University High School Chemistry: The Mathematics of Compounds

The Mathematics of Compounds

Determining Formula and Molar Masses

Lesson Objectives

Given the formula or name of a compound and a periodic table, the student will calculate the formula mass.

Introduction

When atoms of one element chemically combine with atoms of another element, a compound is formed. Compounds have names and they have formulas. Previously, you have learned to properly write formulas and name compounds. The formula of a compound contains chemical symbols that tell us what elements are in the compound and the formula also contains subscripts that tell us the ratios of the atoms of the elements that combined. For example, the formula $MgCl_2$ tells us that this compound is composed of the elements magnesium and chlorine and that they combined in the ratio of two atoms of chlorine for each atom of magnesium. Using the relative masses of elements from the periodic table and the formula, we will calculate the relative formula mass for the compound.

Formula Mass

We have, on the periodic table, the relative masses of all the elements. By looking at the squares for carbon and helium, we can see that a carbon atom has about three times the mass of a helium atom. In this way, we can compare the relative masses of any two atoms in the table. We will now be looking at many chemical formulas and we wish to compare their masses in the same way. We can use the relative atomic masses to do exactly that. The formula mass of a compound is the sum of all the atomic masses in the formula. The formula for water, for example, is $H_2O$ . This formula tells us that water is composed of hydrogen and oxygen and that their ratio is two hydrogen atoms for each oxygen atom. We can determine the formula mass for water by adding up the atomic masses of its components.

Example 1

What is the formula mass of $H_2O$ ?

Solution

Element	Atomic Mass	Number of Atoms per Formula	Product
$H$	$1.0 \;\mathrm{Dalton}$	$2$	$2.0 \;\mathrm{Dalton}$
$O$	$16.0 \;\mathrm{Dalton}$	$1$	$16.0 \;\mathrm{Dalton}$
			$\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}$
			$18.0 \;\mathrm{Dalton}$

The formula mass of $H_2O = 18.0 \;\mathrm{Dalton}$ .

Example 2

What is the formula mass of $Ca(NO_3)_2$ ?

Solution

Element	Atomic Mass	Number of Atoms per Formula	Product
$Ca$	$40.0 \;\mathrm{Dalton}$	$1$	$40.0 \;\mathrm{Dalton}$
$N$	$14.0 \;\mathrm{Dalton}$	$2$	$28.0 \;\mathrm{Dalton}$
$O$	$16.0 \;\mathrm{Dalton}$	$6$	$96.0 \;\mathrm{Dalton}$
			$\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}$
			$164.0 \;\mathrm{Dalton}$

The formula mass of $Ca(NO_3)_2 = 164.0 \;\mathrm{Dalton}$ .

These formula masses are in the same units as atomic masses and therefore are exactly comparable. That is, the atomic mass of an oxygen atom is $16 \;\mathrm{Dalton}$ , the atomic mass of a fluorine atom is $19 \;\mathrm{amu}$ , and the formula mass of a water molecule is $18 \;\mathrm{Dalton}$ – that means that a water molecule is slightly more massive than an oxygen atom and slightly less massive than a fluorine atom.

Terminology

Since ionic compounds do not form molecules, the term formula mass is the only proper term for this relative mass. For covalent compounds, since they do form molecules, the formula mass may also be called the molecular mass. On the other hand, there are professional chemists who occasionally refer to the “molecular mass” of $NaCl$ and so far no degrees in chemistry have been recalled due to such a slip of the tongue.

Lesson Summary

The molecular mass of a molecule is found by adding the atomic masses of all the atoms in one molecule.
Not all substances exist as molecules and therefore, the term molecular mass is not used for all substances. The masses of ionic compounds and empirical formulas are called formula mass.

Review Questions

Calculate the formula mass for each of the following.
1. $K_2SO_4$
2. $CuO$
3. $Mg_3(AsO_4)_2$
4. $Ca_3(PO_4)_2$
5. $Fe_2O_3$
6. $Al(OH)_3$
7. $(NH_4)_2S$
8. $C_{12}H_{22}O_{11}$
How many times heavier are bromine atoms on the average than neon atoms?
An unknown element, $M$ , combines with oxygen to form a compound with a formula of $MO_2$ . If $25.0 \;\mathrm{grams}$ of the unknown element combines with $4.50 \;\mathrm{grams}$ of oxygen, what is the atomic mass of $M$ ?

Vocabulary

formula mass: The sum of the atomic masses of the atoms in a formula.

molecular mass: The mass of a molecule found by adding the atomic masses of the atoms comprising the molecule.

The Mole

Lesson Objectives

The student will express the value of Avogadro's number and explain its relevance to moles.
Given the number of particles of a substance, the student will use Avogadro's number to convert to moles and vice versa.
Given the number of moles of a substance, the student will use the molar mass to convert to grams and vice versa.

Introduction

When objects are very small, it is often inconvenient or inefficient, or even impossible to deal with the objects one at a time. For these reasons, we often deal with very small objects in groups, and have even invented names for various numbers of objects. The most common of these is “dozen” which refers to $12$ objects. We frequently buy objects in groups of $12$ , like doughnuts or pencils. Even smaller objects such as straight pins or staples are usually sold in boxes of $144$ , or a dozen dozen. A group of $144$ is called a “gross.”

This problem of dealing with things that are too small to operate with as single items also occurs in chemistry. Atoms and molecules too small to see, let alone to count or measure. Chemists needed to select a group of atoms or molecules that would be convenient to operate with.

