CHEMISTRY - 2202 .

UNIT 1 : STOICHIOMETRY.

NAME : _____________________________

Introduction.

Matter to represent anything that has mass and occupies space. Chemicals are examples of matter. Chemistry is the study of matter, its properties, and the changes it undergoes.

There are two major branches: pure and applied chemistry. Pure chemistry deals with describing known substances and discovering new compounds. Applied chemistry is the search for uses for chemical substances. New chemicals are being discovered or produced every day.

Isotopes and Atomic Mass .

Investigations conducted as long ago as the late 1700's revealed that atoms of different elements have different masses. Dalton included this idea in his atomic theory in 1804. What he didn't know at the time was that two atoms of the same element may also have different masses. That's because he didn't know about subatomic particles. Today, we know that the number of protons and neutrons in an atom determines the mass of an atom. Remember the terms like electron (e-), proton (p+), and neutron (n) — (see the chart below); however, there are a number of other terms used to describe the atom.

It is important that you be familiar with the following key terms.

1.	Atomic Number (Nuclear Charge) : The atomic number of an element is equal to the number of protons in an atom. Atomic number is also called nuclear charge because it is based on the number of protons (positively charged particles) in the nucleus. For example, all atoms of carbon have six protons, so the atomic number (or nuclear charge) of carbon is six. Atomic numbers are whole numbers.

2.	Mass Number : The mass number of an atom is equal to the sum of its protons and neutrons. For example, an atom of helium has two protons and two neutrons, so its mass number is four. Mass numbers are also whole numbers. Sometimes the mass number of an atom is included in its name; for example, helium - 4.

3.	Isotopes : It is possible for atoms of the same element to have different numbers of neutrons. These atoms are termed isotopes. For example, if a carbon atom has six neutrons and another has eight neutrons, then they are isotopes of carbon. (The prefix iso- means the same.)

Example : Two isotopes of carbon. Carbon - 12 is stable but carbon - 14 is radioactive.

Average Atomic Mass :

When you look up the atomic mass of an element in the periodic table, you will find that its value usually has 2 decimal places; (C is 12.01 amu). The reason for this is that most naturally occurring elements have more than one isotope. This means that when we measure the atomic mass of an element, we must take into account the average mass of the naturally occurring isotopes. The natural relative abundances of the isotopes are usually expressed as percentages.

Example :	Copper, a metal known since ancient times, is used in electrical cables and pennies. The atomic masses of its two stable isotopes, copper- 63 (69.09%) and copper- 65 (30.91%), are 62.93 amu and 64.9278 amu, respectively. Calculate the average atomic mass of copper.

Solution :

Average Atomic Mass for Cu = (0.6909)(62.93 amu) + (0.3091)(64.9278 amu) = 63.55 amu Sample Problem :

The atomic masses of the two stable isotopes of boron, boron-10 (19.78%) and boron-11 (80.22%), are 10.0129 amu and 11.0093 amu, respectively. Calculate the average atomic mass of boron.

The Mole (Avogadro's Number).

The word "mole" comes from the Latin "moles", meaning a mass or pile of something. A mole is a number, 6.022137 x 1023 to be exact. It is sometimes called Avogadro's number in honour of Amedeo Avogadro.(See handout). The mole is a useful amount in chemistry because the basic unit of matter (the atom) is so very, very, very, very, small. Most people have a hard time imagining how big a number like 6.022137 x 1023 really is. Atoms are so small that it takes an incredibly huge number of them to make up a visible amount. For example, a mole of regular marbles spread over the Earth's surface would cover it to a depth of 80 km. When dealing with large numbers of atoms on a consistent basis can get to be tiresome after a while. That's where the mole comes in.

A mole is the number of carbon-12 atoms in exactly 12 grams of carbon-12. This number was determined by dividing the mass of a single carbon-12 atom into a 12 g mass of carbon-12. Since the mass of a single carbon-12 atom is tiny (1.99 x 10-23 g), the value of a mole comes out to be a huge number. Remember that the mole is not an exact number and current estimates place its value at 6.022137 x 1023. The greater precision (number of decimal places) in the current value is a function of the greater precision of the devices currently used to measure the mass. We use the value of 6.02 x 1023 particles (atoms, ions or molecules) in all our calculations for this course; (see the chart that follows). At this point, the most important thing to remember about the mole is that it represents a big number - a very, very, very big number. Chemists use it in the same sort of way that a egg farmer uses the dozen.

Avogadro's Number, 6.02 x 1023

Element	Ionic Compound	Molecular Compound
name	carbon, C	sodium chloride, NaCl	water, H2O
mass	12.01 g	58.44 g	18.02 g
number	6.02 x 1023

atoms	6.02 x 1023 formula units of NaCl

6.02 x 1023 Na+ ions

6.02 x 1023 Cl- ions	6.02 x 1023 molecules of H2O

6.02 x 1023 O atoms

2 (6.02 x 1023) H atoms

VIDEODISC SELECTION :

Sample Problem :

A)	How many particles are there in a mole?

B)	How many atoms of sodium are there in one mole?

C)	What is the mass of one mole of sodium atoms?

The Molar Mass of a Compound.

As you know, few elements exist as pure substances in nature — most of them are found in compounds. Although the periodic table provides you with the molar masses of the elements, it does not provide you with the molar masses of compounds. To find the molar mass of a compound, you have to write the chemical formula, list the number of atoms of each element, and multiply this number by the molar mass of the element.

Example : Calculate the molar mass of water.

To do this, write the chemical formula: H2O. List the number of atoms of each element in the compound: 2 H and 1 O. Multiply each number by the molar mass of the element and add the products. The molar mass of H2O is 18.02 g/mol.

You will be using the molar mass of water frequently, so it is a good idea to memorize it as 18.02 g/mol.

When calculating the molar mass of a compound, it is important to remember what each numerical subscript is indicating. Numbers to the lower right of an element symbol indicate the number of "atoms" of that element in the compound. Numbers outside the parentheses are used to balance a chemical formula. To determine the number of "atoms" in a formula, you multiply terms inside the parentheses by the ones outside the parentheses.

Consider the formula:	(NH4)3PO4.

There are:	3 x 1 nitrogen atoms (a total of 3 N),

3 x 4 hydrogen atoms (a total of 12 H),

1 phosphorus atom,

4 oxygen atoms.

In chemistry, the conversion factor in mole calculations is molar mass (the mass of one mole). This value has been calculated for each element and is usually included in a periodic table. In the Nelson Chemistry periodic table of the elements, the molar mass value is located in the top right hand corner of an element's cell. The value 55.85 g/mol means that a mole of iron atoms (atoms) has a mass of 55.85 grams. See the diagram below.

Look at the following example :

Calculate the molar mass of barium hydroxide octahydrate.

