THE ANSWER OF MID SEMESTER TEST

1. Pure substance x is solid at room temperature. if the substance is heated to 230 C is melted gradually. if then cooled to room temperature, the liquid can not be frozen.

a. is it possible x of an element or a compound. explain it!

 The substance X is a compound because the boiling point of the substance is very high, this possibility is an ionic compound. when frozen at room temperature is around 25 degrees centigrade the substance can not be frozen, because freezing of these compounds are also very high

b. Does it chemical change occured? if so can it be said undergo an endotherm changing, based on the information provided?

yes it can be said of any change in the chemical reaction. because the form of substances and automatically change the atomic structure inside will also change. can also be said that the endothermic reaction occurs because the endothermic reaction heat transfer occurs from the environment to the system or to the reaction heat is needed.

c.can it be said that the liquid is an element, based on the information provided

can be said that the fluid is due as a result of the decomposition element X compound which has been heated

2. When a candle that weighs 10 g is burned in oxigen, carbon dioxide and water vapor formed by combustion the weight more than 10 g. was this case match with the law of conservation of mass. Explain it!

in the event the law does not apply because the system is open to allow for substances other substances to catch fire, causing the mass of a substance produced from the burning candle wax is greater than the mass of the candle

3. When carbon burns in oxygen under limited number, it will form two gaseous compounds. Suggest the way to diffrerentiate the two compounds with one another.

when the carbon was burned then there are two potential outcomes, namely carbon monoxide and carbon dioxide. Carbon monoxide is produced from incomplete combustion of carbon compounds, often occur in internal combustion engines. Carbon monoxide is formed when there is a lack of oxygen in the combustion process. by because it will have a black soot
while the carbon dioxide so that complete combustion occurs because there are no byproducts such as soot

4. after mendeev compiled the periodic table, he concluded that the atomic weights of certain elements was wrong ruling, and this conclusion was apparently correct. How mendeelev was able to predict the several atomic weights were wrong? why his predictions are not always right. Explain it!

Dmitry Ivanovich concluded that the properties of elements are periodic function of the relative atomic mass. That is, if the elements disusunmenurut relative atomic mass increases, then the specific properties will be repeated periodically. Mendeleev placed elements that have a similarity of traits in a vertical column called class. Mendeleev mengkosongkan some places. This was done to establish the similarity of the properties in the class. For example, placing Mendelev Ti (Ar = 48) in group IV and group III left empty because Ti is more similar to C and Si, than with B and Al. Mendeleev predicted the properties of the elements that have not known it. The forecast is based on the known properties unsurlain, adjacent either horizontally or upright. When the elements are predicted was found, teryata are highly in accordance with predictions Mendeleev. One example is a germanium (Ge), which was found in 1886, which by Mendeleev named ekasilikon. The weakness of this theory is still contained elements of a larger mass is located in front of the element whose mass is smaller. Co: Tellurium (te) = 128 in kiriIodin (I) = 127. this is because the elements that have properties kemirpan placed in one class. The weakness of this theory is the atomic mass pemebetulan. Previous atomic mass. In previous atomic mass = 76 to 113. besides Be, from 13.5 to 9. N from 120 to 240.

5. When an aqueous solution of merkuri chloride is added an aqueous solution of silver nitrate, a white solid forms. Identify the white solid and write the balanced equation for the reaction that occurs 

HgCl2 + 2AgNO3 ---> Hg (NO3) 2 + 2AgCl

The resulting white powder is silver nitrate. 

 

Stoichiometry

Stoichiometry ( /ˌstɔɪkiˈɒmɨtri/) is a branch of chemistry that deals with the relative quantities of reactants and products in chemical reactions. In a balanced chemical reaction, the relations among quantities of reactants and products typically form a ratio of whole numbers. For example, in a reaction that forms ammonia (NH₃), exactly one molecule of nitrogen (N₂) reacts with three molecules of hydrogen (H₂) to produce two molecules of NH₃:

N₂ + 3H₂ → 2NH₃

Stoichiometry can be used to calculate quantities such as the amount of products (in mass, moles, volume, etc.) that can be produced with given reactants and percent yield (the percentage of the given reactant that is made into the product). Stoichiometry calculations can predict how elements and components diluted in a standard solution react in experimental conditions. Stoichiometry is founded on the law of conservation of mass: the mass of the reactants equals the mass of the products.

Reaction stoichiometry describes the quantitative relationships among substances as they participate in chemical reactions. In the example above, reaction stoichiometry describes the 1:3:2 ratio of molecules of nitrogen, hydrogen, and ammonia.
Composition stoichiometry describes the quantitative (mass) relationships among elements in compounds. For example, composition stoichiometry describes the nitrogen to hydrogen (mass) relationship in the compound ammonia: i.e., one mole of nitrogren and three moles of hydrogen are in every mole of ammonia.
A stoichiometric amount or stoichiometric ratio of a reagent is the amount or ratio where, assuming that the reaction proceeds to completion:

1. all reagent is consumed,
2. there is no shortfall of reagent, and
3. no residues remain.

