What Does One Mole of H2o Correspond to?

The Mole

Avogadro'south Number

  • Define mole.
  • Ascertain Avogadro's number.

It is often convenient to have a unit for large amounts, such as a mole

Is there an easier fashion to load this truck?

When the atmospheric condition is nice, many people begin to piece of work on their yards and homes. For many projects, sand is needed as a foundation for a walk or to add together to other materials. You lot could order up xx million grains of sand and accept people really stare at you. Yous could social club by the pound, but that takes a lot of time weighing out. The best bet is to guild by the one thousand, significant a cubic yard. The loader can easily scoop upwardly what y'all need and put it direct in your truck.

Avogadro's Number

It certainly is easy to count bananas or to count elephants (as long as you stay out of their way). However, you lot would exist counting grains of sugar from your sugar canister for a long, long time. Atoms and molecules are extremely small – far, far smaller than grains of sugar. Counting atoms or molecules is not only unwise, it is absolutely impossible. Ane drib of water contains nigh 10 22 molecules of water. If you lot counted 10 molecules every 2d for l years without stopping you would have counted only 1.6 × 10 ten molecules. Put some other manner, at that counting rate, information technology would accept yous over 30 trillion years to count the water molecules in i tiny drib.

Chemists needed a name that can stand for a very large number of items. Amedeo Avogadro (1776 – 1856), an Italian scientist, provided but such a number. He is responsible for the counting unit of measure chosen the mole. A mole (mol) is the amount of a substance that contains 6.02 × 10 23 representative particles of that substance. The mole is the SI unit for amount of a substance.  Merely similar the dozen and the gross, information technology is a name that stands for a number. There are therefore 6.02 × x 23 water molecules in a mole of water molecules. There likewise would be 6.02 × ten 23 bananas in a mole of bananas, if such a huge number of bananas ever existed.

Portrait of Amedeo Avogadro

Effigy 10.1

Italian scientist Amedeo Avogadro, whose piece of work led to the concept of the mole every bit a counting unit in chemistry.

The number 6.02 × 10 23 is chosen Avogadro's number , the number of representative particles in a mole. Information technology is an experimentally determined number. A representative particle is the smallest unit in which a substance naturally exists. For the bulk of elements, the representative particle is the atom. Iron, carbon, and helium consist of iron atoms, carbon atoms, and helium atoms, respectively. Seven elements exist in nature as diatomic molecules and they are H 2 , N 2 , O 2 , F ii , Cl ii , Br 2 , and I two . The representative particle for these elements is the molecule. Likewise, all molecular compounds such as H 2 O and CO 2 exist as molecules and so the molecule is their representative particle.  For ionic compounds such every bit NaCl and Ca(NO 3 ) two , the representative particle is the formula unit. A mole of whatever substance contains Avogadro's number (6.02 × x 23 ) of representative particles.

The animal mole is very different from the unit mole

Figure 10.2

The animal mole is very different than the counting unit of the mole. Chemists nonetheless take adopted the mole equally their unofficial mascot. National Mole Day is a commemoration of chemical science that occurs on Oct 23rd (10/23) of each year.

Summary

  • A mole of any substance contains Avogadro'south number (half dozen.02 × 10 23 ) of representative particles.

Practice

Questions

Use the link beneath to reply the following questions:

http://www.scientificamerican.com/article.cfm?id=how-was-avogadros-number

  1. What was Avogadro's hypothesis?
  2. Who first calculated this number?
  3. Who coined the term "Avogadro's number"?
  4. What contribution did Robert Millikan make to the determination for the value for the number?

Review

Questions

  1. What is the SI unit for amount of a substance?
  2. What is the representative particle for an element?
  3. The formula unit is the representative particle for what?
  • Avogadro'due south number: The number of representative particles in a mole, 6.02 × x 23 .
  • mole (mol): The amount of a substance that contains 6.02 × 10 23 representative particles of that substance.
  • representative particle : The smallest unit in which a substance naturally exists.

Conversions Between Moles and Atoms

  • Perform calculations involving conversions between number of moles and number of atoms or molecules.

Using moles allows us to avoid superscripts and subscripts

Big numbers or picayune numbers?

Do you detest to blazon subscripts and superscripts? Even with a good discussion-processing plan, having to click on an icon to get a superscript and so remembering to click off later on y'all type the number can be a real hassle.  If nosotros did not know about moles and just knew nearly numbers of atoms or molecules (those large numbers that crave lots of superscripts), life would exist much more complicated and we would brand many more typing errors.

Conversions Between Moles and Atoms

Conversions Between Moles and Number of Particles

Using our unit of measurement conversion techniques, nosotros tin use the mole label to convert back and forth between the number of particles and moles.

Sample Problem 1: Converting Number of Particles to Moles

The chemical element carbon exists in ii primary forms: graphite and diamond.  How many moles of carbon atoms is four.72 × x 24 atoms of carbon?

Stride 1: List the known quantities and program the problem.

Known

  • number of C atoms = 4.72 × 10 24
  • 1 mole = half dozen.02 × 10 23 atoms

Unknown

  • 4.72 × 10 24 = ? mol C

One conversion factor will allow u.s. to convert from the number of C atoms to moles of C atoms.

Stride 2: Calculate.

4.72 times 10^{24} text{atoms C} times frac{1  text{mol C}}{6.02 times 10^{23} text{atoms C}}=7.84  text{mol C}

Stride 3: Think about your outcome.

The given number of carbon atoms was greater than Avogadro'due south number, and then the number of moles of C atoms is greater than 1 mole.  Since Avogadro's number is a measured quantity with three significant figures, the result of the calculation is rounded to three pregnant figures.

Suppose that y'all wanted to know how many hydrogen atoms were in a mole of h2o molecules.  Beginning, you lot would need to know the chemic formula for water, which is H ii O.  There are 2 atoms of hydrogen in each molecule of water.  How many atoms of hydrogen would in that location be in ii water molecules?  There would be ii × 2 = four hydrogen atoms. How near in a dozen?  In that case a dozen is 12 so 12 × 2 = 24 hydrogen atoms in a dozen h2o molecules.  To get the answers, (iv and 24) you had to multiply the given number of molecules by two atoms of hydrogen per molecule.  So to notice the number of hydrogen atoms in a mole of water molecules, the trouble could be solved using conversion factors.

