# استوکیومتری

__Stoichiometry__

__Stoichiometry__

__Stoichiometry is the accounting, or math, behind chemistry. Given enough information, one can use stoichiometry to calculate masses, moles, and percents within a chemical equation.__What is a Chemical Equation

The Mole

Balancing Chemical Equations

Limiting Reagents

Percent Composition

Empirical and Molecular Formulas

Density

Concentrations of Solutions

__What is a chemical equation?__

__What is a chemical equation?__

__In chemistry, we use symbols to represent the various chemicals. Success in chemistry depends upon developing a strong familiarity with these basic symbols. For example, the symbol "C" represents an atom of carbon, and "H" represents an atom of hydrogen. To represent a molecule of table salt, sodium chloride, we would use the notation "NaCl", where "Na" represents sodium and "Cl" represents chlorine. We call chlorine "chloride" in this case because of its connection to sodium. You should have reviewed naming schemes, or nomenclature, in earlier readings. __

__A chemical equation is an expression of a chemical process. For example: __

__AgNO___{3}(aq) + NaCl(aq) ---> AgCl (s) + NaNO_{3}(aq)__In this equation, AgNO _{3} is mixed with NaCl. The equation shows that the __reactants

__(AgNO__products

_{3}and NaCl) react through some process (--->) to form the__(AgCl and NaNO__

_{3}). Since they undergo a chemical process, they are changed fundamentally.__Often __chemical equations__ are written showing the __state__ that each substance is in. The (s) sign means that t he compound is a solid. The (l) sign means the substance is a liquid. The (aq) sign stands for aqueous in water and means the __compound __is dissolved in water. Finally, the (g) sign mean s that the compound is a gas. __

__Coefficients are used in all chemical equations to show the relative amounts of each substance present. This amount can represent either the relative number of __molecules__, or the relative number of moles (described below). If no coefficient is shown, a one (1) is assumed. __

__On some occasions, a variety of information will be written above or below the arrows. This information, such as a value for temperature, shows what conditions need to be present for a reaction to occur. For example, in the graphic below, the notation above and below the arrows shows that we need a chemical Fe _{2}O_{3}, a temperature of 1000 degrees C, and a pressure of 500 atmospheres for this reaction to occur. __

__The graphic below works to capture most of the concepts described above: __

__The Mole____Balancing Chemical Equations____ in the reactants equals the number of atoms of that same element in the products. To do this we have to figure out the relative number of molecules of each term expressed by the term's coefficient. __

__The Mole____Balancing Chemical Equations____in the reactants equals the number of atoms of that same element in the products. To do this we have to figure out the relative number of molecules of each term expressed by the term's coefficient.____Balancing a chemical equation is essentially done by trial and error. There are many different ways and systems of doing this, but for all methods, it is important to know how to count the number of atoms in an equation. For example we will look at the following term. __

__Balancing a chemical equation is essentially done by trial and error. There are many different ways and systems of doing this, but for all methods, it is important to know how to count the number of atoms in an equation. For example we will look at the following term.__

__2Fe___{3}O_{4}

__2Fe___{3}O_{4}__This term expresses two (2) molecules of Fe___{3}O_{4}. In each molecule of this substance there are three (3) Fe atoms. Therefore in two (2) molecules of the substance there must be six (6) Fe atoms. Similarly there are four (4) oxygen a toms in one (1) molecule of the substance so there must be eight (8) oxygen atoms in two (2) molecules.

__This term expresses two (2) molecules of Fe___{3}O_{4}. In each molecule of this substance there are three (3) Fe atoms. Therefore in two (2) molecules of the substance there must be six (6) Fe atoms. Similarly there are four (4) oxygen a toms in one (1) molecule of the substance so there must be eight (8) oxygen atoms in two (2) molecules.__Now let's try balancing the equation mentioned earlier: __

__Now let's try balancing the equation mentioned earlier:__

__Al + Fe___{3}O_{4}---> Al_{2}O_{3}+ Fe

__Al + Fe__

_{3}O_{4}---> Al_{2}O_{3}+ Fe__Developing a strategy can be difficult, but here is one way of approaching a problem like this. __

1) Count the number of each atom on the reactant and on the product side.

2) Determine a term to balance first. When looking at this problem it appears that the oxygen will be the most difficult to balance so we'll try to balance the oxygen first. The simplist way to balance the oxygen terms is:

__Developing a strategy can be difficult, but here is one way of approaching a problem like this.__

1) Count the number of each atom on the reactant and on the product side.

2) Determine a term to balance first. When looking at this problem it appears that the oxygen will be the most difficult to balance so we'll try to balance the oxygen first. The simplist way to balance the oxygen terms is:1) Count the number of each atom on the reactant and on the product side.

