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Can You Color A Mole Chemistry Answers

The Mole

Avogadro's Number

  • Ascertain mole.
  • Define Avogadro's number.

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

Is at that place an easier way to load this truck?

When the weather 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 to other materials. Y'all could guild up twenty 1000000 grains of sand and accept people actually stare at you. You could order by the pound, just that takes a lot of fourth dimension weighing out. The best bet is to order by the 1000, pregnant a cubic one thousand. The loader can hands scoop up what you need and put it directly in your truck.

Avogadro's Number

It certainly is easy to count bananas or to count elephants (as long equally you stay out of their way). However, yous would be counting grains of saccharide from your carbohydrate 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 admittedly impossible. Ane drib of water contains about 10 22 molecules of water. If you counted 10 molecules every 2d for 50 years without stopping you would accept counted only i.half dozen × 10 x molecules. Put another way, at that counting rate, information technology would accept you over 30 trillion years to count the water molecules in one tiny drop.

Chemists needed a name that tin can stand for a very large number of items. Amedeo Avogadro (1776 – 1856), an Italian scientist, provided simply such a number. He is responsible for the counting unit of measurement of measure called the mole. A mole (mol) is the corporeality 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.  Simply like the dozen and the gross, information technology is a name that stands for a number. In that location are therefore six.02 × 10 23 water molecules in a mole of h2o molecules. There also would exist half dozen.02 × 10 23 bananas in a mole of bananas, if such a huge number of bananas ever existed.

Portrait of Amedeo Avogadro

Figure x.1

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

The number six.02 × 10 23 is called Avogadro's number , the number of representative particles in a mole. It is an experimentally determined number. A representative particle is the smallest unit of measurement in which a substance naturally exists. For the majority of elements, the representative particle is the atom. Iron, carbon, and helium consist of fe atoms, carbon atoms, and helium atoms, respectively. Vii elements exist in nature as diatomic molecules and they are H 2 , Due north 2 , O two , F 2 , Cl 2 , Br 2 , and I 2 . The representative particle for these elements is the molecule. Besides, all molecular compounds such every bit H two O and CO 2 exist equally molecules then the molecule is their representative particle.  For ionic compounds such as NaCl and Ca(NO 3 ) 2 , the representative particle is the formula unit. A mole of any substance contains Avogadro's number (vi.02 × 10 23 ) of representative particles.

The animal mole is very different from the unit mole

Figure ten.2

The beast mole is very different than the counting unit of the mole. Chemists nonetheless accept adopted the mole every bit their unofficial mascot. National Mole Day is a celebration of chemistry that occurs on October 23rd (10/23) of each year.

Summary

  • A mole of whatsoever substance contains Avogadro's number (6.02 × 10 23 ) of representative particles.

Practise

Questions

Apply the link below to answer the following questions:

http://www.scientificamerican.com/commodity.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 conclusion for the value for the number?

Review

Questions

  1. What is the SI unit for corporeality of a substance?
  2. What is the representative particle for an element?
  3. The formula unit is the representative particle for what?
  • Avogadro's number: The number of representative particles in a mole, 6.02 × 10 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 Betwixt Moles and Atoms

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

Using moles allows us to avoid superscripts and subscripts

Big numbers or little numbers?

Do yous detest to type subscripts and superscripts? Even with a proficient word-processing program, having to click on an icon to get a superscript then remembering to click off after yous type the number can be a real hassle.  If we did non know well-nigh moles and merely knew about numbers of atoms or molecules (those big numbers that crave lots of superscripts), life would exist much more complicated and we would make many more typing errors.

Conversions Betwixt Moles and Atoms

Conversions Between Moles and Number of Particles

Using our unit conversion techniques, nosotros can apply the mole label to convert dorsum and along between the number of particles and moles.

Sample Trouble 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 4.72 × ten 24 atoms of carbon?

