C, 2.22% H, 18.50% N, and 63.41% O by mass. Determine the empirical formula of the compound.
If we also know the molar mass of the compound, the process used in the pre- vious example can easily be extended to allow the determination of the molecular formula.
The connection between mass and moles can also be applied in other ways.
Materials scientists and engineers often describe the composition of alloys in terms of either weight percentage (wt %) or mole percentage (mol %). Converting between these two units involves the use of molar mass, as shown in Example 3.9.
An alloy is a solution in which both the solute and solvent are solids. Materials engineers can choose or design alloys to have useful properties for a particular application.
An alloy is a solution in which both the solute and solvent are solids. Materials engineers can choose or design alloys to have useful properties for a particular application.
E X A M P L E P RO B L E M 3 . 9
Alloys of palladium and nickel are often used in the manufacture of electronic con- nectors. One such alloy contains 70.8 mol % Pd and 29.2 mol % Ni. Express the composition of this alloy in weight percentage.
Strategy In order to get started, we need to choose an amount of the alloy. Al- though any amount would work, one mole might be convenient. Once we have speci- fi ed a number of moles, we can use molar masses to determine the mass of each metal and the total mass. And from those we can calculate the weight percentages.
Solution The molar masses of palladium and nickel are 106.42 g/mol and 58.69 g/mol, respectively. From the given mole percentages, we know that one mole of alloy would contain 0.708 mole of Pd and 0.292 mole of Ni. We can use these val- ues to determine the mass of each component in one mole of alloy.
mPd= 0.708 mol Pd × 106.42 g Pd
——————
1 mol Pd = 75.4 g Pd mNi= 0.292 mol Ni × 58.6934 g Ni
——————
1 mol Ni = 17.1 g Ni So the total mass of one mole of alloy is just
75.4 g Pd + 17.1 g Ni = 92.5 g alloy Finally, calculate the percentage by mass (weight) of each metal.
wt % (Pd) = 75.4 g Pd
——————
92.5 g alloy × 100% = 81.5 wt % Pd wt % (Ni) = 17.1 g Ni
——————
92.5 g alloy × 100% = 18.5 wt % Ni
Analyze Your Answer It makes sense that the weight percentage of the palladium is higher than the mole percentage because palladium has a higher molar mass than nickel.
Discussion There is nothing special about the assumption of one mole in this problem.
It was chosen because it was likely to provide numbers for the masses that were a “com- fortable” size to think about. If the problem were worked in the other direction (from wt % to mol %), we might choose 100 g as the amount of material to use in the calculation.
An additional unit sometimes used for alloy composition is atomic percentage (at %). The conversion between at % and wt % involves the same steps as the mol % and wt % conversion in this problem.
Check Your Understanding Eighteen carat gold typically contains 75% gold, 16% silver, and 9% copper by weight. Express this composition in mol percentages.
Molarity
Molar mass is useful for chemical calculations because it provides a connection be- tween a quantity that is easy to measure (mass) and one that is conceptually important (moles). Another readily measured quantity is volume. When we work with aqueous solutions, we often use volume in calculations rather than mass. It should not be sur- prising, therefore, that we would want to defi ne quantities that will help us relate a volume measurement to the number of moles.
Many different ways to express the concentration of a solute in a solvent have been developed. To defi ne any unit of concentration, we must know both the amount of solute and the quantity of solvent in the solution. The most commonly used concentration unit
in chemistry is molarity or molar concentration. Molarity, represented by the symbol M, is defi ned as the number of moles of solute per liter of solution:
Molarity (M) = ——————— moles of solute
liter of solution (3.1)
The defi nition of molarity points to the ways in which it can be used. It provides a relationship among three things: molar concentration, moles of solute, and liters of solution. If we know any two of these quantities, we can determine the third one.
We can measure the volume of a solution in the laboratory. Multiplying that volume ( expressed in liters) by molarity readily gives moles.
The same relationship also allows us to fi nd the number of moles of solute pres- ent if we know both the volume and molarity of any solution. If we use n to represent the number of moles, V for volume, and M for molarity, we can simply rearrange the defi nition of molarity:
n = M × V (3.2)
E X A M P L E P RO B L E M 3 .10
Aqueous solutions of sodium hypochlorite (NaClO) can be used in the synthesis of hydrazine (N2H4). Hydrazine has often been used as a rocket fuel, and a derivative of hydrazine is used as a fuel for engines in the orbital maneuvering system of the space shuttle. A solution is prepared by dissolving 45.0 g of NaClO in enough water to produce exactly 750 mL of solution. What is the molarity of the solution?