1  text{mol} text{H}_2text{O} times frac{6.02 times 10^{23} text{molecules} text{H}_2text{O}}{1  text{mol}  text{H}_2text{O}} times frac{2  text{atoms H}}{1  text{molecule} text{H}_2text{O}}=1.20 times 10^{24}  text{atoms H}

The outset conversion factor converts from moles of particles to the number of particles. The second conversion factor reflects the number of atoms contained within each molecule.

Two water molecules

Figure 10.3

2 water molecules contain iv hydrogen atoms and 2 oxygen atoms. A mole of water molecules contains two moles of hydrogen atoms and 1 mole of oxygen atoms.

Sample Trouble 2: Atoms, Molecules, and Moles

Sulfuric acid has the chemical formula H 2 SO 4 .  A certain quantity of sulfuric acid contains 4.89 × 10 25 atoms of oxygen.  How many moles of sulfuric acid is the sample?

Step i: List the known quantities and plan the problem.

Known

  • iv.89 × x 25 = O atoms
  • i mole = 6.02 × 10 23 molecules H 2 SO 4

Unknown

  • mol of H 2 So 4 molecules

Two conversion factors will be used.  Showtime, catechumen atoms of oxygen to molecules of sulfuric acrid.  So, convert molecules of sulfuric acid to moles of sulfuric acid.

Step 2: Calculate.

4.89 times 10^{25} text{atoms O} times frac{1  text{molecule} text{H}_2text{SO}_4}{4  text{atoms O}} times frac{1  text{mol} text{H}_2text{SO}_4}{6.02 times 10^{23} text{molecules} text{H}_2text{SO}4}=20.3  text{mol} text{H}_2text{SO}_4

Step 3: Recall most your result.

The original number of oxygen atoms was about 80 times larger than Avogadro's number.  Since each sulfuric acid molecule contains 4 oxygen atoms, in that location are nigh 20 moles of sulfuric acrid molecules.

Summary

  • Methods are described for conversions between moles, atoms, and molecules.

Practice

Read the relevant portions of the post-obit commodity and practice problems 3, 5, 9, 13, and 18.  Do not worry about the calculations involving conversions dealing with molar mass (that will come adjacent).

http://kinesthesia.rcc.edu/freitas/1AWorksheets/14GramsToMolesToMolecules.pdf

Review

Questions

  1. What important number do we need to know to do these conversions?
  2. I want to convert atoms to moles. My friend tells me to multiply the number of atoms by 6.02 × 10 23 atoms/mole.  Is this correct?
  3. Why should I know the formula for a molecule in order to summate the number of moles of one of the atoms?

Molar Mass

  • Ascertain molar mass.
  • Perform calculations involving tooth mass.

Pile of potassium dichromate

When creating a solution, how do I know how much of each substance to put in?

I want to make a solution that contains 1.8 moles of potassium dichromate.  I don't have a residuum calibrated in molecules, simply I do have ane calibrated in grams.  If I know the relationship between moles and the number of grams in a mole, I can employ my remainder to measure out out the needed corporeality of textile.

Molar Mass

Molar mass is defined equally the mass of one mole of representative particles of a substance. By looking at a periodic table, we can conclude that the molar mass of lithium is half-dozen.94 1000, the molar mass of zinc is 65.38 g, and the molar mass of gold is 196.97 g.  Each of these quantities contains half dozen.02 × x 23 atoms of that particular element.  The units for molar mass are grams per mole or g/mol.

Molar Masses of Compounds

A molecular formula of the compound carbon dioxide is CO two .  Ane molecule of carbon dioxide consists of 1 atom of carbon and 2 atoms of oxygen.  We can calculate the mass of i molecule of carbon dioxide by adding together the masses of 1 atom of carbon and two atoms of oxygen.

12.01  text{amu} + 2(16.00  text{amu})=44.01  text{amu}

The molecular mass of a compound is the mass of one molecule of that compound.  The molecular mass of carbon dioxide is 44.01 amu.

The molar mass of whatever compound is the mass in grams of one mole of that compound.  One mole of carbon dioxide molecules has a mass of 44.01 g, while one mole of sodium sulfide formula units has a mass of 78.04 g.  The tooth masses are 44.01 g/mol and 78.04 k/mol respectively.  In both cases, that is the mass of 6.02 × 10 23 representative particles.  The representative particle of CO ii is the molecule, while for Na 2 S, information technology is the formula unit.

Sample Trouble:  Molar Mass of  a Compound

Calcium nitrate, Ca(NO 3 ) 2 , is used as a component in fertilizer Determine the molar mass of calcium nitrate.

Step 1:  List the known and unknown quantities and programme the problem.

Known

  • formula = Ca(NO 3 ) 2
  • molar mass Ca = 40.08 1000/mol
  • tooth mass North = 14.01 g/mol
  • molar mass O = xvi.00 k/mol

Unknown

  • tooth mass Ca(NO 3 ) 2

First we demand to analyze the formula.  Since the Ca lacks a subscript, there is one Ca cantlet per formula unit.  The 2 exterior the parentheses means that there are two nitrate ions per formula unit and each nitrate ion consists of 1 nitrogen atom and three oxygen atoms.  Therefore, there are a total of 1 × 2 = 2 nitrogen atoms and three × 2 = half-dozen oxygen atoms per formula unit of measurement.  Thus, one mol of calcium nitrate contains 1 mol of Ca atoms, 2 mol of North atoms, and 6 mol of O atoms.

Step ii:  Calculate.

Use the molar masses of each atom together with the number of atoms in the formula and add together.

& 1  text{mol Ca} times frac{40.08  text{g Ca}}{1  text{mol Ca}}=40.08  text{g Ca} \& 2  text{mol N} times frac{14.01  text{g N}}{1  text{mol N}}=28.02  text{g N} \& 6  text{mol O} times frac{16.00  text{g O}}{1  text{mol O}}=96.00  text{g O} \& text{molar mass of } text{Ca(NO}_3)_2=40.08  text{g} + 28.02  text{g} + 96.00  text{g} = 164.10  text{g} / text{mol}

Summary

  • Calculations are described for the determination of tooth mass of an atom or a chemical compound.