Avogadro's Number

In chemistry, it is impossible to deal with a single atom or molecule because we can’t see them or count them or weigh them. Chemists have selected a number of particles with which to work that is convenient. Since molecules are extremely small, you may suspect that this number is going to be very large and you are right. The number of particles in this group is $6.02 \times 10^{23}$ particles and the name of this group is the mole (the abbreviation for mole is mol). One mole of any object is $6.02 \times 10^{23}$ of those objects. There is a very particular reason that this number was chosen and we hope to make that reason clear to you.

When chemists are carrying out chemical reactions, it is important that the relationship between the numbers of particles of each reactant is known. Chemists looked at the atomic masses on the periodic table and understood that the mass ratio of one carbon atom to one sulfur atom was $12 \;\mathrm{amu}$ to $32 \;\mathrm{amu}$ . They realized that if they massed out $12 \;\mathrm{grams}$ of carbon and $32 \;\mathrm{grams}$ of sulfur, they would have the same number of atoms of each element. They didn’t know how many atoms were in each pile but they knew the number in each pile had to be the same. This is the same logic as knowing that if a basketball has twice the mass of a soccer ball and you massed out $100 \;\mathrm{lbs}$ of basketballs and $50 \;\mathrm{lbs}$ of soccer balls, you would have the same number of each ball.

The technique of taking one atomic mass or molecular mass in grams of substances for reactions was used to get equal numbers of particles. This amount of substance (its molecular mass in grams) became known as a gram-molecular mass. One gram-molecular mass of any substance had the same number of particles in it. Many years later, when it became possible to count particles using electrochemical reactions, the number of particles in a gram-molecular mass was counted. That number turned out to be $6.02 \times 10^{23}$ particles. That group of particles continued to be called a gram-molecular mass for many years, but eventually chemists decided to call that number of particles a mole.

In present day, $1.00 \;\mathrm{mole}$ of carbon- $12 \;\mathrm{atoms}$ has a mass of $12.0 \;\mathrm{grams}$ and contains $6.02 \times 10^{23} \;\mathrm{atoms}$ . Likewise, $1.00 \;\mathrm{mole}$ of water has a mass of $18.0 \;\mathrm{grams}$ and contains $6.02 \times 10^{23} \;\mathrm{molecules}$ . $1.00 \;\mathrm{mole}$ of any element or compound has a mass equal to its molecular mass in grams and contains $6.02 \times 10^{23}$ particles. The mass in grams of $6.02 \times 10^{23}$ particles of a substance is now called the molar mass (mass of $1.00 \;\mathrm{mole}$ ).

Converting Molecules to Moles and Vice Versa

The number $6.02 \times 10^{23}$ is called Avogadro’s number and is symbolized as $N$ (the capital letter $N$ ). Avogadro, of course, had no hand in determining this number, rather it was named in honor of Avogadro. The mass of a single molecule of $H_2SO_4$ is $98 \;\mathrm{Dalton}$ and the mass of an Avogadro number of molecules of $H_2SO_4$ is $98 \;\mathrm{grams}$ . We can use this information to find the mass in grams of a single $H_2SO_4 \;\mathrm{molecule}$ because we know that $98 \;\mathrm{grams}$ contains $6.02 \times 10^{23} \;\mathrm{molecules}$ . If we divide $6.02 \times 10^{23} \;\mathrm{molecules}$ into $98 \;\mathrm{grams}$ , we will get the mass of a single molecule of $H_2SO_4$ in grams. The answer is $1.6 \times 10^{-22} \;\mathrm{grams/molecule}$ – tiny, indeed. If we are given a number of molecules of a substance, we can convert it into moles by dividing by Avogadro’s number and vice versa.

Example 3

How many moles are present in $1,000,000,000$ ( $1$ billion or $1 \times 10^9$ ) molecules of water?

Solution

$\text{moles} = (1,000,000,000 \ \text{molecules}) \left (\frac {1\ \text{mole}} {6.02 \times 10^{23}\ \text{molecules}}\right ) = 1.7 \times 10^{-15} \ \text{moles}$

You should note that this amount of water is too small for even our most delicate balances to determine the mass. A very large number of molecules must be present before the mass is large enough to detect with our balances.

Example 4

How many molecules are present in $0.00100 \;\mathrm{mole}$ ?

Solution

$\text{molecules} & = (\text{moles})(6.02 \times 10^{23} \ \text{molecules/mole})\\ \text{molecules} & = (0.00100 \ \text{mole})(6.02 \times 10^{23} \ \text{molecules/mole}) = 6.02 \times 10^{20} \ \text{molecules}$

This is a number of water molecules whose mass we can measure with a balance, $602,000,000,000,000,000,000$ .

Converting Grams to Moles and Vice Versa

We can also convert back and forth between grams of substance and moles. The conversion factor for this is the molar mass of the substance. To convert the grams of a substance into moles, we divide by the molar mass and to convert the moles of a substance into grams, we multiply by the molar mass.

Example 5

How many moles are present in $108 \;\mathrm{grams}$ of water?

Solution

$\text{moles} = \frac {\text{grams}} {\text{molar mass}} = \frac {108\ \text{grams}} {18.0\ \text{grams/mole}} = 6.00 \ \text{moles}$

Example 6

What is the mass of $7.50 \;\mathrm{moles}$ of $CaO$ ?