A nonstoichiometric mixture, where reactions have gone to completion, will have only the limiting reagent consumed completely.
While almost all reactions have integer-ratio stoichiometry in amount of matter units (moles, number of particles), some nonstoichiometric compounds are known that cannot be represented by a ratio of well-defined natural numbers. These materials therefore violate the law of definite proportions that forms the basis of stoichiometry along with the law of multiple proportions.
Gas stoichiometry deals with reactions involving gases, where the gases are at a known temperature, pressure, and volume, and can be assumed to be ideal gases. For gases, the volume ratio is ideally the same by the ideal gas law, but the mass ratio of a single reaction has to be calculated from the molecular masses of the reactants and products. In practice, due to the existence of isotopes, molar masses are used instead when calculating the mass ratio.

Stoichiometry rests upon the very basic laws which help to understand it better i.e law of conservation of mass, the law of definite proportions (i.e., the law of constant composition) and the law of multiple proportions. In general, chemical reactions combine in definite ratios of chemicals. Since chemical reactions can neither create nor destroy matter, nor transmute one element into another, the amount of each element must be the same throughout the overall reaction. For example, the amount of element X on the reactant side must equal the amount of element X on the product side.
Stoichiometry is often used to balance chemical equations (reaction stoichiometry). For example, the two diatomic gases, hydrogen and oxygen, can combine to form a liquid, water, in an exothermic reaction, as described by the following equation:

Reaction stoichiometry describes the 2:1:2 ratio of hydrogen, oxygen, and water molecules in the above equation.
The term stoichiometry is also often used for the molar proportions of elements in stoichiometric compounds (composition stoichiometry). For example, the stoichiometry of hydrogen and oxygen in H₂O is 2:1. In stoichiometric compounds, the molar proportions are whole numbers.
Stoichiometry is not only used to balance chemical equations but also used in conversions, i.e., converting from grams to moles, or from grams to millilitres. For example, to find the number of moles in 2.00 g of NaCl, one would do the following:

In the above example, when written out in fraction form, the units of grams form a multiplicative identity, which is equivalent to one (g/g=1), with the resulting amount of moles (the unit that was needed), is shown in the following equation,

Stoichiometry is also used to find the right amount of reactants to use in a chemical reaction (stoichiometric amounts). An example is shown below using the thermite reaction,

This equation shows that 1 mole of aluminium oxide and 2 moles of iron will be produced with 1 mole of iron(III) oxide and 2 moles of aluminium. So, to completely react with 85.0 g of iron(III) oxide (0.532 mol), 28.7 g (1.06 mol) of aluminium are needed.

[edit] Different stoichiometries in competing reactions

Often, more than one reaction is possible given the same starting materials. The reactions may differ in their stoichiometry. For example, the methylation of benzene (C₆H₆), through a Friedel-Crafts reaction using AlCl₃ as catalyst, may produce singly methylated (C₆H₅CH₃), doubly methylated (C₆H₄(CH₃)₂), or still more highly methylated products, as shown in the following example,

In this example, which reaction takes place is controlled in part by the relative concentrations of the reactants.

[edit] Stoichiometric coefficient

In layman's terms, the stoichiometric coefficient (or stoichiometric number in the IUPAC nomenclature^[1]) of any given component is the number of molecules which participate in the reaction as written.
For example, in the reaction CH₄ + 2 O₂ → CO₂ + 2 H₂O, the stoichiometric coefficient of CH₄ would be 1 and the stoichiometric coefficient of O₂ would be 2.
In more technically-precise terms, the stoichiometric coefficient in a chemical reaction system of the i–th component is defined as

or

where N_i is the number of molecules of i, and ξ is the progress variable or extent of reaction (Prigogine & Defay, p. 18; Prigogine, pp. 4–7; Guggenheim, p. 37 & 62).

The extent of reaction ξ can be regarded as a real (or hypothetical) product, one molecule of which is produced each time the reaction event occurs. It is the extensive quantity describing the progress of a chemical reaction equal to the number of chemical transformations, as indicated by the reaction equation on a molecular scale, divided by the Avogadro constant (it is essentially the amount of chemical transformations). The change in the extent of reaction is given by dξ = dn_B/n_B, where n_B is the stoichiometric number of any reaction entity B (reactant or product) an dn_B is the corresponding amount.^[2]