SOLUTION :

Begin by writing the chemical formula for barium hydroxide octahydrate.

Barium forms a 2+ ion because it is a Group 2 element ; Ba2+.

Hydroxide is a polyatomic ion with a 1- charge ; OH -.

Octahydrate means 8 H2O.

It takes two hydroxide ions to balance the charge on one barium ion, so the chemical formula is Ba(OH)2 8 H2 O .

Count and list the atoms/ions of each element, but treat water as a unit.

1 Ba

2 O

2 H

8 H2O

Next, look-up the molar mass of each item in the list and multiply. (Remember the molar mass of water is 18.02 g/mol.)

Answer :	The molar mass of barium hydroxide octahydrate is 315.51 g/mol.

As you get comfortable with molar mass calculations, you can combine the steps into a single line like this:

(1 Ba x 137.33) + (2 O x 16.00) + (2 H x 1.01) + (8 H2O x 18.02) = 315.51 g/mol
Always show the unit for molar mass (g/mol) in your final answer.

Note that in this course, molar mass values (calculated or listed in tables) ALWAYS show precision to two decimal places.

Summary :

Here is a summary of the steps to complete in molar mass calculations.

1.	Write the chemical formula for the compound.

2.	List the number of particles of each element in the compound.

3.	Multiply each number in the list by the molar mass.

4.	Add the products. ALL molar mass values must show precision to TWO decimal places in this course.

5.	Communicate your answer.

Sample Problem :	Calculate the molar mass for each compound.

A)	NaHCO3	E)	FeSO4 3H2O

B)	Sr(NO3)2	F)	lead(II) acetate

C)	Al2(SO4)3	G)	magnesium chloride

D)	CuSO4

Molar Volume.

Another way of measuring amounts of substances is by volume. The SI unit of volume is the litre, (L). Amounts of gases like hydrogen, helium, nitrogen and oxygen can be measured and compared in litres if the samples are at the same temperature and pressure. The reason for this is that changes in pressure and/or temperature can effect the volume of a gas. Chemists compare gas volumes at 0° C and 101.3 kPa. These conditions are known as Standard Temperature and Pressure or STP.

The reason why gases appear to be lighter than liquids or solids is that they are less dense. As a comparison, let us look at the density of iron and hydrogen; (pay attention to the density units for solids versus gases in your periodic table)! Hydrogen is much less dense (0.0899 g/L versus 7.87 g/mL)!

What does all of this mean? Well in a gas sample, the particles are really spread out; (ie.) there is a lot of empty space between them. The same is not true for solids and liquids. It turns out that gas volumes contain a consistent number of gas particles regardless of the type of gas. This idea was proposed by Amadeo Avogadro in 1811. (Nelson p. 205)

Based on what was known about how gas volumes combined in chemical reactions, Amadeo Avogadro concluded that equal volumes of different gases at the same temperature and temperature, have the same number of particles (ie.) same number of moles. This very significant discovery led to the concept of molar volume : 1 mol of any gas at STP has a volume of 22.4 L.

You should memorize the value 22.4 L/mol as molar volume at STP, but in the event that you forget it, you can look it up in the table on the data side of your periodic table under the heading "Measured (Uncertain) Quantities". Why "uncertain"? Well it turns out that there is a little variance in molar volume from one gas to another, but not that much — so we can apply the value 22.4 L/mol for all gases discussed in this course.

Sample Problems :

A)	What does STP mean?

B)	What is molar volume at STP?

C)	What is the volume of one mole of fluorine gas at STP?

D)	What is the volume of one mole of chlorine gas at STP?

E)	Compare the answers for items (B) and (C).

A point to consider : In chemistry, you are often required to compare amounts of substances. Comparisons can be made by mass, but do equal masses represent the same amounts? For example, do 10.0 g samples of iron and aluminum contain the same number of atoms? The answer is no. An atom of aluminum has a lower mass than an atom of iron because it has fewer protons and neutrons. Since aluminum atoms are less massive, it takes more of them to make up a 10.0 g sample. Therefore, amounts of chemical substances are not compared by mass, they are compared in moles. So if amounts of chemicals are compared in moles, why not measure amounts of chemicals in moles directly? Well, because moles cannot be measured directly. We measure amounts by mass and volume and then convert them to moles using conversion factors.

Mole Calculations :

Conversion Factor Method :

Formulas :

Mass to Mole Conversions : The mass of a substance can be found using a balance. If you know the molar mass of the substance, then you can convert the mass to a mole amount.

Use this problem solving approach to convert a mass to moles:

1.	List the given mass.

2.	Look-up or calculate the molar mass of the pure substance using data from a periodic table.

3.	Apply the mole formula, n = m/M , and round-off the answer to the appropriate number of significant digits using the rule for multiplication and division.

4.	Communicate your answer.

Here's an example : Let's say you want to determine the number of moles in a 56.83 g piece of aluminum.

To solve this problem, you can apply a simple formula, , where n is the number of moles, m is the mass, and M is the molar mass (in g/mol). You can find the molar mass of aluminum in a periodic table and substitute it into the mole formula along with the given mass.

There are two things you should notice about the calculation:

• First, the unit for grams cancels out leaving you with the unit for moles.

• Second, the answer is rounded-off to four significant digits.

Moles to Mass Conversions : Let's say you are asked to measure out 0.350 mol of potassium chloride. Is there a device that allows you to obtain this amount directly? No there isn't. You would have to measure the amount as a mass using a balance. Obviously, there you need to be able to convert a mole amount to a mass first. A simple rearrangement of the mole formula allows you to solve for mass given moles. The rearranged formula is:, where once again, m is mass, n is number of moles and M is molar mass.

The steps to convert from moles to mass are :

1.	List the given number of moles.

2.	Look up or calculate the molar mass of the substance or compound.

3.	Substitute the values into the formula, m = nM , and round-off the answer to the correct number of significant digits.

4.	Communicate the answer.

Example : A sample of neon gas contains 2.00 mol of atoms. Calculate the mass of the sample.

Solution :	1. List the given amount : n = 2.00 mol

2. Look up the molar mass of neon : M = 20.18 g/mol

3. Apply the rearranged mole formula and solve for mass:

4. Communicate the answer : The mass of the neon sample is 40.4 g.

Notice that the unit mol cancelled out and that the answer was rounded-off to three significant digits or the lower of the two numbers of significant digits in the measured values.

Another Example : Calculate the mass of 2.40 mol of aluminum nitrate, Al(NO3)3.

Molar Mass (M) for Al(NO3)3 :

(1 Al x 26.98) + (3 N x 14.01) + (9 O x 16.00) = 213.01 g/mol

Apply the rearranged mole formula and solve for mass:

Answer :	The mass of 2.40 mol of aluminum nitrate is 511 g.