Do

Read the material at the link below and work the problems at the end:

http://misterguch.brinkster.net/molarmass.html

Review

Questions

  1. What is the molar mass of Pb?
  2. Why do we need to include the units in our answer?
  3. I want to summate the molar mass of CaCl two .  How many moles of Cl are in 1 mole of the compound?
  4. How many moles of H are in the compound (NH 4 ) iii PO 4 ?
  • tooth mass : The mass of one mole of representative particles of a substance.
  • molecular mass : The mass of one molecule of that chemical compound.

Conversions betwixt Moles and Mass

  • Perform calculations dealing with conversions between moles and mass.

Chemical plants need to know how to convert between moles and mass

How tin can we go more than product?

Chemical manufacturing plants are always seeking to improve their processes.  One of the ways this improvement comes about is through measuring the corporeality of material produced in a reaction.  By knowing how much is made, the scientists and engineers can endeavour different ways of getting more than production at less price.

Conversions Between Moles and Mass

The molar mass of whatever substance is the mass in grams of one mole of representative particles of that substance.  The representative particles can be atoms, molecules, or formula units of ionic compounds.  This relationship is frequently used in the laboratory.  Suppose that for a certain experiment you demand 3.00 moles of calcium chloride (CaCl 2 ).  Since calcium chloride is a solid, it would exist convenient to use a rest to measure the mass that is needed.  The molar mass of CaCl ii is 110.98 g/mol.  The conversion factor that can be used is then based on the equality that ane mol = 110.98 chiliad CaCl two .  Dimensional analysis volition allow yous to summate the mass of CaCl 2 that you should measure.

3.00  text{mol}  text{CaCl}_2 times frac{110.98  text{g}  text{CaCl}_2}{1  text{mol}  text{CaCl}_2}=333  text{g}  text{CaCl}_2

When you mensurate the mass of 333 g of CaCl ii , y'all are measuring iii.00 moles of CaCl 2 .

Calcium chloride is used as a drying agent and as a road deicer

Effigy 10.4

Calcium chloride is used equally a drying amanuensis and as a road deicer.

Sample Problem: Converting Moles to Mass

Chromium metallic is used for decorative electroplating of automobile bumpers and other surfaces.  Find the mass of 0.560 moles of chromium.

Step 1: List the known quantities and plan the problem.

Known

  • molar mass of Cr = 52.00 g mol
  • 0.560 mol Cr

Unknown

  • 0.560 mol Cr = ? g

One conversion cistron volition allow us to convert from the moles of Cr to mass.

Step 2: Calculate.

0.560  text{mol Cr} times frac{52.00  text{g}  text{Cr}}{1  text{mol Cr}}=29.1  text{g Cr}

Step 3: Recall most your event.

Since the desired corporeality was slightly more than than ane half of a mole, the mass should exist slightly more than one half of the tooth mass.  The reply has three significant figures considering of the 0.560 mol.

A similar conversion factor utilizing molar mass can be used to convert from the mass of an substance to moles.  In a laboratory situation, you may perform a reaction and produce a certain corporeality of a product which tin be massed.  It will often then be necessary to make up one's mind the number of moles of the product that was formed.  The next problem illustrates this situation.

Sample Problem: Converting Mass to Moles

A certain reaction produces 2.81 thou of copper(II) hydroxide, Cu(OH) two .  Determine the number of moles produced in the reaction.

Step i: List the known quantities and plan the problem.

Known

  • mass = ii.81 g

Unknown

  • mol Cu(OH) two

One conversion cistron will allow us to catechumen from mass to moles.

Step 2: Calculate.

Beginning, it is necessary to calculate the tooth mass of Cu(OH) 2 from the molar masses of Cu, O, and H.  The molar mass is 97.57 g/mol.

2.81  text{g}  text{Cu(OH)}_2 times frac{1   text{mol}  text{Cu(OH)}_2}{97.57  text{g}  text{Cu(OH)}_2}=0.0288  text{mol}  text{Cu(OH)}_2

Step 3: Think about your result.

The relatively small mass of product formed results in a modest number of moles.

Summary

  • Calculations involving conversions between moles of a material and the mass of that material are described.

Practice

Read the material in the link beneath and work the issues at the end.

http://www.occc.edu/kmbailey/chem1115tutorials/Stoichiometry_Molar_Mass_Calculations.htm

Review

Questions

  1. Why would you want to calculate the mass of a textile?
  2. Why would you want to decide how many moles of material you produced in a reaction?
  3. You lot have 19.7 grams of a material and wonder how many moles were formed. Your friend tells you to multiply the mass by grams/mole.  Is your friend correct?

Conversions between Mass and Number of Particles

  • Perform calculations involving conversions of mass and number of particles.

Equal volumes of gas under the same conditions contain the same number of particles

How much gas is there?

Avogadro was interested in studying gases.  He theorized that equal volumes of gases under the same weather contained the aforementioned number of particles.  Other researchers studied how many gas particles were in a specific volume of gas. Somewhen, scientists were able to develop the relationship between number of particles and mass using the idea of moles.

Conversions Between Mass and Number of Particles

In "Conversions between Moles and Mass", y'all learned how to catechumen back and along betwixt moles and the number of representative particles.  At present yous have seen how to convert back and forth between moles and mass of a substance in grams.  Nosotros can combine the two types of problems into ane.  Mass and number of particles are both related to grams.  In society to convert from mass to number of particles or vice-versa, information technology volition kickoff crave a conversion to moles.

Flowchart of conversion between number of particles, moles, and mass

Effigy x.5

Conversion from number of particles to mass or from mass to number of particles requires 2 steps

Sample Problem: Converting Mass to Particles

How many molecules is 20.0 g of chlorine gas, Cl ii ?

Step ane: Listing the known quantities and plan the problem.

Known

  • molar mass Cl ii = 70.90 yard/mol
  • 20.0 thousand Cl two

Unknown

  • number of molecules of Cl 2

Utilise two conversion factors.  The first converts grams of Cl 2 to moles.  The 2nd converts moles of Cl 2 to the number of molecules.