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

Known

  • number of C atoms = four.72 × 10 24
  • 1 mole = half dozen.02 × 10 23 atoms

Unknown

  • 4.72 × x 24 = ? mol C

1 conversion cistron volition allow united states to convert from the number of C atoms to moles of C atoms.

Pace 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}

Step 3: Call up about your outcome.

The given number of carbon atoms was greater than Avogadro'southward number, so the number of moles of C atoms is greater than 1 mole.  Since Avogadro's number is a measured quantity with three meaning figures, the effect of the calculation is rounded to three pregnant figures.

Suppose that yous wanted to know how many hydrogen atoms were in a mole of water molecules.  First, yous would need to know the chemic formula for water, which is H 2 O.  In that location are two atoms of hydrogen in each molecule of water.  How many atoms of hydrogen would there exist in 2 h2o molecules?  There would be 2 × 2 = 4 hydrogen atoms. How nigh in a dozen?  In that case a dozen is 12 and so 12 × 2 = 24 hydrogen atoms in a dozen water molecules.  To go the answers, (4 and 24) you lot had to multiply the given number of molecules by two atoms of hydrogen per molecule.  Then to notice the number of hydrogen atoms in a mole of water molecules, the trouble could exist 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 get-go conversion factor converts from moles of particles to the number of particles. The second conversion cistron reflects the number of atoms independent within each molecule.

Two water molecules

Effigy ten.3

Two water molecules contain 4 hydrogen atoms and 2 oxygen atoms. A mole of water molecules contains 2 moles of hydrogen atoms and i mole of oxygen atoms.

Sample Problem ii: Atoms, Molecules, and Moles

Sulfuric acid has the chemic formula H 2 And then four .  A certain quantity of sulfuric acrid contains 4.89 × x 25 atoms of oxygen.  How many moles of sulfuric acid is the sample?

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

Known

  • four.89 × ten 25 = O atoms
  • one mole = 6.02 × 10 23 molecules H two SO iv

Unknown

  • mol of H 2 And then four molecules

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

Footstep 2: Summate.

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 iii: Call back virtually your result.

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

Summary

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

Practise

Read the relevant portions of the following article and do problems 3, 5, 9, 13, and 18.  Do non worry virtually the calculations involving conversions dealing with tooth mass (that will come side by side).

http://faculty.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 catechumen atoms to moles. My friend tells me to multiply the number of atoms past half-dozen.02 × ten 23 atoms/mole.  Is this right?
  3. Why should I know the formula for a molecule in lodge to calculate the number of moles of one of the atoms?

Molar Mass

  • Define molar mass.
  • Perform calculations involving molar mass.

Pile of potassium dichromate

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

I want to make a solution that contains 1.eight moles of potassium dichromate.  I don't have a balance calibrated in molecules, simply I do have one calibrated in grams.  If I know the relationship between moles and the number of grams in a mole, I tin use my balance to measure out the needed amount of material.

Molar Mass

Molar mass is defined as the mass of i mole of representative particles of a substance. By looking at a periodic table, we can conclude that the tooth mass of lithium is six.94 g, the molar mass of zinc is 65.38 g, and the tooth mass of gilded is 196.97 g.  Each of these quantities contains 6.02 × 10 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 ii .  One molecule of carbon dioxide consists of 1 cantlet of carbon and 2 atoms of oxygen.  We can calculate the mass of one molecule of carbon dioxide by adding together the masses of ane atom of carbon and ii 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 chemical compound is the mass in grams of i mole of that compound.  1 mole of carbon dioxide molecules has a mass of 44.01 chiliad, while i mole of sodium sulfide formula units has a mass of 78.04 g.  The molar masses are 44.01 g/mol and 78.04 g/mol respectively.  In both cases, that is the mass of 6.02 × 10 23 representative particles.  The representative particle of CO two is the molecule, while for Na 2 S, it is the formula unit.

