Chemistry Packet 2 Mole Concept Name ______
When the weather is nice, many people begin to work on their yards and homes. For many projects, sand is needed as a foundation for a walk or to add to other materials. You could order up twenty million grains of sand and have people really stare at you. You could order by the pound, but that takes a lot of time weighing out. The best bet is to order by the yard, meaning a cubic yard. The loader can easily scoop up what you need and put it directly in your truck.
Avogadro’s Number
It certainly is easy to count bananas or to count elephants (as long as you stay out of their way). However, you would be counting grains of sugar from your sugar canister for a long, long time. Atoms and molecules are extremely small – far, far smaller than grains of sugar. Counting atoms or molecules is not only unwise, it is absolutely impossible. One drop of water contains about 1022 molecules of water. If you counted 10 molecules every second for 50 years without stopping you would have counted only 1.6 × 1010 molecules. Put another way, at that counting rate, it would take you over 30 trillion years to count the water molecules in one tiny drop.
Chemists needed a name that can stand for a very large number of items. Amedeo Avogadro (1776 - 1856), an Italian scientist, provided just such a number. He is responsible for the counting unit of measure called the mole. A mole (mol) is the amount of a substance that contains 6.02 × 1023 representative particles of that substance. The mole is the SI unit for amount of a substance. Just like the dozen and the gross, it is a name that stands for a number. There are therefore 6.02 × 1023 water molecules in a mole of water molecules. There also would be 6.02 × 1023 bananas in a mole of bananas, if such a huge number of bananas ever existed.
Italian scientist Amedeo Avogadro, whose work led to the concept of the mole as a counting unit in chemistry.
The number 6.02 × 1023 is called Avogadro’s number, the number of representative particles in a mole. It is an experimentally determined number. A representative particle is the smallest unit in which a substance naturally exists. For the majority of elements, the representative particle is the atom. Iron, carbon, and helium consist of iron atoms, carbon atoms, and helium atoms, respectively. Seven elements exist in nature as diatomic molecules and they are H2, N2, O2, F2, Cl2, Br2, and I2. The representative particle for these elements is the molecule. Likewise, all molecular compounds such as H2O and CO2 exist as molecules and so the molecule is their representative particle. For ionic compounds such as NaCl and Ca(NO3)2, the representative particle is the formula unit. A mole of any substance contains Avogadro’s number (6.02 × 1023) of representative particles.
The animal mole is very different than the counting unit of the mole. Chemists nonetheless have adopted the mole as their unofficial mascot. National Mole Day is a celebration of chemistry that occurs on October 23rd (10/23) of each year.
Summary
A mole of any substance contains Avogadro’s number (6.02 × 1023) of representative particles.
Review
1. What is the SI unit for amount of a substance?
2. What is the representative particle for an element?
3. The formula unit is the representative particle for what?
Big numbers or little numbers?
Do you hate to type subscripts and superscripts? Even with a good word-processing program, having to click on an icon to get a superscript and then remembering to click off after you type the number can be a real hassle. If we did not know about moles and just knew about numbers of atoms or molecules (those big numbers that require lots of superscripts), life would be much more complicated and we would make many more typing errors.
Conversions Between Moles and Atoms
Conversions Between Moles and Number of Particles
Using our unit conversion techniques, we can use the mole label to convert back and forth between the number of particles and moles.
Sample Problem 1: Converting Number of Particles to Moles
The element carbon exists in two primary forms: graphite and diamond. How many moles of carbon atoms is 4.72 × 1024 atoms of carbon?
Step 1: List the known quantities and plan the problem.
Known
number of C atoms = 4.72 × 1024
1 mole = 6.02 × 1023atoms
Unknown 4.72 × 1024 = ? mol C
One conversion factor will allow us to convert from the number of C atoms to moles of C atoms.
Step 2: Calculate.
Step 3: Think about your result.
The given number of carbon atoms was greater than Avogadro’s number, so the number of moles of C atoms is greater than 1 mole. Since Avogadro’s number is a measured quantity with three significant figures, the result of the calculation is rounded to three significant figures.