2) Determine a term to balance first. When looking at this problem it appears that the oxygen will be the most difficult to balance so we'll try to balance the oxygen first. The simplist way to balance the oxygen terms is:

__Al +3 Fe___{3}O_{4}---> 4Al_{2}O_{3}+Fe

__Al +3 Fe___{3}O_{4}---> 4Al_{2}O_{3}+Fe__Be sure to notice that the subscript times the coefficient will give the number of atoms of that element. On the reactant side, we have a coefficient of three (3) multiplied by a subscript of four (4), giving 12 oxygen atoms. On the product side, we ha ve a coefficient of four (4) multiplied by a subscript of three (3), giving 12 oxygen atoms. Now, the oxygens are balanced.__

3) Choose another term to balance. We'll choose iron, Fe. Since there are nine (9) iron atoms in the term in which the oxygen is balanced we add a nine (9) coefficient in front of the Fe. We now have:

__Be sure to notice that the subscript times the coefficient will give the number of atoms of that element. On the reactant side, we have a coefficient of three (3) multiplied by a subscript of four (4), giving 12 oxygen atoms. On the product side, we ha ve a coefficient of four (4) multiplied by a subscript of three (3), giving 12 oxygen atoms. Now, the oxygens are balanced.__

3) Choose another term to balance. We'll choose iron, Fe. Since there are nine (9) iron atoms in the term in which the oxygen is balanced we add a nine (9) coefficient in front of the Fe. We now have:3) Choose another term to balance. We'll choose iron, Fe. Since there are nine (9) iron atoms in the term in which the oxygen is balanced we add a nine (9) coefficient in front of the Fe. We now have:

__Al +3 Fe___{3}O_{4}---> 4Al_{2}O_{3}+9Fe

__Al +3 Fe___{3}O_{4}---> 4Al_{2}O_{3}+9Fe__4) Balance the last term. In this case, since we had eight (8) aluminum atoms on the product side we need to have eight (8) on the reactant side so we add an eight (8) in front of the Al term on the reactant side. __

Now, we're done, and the balanced equation is:

__4) Balance the last term. In this case, since we had eight (8) aluminum atoms on the product side we need to have eight (8) on the reactant side so we add an eight (8) in front of the Al term on the reactant side.__

Now, we're done, and the balanced equation is:Now, we're done, and the balanced equation is:

__8Al + 3Fe___{3}O_{4} ---> 4Al_{2}O_{3} + 9 Fe

__8Al + 3Fe___{3}O_{4}---> 4Al_{2}O_{3}+ 9 Fe

__Limiting Reagents__

__Limiting Reagents____Sometimes when reactions occur between two or more substances, one reactant runs out before the other. That is called the "limiting reagent". Often, it is necessary to identify the limiting reagent in a problem. __

Example: A chemist only has 6.0 grams of C_{2}H_{2} and an unlimitted supply of oxygen and desires to produce as much CO_{2} as possible. If she uses the equation below, how much oxygen should she add to the reaction?

__Sometimes when reactions occur between two or more substances, one reactant runs out before the other. That is called the "limiting reagent". Often, it is necessary to identify the limiting reagent in a problem.__

Example: A chemist only has 6.0 grams of CExample: A chemist only has 6.0 grams of C

_{2}H_{2}and an unlimitted supply of oxygen and desires to produce as much CO_{2}as possible. If she uses the equation below, how much oxygen should she add to the reaction?

__2C___{2}H_{2}(g) + 5O_{2}(g) ---> 4CO_{2}(g) + 2 H_{2}O(l)

__2C___{2}H_{2}(g) + 5O_{2}(g) ---> 4CO_{2}(g) + 2 H_{2}O(l)__To solve this problem, it is necessary to determine how much oxygen should be added if all of the reactants were used up (this is the way to produce the maximum amount of CO___{2}).

First, we calculate the number of moles of C_{2}H_{2} in 6.0 grams of C_{2}H_{2}. To be able to calculate the moles we need to look at a periodic table and see that 1 mole of C weighs 12.0 grams and H weighs 1.0 gram. Therefore we know that 1 mole of C_{2}H_{2} weighs 26 grams (2*12 grams + 2*1 gram).

__To solve this problem, it is necessary to determine how much oxygen should be added if all of the reactants were used up (this is the way to produce the maximum amount of CO__

First, we calculate the number of moles of C_{2}).First, we calculate the number of moles of C

_{2}H_{2}in 6.0 grams of C_{2}H_{2}. To be able to calculate the moles we need to look at a periodic table and see that 1 mole of C weighs 12.0 grams and H weighs 1.0 gram. Therefore we know that 1 mole of C_{2}H_{2}weighs 26 grams (2*12 grams + 2*1 gram).