The stoichiometric coefficient ν_i represents the degree to which a chemical species participates in a reaction. The convention is to assign negative coefficients to reactants (which are consumed) and positive ones to products. However, any reaction may be viewed as "going" in the reverse direction, and all the coefficients then change sign (as does the free energy). Whether a reaction actually will go in the arbitrarily-selected forward direction or not depends on the amounts of the substances present at any given time, which determines the kinetics and thermodynamics, i.e., whether equilibrium lies to the right or the left.
If one contemplates actual reaction mechanisms, stoichiometric coefficients will always be integers, since elementary reactions always involve whole molecules. If one uses a composite representation of an "overall" reaction, some may be rational fractions. There are often chemical species present that do not participate in a reaction; their stoichiometric coefficients are therefore zero. Any chemical species that is regenerated, such as a catalyst, also has a stoichiometric coefficient of zero.
The simplest possible case is an isomerism

in which ν_B = 1 since one molecule of B is produced each time the reaction occurs, while ν_A = −1 since one molecule of A is necessarily consumed. In any chemical reaction, not only is the total mass conserved, but also the numbers of atoms of each kind are conserved, and this imposes corresponding constraints on possible values for the stoichiometric coefficients.
There are usually multiple reactions proceeding simultaneously in any natural reaction system, including those in biology. Since any chemical component can participate in several reactions simultaneously, the stoichiometric coefficient of the i–th component in the k–th reaction is defined as

so that the total (differential) change in the amount of the i–th component is

.

Extents of reaction provide the clearest and most explicit way of representing compositional change, although they are not yet widely used.
With complex reaction systems, it is often useful to consider both the representation of a reaction system in terms of the amounts of the chemicals present { N_i } (state variables), and the representation in terms of the actual compositional degrees of freedom, as expressed by the extents of reaction { ξ_k }. The transformation from a vector expressing the extents to a vector expressing the amounts uses a rectangular matrix whose elements are the stoichiometric coefficients [ ν_i k ].
The maximum and minimum for any ξ_k occur whenever the first of the reactants is depleted for the forward reaction; or the first of the "products" is depleted if the reaction as viewed as being pushed in the reverse direction. This is a purely kinematic restriction on the reaction simplex, a hyperplane in composition space, or N‑space, whose dimensionality equals the number of linearly-independent chemical reactions. This is necessarily less than the number of chemical components, since each reaction manifests a relation between at least two chemicals. The accessible region of the hyperplane depends on the amounts of each chemical species actually present, a contingent fact. Different such amounts can even generate different hyperplanes, all of which share the same algebraic stoichiometry.
In accord with the principles of chemical kinetics and thermodynamic equilibrium, every chemical reaction is reversible, at least to some degree, so that each equilibrium point must be an interior point of the simplex. As a consequence, extrema for the ξ's will not occur unless an experimental system is prepared with zero initial amounts of some products.
The number of physically-independent reactions can be even greater than the number of chemical components, and depends on the various reaction mechanisms. For example, there may be two (or more) reaction paths for the isomerism above. The reaction may occur by itself, but faster and with different intermediates, in the presence of a catalyst.
The (dimensionless) "units" may be taken to be molecules or moles. Moles are most commonly used, but it is more suggestive to picture incremental chemical reactions in terms of molecules. The N's and ξ's are reduced to molar units by dividing by Avogadro's number. While dimensional mass units may be used, the comments about integers are then no longer applicable.

[edit] Stoichiometry matrix

In complex reactions, stoichiometries are often represented in a more compact form called the stoichiometry matrix. The stoichiometry matrix is denoted by the symbol, .
If a reaction network has n reactions and m participating molecular species then the stoichiometry matrix will have corresponding m rows and n columns.
For example, consider the system of reactions shown below:

S₁ → S₂

5S₃ + S₂ → 4S₃ + 2S₂

S₃ → S₄

S₄ → S₅.

This systems comprises four reactions and five different molecular species. The stoichiometry matrix for this system can be written as:

where the rows correspond to S₁, S₂, S₃, S₄ and S₅, respectively. Note that the process of converting a reaction scheme into a stoichiometry matrix can be a lossy transformation, for example, the stoichiometries in the second reaction simplify when included in the matrix. This means that it is not always possible to recover the original reaction scheme from a stoichiometry matrix.
Often the stoichiometry matrix is combined with the rate vector, v to form a compact equation describing the rates of change of the molecular species:

[edit] Gas stoichiometry

Gas stoichiometry is the quantitative relationship (ratio) between reactants and products in a chemical reaction with reactions that produce gases. Gas stoichiometry applies when the gases produced are assumed to be ideal, and the temperature, pressure, and volume of the gases are all known. The ideal gas law is used for these calculations. Often, but not always, the standard temperature and pressure (STP) are taken as 0°C and 1 bar and used as the conditions for gas stoichiometric calculations.
Gas stoichiometry calculations solve for the unknown volume or mass of a gaseous product or reactant. For example, if we wanted to calculate the volume of gaseous NO₂ produced from the combustion of 100 g of NH₃, by the reaction:

4NH₃ (g) + 7O₂ (g) → 4NO₂ (g) + 6H₂O (l)

we would carry out the following calculations:

There is a 1:1 molar ratio of NH₃ to NO₂ in the above balanced combustion reaction, so 5.871 mol of NO₂ will be formed. We will employ the ideal gas law to solve for the volume at 0 °C (273.15 K) and 1 atmosphere using the gas law constant of R = 0.08206 L · atm · K⁻¹ · mol⁻¹ :