Suppose that you wanted to know how many hydrogen atoms were in a mole of water molecules. First, you would need to know the chemical formula for water, which is H2O. There are two atoms of hydrogen in each molecule of water. How many atoms of hydrogen would there be in two water molecules? There would be 2 × 2 = 4 hydrogen atoms. How about in a dozen? In that case a dozen is 12 so 12 × 2 = 24 hydrogen atoms in a dozen water molecules. To get the answers, (4 and 24) you had to multiply the given number of molecules by two atoms of hydrogen per molecule. So to find the number of hydrogen atoms in a mole of water molecules, the problem could be solved using conversion factors.
The first conversion factor converts from moles of particles to the number of particles. The second conversion factor reflects the number of atoms contained within each molecule.
Two water molecules contain 4 hydrogen atoms and 2 oxygen atoms. A mole of water molecules contains 2 moles of hydrogen atoms and 1 mole of oxygen atoms.
Sample Problem 2: Atoms, Molecules, and Moles
Sulfuric acid has the chemical formula H2SO4. A certain quantity of sulfuric acid contains 4.89 × 1025 atoms of oxygen. How many moles of sulfuric acid is the sample?
Step 1: List the known quantities and plan the problem.
Known 4.89 × 1025 = O atoms
23 1 mole = 6.02 × 10 molecules H2SO4
Unknown
mol of H2SO4 molecules
Two conversion factors will be used. First, convert atoms of oxygen to molecules of sulfuric acid. Then, convert molecules of sulfuric acid to moles of sulfuric acid.
Step 2: Calculate.
Step 3: Think about your result.
The original number of oxygen atoms was about 80 times larger than Avogadro’s number. Since each sulfuric acid molecule contains 4 oxygen atoms, there are about 20 moles of sulfuric acid molecules.
Summary
Methods are described for conversions between moles, atoms, and molecules.
Review
1. What conversion factor would we need to convert moles of helium to atoms of helium?
2. I want to convert atoms to moles. My friend tells me to multiply the number of atoms by 6.02 × 1023 atoms/mole. Is my friend correct?
3. Why do you need to know the formula for a molecule in order to calculate the number of moles of one of the atoms?
22 4. How many atoms of fluorine are in 5.6×10 molecules of MgF2?
When creating a solution, how do I know how much of each substance to put in?
I want to make a solution that contains 1.8 moles of potassium dichromate. I don’t have a balance calibrated in molecules, but I do have one calibrated in grams. If I know the relationship between moles and the number of grams in a mole, I can use my balance to measure out the needed amount of material.
Molar Mass
Molar mass is defined as the mass of one mole of representative particles of a substance. By looking at a periodic table, we can conclude that the molar mass of lithium is 6.94 g, the molar mass of zinc is 65.38 g, and the molar mass of gold is 196.97 g. Each of these quantities contains 6.02 × 1023 atoms of that particular element. The units for molar mass are grams per mole or g/mol.
Molar Masses of Compounds
A molecular formula of the compound carbon dioxide is CO2. One molecule of carbon dioxide consists of 1 atom of carbon and 2 atoms of oxygen. We can calculate the mass of one molecule of carbon dioxide by adding together the masses of 1 atom of carbon and 2 atoms of oxygen.
12.01 amu+2(16.00 amu)=44.01 amu
The molecular mass of a compound is the mass of one molecule of that compound. The molecular mass of carbon dioxide is 44.01 amu.
The molar mass of any compound is the mass in grams of one mole of that compound. One mole of carbon dioxide molecules has a mass of 44.01 g, while one mole of sodium sulfide formula units has a mass of 78.04 g. The molar masses are 44.01 g/mol and 78.04 g/mol respectively. In both cases, that is the mass of 6.02 × 1023 representative particles. The representative particle of CO2 is the molecule, while for Na2S, it is the formula unit.
Sample Problem: Molar Mass of a Compound
Calcium nitrate, Ca(NO3)2, is used as a component in fertilizer Determine the molar mass of calcium nitrate.
Step 1: List the known and unknown quantities and plan the problem.