Mole-Volume Conversions : Gases are produced in many chemical reactions. Gas amounts are often measured as volumes because their low densities make obtaining accurate mass measurements difficult. You can use the molar volume constant for gases at STP to convert between a measured volume and a mole amount.

Converting a gas volume to a number of moles requires use of this formula :

, where n = number of moles, v = volume in litres, and V = molar volume.

Examples :

A)	Calculate the number of moles of neon in a 6.81 L sample at STP conditions.

Solution :

Answer : The number of moles of neon in a 6.81 L sample at STP is 0.304 mol.

B)	Calculate the volume that 0.200 mol of hydrogen gas occupies at STP.

Solution :

Answer : The volume of hydrogen gas in a 0.200 mol sample at STP is 4.48 L.

C)	Calculate the mass of oxygen gas present in a volume of 896 L at STP.

Solution :

Answer : The mass of oxygen gas in a 896 L mol sample at STP is 1.28 x 10 3 g.

Summary for all mole-mass and mole-volume inter-conversions :

1.	Identifying the given and wanted information.

2.	Finding or calculating values like molar mass.

3.	Choosing a formula suited to finding the wanted information.

4.	Communicate the answer.

The Mole and Quantitative Analysis :

Qualitative analysis is carried out to identify the composition of a substance or a mixture. The focus is on determining which chemical species are present in a sample. A quantitative analysis involves identifying the quantity of each species that is present in a sample.

One type of quantitative analysis that is particularly important in chemistry is percentage composition - finding the percent by mass of each element in a substance. Percentage composition analysis is a key step in the determination of the compound's chemical formula.

Analysis begins with decomposition of the compound into its elements. Once the elements are identified, their masses are determined and expressed as a percentage of the mass of the sample. This information is used in conjunction with the molar mass of the compound to determine the chemical formula.

Calculating Percentage Composition :

The steps involved in calculating percentage composition would be...

1. Interpret the subscripts in terms of moles.

2. Convert moles to mass.

3. Calculate the molar mass of the compound.

4. Calculate the percentage composition of each element.

Example :	Calculate the percent composition by mass of each element in sulfuric acid (H2SO4).

Mass of one mole of H2SO4 =	2 • 1.01 g + 1 • 32.06 g + 4 • 16.00 g

=	2.02 g + 32.06 g + 64.00 g

=	98.08 g

% H = x 100%	=	2.06 %

% S =	x 100%	=	32.7 %

% O =	x 100%	=	65.3 %

Answer :	The percentage composition of sulfuric acid is 2.06 % hydrogen, 32.7 % sulfur and 65.3 % oxygen.

Empirical and Molecular Formulas :

When a new substance is discovered, the first thing that has to be done is an analysis of its composition. This is done by experiment, using sophisticated equipment like combustion analysers and mass spectrometers. Once percentage composition is known, then a chemical formula for the compound can be determined. This is a really important point. Whether the substance is one that is known to exist or one that as never been discovered, a key step in identifying it (i.e. coming up with a chemical formula) is to determine its percent composition.

Essentially, there are just two kinds of chemical formulas: empirical formulas and molecular formulas. An empirical formula is a lowest whole number ratio of the elements in a compound. A molecular formula shows the number of atoms of element in the compound; it is not necessarily a lowest whole number ratio. For example, NaCl, CH4, and H2O are all empirical formulas because they are lowest whole number ratios: 1:1, 1:4, and 1:2 respectively. These ratios cannot be simplified any further. Molecular formulas are not necessarily lowest whole number ratios. For example, C3H8, C6H6, and C2H2 are molecular formulas — they all show the exact numbers of atoms in the molecules; however, C6H6 and C2H2 can be reduced to the empirical formula CH — they are not lowest whole number ratios.

Finding an Empirical Formula : An empirical formula is the lowest whole number ratio of the atoms or ions in a compound. To determine the empirical formula of a compound, you need to know the results of a percentage composition analysis. Devices like the combustion analyzer (see page 269 Nelson Chemistry) can provide this kind of information, but in this course, you will use simpler methods to obtain this kind of data.

The steps in determining an empirical formula are:

1.	Convert the percentage to a mass.

2.	Convert mass to moles.

3.	Divide each mole value by the lowest of the mole values.

4.	Multiply the mole values by a factor to convert them to whole numbers (if necessary).

Example :	A prospector finds a rock that may contain rutile - an ore of titanium. A crystal from the sample is analyzed and found to be 59.94% titanium and 40.06% oxygen by mass. Determine the empirical formula of the crystalline substance.

Assume the mass of the sample is 100.0 g and multiply it by the percentage composition for each element:

mtitanium = 100.0 g x 59.94 % = 59.94 g

moxygen	= 100.0 g x 40.06 % = 40.06 g

Use the mole formula to convert each mass to moles:

Write a ratio of the mole values and divide the larger mole value by the smaller mole value:

Express the empirical formula:

▪The ratio of titanium to oxygen (1 Ti : 2 O ) is a whole number ratio, so the empirical formula for the compound is TiO2.

A Point to Remember : Sometimes the division of the ratio by the smaller value does not produce a neat whole number ratio. In such cases you have to either round off the value to a whole number (e.g. 1.997 rounds up to 2) or use a multiplier to convert the ratio to whole numbers.

Finding a Molecular Formula : A molecular formula shows the actual number of atoms of each element in a compound. It may or may not be a lowest whole number ratio. In order to find the molecular formula of an unknown, the molar mass of the compound must be known. Nowadays, molar mass is usually determined by mass spectroscopy (see page 272 in Nelson Chemistry). In this course, you are provided with the molar mass values of unknowns; that is, you are not required to determine molar masses in the lab. In the molecular formula calculations, it is convenient to use the mass of a one mole sample of the unknown.

The steps in determining a molecular formula are:

1.	Assume a sample mass equal to one mole of the compound and convert the percentage composition data to mass amounts.

2.	Convert the mass amount for each element to moles.

3.	Roundoff the mole amounts and use them as numerical subscripts in the chemical formula.

Example :	An unknown was subjected to a combustion analysis and found to be 85.60% carbon and 14.40% hydrogen. Mass spectrometry indicated that its molar mass was 28.06 g/mol. Determine the molecular formula of the compound.

Assume a one mole sample is analyzed, and multiply it by the percentage composition for each element:

Convert the mass of each element to moles:

Write a mole ratio for the pair of elements in the compound:

2.000 C : 4.00 H

Round-off each mole amount to a whole number and use it as a numerical subscript in the molecular formula. The molecular formula is C2H4.

STOICHIOMETRY :

A chemical equation tells you the chemical species that react and the chemical species produced. A balanced chemical equation tells you the relative amounts of the species involved in a reaction. The number in front of each species (ie.) the coefficients in chemical equations are interpreted as counted numbers and represent the number of moles of that substance in a balanced chemical equation.