PV	= nRT
V

Gas stoichiometry often involves having to know the molar mass of a gas, given the density of that gas. The ideal gas law can be re-arranged to obtain a relation between the density and the molar mass of an ideal gas:

    and    

and thus:

where:
P	= absolute gas pressure
V	= gas volume
n	= number of moles
R	= universal ideal gas law constant
T	= absolute gas temperature
ρ	= gas density at T and P
m	= mass of gas
M	= molar mass of gas

[edit] Stoichiometric air-fuel ratios of common fuels

Fuel	By mass [3]	By volume [4]	Percent fuel by mass
Gasoline	14.6 : 1	—	6.8%
Natural gas	14.5 : 1	9.7  : 1	5.8%
Propane (LP)	15.67 : 1	23.9 : 1	6.45%
Ethanol	9 : 1	—	11.1%
Methanol	6.47 : 1	—	15.6%
Hydrogen	34.3 : 1	2.39 : 1	2.9%
Diesel	14.5 : 1	0.094 : 1	6.8%

Gasoline engines can run at stoichiometric air-to-fuel ratio, because gasoline is quite volatile and is mixed (sprayed or carburetted) with the air prior to ignition. Diesel engines, in contrast, run lean, with more air available than simple stoichiometry would require. Diesel fuel is less volatile and is effectively burned as it is injected, leaving less time for evaporation and mixing. Thus, it would form soot (black smoke) at stoichiometric ratio.

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:
AgNO₃(aq) + NaCl(aq) ---> AgCl(s) + NaNO₃(aq) In this equation, AgNO₃ is mixed with NaCl. The equation shows that the reactants (AgNO₃ and NaCl) react through some process (--->) to form the products (AgCl and NaNO₃). 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 the 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 means 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 Fe₂O₃, a temperature of 1000° 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

Given the equation above, we can tell the number of moles of reactants and products. A mole simply represents Avogadro's number (6.022 x 10²³) of molecules. A mole is similar to a term like a dozen. If you have a dozen carrots, you have twelve of them. Similarly, if you have a mole of carrots, you have 6.022 x 10²³ 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 AgNO₃, NaCl, AgCl, NaNO₃. Converting between moles and grams of a substance is often important. This conversion can be easily done when the atomic and/or molecular mass of the substance(s) are known. Given the atomic or molecular mass of a substance, that mass in grams makes a mole of the substance. For example, calcium has an atomic mass of 40 atomic mass units. So, 40 grams of calcium makes one mole, 80 grams makes two moles, etc.

Balancing Chemical Equations

Sometimes, however, we have to do some work before using the coefficients of the terms to represent the relative number of molecules of each compound. This is the case when the equations are not properly balanced. We will consider the following equation:
Al + Fe₃O₄---> Al₂O₃
+ Fe Since no coefficients are in front of any of the terms, it is easy to assume that one (1) mole of Al and one (1) mole of Fe₃O₄ react to form one (1) mole of Al₂O₃. If this were the case, the reaction would be quite spectacular: an aluminum atom would appear out of nowhere, and two (2) iron atoms and one (1) oxygen atom would magically disappear. We know from the Law of Conservation of Mass (which states that matter can neither be created nor destroyed) that this simply cannot occur. We have to make sure that the number of atoms of each particular element 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 simple 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.

2Fe₃O₄ This term expresses two (2) molecules of Fe₃O₄. 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 atoms 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:

Al + Fe₃O₄---> Al₂O₃+ 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 simplest way to balance the oxygen terms is:
   Al + 3 Fe₃O₄---> 4 Al₂O₃+ 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 have 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:
   Al +3 Fe₃O₄---> 4Al₂O₃+ 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:

8Al + 3Fe₃O₄ ---> 4Al₂O₃ + 9 Fe

Limiting Reagents

in a chemical reaction, the limiting reagent, also known as the "limiting reactant", is the substance which is totally consumed when the chemical reaction is complete. The amount of product formed is limited by this reagent since the reaction cannot proceed further without it. The other reagents may be present in excess of the quantities required to react with the limiting reagent.

Method 1: Comparison of reactant amounts

This method is most useful when there are only two reactants. One reactant (A) is chosen, and the balanced chemical equation is used to determine the amount of the other reactant (B) necessary to react with A. If the amount of B actually present exceeds the amount required, then B is in excess and A is the limiting reagent. If the amount of B present is less than required, then B is the limiting reagent.

Example for two reactants 

Consider the combustion of benzene, represented by the following chemical equation:

This means that 15 mol molecular oxygen (O₂) is required to react with 2 mol benzene (C₆H₆).
The amount of oxygen required for other quantities of benzene can be calculated using cross-multiplication (the rule of three). For example, if 1.5 mol C₆H₆ is present, 11.25 mol O₂ is required:

If in fact 18 mol O₂ are present, there will be an excess of (18 - 11.25) = 6.75 mol of unreacted oxygen when all the benzene is consumed. Benzene is then the limiting reagent.
This conclusion can be verified by comparing the mole ratio of O₂ and C₆H₆ required by the balanced equation with the mole ratio actually present:

required: =

actual: =

Since the actual ratio is larger than required, O₂ is the reagent in excess, which confirms that benzene is the limiting reagent.