Known
formula = Ca(NO3)2
molar mass Ca = 40.08 g/mol molar mass N = 14.01 g/mol
molar mass O = 16.00 g/mol
Unknown
molar mass Ca(NO3)2
First we need to analyze the formula. Since the Ca lacks a subscript, there is one Ca atom per formula unit. The 2 outside the parentheses means that there are two nitrate ions per formula unit and each nitrate ion consists of one nitrogen atom and three oxygen atoms. Therefore, there are a total of 1 × 2 = 2 nitrogen atoms and 3 × 2 = 6 oxygen atoms per formula unit. Thus, 1 mol of calcium nitrate contains 1 mol of Ca atoms, 2 mol of N atoms, and 6 mol of O atoms.
Step 2: Calculate.
Use the molar masses of each atom together with the number of atoms in the formula and add together.
Summary
Calculations are described for the determination of molar mass of an atom or a compound.
Review
1. What is the molar mass of Pb?
2. Where do you find the molar mass of an element?
3. How many moles of Cl are in one mole of the CaCl2? 4. How many moles of H are in one mole of the compound (NH4)3PO4?
5. Calculate the molar mass of CaCl2.
How can we get more product?
Chemical manufacturing plants are always seeking to improve their processes. One of the ways this improvement comes about is through measuring the amount of material produced in a reaction. By knowing how much is made, the scientists and engineers can try different ways of getting more product at less cost.
Conversions Between Moles and Mass
The molar mass of any substance is the mass in grams of one mole of representative particles of that substance. The representative particles can be atoms, molecules, or formula units of ionic compounds. This relationship is frequently used in the laboratory. Suppose that for a certain experiment you need 3.00 moles of calcium chloride (CaCl2). Since calcium chloride is a solid, it would be convenient to use a balance to measure the mass that is needed. The molar mass of CaCl2 is 110.98 g/mol. The conversion factor that can be used is then based on the equality that 1 mol = 110.98 g CaCl2. Dimensional analysis will allow you to calculate the mass of CaCl2 that you should measure.
When you measure the mass of 333 g of CaCl2, you are measuring 3.00 moles of CaCl2.
Calcium chloride is used as a drying agent and as a road deicer.
Sample Problem: Converting Moles to Mass
Chromium metal is used for decorative electroplating of car bumpers and other surfaces. Find the mass of 0.560 moles of chromium.
Step 1: List the known quantities and plan the problem.
Known
molar mass of Cr = 52.00 g mol
0.560 mol Cr
Unknown
0.560 mol Cr = ? g
One conversion factor will allow us to convert from the moles of Cr to mass.
Step 2: Calculate.
Step 3: Think about your result.
Since the desired amount was slightly more than one half of a mole, the mass should be slightly more than one half of the molar mass. The answer has three significant figures because of the 0.560 mol.
A similar conversion factor utilizing molar mass can be used to convert from the mass of an substance to moles. In a laboratory situation, you may perform a reaction and produce a certain amount of a product which can be massed. It will often then be necessary to determine the number of moles of the product that was formed. The next problem illustrates this situation.
Sample Problem: Converting Mass to Moles
A certain reaction produces 2.81 g of copper(II) hydroxide, Cu(OH)2. Determine the number of moles produced in the reaction.
Step 1: List the known quantities and plan the problem.
Known
mass = 2.81 g
Unknown
mol Cu(OH)2
One conversion factor will allow us to convert from mass to moles.
Step 2: Calculate.
First, it is necessary to calculate the molar mass of Cu(OH)2 from the molar masses of Cu, O, and H. The molar mass is 97.57 g/mol.
Step 3: Think about your result.
The relatively small mass of product formed results in a small number of moles.
Learn how the molar mass of an object will determine if it will sink or float in this simulation:
Summary
Calculations involving conversions between moles of a material and the mass of that material are described.
Review 1. You have 19.7 grams of a material and wonder how many moles were formed. Your friend tells you to multiply the mass by grams/mole. Is your friend correct?
2. How many grams of MgO are in 3.500 moles?
3. How many moles of H2O are in 15.2 grams of pure ice?
How much gas is there?
Avogadro was interested in studying gases. He theorized that equal volumes of gases under the same conditions contained the same number of particles. Other researchers studied how many gas particles were in a specific volume of gas. Eventually, scientists were able to develop the relationship between number of particles and mass using the idea of moles.
Conversions Between Mass and Number of Particles
In "Conversions between Moles and Mass," you learned how to convert back and forth between moles and the number of representative particles. Now you have seen how to convert back and forth between moles and mass of a substance in grams. We can combine the two types of problems into one. Mass and number of particles are both related to grams. In order to convert from mass to number of particles or vice-versa, it will first require a conversion to moles.