Sample Problem:  Molar Mass of  a Compound

Calcium nitrate, Ca(NO three ) ii , is used equally a component in fertilizer Determine the molar mass of calcium nitrate.

Pace i:  List the known and unknown quantities and plan the problem.

Known

  • formula = Ca(NO three ) 2
  • molar mass Ca = 40.08 1000/mol
  • molar mass N = xiv.01 yard/mol
  • molar mass O = 16.00 chiliad/mol

Unknown

  • molar mass Ca(NO 3 ) 2

First we need to analyze the formula.  Since the Ca lacks a subscript, in that location is one Ca atom per formula unit.  The 2 outside the parentheses means that there are two nitrate ions per formula unit of measurement and each nitrate ion consists of 1 nitrogen atom and three oxygen atoms.  Therefore, at that place are a full of 1 × 2 = 2 nitrogen atoms and 3 × ii = 6 oxygen atoms per formula unit.  Thus, one mol of calcium nitrate contains 1 mol of Ca atoms, ii mol of N atoms, and 6 mol of O atoms.

Step ii:  Summate.

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

Practice

Read the cloth at the link below and piece of work the problems at the end:

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

Review

Questions

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

Conversions between Moles and Mass

  • Perform calculations dealing with conversions betwixt moles and mass.

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

How can we get more product?

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

Conversions Between Moles and Mass

The molar mass of whatsoever 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 oft used in the laboratory.  Suppose that for a sure experiment you need 3.00 moles of calcium chloride (CaCl 2 ).  Since calcium chloride is a solid, it would be user-friendly to utilize a balance to measure the mass that is needed.  The molar mass of CaCl 2 is 110.98 g/mol.  The conversion factor that can be used is and so based on the equality that 1 mol = 110.98 g CaCl 2 .  Dimensional analysis will let you to calculate the mass of CaCl 2 that yous should mensurate.

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 measure the mass of 333 g of CaCl 2 , y'all are measuring three.00 moles of CaCl ii .

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

Effigy ten.4

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

Sample Problem: Converting Moles to Mass

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

Step ane: List the known quantities and program the problem.

Known

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

Unknown

  • 0.560 mol Cr = ? g

One conversion cistron will 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: Recollect about your consequence.

Since the desired amount was slightly more than one one-half of a mole, the mass should exist slightly more than than one half of the molar mass.  The answer has three significant figures because of the 0.560 mol.

A like conversion gene utilizing molar mass can exist used to catechumen from the mass of an substance to moles.  In a laboratory state of affairs, y'all may perform a reaction and produce a certain corporeality of a production which can be massed.  Information technology will oftentimes 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 sure reaction produces 2.81 chiliad of copper(II) hydroxide, Cu(OH) 2 .  Determine the number of moles produced in the reaction.

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

Known

  • mass = 2.81 yard

Unknown

  • mol Cu(OH) ii

Ane conversion factor will allow us to convert from mass to moles.

Step 2: Calculate.

First, it is necessary to calculate the molar mass of Cu(OH) ii from the molar masses of Cu, O, and H.  The molar mass is 97.57 chiliad/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

Footstep 3: Recollect nearly your result.

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

Summary

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

Practice

Read the material in the link below and work the problems at the cease.

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

Review

Questions

  1. Why would you want to calculate the mass of a cloth?
  2. Why would yous want to decide how many moles of cloth you produced in a reaction?
  3. You have nineteen.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 at that place?

Avogadro was interested in studying gases.  He theorized that equal volumes of gases under the same conditions contained the same number of particles.  Other researchers studied how many gas particles were in a specific volume of gas. Eventually, 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 betwixt Moles and Mass", you lot learned how to convert back and forth betwixt moles and the number of representative particles.  Now you have seen how to catechumen back and forth between moles and mass of a substance in grams.  We tin can combine the two types of problems into one.  Mass and number of particles are both related to grams.  In order to convert from mass to number of particles or vice-versa, it will first require a conversion to moles.