Solution

$\text{Grams} = (\text{moles})(\text{molar mass}) = (7.50 \ \text{moles})(56.0 \ \text{grams/mole}) = 420. \ \text{Grams}$

Using Avogadro's law, the mass of a substance can be related to the number of particles contained in that mass. The Mole (http://www.learner.org/vod/vod_window.html?pid=803)

Lesson Summary

There are $6.02 \times 10^{23}$ particles in $1.00 \;\mathrm{mole}$ . This number is called Avogadro’s number.
The number of moles in a given number of molecules of a substance can be found by dividing the number of molecules by Avogadro’s number.
The number of moles in a given mass of substance can be found by dividing the mass by the formula mass expressed in $\;\mathrm{g/mol}$ .

Review Questions

Convert the following to moles.
1. $60.0 \;\mathrm{grams \ of \ NaOH}$
2. $2.73 \;\mathrm{grams \ of} \ NH_4Cl$
3. $5.70 \;\mathrm{grams \ of} \ ZrF_4$
4. $10.0 \;\mathrm{grams \ of} PbO_2$
Convert the following to grams.
1. $0.100 \;\mathrm{moles \ of} \ CO_2$
2. $0.437 \;\mathrm{moles \ of \ NaOH}$
3. $0.500 \;\mathrm{moles \ of} \ (NH_4)_2CO_3$
4. $3.00 \;\mathrm{moles \ of} \ ZnCl_2$
How many molecules are present in the following quantities?
1. $0.250 \;\mathrm{mole \ of} \ H_2O$
2. $6.00 \;\mathrm{moles \ of} \ H_2SO_4$
3. $0.00450 \;\mathrm{mole \ of} \ Al_2(CO_3)_3$
How many moles are present in the following quantities?
1. $1.00 \times 10^{20} \;\mathrm{molecules \ of} \ H_2O$
2. $1.00 \times 10^{25} \;\mathrm{molecules \ of} \ H_2$
3. $5,000,000,000,000 \;\mathrm{atoms \ of \ carbon}$
How many molecules (or atoms) are present in the following masses?
1. $1.00 \;\mathrm{gram \ of} \ Na_2CO_3$
2. $8.00 \;\mathrm{grams \ of \ helium}$
3. $1000. \;\mathrm{grams \ of} \ H_2O$
Convert the following to grams.
1. $1.00 \times 10^{23} \;\mathrm{molecules \ of} \ H_2$
2. $1.00 \times 10^{24} \;\mathrm{molecules \ of} \ AlPO_4$
3. $1.00 \times 10^{22} \;\mathrm{molecules \ of \ NaOH}$
What is the mass of a single atom of silver, $Ag$ , in grams?
If you had a silver bar that exactly fit inside a $100. \;\mathrm{mL}$ graduated cylinder and filled it precisely to the $100. \;\mathrm{mL}$ mark, how many atoms of silver would be in the bar? The density of silver is $10.5 \;\mathrm{g/mL}$ .

Vocabulary

Avogadro's number: The number of objects in a mole; equal to $6.02 \times 10^{23}$ .

mole: An Avogadro’s number of objects.

Percent Composition

Lesson Objectives

Given either masses of elements in a compound, the student will calculate the percent composition by mass.
Given the formula or name of a compound, the student will calculate the percent composition by mass.

Introduction

Metals that are useful to man usually begin as ore in a mine. The ore is removed from the mine and partially purified by simply washing away the dirt and other non-chemically combined materials. This partially purified ore is then treated chemically in smelters or other purifying processes to separate pure metal from the other elements. The value of the original ore is very dependent on how much pure metal can eventually be separated from it. High grade ore and low grade ore command a significantly different price. The ore can be evaluated before it is mined or smelted to determine what percent of the ore can eventually become pure metal. The process involves determining what percent of the ore is metal compounds and then determining what percent of the metal compounds is pure metal.

Percent Composition from Masses

Compounds are made up of two or more elements. The Law of Definite Proportions tells us that the proportion, by mass, of the elements in a compound is always the same. Water, for example, is always $11\%$ hydrogen and $89\%$ oxygen by mass. The percentage composition of a compound is the percentage by mass of each of the elements in the compound.

Percentage composition can be determined by experiment. To do this, a known quantity of a compound is decomposed in the laboratory and the mass of each of the elements is measured. Then the mass of each element is divided by the total mass of the original compound. This tells us what fraction of the compound is made up of that element. This fraction can then be multiplied by $100$ to convert it to a percent.

Example 7

Laboratory procedures show that $50.0 \;\mathrm{grams}$ of ammonia, $NH_3$ , yields $41.0 \;\mathrm{grams}$ of nitrogen and $9.00 \;\mathrm{grams}$ of hydrogen upon decomposition. What is the percent composition of ammonia?

Solution

$\% \ \text{Nitrogen} & = \left (\frac {41.0\ \text{grams}} {50.0\ \text{grams}}\right ) (100) = 82\%\\ \% \ \text{Hydrogen} & = \left (\frac {9.00\ \text{grams}} {50.0\ \text{grams}}\right ) (100) = 18\%$

Example 8

The decomposition of $25.0 \;\mathrm{grams}$ of $Ca(OH)_2$ in the lab produces $13.5 \;\mathrm{grams}$ of calcium, $10.8 \;\mathrm{grams}$ of oxygen and $0.68 \;\mathrm{grams}$ of hydrogen. What is the percent composition of calcium hydroxide?