Stoichiometry is the prediction of how much of one substance will react or be produced in a chemical reaction relative to the amount of another substance in the reaction.

Note: The amount of substances may be expressed in: (i) Moles or (ii) Mass (grams).

Gravimetric Stoichiometry refers to stoichiometry involving the measurement of gravity or mass. It's the procedure for calculating the masses of reactants or products in a chemical reaction.

Practical Importance of stoichiometry:

1.	The carburetor in a car controls the gasoline to air mixture. Proper proportions of gasoline and air are necessary for maximum power and gas mileage.

2.	Cooking food involves chemical reactions and the proper proportions are necessary in order to produce a complete reaction.. An excess of one substance (reactant) may adversely affect the cooked product.

3.	Antacid tablets may be harmful if taken in excess. If too many tablets are taken, too much stomach acid is destroyed and proper digestion can not occur.

Law of Conservation of Mass : In any chemical reaction, matter is neither created or destroyed. (ie.) The mass of the reactants equals the mass of the products.

Show that the total mass on the reactant side of the given equation is equal to total mass on the products side.

4 Fe(s)	+	3 O2(g)	2 Fe2O3(s)

Reactants:

4 mol Fe = 4 x 55.85 g/mol = 223.4 g

3 mol O2 = 3 x ( 2 x 16.00 g/mol) = 96 g

Products:

2 mol Fe2O3 = 2 [ (2 x 55.85 g/mol ) + ( 3 x 16.00 g/mol)]

= 2 [ 111.7 g + 48.00 g ]

= 2 [ 159.7 g ]

= 319.4 g

Now,	4 Fe(s)	+	3 O2(g)	2 Fe2O3(s)

223.4 g +	96.00 g =	319.4 g

Hence, mass is conserved.

Antoine Laurent Lavoisier (1743 - 1794) was known as the father of modern chemistry. His emphasis was on quantitative experimentation (such as the quantity of matter in a substance, etc.).

Gravimetric Stoichiometry (Important formulas to recal)l:

Mole to Mole Stoichiometry :

Example 1: Henry Cavendish proved in 1781 that water was the only product resulting from the combustion of hydrogen. How many moles of hydrogen are required to react exactly with 4.13 mol of oxygen?

Step 1:	Write a balanced chemical equation for the reaction. Identify the required and given.

2 H2(g) + O2(g)	2 H2O(g)

?	4.13 mol

Step 2 :	Use the mole ratio from the balanced equation to calculate the number of moles of the required substance.

There are two moles of hydrogen to every one mole of oxygen.

Therefore, nR = x mole ratio (MR)

= 4.13 mol O2 x 2 mol H2 = 8.26 mol H2

1 mol O2

Example 2: Determine the number of moles of natural gas(methane, CH4) which must undergo combustion to produce 1.62 mol of water vapor.

CH4(g)	+	2 O2(g)	CO2(g) +	2 H2O(g)

?	(1.62 mol )

= 1.62 mol H2O x

= 0.810 mol CH4

Mole to Mass Stoichiometry :

Example 3:	Zinc ore (ZnS) is converted into an oxide by roasting before it is reduced into the pure metal. What mass of zinc sulfide could react with 21.0 mol of oxygen to produce zinc oxide and sulfur dioxide?

Step 1:	Write a balanced chemical equation. Identify the required and given.

2 ZnS(s) + 3 O2	2 ZnO(s) + 2 SO2(g)

?	21.0 mol

Step 2:	Use the mole ratio from the balanced equation to determine the number of moles of the required substance.

n ZnS	= 21.0 mol O2 x = 14.0 mol ZnS

Step 3:	Determine the mass of the required substance.

m = n M

= 14.0 mol x 97.44 g/mol

= 1.36 x 103 g

= 1.36 kg

Mass to Mole Stoichiometry :

Example 4: Phosphoric acid is produced by reacting rock phosphorus with sulfuric acid. Determine the number of moles of sulfuric acid required to react with 62.0 g of calcium phosphate.

Step 1:	Write a balanced chemical equation. Identify the required and given.

3 H2SO4(aq)	+	Ca3(PO4)2(s)	2 H3PO4(aq) + 3 CaSO4(s)

?	62.0 g

Step 2:	Convert mass of given to moles of given.

Step 3:	Use the mole ratio from the balanced equation to calculate the number of moles of the required substance.

= 0.600 mol H2SO4

Mass to Mass Stoichiometry :

Example 5: What mass of calcium carbonate must react with 70.5 g of aluminum silicate to produce a cement mixture of calcium oxide, carbon dioxide, aluminum oxide, and calcium silicate?

Step 1:	Write the balanced chemical equation. Identify the required and given.

4 CaCO3(s) + Al2(SiO3)3(s) CaO(s) + 4 CO2(g) + Al2 O3(s) + 3 Ca SiO3(s)

?	70.5 g

Step 2:	Convert mass of given to moles of given.

Step 3:	Determine the number of moles of the required substance.

= 0.250 mol Al2(Si O3)3 x 4 mol CaCO3

1 mol Al2(Si O3)3

= 1.00 mol CaCO3

Step 4:	Determine the mass of the required substance.

m = n M	= 1.00 mol x 100.09 g/mol

= 100 g

Solution Chemistry :

Reference — Nelson : Chapter 5 : p. 166 - 195 ; Chapter 4 : p. 151 - 152.

What is a Solution?

A solution is a homogeneous (one phase) mixture. A solution forms when two substances are mixed together so evenly that they appear to be a single phase. The components of a solution are the solute - the substance that dissolves, and the solvent - the substance in which the solute is dissolved. Any solid, liquid, or gas which is evenly distributed throughout another solid, liquid or gas is said to be in solution. There are nine possible combinations of these states of matter.

Of the nine possible types of solutions, you are probably most familiar with those in which the solvent is a liquid, especially those in which the solvent is water.

Aqueous or water-based solutions are important in living and nonliving systems. Your body is about 60% water. Dissolved in that water are numerous solutes ranging from simple ionic species like chlorides, carbonates, and phosphates to complex molecular compounds and acids. For example, the chemical reactions which join amino acids together to form proteins in your cells, occur in water.

In industrial processes like paper making, ore processing, and chemical manufacturing, water is the carrier for atoms, ions and molecules that are undergoing chemical change to form new and different substances. For example, gold ions are dissolved in water before they are converted to gold atoms in a single replacement reaction with zinc metal.