Footstep 2: Calculate.

20.0  text{g}  text{Cl}_2 times frac{1  text{mol}  text{Cl}_2}{70.90  text{g}  text{Cl}_2} times frac{6.02 times 10^{23}  text{molecules Cl}_2}{1  text{mol}  text{Cl}_2}=1.70 times 10^{23}  text{molecules Cl}_2

The problem is washed using two sequent conversion factors. There is no need to explicitly calculate the moles of Cl 2 .

Pace iii: Think about your result.

Since the given mass is less than one-half of the tooth mass of chlorine, the resulting number of molecules is less than half of Avogadro's number.

Summary

  • Calculations are illustrated for conversions between mass and number of particles.

Practice

Read the material at the link below and then do practice problems on folio ix and the problem on page 17 (don't acme at the answers until you have tried the problems).

http://schools.fwps.org/decatur-old/staff/adewaraja/chemistry/curriculum%20units/chapter%2010/Lesson5/particle_mole_mass_calcu.pdf

Review

Questions

  1. Why tin't we convert directly from number of particles to grams?
  2. How many atoms of chlorine are present in the trouble above?
  3. The periodic tabular array says the atomic weight of chlorine is 35.5. Why can't I utilize that value in my calculations?

Avogadro'south Hypothesis and Molar Volume

  • Land Avogadro'southward hypothesis.
  • Ascertain standard temperature and pressure.
  • Define molar book.

Gas laws are used to determine the amount of air in a scuba tank

How exercise scuba divers know if they will run out of gas?

Knowing how much gas is available for a dive is crucial to the survival of the diver.  The tank on the diver's back is equipped with gauges to tell how much gas is present and what the pressure is.  A basic knowledge of gas beliefs allows the diver to assess how long to stay nether water without developing issues.

Avogadro's Hypothesis and Molar Volume

Volume is a 3rd mode to measure the amount of matter, after particular count and mass.  With liquids and solids, volume varies greatly depending on the density of the substance.  This is because solid and liquid particles are packed shut together with very footling space in between the particles.  However, gases are largely composed of empty space between the bodily gas particles (see Figure below ).

Gas particles are small compared to the empty space between them

Figure 10.6

Gas particles are very small compared to the big amounts of empty space between them.

In 1811, Amedeo Avogadro explained that the volumes of all gases can exist easily determined. Avogadro's hypothesis states that equal volumes of all gases at the same temperature and force per unit area incorporate equal numbers of particles.  Since the full volume that a gas occupies is made up primarily of the empty space between the particles, the actual size of the particles themselves is nearly negligible.  A given volume of a gas with small calorie-free particles such as hydrogen (H 2 ) contains the aforementioned number of particles every bit the same book of a heavy gas with large particles such as sulfur hexafluoride, SF 6 .

Gases are compressible, meaning that when put nether high pressure, the particles are forced closer to i some other.  This decreases the corporeality of empty space and reduces the book of the gas.  Gas volume is too affected by temperature.  When a gas is heated, its molecules motility faster and the gas expands.  Considering of the variation in gas volume due to pressure and temperature changes, the comparison of gas volumes must be done at ane standard temperature and pressure. Standard temperature and pressure (STP) is divers as 0°C (273.15 Grand) and 1 atm force per unit area. The molar book of a gas is the volume of one mole of a gas at STP.  At STP, ane mole (6.02 × 10 23 representative particles) of any gas occupies a volume of 22.4 L ( Figure below ).

Avogadro's hypothesis states that one mole of gas at STP occupies 22.4 liters

Figure ten.7

A mole of any gas occupies 22.4 50 at standard temperature and pressure (0°C and ane atm).

The Effigy beneath illustrates how molar volume can be seen when comparing different gases. Samples of helium (He), nitrogen (N 2 ), and methane (CH iv ) are at STP.  Each contains 1 mole or 6.02 × 10 23 particles.  However, the mass of each gas is different and corresponds to the tooth mass of that gas: iv.00 g/mol for He, 28.0 grand/mol for North 2 , and 16.0 grand/mol for CH 4 .

Avogadro's hypothesis means that  one mole of neon, nitrogen, or methane occupies 22.4 liters at STP

Figure 10.8

Avogadro'south hypothesis states that equal volumes of any gas at the aforementioned temperature and force per unit area incorporate the aforementioned number of particles. At standard temperature and pressure, 1 mole of any gas occupies 22.four 50.

Summary

  • Equal volumes of gases at the same weather contain the same number of particles.
  • Standard temperature and pressure are defined.

Practice

Questions

Use the link below to answer the following questions:

http://chemed.chem.purdue.edu/demos/main_pages/4.6.html

  1. What was the volume of each gas that was weighed?
  2. What did the experiment find?
  3. What was the relationship betwixt gas weight and molecular weight?

Review

Questions

  1. What practice nosotros know most the space actually taken upwardly by a gas?
  2. Why practice we demand to exercise all our comparisons at the same temperature and force per unit area?
  3. How can we use this information?
  • Avogadro'southward hypothesis: Equal volumes of all gases at the same temperature and pressure incorporate equal numbers of particles.
  • molar volume: The volume of one mole of a gas at STP.
  • standard temperature and pressure (STP): 0°C (273.15 M) and 1 atm pressure.

Conversions between Moles and Gas Volume

  • Make conversions between the volume of a gas and the number of moles of that gas.

Gas volumes can be used to determine the moles of gas in these tanks

How tin can you tell how much gas is in these containers?

Small gas tanks are often used to supply gases for chemistry reactions.  A gas gauge will give some information well-nigh how much is in the tank, only quantitative estimates are needed and so the reaction will exist able to proceed to completion.  Knowing how to calculate needed parameters for gases is very helpful to avoid running out too early on.

Conversions Between Moles and Gas Volume

Molar volume at STP tin be used to convert from moles to gas volume and from gas volume to moles.  The equality of i mole = 22.four 50 is the footing for the conversion factor.

Sample Problem 1: Converting Gas Volume to Moles

Many metals react with acids to produce hydrogen gas.  A certain reaction produces 86.5 L of hydrogen gas at STP.  How many moles of hydrogen were produced?