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

Figure 10.5

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

Sample Trouble: Converting Mass to Particles

How many molecules is 20.0 g of chlorine gas, Cl 2 ?

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

Known

  • molar mass Cl 2 = 70.ninety 1000/mol
  • 20.0 thousand Cl two

Unknown

  • number of molecules of Cl ii

Use 2 conversion factors.  The get-go converts grams of Cl 2 to moles.  The second converts moles of Cl two to the number of molecules.

Stride ii: 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 done using two consecutive conversion factors. In that location is no demand to explicitly calculate the moles of Cl ii .

Step 3: Think about your effect.

Since the given mass is less than half of the molar mass of chlorine, the resulting number of molecules is less than half of Avogadro'southward number.

Summary

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

Practice

Read the material at the link below and then do do problems on page 9 and the problem on page 17 (don't meridian 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 can't we catechumen 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 diminutive weight of chlorine is 35.5. Why tin't I utilize that value in my calculations?

Avogadro'due south Hypothesis and Molar Volume

  • State Avogadro's hypothesis.
  • Define standard temperature and pressure level.
  • Define tooth volume.

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

How do scuba divers know if they volition run out of gas?

Knowing how much gas is available for a swoop is crucial to the survival of the diver.  The tank on the diver'due south back is equipped with gauges to tell how much gas is present and what the pressure level is.  A bones knowledge of gas behavior allows the diver to assess how long to stay under water without developing issues.

Avogadro'due south Hypothesis and Molar Book

Volume is a third style to measure the amount of matter, after item count and mass.  With liquids and solids, volume varies profoundly depending on the density of the substance.  This is because solid and liquid particles are packed close together with very little infinite in betwixt the particles.  However, gases are largely equanimous of empty space between the actual gas particles (see Figure beneath ).

Gas particles are small compared to the empty space between them

Figure 10.6

Gas particles are very small-scale compared to the large amounts of empty infinite 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 total volume that a gas occupies is made upward 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 light particles such as hydrogen (H 2 ) contains the same number of particles as the aforementioned volume of a heavy gas with big particles such as sulfur hexafluoride, SF 6 .

Gases are compressible, meaning that when put under high pressure, the particles are forced closer to one another.  This decreases the amount of empty space and reduces the volume of the gas.  Gas volume is besides affected past temperature.  When a gas is heated, its molecules movement faster and the gas expands.  Because of the variation in gas volume due to pressure and temperature changes, the comparison of gas volumes must be done at i standard temperature and pressure. Standard temperature and pressure (STP) is defined every bit 0°C (273.15 K) and i atm pressure. The molar volume of a gas is the volume of one mole of a gas at STP.  At STP, i mole (half-dozen.02 × 10 23 representative particles) of whatsoever gas occupies a book of 22.4 L ( Figure beneath ).

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

Figure 10.vii

A mole of any gas occupies 22.4 Fifty at standard temperature and pressure level (0°C and one atm).

The Figure below illustrates how molar volume can exist seen when comparison different gases. Samples of helium (He), nitrogen (N 2 ), and methyl hydride (CH four ) are at STP.  Each contains ane mole or 6.02 × ten 23 particles.  Withal, the mass of each gas is dissimilar and corresponds to the molar mass of that gas: 4.00 k/mol for He, 28.0 g/mol for N 2 , and xvi.0 chiliad/mol for CH 4 .

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

Figure x.eight

Avogadro's hypothesis states that equal volumes of any gas at the aforementioned temperature and pressure contain the same number of particles. At standard temperature and pressure, one mole of whatsoever gas occupies 22.4 Fifty.