Conversion from number of particles to mass or from mass to number of particles requires two steps
Sample Problem: Converting Mass to Particles
How many molecules is 20.0 g of chlorine gas, Cl2?
Step 1: List the known quantities and plan the problem.
Known
molar mass Cl2 = 70.90 g/mol
20.0 g Cl2 Unknown
number of molecules of Cl2
Use two conversion factors. The first converts grams of Cl2 to moles. The second converts moles of Cl2 to the number of molecules.
Step 2: Calculate.
The problem is done using two consecutive conversion factors. There is no need to explicitly calculate the moles of Cl2.
Step 3: Think about your result.
Since the given mass is less than half of the molar mass of chlorine, the resulting number of molecules is less than half of Avogadro’s number.
Summary
Calculations are illustrated for conversions between mass and number of particles.
Review
1. Why can’t we convert directly from number of particles to grams?
23 2. How many atoms of chlorine are present in 1.70×10 molecules Cl2?
3. How many molecules of BH3 are in 14.32 grams BH3?
How do scuba divers know if they will run out of gas?
Knowing how much gas is available for a dive is crucial to the survival of the diver. The tank on the diver’s back is equipped with gauges to tell how much gas is present and what the pressure is. A basic knowledge of gas behavior allows the diver to assess how long to stay under water without developing problems.
Avogadro’s Hypothesis and Molar Volume
Volume is a third way to measure the amount of matter, after item count and mass. With liquids and solids, volume varies greatly depending on the density of the substance. This is because solid and liquid particles are packed close together with very little space in between the particles. However, gases are largely composed of empty space between the actual gas particles (see Figure below).
Gas particles are very small compared to the large amounts of empty space between them.
In 1811, Amedeo Avogadro explained that the volumes of all gases can be easily determined. Avogadro’s hypothesis states that equal volumes of all gases at the same temperature and pressure contain equal numbers of particles. Since the total volume that a gas occupies is made up primarily of the empty space between the particles, the actual size of the particles themselves is nearly negligible. A given volume of a gas with small light particles such as hydrogen (H2) contains the same number of particles as the same volume of a heavy gas with large particles such as sulfur hexafluoride, SF6.
Gases are compressible, meaning that when put under high pressure, the particles are forced closer to one another. This decreases the amount of empty space and reduces the volume of the gas. Gas volume is also affected by temperature. When a gas is heated, its molecules move faster and the gas expands. Because of the variation in gas volume due to pressure and temperature changes, the comparison of gas volumes must be done at one standard temperature and pressure. Standard temperature and pressure (STP) is defined as 0°C (273.15 K) and 1 atm pressure. The molar volume of a gas is the volume of one mole of a gas at STP. At STP, one mole (6.02 × 1023 representative particles) of any gas occupies a volume of 22.4 L (Figure below).
A mole of any gas occupies 22.4 L at standard temperature and pressure (0°C and 1 atm).
Figure below illustrates how molar volume can be seen when comparing different gases. Samples of helium (He), nitrogen (N2), and methane (CH4) are at STP. Each contains 1 mole or 6.02 × 1023 particles. However, the mass of each gas is different and corresponds to the molar mass of that gas: 4.00 g/mol for He, 28.0 g/mol for N2, and 16.0 g/mol for CH4.
Avogadro’s hypothesis states that equal volumes of any gas at the same temperature and pressure contain the same number of particles. At standard temperature and pressure, 1 mole of any gas occupies 22.4 L.
Summary
Equal volumes of gases at the same conditions contain the same number of particles.
Standard temperature and pressure is abbreviated (STP).
Standard temperature is 0°C (273.15 K) and standard pressure is 1 atm.
At STP, one mole of any gas occupies a volume of 22.4 L
Review
1. A container is filled with gas, what do we know about the space actually taken up by a gas?
2. Why do we need to do all our comparisons at the same temperature and pressure?
3. At standard temperature and pressure, 1 mole of gas is always equal to how many liters?
How can you tell how much gas is in these containers?