Step 1: List the known quantities and plan the problem.

Known

  • 86.5 L H ii
  • 1 mol = 22.iv L

Unknown

  • moles of H 2

Employ a conversion factor to convert from liters to moles.

Step 2: Calculate.

86.5  text{L H}_2 times frac{1  text{mol H}_2}{22.4  text{L H}_2}=3.86  text{mol H}_2

Step three: Think about your result.

The book of gas produced is nearly iv times larger than the molar volume.  The fact that the gas is hydrogen plays no role in the adding.

Sample Trouble Two: Converting Moles to Gas Volume

What volume does 4.96 moles of O 2 occupy at STP?

Stride one: List the known quantities and plan the problem.

Known

  • 4.96 moles O 2
  • 1 mol = 22.4 50

Unknown

  • book of O 2

Step 2:  Calculate.

4.96  text{moles} times   {22.4  text{liters/mole}}=111.1  text{liters}

Footstep 3: Remember nearly your result.

The volume seems correct given the number of moles.

Sample Problem 3: Converting Volume to Mass

If we know the volume of a gas sample at STP, we tin can determine how much mass is nowadays.  Presume we have 867 liters of N 2 at STP.  What is the mass of the nitrogen gas?

Step 1: List the known quantities and plan the trouble.

Known

  • 867 50 Due north 2
  • 1 mol = 22.4 L
  • tooth mass of N ii = 28.02 1000/mol

Unknown

  • mass of N 2

Step 2: Calculate.

We offset by determining the number of moles of gas present. Nosotros know that 22.four liters of a gas at STP equals ane mole, so:

867  text{litres} times frac{1  text{mole}}{22.4  text{liters}}=3.87  text{moles}

We also know the molecular weight of N two (28.0 grams/mole), so nosotros can then summate the weight of nitrogen gas in 867 liters:

38.7  text{moles} times frac{28  text{grams}}{text{mole}}=1083.6  text{grams N}_2

Step iii: Think nearly your result.

In a multi-step trouble, be certain that the units bank check.

Summary

  • Conversions between moles and book of a gas are shown.

Exercise

Piece of work the practice issues at the link beneath. Focus on conversions between volume and moles, merely try some of the others:

http://www.sciencegeek.cyberspace/Chemistry/taters/Unit4GramMoleVolume.htm

Review

Questions

  1. Why do the gases need to exist at STP?
  2. When does the identity of the gas go of import?

Gas Density

  • Brand calculations dealing with molar mass and density of a gas.

Carbon dioxide sinks in air

Why does carbon dioxide sink in air?

When we run a reaction to produce a gas, we expect it to rise into the air.  Many students take done experiments where gases such as hydrogen are formed.  The gas can exist trapped in a exam tube held upside-down over the reaction.  Carbon dioxide, on the other manus, sinks when it is released.  Carbon dioxide has a density greater that air, so information technology will not rising like these other gases would.

Gas Density

Equally you know, density is defined as the mass per unit volume of a substance.  Since gases all occupy the same volume on a per mole ground, the density of a particular gas is dependent on its molar mass.  A gas with a minor molar mass will accept a lower density than a gas with a large molar mass.  Gas densities are typically reported in one thousand/50.  Gas density can be calculated from molar mass and molar volume.

Balloons float because they contain helium, which is lighter than air

Figure x.ix

Balloons filled with helium gas float in air because the density of helium is less than the density of air.

Sample Problem One: Gas Density

What is the density of nitrogen gas at STP?

Pace 1: List the known quantities and programme the trouble.

Known

  • Due north ii = 28.02 g/mol
  • 1 mol = 22.four L

Unknown

  • density = ? g/L

Molar mass divided by molar volume yields the gas density at STP.

Pace 2: Calculate.

frac{28.02  text{g}}{1  text{mol}} times frac{1  text{mol}}{22.4  text{L}}=1.25  text{g} / text{L}

When prepare upwards with a conversion factor, the mol unit cancels, leaving g/50 as the unit in the result.

Step 3: Think about your result.

The molar mass of nitrogen is slightly larger than molar book, so the density is slightly greater than 1 g/L.

Alternatively, the tooth mass of a gas can be determined if the density of the gas at STP is known.

Sample Problem Ii: Molar Mass from Gas Density

What is the tooth mass of a gas whose density is 0.761 thousand/L at STP?

Step i: List the known quantities and plan the problem.

Known

  • Due north 2 = 28.02 g/mol
  • 1 mol = 22.4 L

Unknown

  • molar mass = ? m/Fifty

Tooth mass is equal to density multiplied past molar volume.

Step 2: Summate.

frac{0.761  text{g}}{1  text{L}} times frac{22.4  text{L}}{1  text{mol}}=17.0  text{g} / text{mol}

Stride 3: Remember about your issue.

Because the density of the gas is less than one k/Fifty, the tooth mass is less than 22.four.

Summary

  • Calculations are described showing conversions between molar mass and density for gases.

Do

Questions

Use the link beneath to answer the following questions:

http://employees.oneonta.edu/viningwj/sims/gas_density_s.html

  1. Which of the gases has the highest density?
  2. Which gas has the lowest density?
  3. Would you await nitrogen to have a higher or lower density that oxygen? Why?

Review

Questions

  1. How is density calculated?
  2. How is molar mass calculated?
  3. What would be the volume of three.five moles of a gas?

Mole Road Map

  • Perform calculations involving interconversions of mass, moles, and volume of a gas.

Chemistry road maps are similar to actual maps

How do I get from here to there?

If I want to visit the town of Manteo, Due north Carolina, out on the declension, I volition need a map of how to get in that location.  I may have a printed map or I may download directions from the internet, but I demand something to get me going in the correct direction.  Chemistry road maps serve the same purpose.  How exercise I handle a certain blazon of calculation? There is a process and a set of directions to help.

Mole Road Map

Previously, we saw how the conversions between mass and number of particles required 2 steps, with moles equally the intermediate.  This concept tin can now be extended to also include gas volume at STP.  The resulting diagram is referred to every bit a mole road map (come across Figure below ).