Solution

$\% \ \text{Calcium} & = \left (\frac {13.5\ \text{grams}} {25.0\ \text{grams}}\right ) (100) = 54.0\%\\ \% \ \text{Oxygen} & = \left(\frac {10.8\ \text{grams}} {25.0\ \text{grams}}\right ) (100) = 43.2\%\\ \% \ \text{Hydrogen} & = \left (\frac {0.68\ \text{grams}} {25.0\ \text{grams}}\right ) (100) = 2.8\%$

You should note that the sum of the percentages always adds to $100 \%$ . Sometimes, it may total $99 \%$ or $101 \%$ due to rounding, but if it totals to $96 \%$ or $103 \%$ , you have an error.

Percent Composition from the Formula

Percent composition can also be calculated from the formula of a compound. Consider the formula for the compound, iron (III) oxide, $Fe_2O_3$ . The percent composition of the elements in this compound can be calculated by dividing the total atomic mass of the atoms of each element in the formula by the formula mass

Example 9

What is the percent composition of iron (III) oxide, $Fe_2O_3$ ?

Solution

Element	Atomic Mass	Number of Atoms per Formula	Product
$Fe$	$55.8 \;\mathrm{daltons}$	$2$	$111.6 \;\mathrm{daltons}$
$O$	$16.0 \;\mathrm{daltons}$	$3$	$48.0 \;\mathrm{daltons}$
			$\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}$
		The formula mass of $Fe_2O_3 =$	$159.6 \;\mathrm{daltons}$

$\% \ \text{iron} & = \left (\frac {111.6} {159.6}\right ) (100) = 69.9 \%\\ \% \ \text{oxygen} & = \left (\frac {48.0} {159.6}\right ) (100) = 30.1 \%$

Example 10

What is the percent composition of aluminum sulfate, $Al_2(SO_4)_3$ ?

Solution

Adding up the formula $\;\mathrm{mass} = 2(27.0 \;\mathrm{daltons}) + 3(32.0 \;\mathrm{daltons}) + 12(16.0 \;\mathrm{daltons}) = 342.0 \;\mathrm{daltons}$

$\% \ \text{aluminum} & = \left (\frac {54.0} {342}\right ) (100) = 15.8 \%\\ \% \ \text{sulfur} & = \left (\frac {96.0} {342}\right ) (100) = 28.1 \%\\ \% \ \text{oxygen} & = \left (\frac {192} {342}\right ) (100) = 56.1 \%$

Lesson Summary

The percent composition of a compound is the percent of the total mass contributed by each of the elements in the compound.

Percent composition can be determined either from the masses of each element in the compound or from the formula of the compound.

Review Questions

Determine the percent composition of the following compounds.

$BF_3$
$Ca(C_2H_3O_2)_2$
$FeF_3$
$CrCl_3$
$(NH_4)_3PO_4$

Vocabulary

percent composition: The proportion of an element present in a compound found by dividing the mass of the element by the mass of the whole compound and multiplying by $100$ .

Empirical and Molecular Formulas

Lesson Objectives

The student will reduce molecular formulas to empirical formulas.
Given either masses or percent composition of a compound, the student will determine the empirical formula.
Given either masses or percent composition of a compound and the molar mass, the student will determine the molecular formula.

Introduction

The empirical formula is the simplest ratio of atoms in a compound. Formulas for ionic compounds are always empirical formulas but for covalent compounds, the empirical formula is not always the actual formula for the molecule. Molecules such as benzene, $C_6H_6$ , would have an empirical formula of $CH$ .

Finding Empirical Formula from Experimental Data

Empirical formulas can be determined from experimental data or from percent composition. Consider the following example.

Example 11

We find that a $2.50 \;\mathrm{gram}$ sample of a compound contains $0.900 \;\mathrm{grams}$ of calcium and $1.60 \;\mathrm{grams}$ of chlorine. The compound contains only these two elements. We can calculate the number of moles of calcium atoms and the number of moles of chlorine atoms in the compound. We can then find the ratio of moles of calcium atoms to moles of chlorine atoms and from this; we can determine the empirical formula.

Solution

First, we convert the mass of each element into moles.

$\text{moles of Ca} & = \frac {0.900\ \text{g}} {40.1\ \text{g/mol}} = 0.0224 \ \text{mole Ca}\\ \text{moles of Cl atoms} & = \frac {1.60\ \text{g}} {35.5\ \text{g/mol}} = 0.0451 \ \text{mole Cl}$

At this point, we have the correct ratio for the atoms in the compound, $Ca_{0.0224} \ Cl_{0.0451}$ , except that this isn’t an acceptable formula. We need to find the simplest whole number ratio. To find a simple whole number ratio for these numbers, we divide each of them by the smallest of them.

$\text{moles of Ca} & = \frac {0.0224} {0.0224} = 1.00 \ \text{Ca}\\ \ \text{moles of Cl} & = \frac {0.0451} {0.0224} = 2.01 \ \text{Cl}$

Now, we can see the correct empirical formula for this compound is $CaCl_2$ .

It is important to note that when solving for empirical formulas, we are determining the number of atoms of each element in the compound. Therefore, those substances which occur in nature as diatomic molecules such as $Cl_2, \ O_2, \ H_2, \ N_2$ , and so on, are dealt with as atoms in this procedure.