__Then, because there are five (5) molecules of oxygen to every two (2) molecules of C___{2}H_{2}, we need to multiply the result by 5/2 to get the total molecules of oxygen. Then we convert to grams to find the amount of oxygen that needs to be added:

__Then, because there are five (5) molecules of oxygen to every two (2) molecules of C___{2}H_{2}, we need to multiply the result by 5/2 to get the total molecules of oxygen. Then we convert to grams to find the amount of oxygen that needs to be added:

__Percent Composition__

__Percent Composition____It is possible to calculate the mole ratios (also called mole fractions) between terms in a chemical equation when given the percent by mass of products or reactants. __

__It is possible to calculate the mole ratios (also called mole fractions) between terms in a chemical equation when given the percent by mass of products or reactants.__

__percentage by mass = mass of part/ mass of whole__

__percentage by mass = mass of part/ mass of whole__

__There are two types of __percent composition__ problems-- problems in which you are given the formula (or the weight of each part) and asked to calculate the percentage of each element and problems in which you are given the percentages and asked to calculate the formula.__

__In percent composition problems, there are many possible solutions. It is always possible to double the answer. For example, CH and C___{2}H_{2} have the same proportions, but they are different compounds. It is standard to give compounds in their simplest form, where the ratio between the elements is as reduced as it can be-- called the empirical formula__. When calculating the empirical formula from percent composition, one can convert the percentages to grams. For example, it is usually the easiest to assume you have 100 grams so 54.3% would become 54.3 grams. Then we can convert the masses to moles which gives us mole ratios. It is necessary to reduce to whole numbers. A good technique is to divide all the terms by the smallest number of moles. Then the ratio of the moles can be transfered to write the empirical formula. __

__Example: If a compound is 47.3% C (carbon), 10.6% H (hydrogen) and 42.0% S (sulfur), what is its empirical formula?__

To do this problem we need to transfer all of our percents to masses. We assume that we have 100 g of this substance. Then we convert to moles:

__Now we try to get an even ratio between the elements so we divide by the number of moles of sulfur, because it is the smallest number: __

__So we have: C___{3}H_{8 }S

__Example: Figure out the percentage by mass of hydrogen sulfate, H___{2}SO_{4}.

In this problem we need to first calculate the total weight of the compound by looking at the periodic table. This gives us:

2(1.008) + 32.07 + 4(16.00) grams/mol = 98.09 g/mol

Now, we need to take the weight fraction of each element over the total mass (which we just found) and multiply by 100 to get a percentage.

__Now, we can check that the percentages add up to 100% __

__65.2 + 2.06 + 32.7 = 99.96__

__This is essentially 100 so we know that everything has worked, and we probably have not made any careless errors. __

So the answer is that H_{2}SO_{4} is made up of 2.06% H, 32.7% S, and 65.2% O by mass.

__Empirical Formula and Molecular Formula__

__While the empirical formula is the simplest form of a compound, the molecular formula is the form of the term as it would appear in a chemical equation. The empirical formula and the molecular formula can be the same, or the molecular formula can be any positive integer multiple of the empirical formula. Examples of empirical formulas: AgBr, Na___{2}S, C_{6}H_{10}O_{5}. Examples of molecular formulas: P_{2}, C_{2}O_{4}, C_{6}H_{14}S_{2}, H_{2}, C_{3}H_{9}.

__One can calculate the empirical formula from the masses or percentage composition of any compound. We have already discussed percent composition in the section above. If we only have mass, all we are doing is essentially eliminating the step of converting from percentage to mass. __

__Example: Calculate the empirical formula for a compound that has 43.7 g P (phosphorus) and 56.3 grams of oxygen. First we convert to moles: __

__Next we divide the moles to try to get a even ratio. __

__When we divide, we did not get whole numbers so we must multiply by two (2). The answer=P___{2}O_{5}

__Calculating the molecular formula once we have the empirical formula is easy. If we know the empirical formula of a compound, all we need to do is divide the molecular mass of the compound by the mass of the empirical formula. It is also possible to do this with one of the elements in the formula; simply divide the mass of that element in one mole of compound by the mass of that element in the empirical formula. The result should always be a natural number. __

__Example: if we know that the empirical formula of a compound is HCN and we are told that a 2.016 grams of hydrogen are necesary to make the compound, what is the molecular formula? In the empirical formula hydrogen weighs 1.008 grams. Dividing 2.016 by 1.008 we see that the amount of hydrogen needed is twice as much. Therefore the empirical formula needs to be increased by a factor of two (2). The answer is: __

H_{2}C_{2}N_{2}.

__Density__

__Density____ refers to the mass per unit volume of a substance. It is a very common term in chemistry.__

__Concentrations of Solutions____The __concentration__ of a solution is the "strength" of a solution. A solution typically refers to the dissolving of some solid substance in a liquid, such as dissolving salt in water. It is also often necessary to figure out how much water to add to a solution to change it to a specific concentration. __

__The concentration of a solution is typically given in __molarity__. Molarity is defined as the number of moles of __solute__ (what is actually dissolved in the solution) divided by the liters of __solution __(the total volume of what is dissolved and what it has been dissolved in). __

__ ____Example: If 5.00 grams of NaOH are dissolved in 5000 mL of water, what is the molarity of the solution?__

One of our first steps is to convert the amount of NaOH given in grams into moles:

__ __

__So the molarity (M) of the solution is 0.025 mol/L. __

__Molality____ is another common measurement of concentration. Molality is defined as moles of solute divided by kilograms of __solvent __(the substance in which it is dissolved, like water). __

__ ____Example: If the molality of a solution of C___{2}H_{5}OH dissolved in water is 1.5 and the weight of the water is 11.7 kg, figure out how much C_{2}H_{5}OH must have been added in grams to the solution?