Summary : Possible Types of Solutions.

solid in liquid

(eg.) brass	solid in liquid

(eg.) sugar water	solid in gas

(eg.) mothball in air
liquid in solid

(eg.) dental amalgam	liquid in liquid

(eg.) ethanol in water	liquid in gas

(eg.) water in air
gas in solid

(eg.) hydrogen in palladium	gas in liquid

(eg.) O2 in water	gas in gas

(eg.) oxygen in nitrogen

Solubility and Miscibility :

Solubility refers to the maximum amount of a solute that can be dissolved in an amount of solvent under specific temperature and pressure conditions. A substance that cannot be dissolved in another (or does so to a very limited extent) is said to be insoluble.

Miscibility refers to the ability of a liquid to dissolve in another in all proportions. Alcohols like methanol and ethanol are miscible with water. There is no limit to the amount of these alcohols that can be dissolved in water - they dissolve in all proportions. When the amount of one liquid exceeds the other, their roles reverse. For example if you add alcohol to water, alcohol is said to be dissolved in water; however, if you add alcohol to the point where its volume is greater than the volume of water, then water becomes the solute and alcohol the solvent. When a liquid does not dissolve in another to any extent, the liquids are said to be immiscible. Oil and water do not mix and are immiscible.

Formation of an Aqueous Solution :

Aqueous solutions are those in which the solvent is water; such as sugar (solute) dissolving in water (solvent) or salt (solute) dissolving in water (solvent). They form in at least three ways depending on the nature of the solute: 1) by molecular solvation ; 2) by dissociation or 3) by ionization. Each one will be discussed a below in behaviour of substances in solution.

Classifying Solutions :

1. Solutions can be classified as electrolytes and nonelectrolytes. Electrolytic solutions will conduct electricity. Nonelectrolytic solutions do not conduct electricity. An electric current involves the movement of charged particles. Solutions containing ions will conduct electricity.

The diagnostic test of using a conductivity tester can be used to tell if the solution conducts electricity when a light on the tester glows or the needle on the meter moves.

Household solutions (aq) such as fruit juices and cleaning solutions contain electrolytes (ie.) they conduct.

2. Acidic, Basic, or Neutral Solutions.

Diagnostic Test: Litmus Paper test

Type of solute	Effect on litmus paper

Acidic	turns blue litmus paper red

Basic	turns red litmus paper blue

Neutral	no change in colour

Four Possible Solutions :	1. Neutral Molecular, Non-electrolyte : CH3OH(aq)

2. Acidic, Electrolyte : HBr(aq)

3. Neutral Ionic, Electrolyte : (NH4)2SO4(aq)

4. Basic, Electrolyte : LiOH(aq)

Summary of Two Important Diagnostic Tests:

1.	Litmus test

- blue to red - acid ; therefore H+ present.

- red to blue - base ; therefore OH- present.

2.	Conductivity test

- conducts (acid or ionic compound) ( ie.) Electrolytes.

- doesn't conduct (neutral molecular) (ie.) Non-electrolytes.

Behaviour of substances in solution :

Have you ever noticed that certain substances like table salt dissolve in water easily while other like caulk do not? Different substances have different solubilities. Arrhenius (1887) hypothesized that particles of a substance, when dissolving, separate from each other and disperse into the solution. (See Figures 5.5 and 5.6 on p.169.)

Non-electrolytes will disperse electrically neutral particles in solution; (they do not conduct). They include neutral molecular substances.

eg. C12H22O11(s) C12H22O11(aq) (p.169)

Individual neutral sucrose molecules separate from each other and are surrounded by water molecules.( They do not break apart into individual atoms or ions). This process is called dispersion. It can also be referred to as molecular solvation.

Molecular solvation is a process whereby water molecules attract, surround and disperse solute molecules. The key term in this description is "attract". The potential solute and solvent molecules are more likely to form a solution if they have some common structural features. This idea is summarized in the generalization "like dissolves like" which will make more sense to you after you complete the unit on chemical bonding. For now it is sufficient to know that most molecular solutes do not undergo any change in their chemical composition or in the way atoms are held together in molecules when they dissolve.

Electrolytes will disperse electrically charged particles in solution; (they conduct). They include acids and ionic compounds (neutral ionic and bases).

IONIC COMPOUNDS :

When table salt dissolves in water the NaCl molecule separates into Na+ and Cl- ions. (Fig.5.6 p.169). This process of separating ions in an ionic compound is called dissociation.

Examples :	1)	NaCl(s) Na+(aq) + Cl-(aq)

(Individual ions, Na+ and Cl- , are surrounded by water molecules).

2)	Ba(OH)2(s) Ba2+(aq) + 2 OH-(aq)

ACIDS:

Acids are molecular compounds and therefore have neutral molecules. How do they become charged? When acids are placed in water, the acid separates into individual molecules. The individual molecules become ionized by the water and form positive H+ and negative ions.

Ionization is the reaction of neutral atoms or molecules of molecular compounds to form ions.

Example : HCl(g)	H+(aq) + Cl-(aq)

(Hydrogen chloride gas dissolving in water to form hydrochloric acid)

Acids are defined theoretically as species that ionize in water to produce hydrogen ions and negative ions (anions). Empirically, they are defined as species that turn blue litmus red, neutralize bases, and react with active metals and carbonates to produce gases. Acids form electrically conductive solutions; therefore they are electrolytes. However, only a few of them are strong electrolytes. When certain acids are added to water, only some of the molecules separate into hydrogen ions (the cations) and anions. The remainder of the acid molecules remain intact (do not form ions) and therefore cannot carry electrical current. Solutions containing some ionized some unionized molecules are termed weakly electrolytic. Solutions in which all the acid molecules ionize are termed strongly electrolytic.

In Summary :

Acidic solutions vary in their ability to conduct electricity. This difference is due to the degree in which ionization occurs in water.

1.	Strong Acids (SA) are very good conductors and almost all of the molecules are ionized in the solution.

(eg.). HCl(g) H+(aq) + Cl-(aq)

Six strong acids : HBr, HI, H2SO4, HClO4, HNO3, HCl .

2.	Weak Acids (WA) are fair to poor conductors. Their degree of ionization varies.

(eg.) CH3COOH(l) CH3COOH(aq)

Stays intact

Or: CH3COO-(aq) + H+(aq)

Some molecules are ionized.

Examples of the ways of forming solutions :

1.	NaF(s)	Na+ + F-(aq)	(Dissociation)

2.	(NH4)2SO4(s)	2 NH4+(aq) +	SO42-(aq)	(Dissociation)

3.	HBr(g)	H+(aq) + Br -	(Ionization)

4.	C12 H22O11(s) C12H22O11(aq)	(Dispersion or Molecular Solvation)

Solubility of Substances in Water : (Tables 5.2 and 5.5 )

1.	Elements have usually low solubility; (Some exceptions are Cl2, Br2 , O2).

2.	Ionic Compounds are from low to high solubility. (See Solubility Table).

3.	Most Acids have high solubility (if strong); weak acids have low solubility.

4.	Molecular Compounds such as hydrocarbons have low solubility; whereas alcohols, sugars, and ammonia have high solubility.