Finding Empirical Formula from Percent Composition

When finding the empirical formula from percent composition, the first thing we do is to convert the percentages into masses. For example, suppose we are given the percent composition of a compound as $40.0\%$ carbon, $6.71\%$ hydrogen, and $53.3\%$ oxygen. Since every sample of this compound regardless of size will have the same composition in terms of ratio of atoms, we could choose a sample of any size. Suppose we choose a sample size of $100. \;\mathrm{grams}$ . The masses of each of the elements in this sample will be $40.0 \;\mathrm{grams}$ of carbon, $6.71 \;\mathrm{grams}$ of hydrogen, and $53.3 \;\mathrm{grams}$ of oxygen. These masses can then be used to find the empirical formula. You should note that you could use any size sample. You could choose a sample size of $167.8 \;\mathrm{grams}$ and take the percentages of this sample to get the masses of the individual elements. We choose a sample size of $100. \;\mathrm{grams}$ because it makes the arithmetic simple.

Example 12

Find the empirical formula of a compound whose percent composition is $40.0 \%$ carbon, $6.71 \%$ hydrogen, and $53.3 \%$ oxygen.

Solution

We choose a sample size of $100. \;\mathrm{grams}$ and multiply this $100. \;\mathrm{gram}$ sample by each of the percentages to get masses for each element. This would yield $40.0 \;\mathrm{grams}$ of carbon, $6.71 \;\mathrm{grams}$ of hydrogen, and $53.3 \;\mathrm{grams}$ of oxygen. The next step is to convert the mass of each element into moles.

$\text{moles of C} & = \frac {40.0\ \text{g}} {12.0\ \text{g/mol}} = 3.33 \ \text{moles C}\\ \text{moles of H} & = \frac {6.71\ \text{g}} {1.01\ \text{g/mol}} = 6.64 \ \text{moles H}\\ \ \text{moles of O} & = \frac {53.3\ \text{g}} {16.0\ \text{g/mol}} = 3.33 \ \text{mole Ca}$

Then, we divide all three numbers by the smallest one to get simple whole number ratios:

$C = \frac {3.33} {3.33} = 1 && H = \frac {6.64} {3.33} = 2 && O = \frac {3.33} {3.33} = 1$

and finally, we can write the empirical formula, $CH_2O$ .

Sometimes, the technique of dividing each of the moles by the smallest to get a whole number ratio does not yield whole numbers. Whenever the subscript for any element in the empirical formula is “1,” dividing each of the moles by the smallest will yield a simple whole number ratio but if none of the elements in the empirical formula has a subscript of “1,” then this technique will not yield a simple whole number ratio. In those cases, a little more work is required.

Example 13

Determine the empirical formula for a compound that is $66.0 \%$ calcium and $34.0 \%$ phosphorus.

Solution

We choose a sample size of $100. \;\mathrm{grams}$ and multiply the $100. \;\mathrm{grams}$ by the percentage of each element to get masses. This yields $66.0 \;\mathrm{grams}$ of calcium and $34.0 \;\mathrm{grams}$ of phosphorus. We then divide each of these masses by their molar mass to convert to moles.

$\text{moles of Ca} & = \frac {66.0\ \text{g}} {40.1\ \text{g/mol}} = 1.65 \ \text{moles Ca}\\ \text{moles of P} & = \frac {34.0\ \text{g}} {31.0\ \text{g/mol}} = 1.10 \ \text{moles P}$

We then divide each of these moles by the smallest.

$Ca = \frac {1.65} {1.10} = 1.50 && P = \frac {1.10} {1.10} = 1.00$

In this case, dividing each of the numbers by the smallest one does not yield a simple whole number ratio. In such a case, we must multiply both numbers by some factor that will produce a whole number ratio. If we multiply each of these by $2$ , we get a whole number ratio of $3$ $Ca$ to $2 \ P$ . Therefore, the empirical formula is $Ca_3P_2$ .

Finding Molecular Formulas

Empirical formulas show the simplest whole number ratio of the atoms in a compound. Molecular formulas show the actual number of atoms of each element in a compound. When you find the empirical formula from either masses of elements or from percent composition, as demonstrated in the previous section, for the compound $N_2H_4$ , you will get an empirical formula of $NH_2$ and for $C_3H_6$ , you will get $CH_2$ . If we want to determine the actual molecular formula, we need one additional piece of information. The molecular formula is always a whole number multiple of the empirical formula. That is, in order to get the molecular formula for $N_2H_4$ , you must double each of the subscripts in the empirical formula. Since the molecular formula is a whole number multiple of the empirical formula, the molecular mass will be the same whole number multiple of the formula mass. The formula mass for $NH_2$ is $14 \;\mathrm{g/mol}$ and the molecular mass for $N_2H_4$ is $28 \;\mathrm{g/mol}$ . When we have the empirical formula and the molecular mass for a compound, we can divide the formula mass into the molecular mass and find the whole number that we need to multiply by each of the subscripts in the empirical formula.

Example 14

Suppose we have the same problem as in example 12 except that we are also given the molecular mass of the compound as $180 \;\mathrm{grams/mole}$ and we are asked for the molecular formula. In example 12, we determined the empirical formula to be $CH_2O$ . This empirical formula has a formula mass of $30.0 \;\mathrm{g/mol}$ . In order to find the molecular formula for this compound, we divide the formula mass into the molecular mass ( $180$ divided by $30$ ) and find the multiplier for the empirical formula to be $6$ . As a result, the molecular formula for this compound will be $C_6H_{12}O_6$ .