Our first step is to substitute what we know into the equation. Then we try to solve for what we don't know: moles of solute. Once we know the moles of solute we can look at the periodic table and figure out the conversion from moles to grams.

__ __

__Example: If the molarity of a solution is 0.30 M, calculate the molality of the solution knowing that the density is 3.25 g/mL.__

To do this problem we can assume one (1) liter of solution to make the numbers easier. We need to get from the molarity units of mols/Liter to the molality units of mols/kg. We work the problem as follows, remembering that there are 1000 mL in a Liter and 1000 grams in a kg. This conversion will only be accurate at small molarities and molalities.

__ __

__Change in T= K * m____Where: T is temperature depression (for freezing point) or temperature expansion (for boiling point) (°C)__

K is the freezing point constant (kg °C/moles)

m is molality in moles/kg

__Example: If the freezing point of the salt water put on roads is -5.2 C, what is the molality of the solution? (The K___{f} for water is 1.86 C/m.)

This is a simple problem where we just plug in numbers into the equation. One piece of information we do have to know is that water usually freezes at 0^{0}C.

__T=K * m __

T/K= m

m = 5.2/1.86

m = 2.8 mols/kg

__It is also possible to calculate __colligative properties__, such as boiling point depression, using molality. The equation for temperature depression or expansion is ____It is possible to convert between molarity and molality. The only information needed is density. __

__Molality is sometimes used in place of molarity at extreme temperatures because the volume can contract or expand. ____Now we simply use the definition of molarity: moles/liters to get the answer__

__There are two types of__percent composition__problems-- problems in which you are given the formula (or the weight of each part) and asked to calculate the percentage of each element and problems in which you are given the percentages and asked to calculate the formula.__

__In percent composition problems, there are many possible solutions. It is always possible to double the answer. For example, CH and C___{2}H_{2} have the same proportions, but they are different compounds. It is standard to give compounds in their simplest form, where the ratio between the elements is as reduced as it can be-- called the empirical formula__. When calculating the empirical formula from percent composition, one can convert the percentages to grams. For example, it is usually the easiest to assume you have 100 grams so 54.3% would become 54.3 grams. Then we can convert the masses to moles which gives us mole ratios. It is necessary to reduce to whole numbers. A good technique is to divide all the terms by the smallest number of moles. Then the ratio of the moles can be transfered to write the empirical formula. __

_{2}H

_{2}have the same proportions, but they are different compounds. It is standard to give compounds in their simplest form, where the ratio between the elements is as reduced as it can be-- called the

__Example: If a compound is 47.3% C (carbon), 10.6% H (hydrogen) and 42.0% S (sulfur), what is its empirical formula?__

To do this problem we need to transfer all of our percents to masses. We assume that we have 100 g of this substance. Then we convert to moles:

To do this problem we need to transfer all of our percents to masses. We assume that we have 100 g of this substance. Then we convert to moles:

__Now we try to get an even ratio between the elements so we divide by the number of moles of sulfur, because it is the smallest number: __

__Now we try to get an even ratio between the elements so we divide by the number of moles of sulfur, because it is the smallest number:__

__So we have: C___{3}H_{8 }S

__So we have: C___{3}H_{8 }S__Example: Figure out the percentage by mass of hydrogen sulfate, H___{2}SO_{4}.

In this problem we need to first calculate the total weight of the compound by looking at the periodic table. This gives us:

2(1.008) + 32.07 + 4(16.00) grams/mol = 98.09 g/mol

Now, we need to take the weight fraction of each element over the total mass (which we just found) and multiply by 100 to get a percentage.

__Example: Figure out the percentage by mass of hydrogen sulfate, H__

In this problem we need to first calculate the total weight of the compound by looking at the periodic table. This gives us:

2(1.008) + 32.07 + 4(16.00) grams/mol = 98.09 g/mol

Now, we need to take the weight fraction of each element over the total mass (which we just found) and multiply by 100 to get a percentage._{2}SO_{4}.In this problem we need to first calculate the total weight of the compound by looking at the periodic table. This gives us:

2(1.008) + 32.07 + 4(16.00) grams/mol = 98.09 g/mol

Now, we need to take the weight fraction of each element over the total mass (which we just found) and multiply by 100 to get a percentage.

__Now, we can check that the percentages add up to 100% __

__65.2 + 2.06 + 32.7 = 99.96__

__This is essentially 100 so we know that everything has worked, and we probably have not made any careless errors. __

So the answer is that H_{2}SO_{4} is made up of 2.06% H, 32.7% S, and 65.2% O by mass.