See Table 5.2 (p.172) for major entities present in water environment. Also check naming of acids as derived from the negative ions;(p. 173).

Points to consider:

1.	If high solubility, then it is aqueous (aq).

2.	If also ionic, then the compound dissociates into separate ions.

NaBr(s) + water	Na+(aq) , Br -(aq) , H2O(l)

3.	If molecular, then the compound disperses into separate molecules.

CH3OH(l) + water CH3OH(aq), H2O(l)

4.	If a strong acid, then the substance ionizes.

5.	If a weak acid, then the molecules disperses.

6.	If low solubility: elements such as Zn(s) or O2(g) will remain the same in H2O(l). Some ionic compounds (check solubility table) will remain as solids in water.

Ca3(PO4)2(s) + water	Ca3(PO4)2(s) , H2O(l)

7.	Remember that high solubility will form an aqueous solution; while low Solubility will form a precipitate.

Identifying Ions : The solubility table can be used in another way. Let's say you wanted to see if a solution contains hydroxide ions and sulfate ions. What tests can you perform to identify these ions? Well, according to the solubility table, at least two ions will form a precipitate with sulfate but not with hydroxide. Once sulfate has been precipitated out of the solution, you can separate the precipitate and the solution by filtration and try some tests to identify the hydroxide ion.

Solubility — Types of solutions : (Text : p. 187 - 192)

Solubility refers to the concentration of solute in a saturated solution, at a given temperature. A saturated solution is a solution in which no more solute will dissolve at a given temperature. If the temperature and pressure conditions change, so does the solubility of the solute. For example, 100.0 mL of water at 25° C dissolves 36.2 g of sodium chloride. Any additional sodium chloride settles to the bottom of the container. Another example would be the solubility of Na2SO4 in water is 4.67 g/100 mL at 0 oC. (NOTE : Solubility is temperature and pressure dependent.)

Other solution types possible are unsaturated solutions and supersaturated solutions. A solution that contains less than the maximum amount of solute at specific temperature and pressure conditions is said to be unsaturated. A solution that contains more than the maximum amount of solute at specific temperature and pressure conditions is a supersaturated solution. Supersaturated solutions are produced by heating a saturated solution to dissolve more solute and then cooling the solution down to a lower temperature. These solutions are not stable and crystals eventually form in them. This is the method used to grow perfect ionic crystals.

Crystallization is the removal of the solvent from a solution to yield the solute. When water evaporates from a saturated solution, solids crystallize out of the solution.

Factors that Effect Solubility: Generally, increasing the temperature of the solvent increases the solubility of solid solutes (there are exceptions) and decreases the solubility of gaseous solutes. Pressure changes have little or no effect on solid solutes; however, an increase in pressure increases the amount of a gas that can be dissolved in a solvent.

Solubility Rules (Generalizations) :

Temperature:

A) Solids (in general): The higher the temperature, the greater the solubility of solids in water.

B) Gases (in general): The higher the temperature, the lower the solubility in water.

Pressure:

A) Solids and liquids - pressure has no effect on the solubility of solids and liquids.

B) Gases - the greater the pressure, the higher the solubility of gases in water ( and vice versa)

Miscibility:

A) Liquids such as methanol and ethanol which dissolve in water, in any proportion, are miscible with water.

B) Liquids ,such as oil, which do not dissolve in water are immiscible with water.

Elements:

Most elements have low solubility in water; exceptions are some halogens such as Cl2 and Br2 and O2 .

Dynamic Equilibrium :

Equilibrium is a condition where opposing processes occur at the same rate. A solubility equilibrium can occur in a saturated solution that contains excess solute. Particles constantly leave the solid solute to enter the solution (dissolving) while dissolved particles return to the crystal (recrystallization). For every particle that dissolves, one recrystallizes. The rates of change are equal so the appearance of the system is static (unchanging). Equilibrium does not necessarily mean that the numbers of things in one condition or another are the same. For example, in a solution containing excess solute, the amount of dissolved solute may be greater, less than, or rarely equal to the amount of undissolved solute. The only things that have to be equal in an equilibrium are rates of change of the opposing processes.

When a solute is first added to a solvent, the rate of dissolving is much greater than the rate of recrystallization. In fact, the rate of recrystallization is probably zero. However as the solution approaches saturation levels, the rate of dissolving slows down and the rate of recrystallization increases. The point at which the rate of particles leaving the crystal equals the rate of particles returning to the crystal is called an equilibrium. It is a dynamic equilibrium because the opposing changes continue to occur, but since they occur at the same rate the system displays no visible changes.

A solubility equilibrium involving solid and dissolved road salt can be represented by an equation:

The double arrows indicate that the dissolving process is reversible and that the rates of change between the two states of CaCl2 are the same.

Here's a similar equation for an equilibrium between solid and dissolved sucrose:

RATE OF DISSOLVING = RATE OF CRYSTALLIZING

It requires a closed system where no substance can enter or leave the system. The rates of dissolving and crystallizing are equal, but the total amount of solute and solvent are not necessarily equal.

Concentration of a Solution :

Concentration is the ratio of the quantity of solute to the quantity of solution. Generally, concentrated means a high amount of solute relative to the amount of solvent and dilute means a low amount of solute relative to the amount of solvent. These terms are useful when comparisons are being made but they are not absolute. A solution that is described as dilute when compared to one solution may be described as concentrated when compared to another. In chemistry, quantitative descriptions of solutions are preferred. In SI (International System of Units), concentration is defined as the number of moles of solute dissolved in one litre of a solution. We use the term molar concentration to distinguish it from other ways of expressing concentration. The unit for molar concentration is mol/L or mol L-1 (e.g. 1.0 mol/L). The symbol M is sometimes used to represent this unit (e.g. 1.0 M).

There are many other ways to express concentration.

Method of expression	Definition	Formula	Unit
Molarity (molar concentration, C)	number of moles of solute (n) per litre of solution (v)

mol/L
Percent by Mass	mass of solute (m) per total mass of solution (m)

%g / g
Percent by Volume	volume of solute (v) per total volume of solution (v)

% mL / mL
Percent Mass by Volume	mass of solute (m) per total volume of solution (n)

% g / mL
Mole Fraction,	number of moles of solute (n) per total number of moles of solution (n)

unitless
Molality	number of moles of solute (n) per mass of solvent (m in kg)

mol / kg

Here are some other examples of expressing concentration :

1.	Percent:

(eg.)	A)	Vinegar contains 5 % acetic acid, by volume

- 5 ml of pure acetic acid dissolved in every 100 mL of the vinegar solution.