Example 15

Find the molecular formula for a compound with percent composition of $85.6\%$ carbon and $14.5 \%$ hydrogen. The molecular mass of the compound is $42.1 \;\mathrm{g/mol}$ .

Solution

We choose a sample size of $100. \;\mathrm{g}$ and multiply each element percentage to get masses for the elements in this sample.

This yields $85.6 \;\mathrm{g}$ of $C$ and $14.5 \;\mathrm{g}$ of $H$ .

Dividing each of these by their atomic mass yields $7.13 \;\mathrm{moles}$ of $C$ and $14.4 \;\mathrm{moles}$ of $H$ .

Dividing each of these by the smallest yields a whole number ratio of $1$ carbon to $2$ hydrogen.

Thus, the empirical formula will be $CH_2$ .

The formula mass of $CH_2$ is $14 \;\mathrm{g/mol}$ .

Dividing $14 \;\mathrm{g/mol}$ into the molecular mass of $42.1 \;\mathrm{g/mol}$ yields a multiplier of 3.

The molecular formula will be $C_3H_6$ .

Lesson Summary

The empirical formula of a compound indicates the simplest whole number ratio of atoms present in the compound.
The empirical formula of a compound can be calculate from the masses of the elements in the compound or from the percent composition.
The molecular formula of a compound is some whole number multiple of the empirical formula.

Review Questions

What is the empirical formula for $C_8H_{18}$ ?
What is the empirical formula for $C_6H_6$ ?
What is the empirical formula for $WO_2$ ?
A compound has the empirical formula $C_2H_8N$ and a molar mass of $46 \;\mathrm{grams/mol}$ . What is the molecular formula of this compound?
A compound has the empirical formula $C_2H_4NO$ . If its molar mass is $116.1 \;\mathrm{grams/mole}$ , what is the molecular formula of the compound?
A sample of pure indium chloride with a mass of $0.5000 \;\mathrm{grams}$ is found to contain $0.2404 \;\mathrm{grams}$ of chlorine. What is the empirical formula of this compound?
Determine the empirical formula of a compound that contains $63.0 \;\mathrm{grams}$ of rubidium and $5.90 \;\mathrm{grams}$ of oxygen.
Determine the empirical formula of a compound that contains $58.0 \%$ $Rb$ , $9.50 \%$ $N$ , and $32.5 \%$ $O$ .
Determine the empirical formula of a compound that contains $33.3 \%$ $Ca$ , $40.0 \%$ $O$ , and $26.7 \%$ $S$ .
Find the molecular formula of a compound with percent composition $26.7\%$ $P$ , $12.1\%$ $N$ , and $61.2\%$ $Cl$ and with a molecular mass of $695 \;\mathrm{g/mol}$ .

Vocabulary

empirical formula: The formula giving the simplest ratio between the atoms of the elements present in a compound.

molecular formula: A formula indicating the actual number of each kind of atom contained in a molecule.

Labs and Demonstrations for The Mathematics of Compounds

Teacher's Pages for Empirical Formula of Magnesium Oxide

Investigation and Experimentation Objectives

In this activity, the student will collect and interpret data, use mathematics to determine a solution, and communicate the result.

Lab Notes

It is not crucial that the magnesium ribbon be exactly $35 \ cm$ , but it should be clean. A good way to clean the ribbon is to dip it in $0.1 \ M$ $HCl$ for a couple of seconds, then rinse it in distilled water, and dry it in alcohol or acetone.

The more finely divided the ribbon is, the faster it will react. $0.5 \ cm$ to $1.0 \ cm$ pieces seems to work best. The crucibles often react with the magnesium during this process. This can cause greenish-black discoloration to the crucibles. This does not really affect chemical behavior for later reactions, as long as they are cleaned. However, the crucibles will often crack during this procedure. If a crucible cracks, discard it. Stress to the students that the crucibles get extremely hot. A hot crucible can cause a very serious burn. Show the students how to handle a crucible properly by using crucible tongs. A common source of error in this experiment is not to react the nitride. Make sure the students do this portion of the lab – it is not washing.

Answers to Pre-Lab Questions

1. Assume you have $100 \ grams$ of the compound. This then changes the percent composition to grams of each element.

Find the number of moles of each element present. (Divide the grams of each element by its molar mass.)

Divide each of these answers by the smallest answer. This will give the empirical formula.

2. The molecular weight of the compound is needed.

3. Yes. The charge of magnesium increases from $0$ to $+2$ , and the charge of the oxygen is reduced from $0$ to $-2.$

4. $3.93 \ g$ for $MgO$ , $3.28 \ g$ for $Mg_3N_2$

Lab - Empirical Formula of Magnesium Oxide

Background Information

In this lab, magnesium metal (an element) is oxidized by oxygen gas to magnesium oxide (a compound). Magnesium reacts vigorously when heated in the presence of air. The $Mg - O_2$ reaction is energetic enough to allow some $Mg$ to react with gaseous $N_2$ . Although there is a higher percentage of $N_2$ gas in the atmosphere than $O_2$ , $O_2$ is more reactive, and the magnesium oxide forms in a greater amount than the nitride. The small amount of nitride that forms can be removed with the addition of water, which converts the nitride to magnesium hydroxide and ammonia gas. Heating the product again causes the loss of water and conversion of the hydroxide to the oxide.