__Empirical Formula and Molecular Formula__

__While the empirical formula is the simplest form of a compound, the molecular formula is the form of the term as it would appear in a chemical equation. The empirical formula and the molecular formula can be the same, or the molecular formula can be any positive integer multiple of the empirical formula. Examples of empirical formulas: AgBr, Na___{2}S, C_{6}H_{10}O_{5}. Examples of molecular formulas: P_{2}, C_{2}O_{4}, C_{6}H_{14}S_{2}, H_{2}, C_{3}H_{9}.

__One can calculate the empirical formula from the masses or percentage composition of any compound. We have already discussed percent composition in the section above. If we only have mass, all we are doing is essentially eliminating the step of converting from percentage to mass. __

__Example: Calculate the empirical formula for a compound that has 43.7 g P (phosphorus) and 56.3 grams of oxygen. First we convert to moles: __

__Next we divide the moles to try to get a even ratio. __

__When we divide, we did not get whole numbers so we must multiply by two (2). The answer=P___{2}O_{5}

__Calculating the molecular formula once we have the empirical formula is easy. If we know the empirical formula of a compound, all we need to do is divide the molecular mass of the compound by the mass of the empirical formula. It is also possible to do this with one of the elements in the formula; simply divide the mass of that element in one mole of compound by the mass of that element in the empirical formula. The result should always be a natural number. __

__Example: if we know that the empirical formula of a compound is HCN and we are told that a 2.016 grams of hydrogen are necesary to make the compound, what is the molecular formula? In the empirical formula hydrogen weighs 1.008 grams. Dividing 2.016 by 1.008 we see that the amount of hydrogen needed is twice as much. Therefore the empirical formula needs to be increased by a factor of two (2). The answer is: __

H_{2}C_{2}N_{2}.

__Density__

__Density____ refers to the mass per unit volume of a substance. It is a very common term in chemistry.__

__Concentrations of Solutions____The __concentration__ of a solution is the "strength" of a solution. A solution typically refers to the dissolving of some solid substance in a liquid, such as dissolving salt in water. It is also often necessary to figure out how much water to add to a solution to change it to a specific concentration. __

__The concentration of a solution is typically given in __molarity__. Molarity is defined as the number of moles of __solute__ (what is actually dissolved in the solution) divided by the liters of __solution __(the total volume of what is dissolved and what it has been dissolved in). __

__ ____Example: If 5.00 grams of NaOH are dissolved in 5000 mL of water, what is the molarity of the solution?__

One of our first steps is to convert the amount of NaOH given in grams into moles:

__ __

__So the molarity (M) of the solution is 0.025 mol/L. __

__Molality____ is another common measurement of concentration. Molality is defined as moles of solute divided by kilograms of __solvent __(the substance in which it is dissolved, like water). __

__ ____Example: If the molality of a solution of C___{2}H_{5}OH dissolved in water is 1.5 and the weight of the water is 11.7 kg, figure out how much C_{2}H_{5}OH must have been added in grams to the solution?

Our first step is to substitute what we know into the equation. Then we try to solve for what we don't know: moles of solute. Once we know the moles of solute we can look at the periodic table and figure out the conversion from moles to grams.

__ __

__Example: If the molarity of a solution is 0.30 M, calculate the molality of the solution knowing that the density is 3.25 g/mL.__

To do this problem we can assume one (1) liter of solution to make the numbers easier. We need to get from the molarity units of mols/Liter to the molality units of mols/kg. We work the problem as follows, remembering that there are 1000 mL in a Liter and 1000 grams in a kg. This conversion will only be accurate at small molarities and molalities.

__ __

__Change in T= K * m____Where: T is temperature depression (for freezing point) or temperature expansion (for boiling point) (°C)__

K is the freezing point constant (kg °C/moles)

m is molality in moles/kg

__Example: If the freezing point of the salt water put on roads is -5.2 C, what is the molality of the solution? (The K___{f} for water is 1.86 C/m.)

This is a simple problem where we just plug in numbers into the equation. One piece of information we do have to know is that water usually freezes at 0^{0}C.

__T=K * m __

T/K= m

m = 5.2/1.86

m = 2.8 mols/kg

__It is also possible to calculate __colligative properties__, such as boiling point depression, using molality. The equation for temperature depression or expansion is ____It is possible to convert between molarity and molality. The only information needed is density. __

__Molality is sometimes used in place of molarity at extreme temperatures because the volume can contract or expand. ____Now we simply use the definition of molarity: moles/liters to get the answer__

__Now, we can check that the percentages add up to 100%__

__65.2 + 2.06 + 32.7 = 99.96__

__65.2 + 2.06 + 32.7 = 99.96__

__This is essentially 100 so we know that everything has worked, and we probably have not made any careless errors. __

So the answer is that H_{2}SO_{4} is made up of 2.06% H, 32.7% S, and 65.2% O by mass.