Summary

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

Practice

Questions

Utilize the link beneath to respond the following questions:

http://chemed.chem.purdue.edu/demos/main_pages/4.half-dozen.html

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

Review

Questions

  1. What practice we know nigh the infinite actually taken upwards by a gas?
  2. Why do we need to do all our comparisons at the same temperature and force per unit area?
  3. How tin can nosotros use this information?
  • Avogadro's hypothesis: Equal volumes of all gases at the aforementioned temperature and pressure contain equal numbers of particles.
  • molar volume: The book of one mole of a gas at STP.
  • standard temperature and pressure level (STP): 0°C (273.15 K) and 1 atm force per unit area.

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 can you tell how much gas is in these containers?

Small gas tanks are often used to supply gases for chemistry reactions.  A gas judge volition give some information about how much is in the tank, but quantitative estimates are needed so the reaction will be able to continue 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

Tooth book at STP can be used to catechumen from moles to gas book and from gas volume to moles.  The equality of 1 mole = 22.iv 50 is the ground for the conversion gene.

Sample Trouble 1: Converting Gas Volume to Moles

Many metals react with acids to produce hydrogen gas.  A certain reaction produces 86.5 50 of hydrogen gas at STP.  How many moles of hydrogen were produced?

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

Known

  • 86.5 L H 2
  • i mol = 22.4 Fifty

Unknown

  • moles of H 2

Apply a conversion factor to catechumen 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

Footstep three: Recollect about your result.

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

Sample Trouble Two: Converting Moles to Gas Volume

What volume does four.96 moles of O two occupy at STP?

Step ane: 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 two:  Calculate.

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

Step 3: Call back about your result.

The book seems correct given the number of moles.

Sample Problem Three: Converting Book to Mass

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

Footstep ane: List the known quantities and plan the problem.

Known

  • 867 L N two
  • 1 mol = 22.4 L
  • molar mass of Due north 2 = 28.02 g/mol

Unknown

  • mass of N 2

Footstep ii: Calculate.

We start past determining the number of moles of gas present. Nosotros know that 22.iv liters of a gas at STP equals one mole, and so:

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

Nosotros also know the molecular weight of N 2 (28.0 grams/mole), so nosotros can so calculate 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 3: Recollect near your upshot.

In a multi-stride problem, be sure that the units check.

Summary

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

Practice

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

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

Review

Questions

  1. Why do the gases demand to be at STP?
  2. When does the identity of the gas become important?

Gas Density

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

Carbon dioxide sinks in air

Why does carbon dioxide sink in air?

When nosotros run a reaction to produce a gas, we expect it to ascent into the air.  Many students have done experiments where gases such as hydrogen are formed.  The gas can be trapped in a test tube held upside-down over the reaction.  Carbon dioxide, on the other paw, sinks when it is released.  Carbon dioxide has a density greater that air, and then it will not rise like these other gases would.

Gas Density

As you know, density is defined equally the mass per unit volume of a substance.  Since gases all occupy the same book on a per mole basis, the density of a item gas is dependent on its tooth mass.  A gas with a pocket-sized molar mass will have a lower density than a gas with a large molar mass.  Gas densities are typically reported in g/Fifty.  Gas density tin can be calculated from molar mass and tooth book.

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

Figure x.9

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

Sample Trouble Ane: Gas Density

What is the density of nitrogen gas at STP?

Stride 1: Listing the known quantities and programme the trouble.

Known

  • N two = 28.02 yard/mol
  • 1 mol = 22.4 L

Unknown

  • density = ? g/Fifty

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

Step 2: Summate.

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

When gear up upwardly with a conversion factor, the mol unit cancels, leaving g/50 as the unit in the result.

Step three: Call up nearly your result.

The tooth mass of nitrogen is slightly larger than molar volume, so the density is slightly greater than 1 g/50.

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

Sample Problem 2: Tooth Mass from Gas Density

What is the molar mass of a gas whose density is 0.761 g/L at STP?

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

Known

  • North 2 = 28.02 thou/mol
  • 1 mol = 22.4 L

Unknown

  • tooth mass = ? one thousand/L

Molar mass is equal to density multiplied by 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 iii: Think about your consequence.