The unbalanced equations are:

$Mg_{(s)} + N_{2(g)} + O_{2(g)} & \rightarrow MgO_{(s)} + Mg_3N_{2(s)}\\ MgO_{(s)} + Mg_3N_{2(s)} + H_2O_{(L)} & \rightarrow MgO_{(s)} + Mg(OH)_{2(s)} + NH_{3(g)}\\ MgO_{(s)} + Mg(OH)_{2(s)} & \rightarrow MgO_{(s)} + H_2O_{(g)}$

Pre-Lab Questions

If the mass percent of each element in a compound is known, what steps are taken to determine the compound’s empirical formula?
If the empirical formula of a compound is known, what additional information is required to determine the molecular formula of the compound?
Is the reaction of magnesium metal and oxygen gas an oxidation-reduction reaction? If so, what is the change in oxidation number of each type of atom?
What is the theoretical yield in grams of $MgO$ if $2.37 \ g \ Mg$ metal reacts with excess $O_2$ ? What is the theoretical yield of $Mg_3N_2$ if the same amount of $Mg$ reacts with excess $N_2$ ?

Purpose

To determine the empirical formula of magnesium oxide, and to reinforce the concepts of the law of mass conservation and the law of multiple proportions.

Apparatus and Materials

Safety goggles
Magnesium ribbon, $Mg$
Balance (to $0.01 \ g$ or better)
Ring stand
Bunsen burner
Ring support with clay triangle
Crucible with lid
Crucible Tongs
Heat resistant tile or pad

Safety Issues

The crucible and all of the apparatus gets very hot. The ammonia emitted during the secondary reaction is irritating. Open flames can be dangerous. Do not place a hot crucible on an electronic balance. It can damage the electronics. In addition, a hot crucible causes the air above it to become buoyant. If placed on a balance, the buoyant air will cause a mass reading, which is less than the actual mass.

Procedure

1. Heat the empty crucible and lid for about $3 \ minutes$ to remove water, oils, or other contaminants and to make sure there are no cracks. The bottom of the crucible should glow red-hot for about $20 \ seconds$ . Remove the flame and cool the crucible with lid.

2. Record the mass of crucible and lid once it has cooled. Handle the crucible with tongs.

3. Obtain about $0.3 \ g \ (35 \ cm)$ magnesium ribbon (do not handle the ribbon with your hands). Cut the magnesium into $0.5 - 1.0 \ cm$ pieces with scissors.

4. Record the mass of the magnesium ribbon, lid and crucible.

5. Place the crucible securely on the clay triangle. Set the lid slightly off-center on the crucible to allow air to enter but to prevent the magnesium oxide from escaping.

6. Place the Bunsen burner under the crucible, light it, and heat the bottom of the crucible with a gentle flame for about $1 \ minute$ ; then, place the burner under the crucible and heat strongly.

7. Heat until all the magnesium turns into gray-white powder (around $10 \ minutes$ ).

8. Stop heating and allow the crucible, lid and contents to cool.

9. Add about $1 \ ml$ (approx. 10 drops) of distilled water directly to the solid powder. Carefully waft some of the gas that is generated toward your nose, but be very careful. Record any odor.

10. Heat the crucible and contents, with the lid slightly ajar, gently for about $2 \ minutes$ and then strongly for about another $3$ to $5 \ minutes$ .

11. Allow the crucible to cool and then record the mass of the crucible, lid and contents.

12. Follow instructions for oxide disposal given by your teacher. Clean all equipment thoroughly.

Data

Mass of crucible and lid = _______ g

Mass of the crucible, crucible lid, and the magnesium = _______ g

Mass of the crucible, crucible lid, and magnesium oxide = _______ g

Post-Lab Questions

1. Determine the mass of magnesium ribbon used in the experiment by subtracting the mass of the crucible and lid from the mass of the crucible, lid, and magnesium.

Mass of magnesium = _______ g

2. Determine the number of moles of magnesium used. Remember: $\frac{\text{mass}}{\text{atomic weight}} = \text{number of moles}$ . The atomic weight of magnesium is $24.3 \ g/mol$ .

Number of moles of magnesium = _______ mole

3. Determine the mass of magnesium oxide that was formed by subtracting the mass of the mass of the crucible and lid from the mass of the crucible, lid, and magnesium oxide.

Mass of magnesium oxide formed = _______ g

4. Determine the mass of oxygen that combined with the magnesium.

Mass of oxygen = mass of magnesium oxide - mass of magnesium

Mass of oxygen that combined with the magnesium = _______ g

5. Determine the number of moles of oxygen atoms that were used. This is elemental oxygen so use $16.0 \ g/mol$ for the atomic weight.

Number of moles of oxygen atoms that were used = _______ mole

6. Calculate the ratio between moles of magnesium atoms used and moles of oxygen atoms used. Remember, this is simple division. Divide the number of moles of magnesium by the number of moles of oxygen. Round your answer to the nearest whole number, as we do not use part of an atom. This represents the moles (and also atoms) of magnesium. The moles (and also atoms) of oxygen, are represented by 1, because it was on the bottom of the division.

Moles of Magnesium : Moles of Oxygen

_______:_______

Teacher's Pages for Water of Hydration Lab

Investigation and Experimentation Objectives

In this activity, the student will collect and interpret data, use mathematics to determine a solution, and communicate the result.