So the answer is that H

_{2}SO

_{4}is made up of 2.06% H, 32.7% S, and 65.2% O by mass.

__Empirical Formula and Molecular Formula__

__Empirical Formula and Molecular Formula__

__While the empirical formula is the simplest form of a compound, the molecular formula is the form of the term as it would appear in a chemical equation. The empirical formula and the molecular formula can be the same, or the molecular formula can be any positive integer multiple of the empirical formula. Examples of empirical formulas: AgBr, Na___{2}S, C_{6}H_{10}O_{5}. Examples of molecular formulas: P_{2}, C_{2}O_{4}, C_{6}H_{14}S_{2}, H_{2}, C_{3}H_{9}.

__While the empirical formula is the simplest form of a compound, the molecular formula is the form of the term as it would appear in a chemical equation. The empirical formula and the molecular formula can be the same, or the molecular formula can be any positive integer multiple of the empirical formula. Examples of empirical formulas: AgBr, Na__

_{2}S, C_{6}H_{10}O_{5}. Examples of molecular formulas: P_{2}, C_{2}O_{4}, C_{6}H_{14}S_{2}, H_{2}, C_{3}H_{9}.__One can calculate the empirical formula from the masses or percentage composition of any compound. We have already discussed percent composition in the section above. If we only have mass, all we are doing is essentially eliminating the step of converting from percentage to mass. __

__One can calculate the empirical formula from the masses or percentage composition of any compound. We have already discussed percent composition in the section above. If we only have mass, all we are doing is essentially eliminating the step of converting from percentage to mass.__

__Example: Calculate the empirical formula for a compound that has 43.7 g P (phosphorus) and 56.3 grams of oxygen. First we convert to moles: __

__Example: Calculate the empirical formula for a compound that has 43.7 g P (phosphorus) and 56.3 grams of oxygen. First we convert to moles:__

__Next we divide the moles to try to get a even ratio. __

__Next we divide the moles to try to get a even ratio.__

__When we divide, we did not get whole numbers so we must multiply by two (2). The answer=P___{2}O_{5}

__When we divide, we did not get whole numbers so we must multiply by two (2). The answer=P___{2}O_{5}__Calculating the molecular formula once we have the empirical formula is easy. If we know the empirical formula of a compound, all we need to do is divide the molecular mass of the compound by the mass of the empirical formula. It is also possible to do this with one of the elements in the formula; simply divide the mass of that element in one mole of compound by the mass of that element in the empirical formula. The result should always be a natural number. __

__Calculating the molecular formula once we have the empirical formula is easy. If we know the empirical formula of a compound, all we need to do is divide the molecular mass of the compound by the mass of the empirical formula. It is also possible to do this with one of the elements in the formula; simply divide the mass of that element in one mole of compound by the mass of that element in the empirical formula. The result should always be a natural number.____Example: if we know that the empirical formula of a compound is HCN and we are told that a 2.016 grams of hydrogen are necesary to make the compound, what is the molecular formula? In the empirical formula hydrogen weighs 1.008 grams. Dividing 2.016 by 1.008 we see that the amount of hydrogen needed is twice as much. Therefore the empirical formula needs to be increased by a factor of two (2). The answer is: __

H_{2}C_{2}N_{2}.

__Density__

__Density____ refers to the mass per unit volume of a substance. It is a very common term in chemistry.__

__Concentrations of Solutions____The __concentration__ of a solution is the "strength" of a solution. A solution typically refers to the dissolving of some solid substance in a liquid, such as dissolving salt in water. It is also often necessary to figure out how much water to add to a solution to change it to a specific concentration. __

__The concentration of a solution is typically given in __molarity__. Molarity is defined as the number of moles of __solute__ (what is actually dissolved in the solution) divided by the liters of __solution __(the total volume of what is dissolved and what it has been dissolved in). __

__ ____Example: If 5.00 grams of NaOH are dissolved in 5000 mL of water, what is the molarity of the solution?__

One of our first steps is to convert the amount of NaOH given in grams into moles:

__ __

__So the molarity (M) of the solution is 0.025 mol/L. __

__Molality____ is another common measurement of concentration. Molality is defined as moles of solute divided by kilograms of __solvent __(the substance in which it is dissolved, like water). __

__ ____Example: If the molality of a solution of C___{2}H_{5}OH dissolved in water is 1.5 and the weight of the water is 11.7 kg, figure out how much C_{2}H_{5}OH must have been added in grams to the solution?

Our first step is to substitute what we know into the equation. Then we try to solve for what we don't know: moles of solute. Once we know the moles of solute we can look at the periodic table and figure out the conversion from moles to grams.