Strategy To obtain molarity we need two quantities—moles of solute and liters of solution. Neither of these quantities is given directly in the question, but both can be obtained readily from information given. We must use the molar mass of NaClO to convert from mass to moles and convert the volume from milliliters to liters. Then we can use the defi nition of molarity to obtain the molarity of the solution.
Solution First we compute the moles of solute.
45.0 g NaClO × ———————— 1.00 mol NaClO
74.442 g NaClO = 0.604 mol NaClO Then convert the solution volume from mL to L:
750 mL × ————— 1.00 L 1000 mL = 0.750 L Finally, calculate the molarity:
Molarity = ———————— moles of solute
liters of solution = ————— 0.604 mol
0.750 L = 0.806 M NaClO
Analyze Your Answer Our answer is somewhat smaller than, but fairly close to, 1 M. Does this make sense? The amount of NaClO was about 2/3 of a mole and the volume of the solution was 3/4 of a liter. The ratio of 2/3 to 3/4 is 8/9, close to but a little bit smaller than 1, so our result seems reasonable.
Discussion As usual, this problem could also be set up as a single calculation includ- ing the same conversions. Whichever approach you choose, you should always check to be sure that your units work out correctly. Here we have units of mol/L for our an- swer, which is appropriate for molarity. Confi rming this helps avoid careless errors.
Check Your Understanding To produce hydrazine, sodium hypochlorite must react with ammonia. How many moles of NH3 would be in a 4.0-L bottle of 14.8 M ammonia solution?
Molarity is another ratio between otherwise unrelated variables—similar to mass density. It can be used in calculations in the same way as density.
Molarity is another ratio between otherwise unrelated variables—similar to mass density. It can be used in calculations in the same way as density.
Hydrazine, N2H4
Dilution
One of the most common procedures encountered in any science laboratory is dilu- tion, the process in which solvent is added to a solution to decrease the concentration of the solute. In dilution, the amount of solute does not change. The number of moles of solute is the same before and after dilution. Because we already know that the number of moles of solute equals the product of molarity and volume, we can write the following equation, where the subscripts denote the initial (i) and fi nal (f) values of the quanti- ties involved:
Mi× Vi= Mf× Vf (3.3)
The simplicity of equations such as this is alluring. One of the most common errors made by students in chemistry is to use this relationship in situations where it is not valid. The only place to use this relationship is in the process of dilution or concentration of a solution. Example Problem 3.11 illustrates a common laboratory situation where this equation is valid and useful.
E X A M P L E P RO B L E M 3 .11
A chemist requires 1.5 M hydrochloric acid, HCl, for a series of reactions. The only solution available is 6.0 M HCl. What volume of 6.0 M HCl must be diluted to obtain 5.0 L of 1.5 M HCl?
Strategy First we need to recognize that this is the dilution of the concentrated solution to prepare the desired one. The underlying concept we must use is that the number of moles of HCl will be the same before and after dilution, and this means we can use Equation 3.3. We know the desired fi nal molarity and fi nal volume, as well as the initial molarity. So we can solve for the needed initial volume.
Solution
Initial concentration of HCl: Mi= 6.0 M Final concentration of HCl: Mf= 1.5 M Final volume of solution: Vf= 5.0 L
The unknown is the initial volume, Vi. Rearranging Equation 3.3, Vi= Mf× Vf
————
Mi Inserting the known quantities on the right-hand side,
Vi= 1.5 M × 5.0 L
———————
6.0 M = 1.3 L
To obtain the desired quantity of diluted HCl, the chemist should begin with 1.3 L of the concentrated solution and add enough water to bring the volume up to 5.0 L.
Analyze Your Answer Once we are certain that we are dealing with a dilution prob- lem, one way to make sure the answer makes sense is to think about which solution is concentrated and which is dilute. You will always need smaller volumes of the concen- trated solution. Did we get it right this time? It seems likely that we did: we were asked to calculate how much of the concentrated solution we would need, so the volume we calculate should be the smaller of the two volumes involved in the problem.
Check Your Understanding If 2.70 mL of 12.0 M NaOH is diluted to a volume of 150.0 mL, what is the fi nal concentration?
This dilution formula can be used when there is only one solute present. If a reaction is taking place, the problem must be approached as a stoichiometry problem, which we will take up in Chapter 4.
This dilution formula can be used when there is only one solute present. If a reaction is taking place, the problem must be approached as a stoichiometry problem, which we will take up in Chapter 4.