How to convert between moles, mass, number of particles, and volume of a gas

Effigy 10.10

The mole road map shows the conversion factors needed to interconvert between mass, number of particles, and book of a gas.

The mole is at the center of whatsoever calculation involving amount of a substance.  The sample trouble beneath is i of many different problems that can exist solved using the mole route map.

Sample Trouble One:  Mole Route Map

What is the volume of 79.3 g of neon gas at STP?

Footstep 1:  List the known quantities and programme the trouble.

Known

  • Ne = 20.18 chiliad/mol
  • 1 mol = 22.4 L

Unknown

  • volume = ? Fifty

The conversion factors volition be grams → moles → gas volume.

Step 2:  Summate.

79.3  text{g Ne} times frac{1  text{mol Ne}}{20.18  text{g Ne}} times frac{22.4  text{L Ne}}{1  text{mol Ne}}=88.0  text{L Ne}

Step 3:  Retrieve almost your result.

The given mass of neon is equal to near four moles, resulting in a volume that is about four times larger than molar book.

Summary

  • An overall process is given for calculations involving moles, grams, and gas volume.

Practice

Employ the link beneath to carry out some practice calculations.  Exercise problems i, 2, and five (you can try the others if you are feeling especially dauntless):

http://www.docbrown.info/page04/4_73calcs/MVGmcTEST.htm

Review

Questions

  1. In the problem to a higher place, what is the formula weight of neon?
  2. What value is at the center of all the calculations?
  3. If we had 79.3 grams of Xe, would we expect a volume that is greater than or less than that obtained with neon?

Per centum Limerick

  • Define percent composition.
  • Perform percentage composition calculations.

Nutritional information labels can let you know the percent composition

Is there anything salubrious in this jar?

Packaged foods that you swallow typically have nutritional information provided on the label.  The characterization on a jar of peanut butter (shown above) reveals that one serving size is considered to be 32 chiliad.  The label also gives the masses of various types of compounds that are nowadays in each serving.  Ane serving contains 7 g of poly peptide, 15 chiliad of fat, and 3 g of sugar.  By calculating the fraction of poly peptide, fatty, or sugar in i serving of size of peanut butter and converting to percent values, nosotros can make up one's mind the limerick of the peanut butter on a percent by mass footing.

Percentage Limerick

Chemists ofttimes need to know what elements are nowadays in a chemical compound and in what percent.  The percent composition is the pct by mass of each chemical element in a chemical compound.  It is calculated in a similar mode that nosotros but indicated for the peanut butter.

%  text{by mass}=frac{text{mass of element}}{text{mass of compound}} times 100 %

Percent Limerick from Mass Data

The sample problem below shows the adding of the percent composition of a compound based on mass data.

Sample Trouble 1: Per centum Composition from Mass

A certain newly synthesized compound is known to comprise the elements zinc and oxygen.  When a xx.00 k sample of the sample is decomposed, sixteen.07 g of zinc remains.  Decide the percentage composition of the compound.

Step 1: List the known quantities and plan the problem.

Known

  • mass of chemical compound = 20.00 thou
  • mass of Zn = 16.07 g

Unknown

  • percent Zn = ? %
  • percent O = ? %

Subtract to find the mass of oxygen in the compound.  Divide each element'due south mass by the mass of the compound to find the percentage past mass.

Stride 2: Calculate.

text{Mass of oxygen}&=20.00  text{g} - 16.07  text{g}=3.93  text{g O} \%  text{Zn}&=frac{16.07  text{g Zn}}{20.00  text{g}} times 100 %=80.35 %  text{Zn} \%  text{O}&=frac{3.93  text{g O}}{20.00  text{g}} times 100 %=19.65 %  text{O} \

Footstep 3: Recollect near your result.

The calculations make sense considering the sum of the 2 percentages adds up to 100%.  By mass, the compound is more often than not zinc.

Pct Limerick from a Chemical Formula

The percentage composition of a compound tin can also exist determined from the formula of the compound.  The subscripts in the formula are first used to calculate the mass of each element in one mole of the chemical compound.  That is divided by the molar mass of the compound and multiplied by 100%.

%  text{by mass }=frac{text{mass of element in}  1  text{mol}}{text{molar mass of compound}}times 100 %

The percentage limerick of a given compound is ever the same every bit long every bit the compound is pure.

Sample Trouble Two: Percent Composition from Chemical Formula

Dichlorineheptoxide (Cl 2 O 7 ) is a highly reactive compound used in some organic synthesis reactions.  Calculate the percentage composition of dichlorineheptoxide.

Pace 1: List the known quantities and programme the problem.

Known

  • mass of Cl in ane mol Cl 2 O seven = 70.ninety thousand
  • mass of O in 1 mol Cl two O seven = 112.00 g
  • molar mass of Cl two O 7 = 182.ninety m/mol

Unknown

  • percentage Cl = ? %
  • percent O = ? %

Calculate the percentage past mass of each element by dividing the mass of that element in 1 mole of the compound by the molar mass of the compound and multiplying by 100%.

Step two: Calculate.

%  text{Cl}&=frac{70.90  text{g Cl}}{182.90  text{g}} times 100 %=38.76 %  text{Cl} \%  text{O}&=frac{112.00  text{g O}}{182.90  text{g}} times 100 %=61.24 %  text{O} \

Step 3: Think near your result.

The percentages add together upwardly to 100%.

Percent composition tin also be used to determine the mass of a certain element that is contained in any mass of a compound.  In the previous sample trouble, it was found that the percent limerick of dichlorineheptoxide is 38.76% Cl and 61.24% O.  Suppose that y'all needed to know the masses of chlorine and oxygen present in a 12.fifty g sample of dichlorineheptoxide.  You tin set upwardly a conversion cistron based on the percentage by mass of each element.

12.50  text{g Cl}_2 text{O}_7 times frac{38.76  text{g Cl}}{100  text{g Cl}_2text{O}_7}&=4.845  text{g Cl} \12.50  text{g Cl}_2 text{O}_7 times frac{61.24  text{g O}}{100  text{g Cl}_2text{O}_7}&=7.655  text{g O} \

The sum of the two masses is 12.50 grand, the mass of the sample size.

Summary

  • Processes are described for calculating the pct limerick of a material based on mass or on chemic composition.