__ __

__Example: If the molarity of a solution is 0.30 M, calculate the molality of the solution knowing that the density is 3.25 g/mL.__

To do this problem we can assume one (1) liter of solution to make the numbers easier. We need to get from the molarity units of mols/Liter to the molality units of mols/kg. We work the problem as follows, remembering that there are 1000 mL in a Liter and 1000 grams in a kg. This conversion will only be accurate at small molarities and molalities.

__ __

__Change in T= K * m____Where: T is temperature depression (for freezing point) or temperature expansion (for boiling point) (°C)__

K is the freezing point constant (kg °C/moles)

m is molality in moles/kg

__Example: If the freezing point of the salt water put on roads is -5.2 C, what is the molality of the solution? (The K___{f} for water is 1.86 C/m.)

This is a simple problem where we just plug in numbers into the equation. One piece of information we do have to know is that water usually freezes at 0^{0}C.

__T=K * m __

T/K= m

m = 5.2/1.86

m = 2.8 mols/kg

__It is also possible to calculate __colligative properties__, such as boiling point depression, using molality. The equation for temperature depression or expansion is ____It is possible to convert between molarity and molality. The only information needed is density. __

__Example: if we know that the empirical formula of a compound is HCN and we are told that a 2.016 grams of hydrogen are necesary to make the compound, what is the molecular formula? In the empirical formula hydrogen weighs 1.008 grams. Dividing 2.016 by 1.008 we see that the amount of hydrogen needed is twice as much. Therefore the empirical formula needs to be increased by a factor of two (2). The answer is:__

H

H

_{2}C_{2}N_{2}.__Density__

__Density__

__Density____ refers to the mass per unit volume of a substance. It is a very common term in chemistry.__

__Concentrations of Solutions____The __concentration__ of a solution is the "strength" of a solution. A solution typically refers to the dissolving of some solid substance in a liquid, such as dissolving salt in water. It is also often necessary to figure out how much water to add to a solution to change it to a specific concentration. __

__The concentration of a solution is typically given in __molarity__. Molarity is defined as the number of moles of __solute__ (what is actually dissolved in the solution) divided by the liters of __solution __(the total volume of what is dissolved and what it has been dissolved in). __

__ ____Example: If 5.00 grams of NaOH are dissolved in 5000 mL of water, what is the molarity of the solution?__

One of our first steps is to convert the amount of NaOH given in grams into moles:

__ __

__So the molarity (M) of the solution is 0.025 mol/L. __

__Molality____ is another common measurement of concentration. Molality is defined as moles of solute divided by kilograms of __solvent __(the substance in which it is dissolved, like water). __

__ ____Example: If the molality of a solution of C___{2}H_{5}OH dissolved in water is 1.5 and the weight of the water is 11.7 kg, figure out how much C_{2}H_{5}OH must have been added in grams to the solution?

Our first step is to substitute what we know into the equation. Then we try to solve for what we don't know: moles of solute. Once we know the moles of solute we can look at the periodic table and figure out the conversion from moles to grams.

__ __

__Example: If the molarity of a solution is 0.30 M, calculate the molality of the solution knowing that the density is 3.25 g/mL.__

To do this problem we can assume one (1) liter of solution to make the numbers easier. We need to get from the molarity units of mols/Liter to the molality units of mols/kg. We work the problem as follows, remembering that there are 1000 mL in a Liter and 1000 grams in a kg. This conversion will only be accurate at small molarities and molalities.

__ __

__Change in T= K * m____Where: T is temperature depression (for freezing point) or temperature expansion (for boiling point) (°C)__

K is the freezing point constant (kg °C/moles)

m is molality in moles/kg

__Example: If the freezing point of the salt water put on roads is -5.2 C, what is the molality of the solution? (The K___{f} for water is 1.86 C/m.)

This is a simple problem where we just plug in numbers into the equation. One piece of information we do have to know is that water usually freezes at 0^{0}C.

__T=K * m __

T/K= m

m = 5.2/1.86

m = 2.8 mols/kg

__refers to the mass per unit volume of a substance. It is a very common term in chemistry.__

__Concentrations of Solutions____The __concentration__ of a solution is the "strength" of a solution. A solution typically refers to the dissolving of some solid substance in a liquid, such as dissolving salt in water. It is also often necessary to figure out how much water to add to a solution to change it to a specific concentration. __

__Concentrations of Solutions__

__The__concentration

__of a solution is the "strength" of a solution. A solution typically refers to the dissolving of some solid substance in a liquid, such as dissolving salt in water. It is also often necessary to figure out how much water to add to a solution to change it to a specific concentration.__

__The concentration of a solution is typically given in __molarity__. Molarity is defined as the number of moles of __solute__ (what is actually dissolved in the solution) divided by the liters of __solution __(the total volume of what is dissolved and what it has been dissolved in). __

__The concentration of a solution is typically given in__molarity

__. Molarity is defined as the number of moles of__solute

__(what is actually dissolved in the solution) divided by the liters of__solution

__(the total volume of what is dissolved and what it has been dissolved in).__

__ ____Example: If 5.00 grams of NaOH are dissolved in 5000 mL of water, what is the molarity of the solution?__

One of our first steps is to convert the amount of NaOH given in grams into moles:

__ __

__So the molarity (M) of the solution is 0.025 mol/L. __

__Molality____ is another common measurement of concentration. Molality is defined as moles of solute divided by kilograms of __solvent __(the substance in which it is dissolved, like water). __

__ ____Example: If the molality of a solution of C___{2}H_{5}OH dissolved in water is 1.5 and the weight of the water is 11.7 kg, figure out how much C_{2}H_{5}OH must have been added in grams to the solution?