Because the density of the gas is less than 1 g/L, the molar mass is less than 22.iv.

Summary

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

Practice

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 y'all expect 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 exist the volume of 3.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 exercise I get from here to there?

If I desire to visit the town of Manteo, Northward Carolina, out on the coast, I will need a map of how to get there.  I may have a printed map or I may download directions from the net, but I need something to get me going in the right direction.  Chemistry road maps serve the same purpose.  How do I handle a sure type of adding? There is a process and a set of directions to help.

Mole Road Map

Previously, we saw how the conversions betwixt mass and number of particles required two steps, with moles as the intermediate.  This concept tin at present be extended to also include gas volume at STP.  The resulting diagram is referred to as a mole road map (see Figure beneath ).

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

Effigy 10.ten

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

The mole is at the center of whatsoever calculation involving amount of a substance.  The sample problem beneath is 1 of many different bug that can be solved using the mole road map.

Sample Problem One:  Mole Road Map

What is the volume of 79.iii g of neon gas at STP?

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

Known

  • Ne = 20.18 g/mol
  • ane mol = 22.4 L

Unknown

  • volume = ? L

The conversion factors will be grams → moles → gas book.

Footstep 2:  Calculate.

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:  Think most your effect.

The given mass of neon is equal to about iv moles, resulting in a volume that is well-nigh 4 times larger than tooth book.

Summary

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

Practice

Use the link below to acquit out some practice calculations.  Practice problems 1, 2, and 5 (you lot can try the others if y'all are feeling especially brave):

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

Review

Questions

  1. In the trouble above, 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?

Percentage Composition

  • Define percent limerick.
  • Perform pct composition calculations.

Nutritional information labels can let you know the percent composition

Is at that place anything healthy in this jar?

Packaged foods that you eat typically have nutritional information provided on the label.  The label on a jar of peanut butter (shown in a higher place) reveals that one serving size is considered to be 32 g.  The characterization also gives the masses of diverse types of compounds that are present in each serving.  One serving contains vii one thousand of protein, xv g of fat, and 3 yard of carbohydrate.  By computing the fraction of protein, fat, or sugar in one serving of size of peanut butter and converting to percent values, we tin can make up one's mind the composition of the peanut butter on a pct by mass footing.

Pct Limerick

Chemists often need to know what elements are present in a compound and in what percentage.  The percentage composition is the percentage past mass of each chemical element in a chemical compound.  It is calculated in a similar mode that we just indicated for the peanut butter.

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

Percent Composition from Mass Data

The sample problem below shows the calculation of the pct limerick of a compound based on mass data.

Sample Trouble One: Percent Composition from Mass

A sure newly synthesized compound is known to incorporate the elements zinc and oxygen.  When a 20.00 one thousand sample of the sample is decomposed, 16.07 g of zinc remains.  Decide the percent composition of the compound.

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

Known

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

Unknown

  • percentage Zn = ? %
  • pct O = ? %

Decrease to observe the mass of oxygen in the compound.  Split up each chemical element'south mass by the mass of the compound to find the per centum by mass.

Step 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} \

Step 3: Think near your upshot.

The calculations make sense because the sum of the two percentages adds up to 100%.  By mass, the compound is mostly zinc.

Percent Composition from a Chemical Formula

The percent composition of a compound can also be determined from the formula of the compound.  The subscripts in the formula are first used to summate the mass of each chemical element in one mole of the 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 percent composition of a given chemical compound is always the same as long as the compound is pure.

Sample Problem 2: Percent Composition from Chemical Formula

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

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

Known

  • mass of Cl in 1 mol Cl ii O vii = seventy.90 thousand
  • mass of O in 1 mol Cl 2 O 7 = 112.00 g
  • molar mass of Cl 2 O 7 = 182.90 g/mol

Unknown

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

Calculate the percent 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 ii: 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} \

Pace three: Think most your result.