B)	6 % W/V of hydrogen peroxide ( H2 O2 )

W (weight) = 6 g of solute ( H2 O2 ) dissolved in v (volume ) = 100 mL of solution.

2.	g/L or g/100 m L :

For solutes that are solids in pure form.

(eg.).	Maximum amount of CaCl2 that will dissolve in water at 200C is

74.5g /100mL

3.	Molar Concentration, C , mol/L :

The amount (in moles) of solute in one litre of solution.

(eg.) Concentrated nitric acid C = 15.4 mol/L

4.	Parts per million (ppm):

For solutions in very dilute concentrations.

(eg.) Water in public swimming pools contain about 1 ppm of chlorine.

1 ppm = 1 g /1000 m L or 1 mg/L

Molar Concentration Calculations :

The following example illustrates typical calculations involving molar concentration.

Molar Concentration is the number of moles of solute dissolved to produce one litre of solution.

Formula:

C = concentration in mol/L

n = number of moles (of solute) in mol

v = volume of the solution in L

Sample Problems :

A)	Calculating Concentration from Moles and Volume.

What is the concentration of a salt solution when 0.350 mol of NaCl(s) is dissolved in 1.50 L of water?

C = n = 0. 350 mol = 0. 233 mol/L

v	1.50 L

B)	Calculating Concentration from Mass and Volume.

What is the concentration of a potassium hydroxide solution prepared by dissolving

4.50 g of KOH(s) in 5.00 x 102 mL of water?

First find the molar mass of KOH :	M = 56.11 g/mol

Convert mL to L : 500. mL = 0.500 L

n = m = 4.50 g = 0.0802 mol

M	56.11 g/mol

Now: C = n = 0.0802 mol = 0.160 mol/L

v	0.500 L

C)	Calculating Moles from Volume and Concentration.

Hydrogen chloride gas may be dissolved in water to prepare a solution of hydrochloric acid. Determine the number of moles in 7.42 x 10-2 L of 6.00 mol/L hydrochloric acid solution.

n = Cv = ( 7.42 x 10-2 ) ( 6.00 mol/L )

= 0.445 mol

D)	Calculating Volume from Mass and Concentration.

Sodium hydroxide, commonly known as caustic soda, has many uses in industry. What volume of 0.600 mol/L NaOH(aq) can be prepared from 4.8 g of solute, NaOH(s) ?

n = m = 4.8 g	= 0.12 mol

M	40.00 g/mol

Now: v = n = 0.12 mol = 0.20 L

C 0.600 mol/L

Volume of NaOH(aq) that can be prepared is 0.20 L.

E)	Calculating Mass from Volume and Concentration.

What mass of washing soda Na2CO3 10 H2O is necessary to make 400 mL of a

0.0500 mol/L solution?

Molar mass of Na2CO3 10 H2O is 286.19 g/mol.

n = C v = 0.0500 mol/L x 0.400 L = 0.0200 mol

m = n M = 0.0200 mol x 286.19 g/mol = 5.72 g

F)	Calculating Volume from Moles and Concentration.

Vinegar is a dilute solution of acetic acid, CH3COOH. What volume of 0.800 mol/L vinegar solution contains 1.60 mol of acetic acid?

v = n = 1.60 mol = 2.00 L

C 0.800 mol/L

Solution Preparation : Preparation of a solution from a mass of solid solute, requires these materials: a source container of the solute, a beaker, a scoopula, a balance, a full wash bottle (water), stirring rod , a funnel, a volumetric flask (they come in various sizes) and a cap

To prepare a solution, follow these steps carefully.

1.	Obtain the desired mass of solute in the beaker by: a) Finding the mass of the beaker. b) Adding the mass of the beaker and the mass of solute to be obtained together and resetting the balance to that amount. c) Transferring solute from its container to the beaker until the balance beam levels.

2.	Add less than half of the final volume of water (i.e. about 50 mL) to the solute in the beaker and stir to dissolve.

3.	Place the funnel in the neck of the volumetric flask and, using the stirring rod as a guide, pour the solution into the flask.

4.	Use the wash bottle to rinse leftover solution on the stirring rod and walls of the beaker into the flask, repeat until you are convinced that all the solute residue has been rinsed into the flask. Rinse walls of the funnel into the flask and remove.

5.	Carefully add water to the flask until the water level reaches the etched mark on the neck of the flask. You may want to use a medicine dropper to add the last few drops so that the bottom of the dip in the water level (known as the meniscus) is even with the etched mark on the neck of the flask.

6.	Stopper the flask and invert several times to ensure the mixture is homogeneous.

Molar Dilution :

Acids and sodium hydroxide are common examples of substances that are sold as concentrated solutions. They tend to be called stock solutions. There may be times when you have to dilute a stock solution to a lower molar concentration. When a solution is diluted, the number of moles of solute does not change. The addition of water to the concentrated solution simply disperses the solute among a greater number of solvent molecules. In other words, the number of moles of solute (ni) in the concentrated volume is equal to the number of moles (nf) in the diluted volume:

ni = nf .

Since the molar concentration formula, , can be rearranged to solve for moles for both

the concentrated and the dilute solutions: .

you can conclude that:

Ci = initial concentration (mol/L)

vi = initial volume (L)

Cf = final concentration (mol/L)

vf = final volume (L)

This formula is known as the dilution formula.

Some terms associated with molar dilution are concentrated and dilute. Concentrated refers to a large quantity of solute per small volume of solvent. Dilute refers to a small quantity of solute per large volume of solvent. Dilution is the process of decreasing the concentration of a solution.

Sample problem: What would be the final concentration of cleaning solution made by diluting 2.50 x 102 mL of a 0.500 mol/L NH3(aq) solution to a final volume of 9.50 x 102 mL ?

Solution:	Given:	Ci = 0.500 mol/L

vi = 2.50 x 102 mL = 0.250 L

Cf = ? ?

Vf = 9.50 x 102 mL = 0.950 L

Cf = Ci Vi

Vf

Cf = 0.500 mol/L x 0.250 L = 0.132 mol/L

0.950 L

Preparation of a Solution by Dilution :

These are the steps involved in diluting small quantities of concentrated solutions.

1.	Use a pipette or a graduated cylinder to deliver a specific volume of concentrated solution into a volumetric flask or other suitable container. (If the concentrated solution is an acid, you must already have some water in the volumetric flask/container to prevent splash back.)

A)	A delivery pipette is used for volumes such as 5.00 mL, 10.00 mL, or 25.00 mL.

B)	A graduated pipette is used when incremental volumes like 8.33 mL or 2.35 mL must be obtained.

C)	A graduated cylinder is used for larger initial volumes.