Method 2: Comparison of product amounts which can be formed from each reactant

In this method the chemical equation is used to calculate the amount of one product which can be formed from each reactant in the amount present. This method can be extended to any number of reactants more easily than the first method.

Example

Which reactant is limiting if 20.0 g of iron (III) oxide (Fe₂O₃) are reacted with 8.00 g aluminium (Al) in the followng thermite reaction?

Since the reactant amounts are given in grams, they must be first converted into moles for comparison with the chemical equation, in order to determine how many moles of Fe can be produced from either reactant.
Moles produced of Fe from reactant Fe₂O₃

Moles produced of Fe from reactant Al

There is enough Al to produce 0.297 mol Fe, but only enough Fe₂O₃ to produce 0.250 mol Fe. This means that the amount of Fe actually produced is limited by the Fe₂O₃ present, which is therefore the limiting reagent.

The limiting reagent must be identified in order to calculate the percentage yield of a reaction, since the theoretical yield is defined as the amount of product obtained when the limiting reagent reacts completely.
Given the balanced chemical equation which describes the reaction, there are several equivalent ways to identify the limiting reagent and evaluate the excess quantities of other 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 C₂H₂ and an unlimited supply of oxygen and he desires to produce as much CO₂ as possible. If she uses the equation below, how much oxygen should she add to the reaction?

2C₂H₂(g) + 5O₂(g) ---> 4CO₂(g) + 2 H₂O(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 CO₂).
First, we calculate the number of moles of C₂H₂ in 6.0 g of C₂H₂. 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 g and H weighs 1.0 g. Therefore we know that 1 mole of C₂H₂ weighs 26 g (2 × 12 grams + 2 × 1 gram).

6.0 g C2H2 x

1 mol C2H2 (24.0 + 2.0)g C2H2

= 0.25 mol C2H2

Then, because there are five (5) molecules of oxygen to every two (2) molecules of C₂H₂, 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:

0.25 mol C2H2 x

5 mol O2 2 mol C2H2

x

32.0 g O2 1 mol O2

= 20 g O2

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. 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 C₂H₂ 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 g so 54.3% would become 54.3 g. Then we can convert the masses to moles; this 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 transferred 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:

Carbon:	47.3 grams 1	x	1 mole 12.01 grams	= 3.94 moles
Hyrdrogen:	10.6 grams 1	x	1 mole 1.008 grams	= 10.52 moles
Sulfur:	42.0 grams 1	x	1 mole 32.07 grams	= 1.310 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:

Carbon:	3.94 1.310	= 3
Hydrogen:	10.52 1.310	= 8
Sulfur:	1.310 1.310	= 1

So we have: C₃H₈ S
Example: Figure out the percentage by mass of hydrogen sulfate, H₂SO₄.
In this problem we need to first calculate the total mass of the compound by looking at the periodic table. This gives us:
2(1.008) + 32.07 + 4(16.00) g/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.

hydrogen:	2(1.008) 98.09	=	2.016 98.09	= 0.0206 ∗ 100 = 2.06%
sulfur:	32.07 98.09	= 0.327 ∗ 100 = 32.7%
oxygen:	4(16.00) 98.09	=	64.00 98.09	= 0.652 ∗ 100 = 65.2%

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 H₂SO₄ 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, Na₂S, C₆H₁₀O₅. Examples of molecular formulas: P₂, C₂O₄, C₆H₁₄S₂, H₂, C₃H₉. 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:

43.7 grams P 1	x	1 mol 30.97 grams	= 1.41 moles
56.3 grams O 1	x	1 mol 16.00 grams	= 3.52 moles

Next we divide the moles to try to get an even ratio.

Phosphorus:	1.41 1.41	= 1.00
Oxygen:	3.52 1.41	= 2.50

When we divide, we did not get whole numbers so we must multiply by two (2). The answer = P₂O₅
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 necessary 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:
H₂C₂N₂.

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

Molarity =

moles of solute liters of solution

Molarity is probably the most commonly used term because measuring a volume of liquid is a fairly easy thing to do.
Example: If 5.00 g 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:

5.00g NaOH 1

x

1 mole (22.9 + 16.00 + 1.008)g

= 0.125 moles

Now we simply use the definition of molarity: moles/liters to get the answer

Molarity =

0.125 moles 5.00 L of soln

= 0.025 mol/L

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).

Molality =

moles of solute kg of solvent

Molality is sometimes used in place of molarity at extreme temperatures because the volume can contract or expand.
Example: If the molality of a solution of C₂H₅OH dissolved in water is 1.5 and the mass of the water is 11.7 kg, figure out how much C₂H₅OH 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.

Molality = moles solute kg solvent

Now we simply use the definition of molarity: moles/liters to get the answer

Molality = moles solute kg solvent

1.5 mols kg = moles solute 11.7 kg

1.5 moles kg x 11.7 kg = 17.55 moles

17.55 moles 1 x (2 ∗ 12.01) + (6 ∗ 1.008) + 16 1 moles = 808.5 g C2H5OH

It is possible to convert between molarity and molality. The only information needed is density.
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 mol/L to the molality units of mol/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.