The percentages add together up to 100%.

Per centum composition can also be used to decide the mass of a certain element that is independent in any mass of a compound.  In the previous sample trouble, information technology was found that the percent composition of dichlorineheptoxide is 38.76% Cl and 61.24% O.  Suppose that you needed to know the masses of chlorine and oxygen present in a 12.50 yard sample of dichlorineheptoxide.  Yous can set up a conversion gene based on the percent 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 ii masses is 12.50 thou, the mass of the sample size.

Summary

  • Processes are described for computing the pct composition of a material based on mass or on chemical composition.

Do

Utilise the link below to review material and do calculations. Read both parts of the lesson and do every bit many calculations every bit you have time for.

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

Review

Questions

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

Per centum of Water in a Hydrate

  • Ascertain hydrate.
  • Calculate the per centum h2o in hydrate when give relevant information.

Copper sulfate changes color when hydrated

Why does the colour change?

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

Percent of H2o in a Hydrate

Many ionic compounds naturally contain h2o every bit function of the crystal lattice construction.  A hydrate is a chemical compound that has one or more water molecules bound to each formula unit of measurement.  Ionic compounds that contain a transition element are oftentimes highly colored.  Interestingly, it is common for the hydrated form of a compound to be of a different color than the anhydrous grade, which has no water in its construction.  A hydrate can usually be converted to the anhydrous compound by heating.  For example, the anhydrous chemical 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

Effigy 10.11

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

The hydrated form of cobalt(Ii) chloride contains six water molecules in each formula unit.  The name of the compound is cobalt(II) chloride hexahydrate and its formula is CoCl two •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 water molecules per formula unit.

It is useful to know the percent of h2o contained within a hydrate.  The sample trouble below demonstrates the procedure.

Sample Problem One: Percent of Water in a Hydrate

Find the percentage water in cobalt(2) chloride hexahydrate, CoCl two •6H two O.

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

The mass of water in the hydrate is the coefficient (half-dozen) multiplied by the molar mass of H 2 O.  The molar mass of the hydrate is the tooth mass of the CoCl 2 plus the mass of h2o.

Known

  • mass of H two O in 1 mol hydrate = 108.12 yard
  • molar mass of hydrate = 237.95 chiliad/mol

Unknown

  • percent H 2 O = ? %

Summate the percent by mass of water by dividing the mass of H 2 O in 1 mole of the hydrate by the molar mass of the hydrate and multiplying by 100%.

Step ii: 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}

Step 3: Retrieve about your result.

Most half of the mass of the hydrate is composed of water molecules within the crystal.

Summary

  • The process of computing the percent water in a hydrate is described.

Practise

Use the following link to practice computing per centum water in a hydrate:

http://www.sd84.k12.id.u.s./shs/departments/science/martz/2007_ssem2/Chemistry/hydrate.htm

Review

Questions

  1. What is a hydrate?
  2. How tin you convert a hydrate to an anhydrous chemical compound?
  3. What does hexahydrate mean?
  • Anhydrous: Without water.
  • Hydrate: A compound that has one or more than water 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 picture?

In the early days of chemistry, there were few tools for the detailed report of compounds.  Much of the information 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 non being able to narrate a compound completely.  The relative amounts of elements could be determined, but and then many of these materials had carbon, hydrogen, oxygen, and possibly nitrogen in simple ratios.  We did non know exactly how many of these atoms were actually in a specific molecule.