Small gas tanks are often used to supply gases for chemistry reactions. A gas gauge will give some information about how much is in the tank, but quantitative estimates are needed so the reaction will be able to proceed to completion. Knowing how to calculate needed parameters for gases is very helpful to avoid running out too early.
Conversions Between Moles and Gas Volume
Molar volume at STP can be used to convert from moles to gas volume and from gas volume to moles. The equality of 1 mole = 22.4 L is the basis for the conversion factor.
Sample Problem One: Converting Gas Volume to Moles
Many metals react with acids to produce hydrogen gas. A certain reaction produces 86.5 L of hydrogen gas at STP. How many moles of hydrogen were produced?
Step 1: List the known quantities and plan the problem.
Known 86.5 L H2
1 mol = 22.4 L
Unknown
moles of H2
Apply a conversion factor to convert from liters to moles.
Step 2: Calculate.
Step 3: Think about your result.
The volume of gas produced is nearly four times larger than the molar volume. The fact that the gas is hydrogen plays no role in the calculation.
Sample Problem Two: Converting Moles to Gas Volume
What volume does 4.96 moles of O2 occupy at STP?
Step 1: List the known quantities and plan the problem.
Known
4.96 moles O2
1 mol = 22.4 L
Unknown
volume of O2
Step 2: Calculate.
Step 3: Think about your result. The volume seems correct given the number of moles.
Sample Problem Three: Converting Volume to Mass
If we know the volume of a gas sample at STP, we can determine how much mass is present. Assume we have 867 liters of N2 at STP. What is the mass of the nitrogen gas?
Step 1: List the known quantities and plan the problem.
Known
867 L N2
1 mol = 22.4 L
molar mass of N2 = 28.02 g/mol
Unknown
mass of N2
Step 2: Calculate.
We start by determining the number of moles of gas present. We know that 22.4 liters of a gas at STP equals one mole, so:
867 litres×1 mole22.4 liters=3.87 moles
We also know the molecular weight of N2 (28.0 grams/mole), so we can then calculate the weight of nitrogen gas in 867 liters:
Step 3: Think about your result.
In a multi-step problem, be sure that the units check.
Summary
Conversions between moles and volume of a gas are shown.
Review 1. In the problems above, why was the gas always at standard temperature and pressure?
2. A container contains 45.2 L of N2 gas at STP. How many moles of N2 gas are in the container?
3. If the gas in the previous problem was CH4 gas at STP instead of N2 gas, then how many
moles of CH4 gas would there be?
How do I get from here to there?
If I want to visit the town of Manteo, North Carolina, out on the coast, I will need a map of how to get there. I may have a printed map or I may download directions from the internet, but I need something to get me going in the right direction. Chemistry road maps serve the same purpose. How do I handle a certain type of calculation? There is a process and a set of directions to help.
Mole Road Map
Previously, we saw how the conversions between mass and number of particles required two steps, with moles as the intermediate. This concept can now be extended to also include gas volume at STP. The resulting diagram is referred to as a mole road map (see Figure below).
The mole road map shows the conversion factors needed to interconvert between mass, number of particles, and volume of a gas.
The mole is at the center of any calculation involving amount of a substance. The sample problem below is one of many different problems that can be solved using the mole road map.
Sample Problem One: Mole Road Map What is the volume of 79.3 g of neon gas at STP?
Step 1: List the known quantities and plan the problem.
Known
Ne = 20.18 g/mol
1 mol = 22.4 L
Unknown
volume = ? L
The conversion factors will be grams → moles → gas volume.
Step 2: Calculate.
Step 3: Think about your result.
The given mass of neon is equal to about 4 moles, resulting in a volume that is about 4 times larger than molar volume.
Summary
An overall process is given for calculations involving moles, grams, and gas volume.
Review
1. In the problem above, what is the formula weight of neon?
2. What value is at the center of all the calculations?
3. If we had 79.3 grams of Xe, would we expect a volume that is greater than or less than that obtained with neon?
Packaged foods that you eat typically have nutritional information provided on the label. The label on a jar of peanut butter reveals that one serving size is considered to be 32 g. The label also gives the masses of various types of compounds that are present in each serving. One serving contains 7 g of protein, 15 g of fat, and 3 g of sugar. By calculating the fraction of protein, fat, or sugar in one serving of size of peanut butter and converting to percent values, we can determine the composition of the peanut butter on a percent by mass basis.