Exercise

Use the link beneath to review cloth and do calculations. Read both parts of the lesson and do as many calculations as you take time for.

http://www.chemteam.info/Mole/Percent-Composition-Part1.html

Review

Questions

  1. What is the formula for calculating pct composition?
  2. What information do yous need to calculate pct composition past mass?
  3. What practise subscripts in a chemical formula tell you?
  • per centum composition: The percent by mass of each element in a compound.

Percent of Water in a Hydrate

  • Define hydrate.
  • Summate the percentage h2o in hydrate when give relevant information.

Copper sulfate changes color when hydrated

Why does the color alter?

If y'all look at a typical bottle of copper sulfate, information technology will be a blueish-light-green.  If someone tells you that copper sulfate is white, you lot won't believe them.  You lot are both correct; it just depends on the copper sulfate.  Your blue-greenish copper sulfate has several water molecules attached to information technology while your friend's copper sulfate is anhydrous (no h2o attached).  Why the difference? The h2o molecules interact with some of the d electrons in the copper ion and produce the colour.  When the water is removed, the electron configuration changes and the colour disappears.

Percentage of Water in a Hydrate

Many ionic compounds naturally contain water as part of the crystal lattice construction.  A hydrate is a compound that has i or more than water molecules jump to each formula unit.  Ionic compounds that incorporate a transition element are often highly colored.  Interestingly, it is common for the hydrated class of a chemical compound to be of a different color than the anhydrous class, which has no water in its structure.  A hydrate can usually be converted to the anhydrous compound past heating.  For instance, the anhydrous compound cobalt(II) chloride is blue, while the hydrate is a distinctive magenta color.

Anhydrous cobalt chloride is blue, while hydrated cobalt chloride is red

Figure 10.eleven

On the left is anhydrous cobalt(Ii) chloride, CoCl 2 . On the right is the hydrated form of the compound called cobalt(Two) chloride hexahydrate, CoCl 2 •6H 2 O.

The hydrated form of cobalt(II) chloride contains six water molecules in each formula unit of measurement.  The name of the chemical compound is cobalt(2) chloride hexahydrate and its formula is CoCl 2 •6H 2 O.  The formula for water is set apart at the end of the formula with a dot, followed by a coefficient that represents the number of h2o molecules per formula unit.

Information technology is useful to know the percent of water contained within a hydrate.  The sample problem below demonstrates the procedure.

Sample Trouble One: Percent of Water in a Hydrate

Find the pct h2o in cobalt(Two) chloride hexahydrate, CoCl 2 •6H 2 O.

Step 1: List the known quantities and plan the trouble.

The mass of water in the hydrate is the coefficient (6) multiplied by the tooth mass of H 2 O.  The molar mass of the hydrate is the molar mass of the CoCl 2 plus the mass of water.

Known

  • mass of H 2 O in one mol hydrate = 108.12 1000
  • molar mass of hydrate = 237.95 g/mol

Unknown

  • percentage H 2 O = ? %

Calculate the percent past mass of water by dividing the mass of H two O in i mole of the hydrate by the molar mass of the hydrate and multiplying by 100%.

Step two: Calculate.

%  text{H}_2text{O}=frac{108.12  text{g H}_2 text{O}}{237.95  text{g}} times 100 %=45.44 %  text{H}_2text{O}

Footstep three: Think about your result.

Most half of the mass of the hydrate is composed of h2o molecules inside the crystal.

Summary

  • The process of computing the per centum h2o in a hydrate is described.

Practice

Employ the following link to practice calculating pct h2o in a hydrate:

http://www.sd84.k12.id.us/shs/departments/science/martz/2007_ssem2/Chemical science/hydrate.htm

Review

Questions

  1. What is a hydrate?
  2. How can you convert a hydrate to an anhydrous compound?
  3. What does hexahydrate mean?
  • Anhydrous: Without water.
  • Hydrate: A chemical compound that has ane or more h2o molecules bound to each formula unit

Determining Empirical Formulas

  • Define empirical formula.
  • Calculate the empirical formula for a compound when given the elemental analysis of the compound.

Scientist trying to determine the formula of a compound in the early days of chemistry

What is occuring in this flick?

In the early days of chemistry, at that place were few tools for the detailed study of compounds.  Much of the data regarding the composition of compounds came from the elemental analysis of inorganic materials.  The "new" field of organic chemistry (the study of carbon compounds) faced the challenge of not existence able to characterize a compound completely.  The relative amounts of elements could be adamant, but then many of these materials had carbon, hydrogen, oxygen, and peradventure nitrogen in elementary ratios.  Nosotros did non know exactly how many of these atoms were actually in a specific molecule.

Determining Empirical Formulas

An empirical formula is 1 that shows the lowest whole-number ratio of the elements in a compound.  Because the structure of ionic compounds is an extended iii-dimensional network of positive and negative ions, all formulas of ionic compounds are empirical.  Nonetheless, we tin can also consider the empirical formula of a molecular chemical compound.  Ethene is a small-scale hydrocarbon compound with the formula C two H 4 (see Figure below ).  While C 2 H 4 is its molecular formula and represents its true molecular structure, information technology has an empirical formula of CH 2 .  The simplest ratio of carbon to hydrogen in ethene is 1:2.  At that place are two ways to view that ratio.  Considering one molecule of ethene, the ratio is one carbon atom for every 2 atoms of hydrogen.  Considering one mole of ethene, the ratio is ane mole of carbon for every ii moles of hydrogen.  Then the subscripts in a formula represent the mole ratio of the elements in that formula.

Ball and stick model of ethene

Figure ten.12

Ball-and-stick model of ethene, C ii H 4 .

In a procedure chosen elemental analysis , an unknown compound can be analyzed in the laboratory in club to decide the percentages of each element independent within it.  These percentages can be transformed into the mole ratio of the elements, which leads to the empirical formula.  The steps to be taken are outlined below.