0.3 mol 1 L

x

1 mL 3.25 g

x

1 L 1000 mL

x

1000 g 1 kg

= 0.09 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

ΔT= K_f × m Where:
ΔT is temperature depression (for freezing point) or temperature expansion (for boiling point) (°C)
K_f is the freezing point constant (kg °C/mol)
m is molality in mol/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° C.
ΔT = K_f * m
ΔT/K_f = m
m = 5.2/1.86
m = 2.8 mols/kg

 TITRATION
Titration, also known as titrimetry,^[1] is a common laboratory method of quantitative chemical analysis that is used to determine the unknown concentration of an identified analyte. Because volume measurements play a key role in titration, it is also known as volumetric analysis. A reagent, called the titrant or titrator^[2] is prepared as a standard solution. A known concentration and volume of titrant reacts with a solution of analyte or titrand^[3] to determine concentration.

History and etymology

The word "titration" comes from the Latin word titulus, meaning inscription or title. The French word titre, also from this origin, means rank. Titration, by definition, is the determination of rank or concentration of a solution with respect to water with a pH of 7 (the pH of pure H₂O under standard conditions).^[4]
Volumetric analysis originated in late 18th-century France. Francois Antoine Henri Descroizilles developed the first burette (which was similar to a graduated cylinder) in 1791.^[5] Joseph Louis Gay-Lussac developed an improved version of the burette that included a side arm, and coined the terms "pipette" and "burette" in an 1824 paper on the standardization of indigo solutions. A major breakthrough in the methodology and popularization of volumetric analysis was due to Karl Friedrich Mohr, who redesigned the burette by placing a clamp and a tip at the bottom, and wrote the first textbook on the topic, Lehrbuch der chemisch-analytischen Titrirmethode (Textbook of analytical-chemical titration methods), published in 1855

Procedure

A typical titration begins with a beaker or Erlenmeyer flask containing a precise volume of the titrand and a small amount of indicator placed underneath a calibrated burette or chemistry pipetting syringe containing the titrant. When the endpoint of the reaction is reached, the volume of reactant consumed is measured and used to calculate the concentration of analyte by

where C_a is the concentration of the analyte, typically in molarity; C_t is the concentration of the titrant, typically in molarity; V_t is the volume of the titrant used, typically in dm³; M is the mole ratio of the analyte and reactant from the balanced chemical equation; and V_a is the volume of the analyte used, typically in dm³

Titration curves

Main article: Titration curve

A typical titration curve of a diprotic acid, oxalic acid, titrated with a strong base, sodium hydroxide. Each of the two equivalence points is visible.

A titration curve is a curve in the plane whose x-coordinate is the volume of titrant added since the beginning of the titration, and whose y-coordinate is the concentration of the analyte at the corresponding stage of the titration (in an acid-base titration, the y-coordinate is usually the pH of the solution).^[12]
In an acid-base titration, the titration curve reflects the strength of the corresponding acid and base. For a strong acid and a strong base, the curve will be relatively smooth and very steep near the equivalence point. Because of this, a small change in titrant volume near the equivalence point results in a large pH change and many indicators would be appropriate (for instance litmus, phenolphthalein or bromothymol blue).
If one reagent is a weak acid or base and the other is a strong acid or base, the titration curve is irregular and the pH shifts less with small additions of titrant near the equivalence point. For example, the titration curve for the titration between oxalic acid (a weak acid) and sodium hydroxide (a strong base) is pictured. The equivalence point occurs between pH 8-10, indicating the solution is basic at the equivalence point and an indicator such as phenolphthalein would be appropriate. Titration curves corresponding to weak bases and strong acids are similarly behaved, with the solution being acidic at the equivalence point and indicators such as methyl orange and bromothymol blue being most appropriate.
Titrations between a weak acid and a weak base have titration curves which are incredibly irregular. Because of this, no definite indicator may be appropriate and a pH meter is often used to monitor the reaction

Types of titrations

There are many types of titrations with different procedures and goals. The most common types of qualitative titration are acid-base titrations and redox titrations.

Acid-base titration

Main article: Acid-base titration

Indicator	Color on Acidic Side	Range of Color Change	Color on Basic Side
Methyl Violet	Yellow	0.0 - 1.6	Violet
Bromophenol Blue	Yellow	3.0 - 4.6	Blue
Methyl Orange	Red	3.1 - 4.4	Yellow
Methyl Red	Red	4.4 - 6.3	Yellow
Litmus	Red	5.0 - 8.0	Blue
Bromothymol Blue	Yellow	6.0 - 7.6	Blue
Phenolphthalein	Colorless	8.3 - 10.0	Pink
Alizarin Yellow	Yellow	10.1 - 12.0	Red

Acid-base titrations depend on the neutralization between an acid and a base when mixed in solution. In addition to the sample, an appropriate indicator is added to the titration chamber, reflecting the pH range of the equivalence point. Common indicators, their colors, and the pH range in which they change color are given in the table above.^[14] When more precise results are required, or when the reagents are a weak acid and a weak base, a pH meter or a conductance meter are used.