Percent Composition
Chemists often need to know what elements are present in a compound and in what percentage. The percent composition is the percent by mass of each element in a compound. It is calculated in a similar way that we just indicated for the peanut butter.
% by mass=mass of elementmass of compound×100%
Percent Composition From Mass Data
The sample problem below shows the calculation of the percent composition of a compound based on mass data.
Sample Problem One: Percent Composition from Mass
A certain newly synthesized compound is known to contain the elements zinc and oxygen. When a 20.00 g sample of the sample is decomposed, 16.07 g of zinc remains. Determine the percent composition of the compound.
Step 1: List the known quantities and plan the problem.
Known
mass of compound = 20.00 g
mass of Zn = 16.07 g
Unknown
percent Zn = ? %
percent O = ? %
Subtract to find the mass of oxygen in the compound. Divide each element’s mass by the mass of the compound to find the percent by mass.
Step 2: Calculate. Step 3: Think about your result.
The calculations make sense because the sum of the two percentages adds up to 100%. By mass, the compound is mostly zinc.
Percent Composition From a Chemical Formula
The percent composition of a compound can also be determined from the formula of the compound. The subscripts in the formula are first used to calculate the mass of each element in one mole of the compound. That is divided by the molar mass of the compound and multiplied by 100%.
% by mass =mass of element in 1 molmolar mass of compound×100%
The percent composition of a given compound is always the same as long as the compound is pure.
Sample Problem Two: Percent Composition from Chemical Formula
Dichlorineheptoxide (Cl2O7) is a highly reactive compound used in some organic synthesis reactions. Calculate the percent composition of dichlorineheptoxide.
Step 1: List the known quantities and plan the problem.
Known
mass of Cl in 1 mol Cl2O7 = 70.90 g
mass of O in 1 mol Cl2O7 = 112.00 g
molar mass of Cl2O7 = 182.90 g/mol
Unknown
percent Cl = ? %
percent O = ? % Calculate the percent by mass of each element by dividing the mass of that element in 1 mole of the compound by the molar mass of the compound and multiplying by 100%.
Step 2: Calculate.
Step 3: Think about your result.
The percentages add up to 100%.
Percent composition can also be used to determine the mass of a certain element that is contained in any mass of a compound. In the previous sample problem, it was found that the percent composition of dichlorineheptoxide is 38.76% Cl and 61.24% O. Suppose that you needed to know the masses of chlorine and oxygen present in a 12.50 g sample of dichlorineheptoxide. You can set up a conversion factor based on the percent by mass of each element.
The sum of the two masses is 12.50 g, the mass of the sample size.
Summary
Processes are described for calculating the percent composition of a material based on mass or on chemical composition.
Review
1. What is the formula for calculating percent composition?
2. What information do you need to calculate percent composition by mass?
3. What do subscripts in a chemical formula tell you?
In the early days of chemistry, there were few tools for the detailed study of compounds. Much of the information regarding the composition of compounds came from the elemental analysis of inorganic materials. The “new” field of organic chemistry (the study of carbon compounds) faced the challenge of not being able to characterize a compound completely. The relative amounts of elements could be determined, but so many of these materials had carbon, hydrogen, oxygen, and possibly nitrogen in simple ratios. We did not know exactly how many of these atoms were actually in a specific molecule.
Determining Empirical Formulas
An empirical formula is one that shows the lowest whole-number ratio of the elements in a compound. Because the structure of ionic compounds is an extended three-dimensional network of positive and negative ions, all formulas of ionic compounds are empirical. However, we can also consider the empirical formula of a molecular compound. Ethene is a small hydrocarbon compound with the formula C2H4 (see Figure below). While C2H4 is its molecular formula and represents its true molecular structure, it has an empirical formula of CH2. The simplest ratio of carbon to hydrogen in ethene is 1:2. There are two ways to view that ratio. Considering one molecule of ethene, the ratio is 1 carbon atom for every 2 atoms of hydrogen. Considering one mole of ethene, the ratio is 1 mole of carbon for every 2 moles of hydrogen. So the subscripts in a formula represent the mole ratio of the elements in that formula.