Lab Notes

This lab is fairly straightforward and simple in execution. The biggest hazard is with the crucibles. They become very, very hot – enough to weld your skin on to it. Make sure to handle hot crucibles with crucible tongs, and never to weigh a hot crucible. Teach your students how to gauge the temperature of a piece of glassware by having them approach the hot item with the back of their hand. If they can bring their hand to within a centimeter of the piece of glassware and it is hot, it is too hot to handle! If it is too hot for the back of your hand, it is certainly too hot for the front. Instruct the students on the proper procedure for using a desiccator.

Water of Hydration Lab

Background Information

When ionic crystals form, they often incorporate water molecules within their structure. This water within the crystal is called the water of hydration, and the compounds themselves are called hydrates. This water of hydration can often be removed by simply heating the hydrate, because the water molecules are only weakly attracted to the ions present. When this is done, the resulting leftover substance is called the anhydrous form of the crystal.

The amount of water present is often in a whole-number stoichiometric amount relative to the anhydrous form. Examples include barium chloride-2-hydrate, $BaCl_2 \cdot 2H_2O$ , and cobalt nitrate-6-hydrate, $Co(NO_3)_2 \cdot 6H_2O$ . Hydration numbers are most often integers, but in calcium compounds they are often fractional. Portland cement is an example of an important hydrate. When water is introduced to the anhydrous form, the water incorporates into the structure, and the cement hardens. Since the removal of water requires an input of heat, it should not be surprising that adding water to a hydrate gives off heat, and it gives off as much heat as was put into the system to remove the water in the first place. This is a problem for civil engineers who pour large amounts of cement: the heat given off by the hardening cement can be so great as to break the cement that has already hardened, due to heat stress. Steps must be taken to remove this heat. The Hoover Dam is such a large piece of concrete (cement + aggregate) that the dam is still cooling – and the last of the cement was poured in 1935.

Purpose

To determine the hydration number and empirical formula of copper(II) sulfate hydrate.

Apparatus and Materials

Ring Stand and Ring
Crucible
Clay Triangle
Bunsen burner
Wash bottle
Matches
Electronic Balance
Copper(II)sulfate hydrate (approximately $3 \ g$ per lab group)
desiccator
watch glass

Safety Issues

Always handle crucibles with crucible tongs. Never place a hot crucible on a balance. It can damage the electronics and give a measurement, which is less than the actual mass.

Procedure

1. Clean a porcelain crucible with soap and water. Rinse and dry the crucible by placing the crucible and cover on a clay triangle over a laboratory burner and heating until red-hot.

2. Carefully remove the crucible and cover with crucible tongs and let it cool. Handle the crucible and cover with tongs for the remainder of the experiment.

3. Measure the mass of the empty crucible and cover to the nearest $0.01 \ gram$ .

4. Add about $3 \ g$ of $CuSO_4$ hydrate crystals to the crucible, replace the cover, and measure the mass to the nearest $0.01 \ g$ .

5. Begin heating slowly. Increase the heat until you have heated the crucible strongly for about $10 \ minutes$ .

6. Remove the crucible from the triangle support, let it cool in a desiccator, and measure the mass.

7. Reheat with a hot flame for a few minutes, cool, and measure the mass again. If the mass is different from that recorded in Step 6, continue to heat and measure until the masses agree.

8. Remove the button of anhydrous copper sulfate by tipping it into a watch glass. Add a few drops of water to the anhydrous copper sulfate, and record your observations below.

Data
Number	Object	Mass (grams)
1.	Mass of Crucible + Cover	_______________ grams
2.	Mass of Crucible + Cover + Copper Sulfate Hydrate	_______________ grams
3.	Mass of Crucible + Cover + Anhydrous Copper Sulfate	_______________ grams
4.	Mass of Water $(2 - 3)$	_______________ grams
5.	Mass of Anhydrous Copper Sulfate $(3 - 1)$	_______________ grams
6.	Moles of Anhydrous Copper Sulfate	_______________ moles
7.	Moles of water driven off	_______________ moles
8.	Ratio of moles of water to moles of anhydrous $CuSO_4$ $\left (\frac{7}{6}\right )$

Sunday, December 9, 2012

The Mathematics of Compounds

The Mathematics of Compounds

Determining Formula and Molar Masses

Lesson Objectives

Introduction

Formula Mass

Terminology

Lesson Summary

Review Questions

Vocabulary

The Mole

Lesson Objectives

Introduction

Avogadro's Number

Converting Molecules to Moles and Vice Versa

Converting Grams to Moles and Vice Versa

Lesson Summary

Review Questions

Further Reading / Supplemental Links

Vocabulary

Percent Composition

Lesson Objectives

Introduction

Percent Composition from Masses

Percent Composition from the Formula

Lesson Summary

Further Reading / Supplemental Links

Review Questions

Vocabulary

Empirical and Molecular Formulas

Lesson Objectives

Introduction

Finding Empirical Formula from Experimental Data

Finding Empirical Formula from Percent Composition

Finding Molecular Formulas

Lesson Summary

Review Questions

Further Reading / Supplemental Links

Vocabulary

Labs and Demonstrations for The Mathematics of Compounds

Teacher's Pages for Empirical Formula of Magnesium Oxide

Lab - Empirical Formula of Magnesium Oxide

Teacher's Pages for Water of Hydration Lab

Water of Hydration Lab

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