[edit] Redox titration

Main article: Redox titration

Redox titrations are based on a reduction-oxidation reaction between an oxidizing agent and a reducing agent. A potentiometer or a redox indicator is usually used to determine the endpoint of the titration, as when one of the constituents is the oxidizing agent potassium dichromate. The color change of the solution from orange to green is not definite, therefore an indicator such as sodium diphenylamine is used.^[15] Analysis of wines for sulfur dioxide requires iodine as an oxidizing agent. In this case, starch is used as an indicator; a blue starch-iodine complex is formed in the presence of excess iodine, signalling the endpoint.^[16]
Some redox titrations do not require an indicator, due to the intense color of the constituents. For instance, in permanganometry a slight faint persisting pink color signals the endpoint of the titration because of the color of the excess oxidizing agent potassium permanganate.^[17]

[edit] Gas phase titration

Ozone, as depicted with a molecular model set

Gas phase titrations are titrations done in the gas phase, specifically as methods for determining reactive species by reaction with an excess of some other gas, acting as the titrant. Most commonly the gaseous analyte is ozone, which is titrated with nitrogen oxide according to the reaction

O₃ + NO → O₂ + NO₂.^[18][19]

After the reaction is complete, the remaining titrant and product are quantified (e.g., by FT-IR); this is used to determine the amount of analyte in the original sample.
Gas phase titration has several advantages over simple spectrophotometry. First, the measurement does not depend on path length, because the same path length is used for the measurement of both the excess titrant and the product. Second, the measurement does not depend on a linear change in absorbance as a function of analyte concentration as defined by the Beer-Lambert law. Third, it is useful for samples containing species which interfere at wavelengths typically used for the analyte.^[20]

[edit] Complexometric titration

Main article: Complexometric titration

Complexometric titrations rely on the formation of a complex between the analyte and the titrant. In general, they require specialized indicators that form weak complexes with the analyte. Common examples are Eriochrome Black T for the titration of calcium and magnesium ions, and the chelating agent EDTA used to titrate metal ions in solution.^[21]

[edit] Zeta potential titration

Main article: Zeta potential titration

Zeta potential titrations are titrations in which the completion is monitored by the zeta potential, rather than by an indicator, in order to characterize heterogeneous systems, such as colloids.^[22] One of the uses is to determine the iso-electric point when surface charge becomes zero, achieved by changing the pH or adding surfactant. Another use is to determine the optimum dose for flocculation or stabilization.^[23]

[edit] Assay

Main article: Assay

An assay is a form of biological titration used to determine the concentration of a virus or bacterium. Serial dilutions are performed on a sample in a fixed ratio (such as 1:1, 1:2, 1:4, 1:8, etc.) until the last dilution does not give a positive test for the presence of the virus. This value is known as the titer, and is most commonly determined through enzyme-linked immunosorbent assay (ELISA).^[24]

[edit] Measuring the endpoint of a titration

Main article: Equivalence point

Different methods to determine the endpoint include^[25]:

• Indicator: A substance that changes color in response to a chemical change. An acid-base indicator (e.g., phenolphthalein) changes color depending on the pH. Redox indicators are also used. A drop of indicator solution is added to the titration at the beginning; the endpoint has been reached when the color changes.
• Potentiometer: An instrument that measures the electrode potential of the solution. These are used for redox titrations; the potential of the working electrode will suddenly change as the endpoint is reached.

An elementary pH meter that can be used to monitor titration reactions

• pH meter: A potentiometer with an electrode whose potential depends on the amount of H⁺ ion present in the solution. (This is an example of an ion-selective electrode.) The pH of the solution is measured throughout the titration, more accurately than with an indicator; at the endpoint there will be a sudden change in the measured pH.
• Conductivity: A measurement of ions in a solution. Ion concentration can change significantly in a titration, which changes the conductivity. (For instance, during an acid-base titration, the H⁺ and OH^- ions react to form neutral H₂O.) As total conductance depends on all ions present in the solution and not all ions contribute equally (due to mobility and ionic strength), predicting the change in conductivity is more difficult than measuring it.
• Color change: In some reactions, the solution changes color without any added indicator. This is often seen in redox titrations when the different oxidation states of the product and reactant produce different colors.
• Precipitation: If a reaction produces a solid, a precipitate will form during the titration. A classic example is the reaction between Ag⁺ and Cl^- to form the insoluble salt AgCl. Cloudy precipitates usually make it difficult to determine the endpoint precisely. To compensate, precipitation titrations often have to be done as "back" titrations (see below).
• Isothermal titration calorimeter: An instrument that measures the heat produced or consumed by the reaction to determine the endpoint. Used in biochemical titrations, such as the determination of how substrates bind to enzymes.
• Thermometric titrimetry: Differentiated from calorimetric titrimetry because the heat of the reaction (as indicated by temperature rise or fall) is not used to determine the amount of analyte in the sample solution. Instead, the endpoint is determined by the rate of temperature change.
• Spectroscopy: Used to measure the absorption of light by the solution during titration if the spectrum of the reactant, titrant or product is known. The concentration of the material can be determined by Beer's Law.
• Aperometry: Measures the current produced by the titration reaction as a result of the oxidation or reduction of the analyte. The endpoint is detected as a change in the current. This method is most useful when the excess titrant can be reduced, as in the titration of halides with Ag⁺.