Ball-and-stick model of ethene, C2H4.
In a procedure called elemental analysis, an unknown compound can be analyzed in the laboratory in order to determine the percentages of each element contained within it. These percentages can be transformed into the mole ratio of the elements, which leads to the empirical formula. The steps to be taken are outlined below.
1. Assume a 100 g sample of the compound so that the given percentages can be directly converted into grams.
2. Use each element’s molar mass to convert the grams of each element to moles.
3. In order to find a whole-number ratio, divide the moles of each element by whichever of the moles from step 2 is the smallest. 4. If all the moles at this point are whole numbers (or very close), the empirical formula can be written with the moles as the subscript of each element.
5. In some cases, one or more of the moles calculated in step 3 will not be whole numbers. Multiply each of the moles by the smallest whole number that will convert each into a whole number. Write the empirical formula.
Sample Problem One: Determining the Empirical Formula of a Compound
A compound of iron and oxygen is analyzed and found to contain 69.94% iron and 30.06% oxygen. Find the empirical formula of the compound.
Step 1: List the known quantities and plan the problem.
Known
% of Fe = 69.94%
% of O = 30.06%
Unknown
Empirical formula = Fe?O?
Steps to follow are outlined in the text.
Step 2: Calculate.
1. Assume a 100 g sample.
69.94 g Fe30.06 g O
2. Convert to moles.
3. Divide both moles by the smallest of the results. 4/5. Since the moles of O, is still not a whole number, both moles can be multiplied by 2, while rounding to a whole number.
1 mol Fe×2=2 mol Fe1.501 mol O×2=3 mol O
The empirical formula of the compound is Fe2O3.
Step 3: Think about your result.
The subscripts are whole numbers and represent the mole ratio of the elements in the compound. The compound is the ionic compound iron(III) oxide.
Summary
A process is described for the calculation of the empirical formula for a compound based on the percent composition of that compound.
Review
1. What is an empirical formula?
2. What does an empirical formula tell you?
3. What does it not tell you?
Molecular Formulas
Molecular formulas give the kind and number of atoms of each element present in a molecular compound. In many cases, the molecular formula is the same as the empirical formula. The molecular formula of methane is CH4 and because it contains only one carbon atom, that is also its empirical formula. Sometimes, however, the molecular formula is a simple whole-number multiple of the empirical formula. Acetic acid is an organic acid that is the main component of vinegar. Its molecular formula is C2H4O2. Glucose is a simple sugar that cells use as a primary source of energy. Its molecular formula is C6H12O6. The structures of both molecules are shown in the figure below. They are very different compounds, yet both have the same empirical formula of CH2O.
Acetic acid (left) has a molecular formula of C2H4O2, while glucose (right) has a molecular formula of C6H12O6. Both have the empirical formula CH2O.
Empirical formulas can be determined from the percent composition of a compound. In order to determine its molecular formula, it is necessary to know the molar mass of the compound. Chemists use an instrument called a mass spectrometer to determine the molar mass of compounds. In order to go from the empirical formula to the molecular formula, follow these steps:
1. Calculate the empirical formula mass (EFM), which is simply the molar mass represented by the empirical formula.
2. Divide the molar mass of the compound by the empirical formula mass. The result should be a whole number or very close to a whole number.
3. Multiply all the subscripts in the empirical formula by the whole number found in step 2. The result is the molecular formula.
Sample Problem One: Determining the Molecular Formula of a Compound
The empirical formula of a compound of boron and hydrogen is BH3. Its molar mass is 27.7 g/mol. Determine the molecular formula of the compound.
Step 1: List the known quantities and plan the problem.
Known
empirical formula = BH3
molar mass = 27.7 g/mol
Unknown
molecular formula = ?
Steps to follow are outlined in the text. Step 2: Calculate.
1. The empirical formula mass (EFM) = 13.84 g/mol
The molecular formula of the compound is B2H6.
Step 3: Think about your result.
The molar mass of the molecular formula matches the molar mass of the compound.
Summary
A procedure is described that allows the calculation of the exact molecular formula for a compound.
Review
1. What is the difference between an empirical formula and a molecular formula?
2. In addition to the elemental analysis, what do you need to know to calculate the molecular formula?
3. What does the empirical formula mass tell you?