  1. Assume a 100 g sample of the compound and so that the given percentages can be directly converted into grams.
  2. Use each element's molar mass to convert the grams of each element to moles.
  3. In order to find a whole-number ratio, separate the moles of each element by whichever of the moles from footstep 2 is the smallest.
  4. If all the moles at this point are whole numbers (or very shut), the empirical formula can be written with the moles as the subscript of each element.
  5. In some cases, one or more than of the moles calculated in pace three will not exist whole numbers.  Multiply each of the moles by the smallest whole number that will convert each into a whole number.  Write the empirical formula.

Sample Problem I: Determining the Empirical Formula of a Compound

A compound of atomic number 26 and oxygen is analyzed and found to incorporate 69.94% iron and 30.06% oxygen.  Find the empirical formula of the compound.

Step one: Listing the known quantities and plan the trouble.

Known

  • % of Atomic number 26 = 69.94%
  • % of O = 30.06%

Unknown

  • Empirical formula = Atomic number 26 ? O ?

Steps to follow are outlined in the text.

Step 2:  Summate.

one. Assume a 100 1000 sample.

& 69.94  text{g Fe} \& 30.06  text{g O}

2. Convert to moles.

69.94  text{g Fe} times frac{1  text{mol Fe}}{55.85  text{g Fe}}&=1.252  text{mol Fe} \30.06  text{g O} times frac{1  text{mol O}}{16.00  text{g O}}&=1.879  text{mol O}

3. Split up both moles past the smallest of the results.

frac{1.252  text{mol Fe}}{1.252}=1  text{mol Fe} qquad qquad frac{1.879  text{mol O}}{1.252}=1.501  text{mol O}

iv/5. Since the moles of O, is all the same non a whole number, both moles can be multiplied by 2, while rounding to a whole number.

1  text{mol Fe} times 2=2  text{mol Fe} qquad qquad 1.501  text{mol O} times 2=3  text{mol O}

The empirical formula of the compound is Fe two O 3 .

Footstep three: Think about your result.

The subscripts are whole numbers and correspond the mole ratio of the elements in the compound.  The chemical compound is the ionic compound iron(Three) oxide.

Summary

  • A process is described for the calculation of the empirical formula for a compound based on the percentage composition of that compound.

Practice

Employ the link below to read almost calculating empirical formulas and exercise working some problems:

http://world wide web.chemteam.info/Mole/EmpiricalFormula.html

Review

Questions

  1. What is an empirical formula?
  2. What does an empirical formula tell you?
  3. What does it not tell y'all?
  • elemental analysis: Determines the percentages of each chemical element in a compound.
  • empirical formula: Shows the lowest whole-number ratio of the elements in a compound.

Determining Molecular Formulas

  • Define molecular formula.
  • Determine the molecular formula when give the empirical formula and the molar mass of the compound.

Fischer projection of glucose

Structure of sucrose

How tin can you lot decide the differences between these two molecules?

To a higher place we run across two carbohydrates: glucose and sucrose.  Sucrose is well-nigh exactly twice the size of glucose, although their empirical formulas are very similar.  Some people could distinguish them on the basis of gustation, but information technology's not a practiced idea to become effectually tasting chemicals. The all-time manner is to decide the molecular weights – this approach allows you to easily tell which compound is which.

Molecular Formulas

Molecular formulas give the kind and number of atoms of each element nowadays in a molecular compound.  In many cases, the molecular formula is the aforementioned equally the empirical formula.  The molecular formula of methane is CH 4 and considering it contains but one carbon atom, that is too its empirical formula.  Sometimes, however, the molecular formula is a simple whole-number multiple of the empirical formula.  Acetic acrid  is an organic acrid that is the primary component of vinegar.  Its molecular formula is C 2 H 4 O 2 .  Glucose is  a uncomplicated sugar that cells utilize as a primary source of energy.  Its molecular formula is C vi H 12 O 6 .  The structures of both molecules are shown in the figure below.  They are very different compounds, yet both have the same empirical formula of CH 2 O.

Acetic acid and glucose both the same empirical formula

Effigy 10.13

Acerb acid (left) has a molecular formula of C 2 H four O two , while glucose (correct) has a molecular formula of C 6 H 12 O 6 . Both have the empirical formula CH 2 O.

Empirical formulas can be determined from the pct composition of a compound.  In society to determine its molecular formula, it is necessary to know the tooth mass of the compound.  Chemists use an instrument called a mass spectrometer to determine the molar mass of compounds.  In lodge to become from the empirical formula to the molecular formula, follow these steps:

  1. Calculate the empirical formula mass (EFM) , which is simply the molar mass represented by the empirical formula.
  2. Split the molar mass of the compound by the empirical formula mass. The result should be a whole number or very close to a whole number.
  3. Multiply all the subscripts in the empirical formula by the whole number found in step 2. The result is the molecular formula.

Sample Problem 1: Determining the Molecular Formula of a Compound

The empirical formula of a compound of boron and hydrogen is BH 3 .  Its molar mass is 27.7 one thousand/mol.  Determine the molecular formula of the compound.

Footstep 1: Listing the known quantities and plan the problem.

Known

  • empirical formula = BH three
  • tooth mass = 27.7 g/mol

Unknown

  • molecular formula = ?

Steps to follow are outlined in the text.

Pace 2: Calculate.

1. The empirical formula mass (EFM) = xiii.84 g/mol

two. frac{text{molar mass}}{text{EFM}}=frac{27.7}{13.84}=2

3. text{BH}_3 times 2 =text{B}_2text{H}_6

The molecular formula of the compound is B ii H 6 .

Step three: Think about your result.

The molar mass of the molecular formula matches the molar mass of the compound.

Summary

  • A procedure is described that allows the calculation of the exact molecular formula for a chemical compound.

Exercise

Apply the link below to access practice issues.  Effort as many equally yous accept time for:

http://chemistry.about.com/od/chemistry-examination-questions/tp/Molecular-Formula-Practice-Exam-Questions.htm

Review

Questions

  1. What is the difference between an empirical formula and a molecular formula?
  2. In add-on to the elemental analysis, what do you need to know to summate the molecular formula?
  3. What does the empirical formula mass tell you?
  • empirical formula mass (EFM): The molar mass represented by the empirical formula.
  • molecular formula: Gives the kind and number of atoms of each element present in a molecular compound.

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Source: https://courses.lumenlearning.com/cheminter/chapter/the-mole/

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