[edit] Endpoint and equivalence point

Though equivalence point and endpoint are used interchangeably, they are different terms. Equivalence point is the theoretical completion of the reaction: the volume of added titrant at which the number of moles of titrant is equal to the number of moles of analyte, or some multiple thereof (as in polyprotic acids). Endpoint is what is actually measured, a physical change in the solution as determined by an indicator or an instrument mentioned above.^[26]
There is a slight difference between the endpoint and the equivalence point of the titration. This error is referred to as an indicator error, and it is indeterminate.^[27]

[edit] Back titration

Back titration is a titration is done in reverse; instead of titrating the original sample, a known excess of standard reagent is added to the solution, and the excess titrated. A back titration is useful if the endpoint of the reverse titration is easier to identify than the endpoint of the normal titration, as with precipitation reactions. Back titrations are also useful if the reaction between the analyte and the titrant is very slow, or when the analyte is in a non-soluble solid.^[28]

[edit] Particular uses

Specific examples of titrations include:

Acid-Base Titrations

• In biodiesel: Waste vegetable oil (WVO) must be neutralized before a batch may be processed. A portion of WVO is titrated with a base to determine acidity, so the rest of the batch may be properly neutralized. This removes free fatty acids from the WVO that would normally react to make soap instead of biodiesel.^[29]
• Kjeldahl method: A measure of nitrogen content in a sample. Organic nitrogen is digested into ammonia with sulfuric acid and potassium sulfate. Finally, ammonia is back titrated with boric acid and then sodium carbonate.^[30]
• Acid value: The mass in milligrams of potassium hydroxide (KOH) required to neutralize carboxylic acid in one gram of sample. An example is the determination of free fatty acid content. These titrations are achieved at low temperatures.
• Saponification value: The mass in milligrams of KOH required to saponify carboxylic acid in one gram of sample. Saponification is used to determine average chain length of fatty acids in fat. These titrations are achieved at high temperatures.
• Ester value (or ester index): A calculated index. Ester value = Saponification value – Acid value.
• Amine value: The mass in milligrams of KOH equal to the amine content in one gram of sample.
• Hydroxyl number: The mass in milligrams of KOH required to neutralize hydroxyl groups in one gram of sample. The analyte is acetylated using acetic anhydride then titrated with KOH.

Redox titrations

• Winkler test for dissolved oxygen: Used to determine oxygen concentration in water. Oxygen in water samples is reduced using manganese(II) sulfate, which reacts with potassium iodide to produce iodine. The iodine is released in proportion to the oxygen in the sample, thus the oxygen concentration is determined with a redox titration of iodine with thiosulfate using a starch indicator.^[31]
• Vitamin C: Also known as ascorbic acid, vitamin C is a powerful reducing agent. Its concentration can easily be identified when titrated with the blue dye Dichlorophenolindophenol (DCPIP) which turns colorless when reduced by the vitamin.^[32]
• Benedict's reagent: Excess glucose in urine may indicate diabetes in the patient. Benedict's method is the conventional method to quantify glucose in urine using a prepared reagent. In this titration, glucose reduces cupric ions to cuprous ions which react with potassium thiocyanate to produce a white precipitate, indicating the endpoint.^[33]
• Bromine number: A measure of unsaturation in an analyte, expressed in milligrams of bromine absorbed by 100 grams of sample.
• Iodine number: A measure of unsaturation in an analyte, expressed in grams of iodine absorbed by 100 grams of sample.

Miscellaneous

• Karl Fischer titration: A potentiometric method to analyze trace amounts of water in a substance. A sample is dissolved in methanol, and titrated with Karl Fischer reagent. The reagent contains iodine, which reacts proportionally with water. Thus, the water content can be determined by monitoring the potential of excess iodine

Preparation techniques

Typical titrations require titrant and analyte to be in a liquid (solution) form. Though solids are usually dissolved into an aqueous solution, other solvents such as glacial acetic acid or ethanol are used for special purposes (as in petrochemistry).^[8] Concentrated analytes are often diluted to improve accuracy.
Many non-acid-base titrations require a constant pH throughout the reaction. Therefore a buffer solution may be added to the titration chamber to maintain the pH.^[9]
In instances where two reactants in a sample may react with the titrant and only one is the desired analyte, a separate masking solution may be added to the reaction chamber which masks the unwanted ion.^[10]
Some redox reactions may require heating the sample solution and titrating while the solution is still hot to increase the reaction rate. For instance, the oxidation of some oxalate solutions requires heating to 60 °C (140 °F) to maintain a reasonable rate of reaction.