2.	Use a wash bottle to bring the volume of the diluted solution to the desired volume.

3.	Stopper and invert the volumetric flask/container several times.

Important Note: A graduated pipette has markings for a whole range of volumes (e.g. 0-10.00 mL). Delivery pipettes and volumetric flasks hold just one specific volume (e.g. 10.00 mL and 100.0 mL respectively). Beakers are generally to imprecise to measure such small volumes. Graduated cylinders are good alternatives for graduated pipettes, but are less precise for small volumes. They are best used for initial volumes of greater than 25.0 mL.

Concentration of Ions : (p.183)

Some substances produce aqueous ions when dissolved in water. Ionic compounds (high solubility) and strong acids are substances that behave in this manner. This formation occurs by dissociation in ionic compounds and by ionization in strong acids. You can calculate the concentration of a specific ion in solution using the molar concentration of the solution and a dissociation or ionization equation.

Ionization equation :

HCl(aq)	H+(aq) + Cl-(aq)

Mole ratio: 1 : 1 : 1

Molar concentration ratio will be the same as the mole ratio.

HCl(aq)	H+(aq)	Cl-(aq)

(ie). 2.4 mol/L	2.4 mol/L 2.4 mol/L

Dissociation equation :

Na2 SO4(s)	2 Na+(aq) + SO2-4(aq)

Mole ratio: 1 : 2 : 1

Therefore the molar concentration will have the same ratio:

Na2 SO4(s)	Na+(aq)	SO2-4(aq)

0.50 mol/L 1.0 mol/L 0.50 mol/L

Steps involved in finding ion concentrations:

1.	List the given information.

2.	Identify the wanted information.

3.	Write an equation for the dissociation of an ionic compound or ionization of an acid.

4.	Perform a stoichiometric calculation using a mole ratio from the equation.

5.	Communicate the answer.

Examples of Ionic Concentration Calculations :

1.	In a 0.23 mol/L of Al2(SO4)3 solution, what is the molar concentration of each ion?

Solution : Write a balanced dissociation equation.

Al2(SO4)3(s)	2 Al3+(aq) + 3 SO2-4(aq)

mole ratio:	1	: 2 : 3

Substituting the molar concentration(given value) into this ratio:

0.23 mol/L 0.46 mol/L 0.69 mol/L

2.	A solution contains 9.61 g of (NH4)2CO3 ,(ammonium carbonate), dissolved in water to form 400. mL of solution. What is the concentration of each ion in the solution?

Solution :	n = m	= 9.61 g = 0.100 mol

M	96.11 g/mol

C = n = 0.100 mol = 0.250 mol/L

v	0.400 L

Write a balanced dissociation equation.

(NH4)2CO3(s)	2 NH4+(aq)	+	CO2-3(aq)

Mole ratio :	1	: 2	:	1

Molar	0.250 mol/L	0.500 mol/L	0.250 mol/L

Concentration

3.	In an ammonium dichromate solution where the concentration of the ammonium ion is 0.0466mol/L, what is the concentration of the solute?

Solution :

(NH4)2Cr2O7(s)	2 NH4-(aq)	+	Cr2O72-(aq)

Mole ratio	1	:	2	: 1

Molar

concentration

0.0233 mol/L	:	0.0466 mol/L :	0.0233 mol/L

Solution Stoichiometry :

Single replacement and double replacement reactions typically involve aqueous species. In quantitative analyses for these reactions, the given and/or wanted information includes fluid volumes and molar concentrations. Stoichiometric calculations involving fluid volumes and molar concentrations are classified as solution stoichiometry. It involves molar concentration and volume of a solution to find the number of moles.

Formula:	C = n

v

( C is expressed in mol/L)

Concentration values can be compared in a balanced chemical equation.

4 steps:

1.	Write a balanced chemical equation and identify required and given.

2.	Convert C and v to # of moles.

C = n	, then	n = C v

v

3.	Use mole ratio:	nR = nG x MR( R/G)

4.	Find the given answer for required substance:

i)	C = n	for Concentration

v

ii)	v = n	for Volume

C

iii)	m = n M	for mass

Sample Problems :

A)	In an experiment, 20.0 mL sample of hydrochloric acid reacts completely with a 16.0 mL sample of magnesium hydroxide. The magnesium hydroxide has a concentration of 0.250 mol/L. Calculate the concentration of hydrochloric acid.

Steps:

1.	Balanced chemical equation.

2 HCl(aq)	+	Mg(OH)2(aq)	MgCl2(aq)	+	2 HOH(l)

(R)	(G)

2.	Convert C and v to the # of moles

nG = C v = 0.250 mol/L x 0.0160 L

nG = 0.00400 mol or 4.00 x 10-3 mol

3.	nR = nG x R/G(mole ratio)

= 4.00 x 10-3 mol x 2

1

= 8.00 x 10-3 mol

4.	C = n = 8.00 x 10 -3 mol = 0.400 mol/L

v	0.0200 mol

The concentration of hydrochloric acid is 0.400 mol/L.

B)	If 200 mL of 0.100 mol/L AgNO3 completely reacts with copper, what mass of silver will be produced?

Steps :

1.	Cu(s) +	2 AgNO3(aq)	2 Ag(s) +	CuNO3(aq)

2.	Now:	= C v	= 0.100 mol/L x 0.200 L = 0.0200 mol

3.	Use the mole ratio from the balanced equation to calculate the # of moles of Ag.

= 0.0200 mol x 2

2

= 0.0200 mol

4.	Calculate the mass of Ag produced.

mAg = n M

= 0.0200 mol x 107.87 g/mol

= 2.16 g

The mass of the silver that should be produced is 2.16 g.

C)	What volume of a 0.25 mol/L Pb(NO3)2 solution would have to be used to react with sufficient KI solution to produce 4.61 g of precipitate?

Steps :

1.	Balanced chemical equation:

Pb(NO3)2(aq)	+ 2 KI(aq)	PbI2(s) + 2KNO3(aq)

2.

3.

4.

The volume of 0.25 mol/L Pb(NO3)2 solution required is 0.040 L (or 40. mL).

GAS STOICHIOMETRY :

Have you noticed that many chemical reactions involve species in the gas phase? Amounts of gases are expressed in units of volume like mL, L and kL. Stoichiometry that involves the use of gas volumes is termed gas stoichiometry. The problem solving approach in gas stoichiometry is basically the same as the method you used for problems in gravimetric stoichiometry; that is, the mole ratio is the connection between the given and wanted quantities. The only significant difference is the use of molar volume values instead of molar mass in the conversion steps.

Mass to Volume Stoichiometry

Many chemical systems involve species in two or more different phases. For example, a reaction between sodium metal and water produces a gas and an aqueous solution. Stoichiometry for these chemical systems requires conversion factors from gravimetric and gas stoichiometry.

Volume to Mass Stoichiometry

Rusting is an example of a reaction in which a gas reacts with a solid. The next example illustrates how to predict a mass amount given a gas volume.