Our first step is to substitute what we know into the equation. Then we try to solve for what we don't know: moles of solute. Once we know the moles of solute we can look at the periodic table and figure out the conversion from moles to grams.

__ __

__Example: If the molarity of a solution is 0.30 M, calculate the molality of the solution knowing that the density is 3.25 g/mL.__

To do this problem we can assume one (1) liter of solution to make the numbers easier. We need to get from the molarity units of mols/Liter to the molality units of mols/kg. We work the problem as follows, remembering that there are 1000 mL in a Liter and 1000 grams in a kg. This conversion will only be accurate at small molarities and molalities.

__ __

__Change in T= K * m____Where: T is temperature depression (for freezing point) or temperature expansion (for boiling point) (°C)__

K is the freezing point constant (kg °C/moles)

m is molality in moles/kg

__Example: If the freezing point of the salt water put on roads is -5.2 C, what is the molality of the solution? (The K___{f} for water is 1.86 C/m.)

This is a simple problem where we just plug in numbers into the equation. One piece of information we do have to know is that water usually freezes at 0^{0}C.

__T=K * m __

T/K= m

m = 5.2/1.86

m = 2.8 mols/kg

__Example: If 5.00 grams of NaOH are dissolved in 5000 mL of water, what is the molarity of the solution?One of our first steps is to convert the amount of NaOH given in grams into moles: __

__So the molarity (M) of the solution is 0.025 mol/L. __

__Molality is another common measurement of concentration. Molality is defined as moles of solute divided by kilograms of solvent (the substance in which it is dissolved, like water). __

__Example: If the molality of a solution of C _{2}H_{5}OH dissolved in water is 1.5 and the weight of the water is 11.7 kg, figure out how much C_{2}H_{5}OH must have been added in grams to the solution? __

Our first step is to substitute what we know into the equation. Then we try to solve for what we don't know: moles of solute. Once we know the moles of solute we can look at the periodic table and figure out the conversion from moles to grams.

__ __

__Example: If the molarity of a solution is 0.30 M, calculate the molality of the solution knowing that the density is 3.25 g/mL.To do this problem we can assume one (1) liter of solution to make the numbers easier. We need to get from the molarity units of mols/Liter to the molality units of mols/kg. We work the problem as follows, remembering that there are 1000 mL in a Liter and 1000 grams in a kg. This conversion will only be accurate at small molarities and molalities. __

__Change in T= K * m____Where: T is temperature depression (for freezing point) or temperature expansion (for boiling point) (°C)K is the freezing point constant (kg °C/moles)m is molality in moles/kg__

__Example: If the freezing point of the salt water put on roads is -5.2 C, what is the molality of the solution? (The K _{f} for water is 1.86 C/m.) This is a simple problem where we just plug in numbers into the equation. One piece of information we do have to know is that water usually freezes at 0^{0}C. __

__T=K * m T/K= mm = 5.2/1.86m = 2.8 mols/kg __

__It is also possible to calculate__colligative properties__, such as boiling point depression, using molality. The equation for temperature depression or expansion is__

__It is possible to convert between molarity and molality. The only information needed is density.__

__Molality is sometimes used in place of molarity at extreme temperatures because the volume can contract or expand.__

__Now we simply use the definition of molarity: moles/liters to get the answer__

__Molarity is probably the most commonly used term because measuring a volume of liquid is a fairly easy thing to do. __

__Molarity is probably the most commonly used term because measuring a volume of liquid is a fairly easy thing to do.____Given the equation above, we can tell the number of moles of reactants and products. A mole simply represents Avogadro's number (6.023 x 10 ^{23}) of molecules. A mole is similar to a term like a dozen. If you have a dozen carrots, you have twelve of them. Similarily, if you have a mole of carrots, you have 6.023 x 10^{23} carrots. In the equation above there are no numbers in front of the terms, so each coefficient is assumed to be one (1). Thus, you have the same number of moles of Ag NO_{3}, NaCl, AgCl, NaNO_{3}. __

__Converting between moles and grams of a substance is often important. This conversion can be easily done when the atomic and/or molecular weights of the substance(s) are known. Given the atomic or molecular weight of a substance, that mass in grams makes a mole of the substance. For example, calcium has an atomic weight of 40 atomic mass units. So, 40 grams of calcium makes one mole, 80 grams makes two moles, etc.__

__
__