Determining Empirical Formulas

An empirical formula is one that shows the everyman whole-number ratio of the elements in a compound.  Considering the structure of ionic compounds is an extended three-dimensional network of positive and negative ions, all formulas of ionic compounds are empirical.  However, nosotros can also consider the empirical formula of a molecular compound.  Ethene is a small hydrocarbon chemical compound with the formula C ii H iv (encounter Effigy below ).  While C two H 4 is its molecular formula and represents its true molecular structure, it has an empirical formula of CH ii .  The simplest ratio of carbon to hydrogen in ethene is 1:two.  There are two ways to view that ratio.  Because ane molecule of ethene, the ratio is 1 carbon atom for every ii atoms of hydrogen.  Considering 1 mole of ethene, the ratio is 1 mole of carbon for every ii moles of hydrogen.  And then the subscripts in a formula represent the mole ratio of the elements in that formula.

Ball and stick model of ethene

Effigy 10.12

Ball-and-stick model of ethene, C 2 H four .

In a procedure called elemental assay , an unknown chemical compound can be analyzed in the laboratory in lodge to determine the percentages of each chemical element independent within it.  These percentages tin 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 so that the given percentages tin can be directly converted into grams.
  2. Use each chemical element's tooth mass to catechumen the grams of each element to moles.
  3. In order to find a whole-number ratio, divide the moles of each element past whichever of the moles from pace 2 is the smallest.
  4. If all the moles at this signal are whole numbers (or very close), the empirical formula can exist written with the moles every bit the subscript of each element.
  5. In some cases, one or more than of the moles calculated in step 3 will non be 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 One: Determining the Empirical Formula of a Compound

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

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

Known

  • % of Fe = 69.94%
  • % of O = 30.06%

Unknown

  • Empirical formula = Fe ? O ?

Steps to follow are outlined in the text.

Step ii:  Calculate.

i. Presume a 100 1000 sample.

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

ii. 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}

three. 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}

four/5. Since the moles of O, is still not a whole number, both moles can exist 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 2 O 3 .

Step three: Call back about your result.

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

Summary

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

Practice

Use the link below to read nearly calculating empirical formulas and practice working some problems:

http://www.chemteam.info/Mole/EmpiricalFormula.html

Review

Questions

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

Determining Molecular Formulas

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

Fischer projection of glucose

Structure of sucrose

How can y'all determine the differences between these two molecules?

Above we see 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 taste, but it's not a good idea to go around tasting chemicals. The best fashion is to determine the molecular weights – this approach allows you to easily tell which compound is which.

Molecular Formulas

Molecular formulas requite the kind and number of atoms of each element present in a molecular compound.  In many cases, the molecular formula is the same as the empirical formula.  The molecular formula of methane is CH 4 and considering information technology contains just one carbon atom, that is also its empirical formula.  Sometimes, however, the molecular formula is a simple whole-number multiple of the empirical formula.  Acerb acrid  is an organic acid that is the main component of vinegar.  Its molecular formula is C 2 H 4 O 2 .  Glucose is  a unproblematic sugar that cells use as a chief source of energy.  Its molecular formula is C 6 H 12 O 6 .  The structures of both molecules are shown in the figure below.  They are very different compounds, notwithstanding both have the same empirical formula of CH 2 O.

Acetic acid and glucose both the same empirical formula

Effigy ten.xiii

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

Empirical formulas can be determined from the percent limerick of a compound.  In club to decide its molecular formula, information technology is necessary to know the molar mass of the compound.  Chemists employ an musical instrument chosen a mass spectrometer to determine the molar mass of compounds.  In order to go from the empirical formula to the molecular formula, follow these steps:

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

Sample Trouble One: Determining the Molecular Formula of a Chemical compound

The empirical formula of a compound of boron and hydrogen is BH 3 .  Its molar mass is 27.seven g/mol.  Decide the molecular formula of the compound.

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

Known

  • empirical formula = BH 3
  • molar mass = 27.7 thousand/mol

Unknown

  • molecular formula = ?

Steps to follow are outlined in the text.

Stride 2: Calculate.

i. The empirical formula mass (EFM) = 13.84 g/mol

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

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

The molecular formula of the chemical compound is B two H vi .

Step 3: Think nearly 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 verbal molecular formula for a compound.

Exercise

Use the link below to admission do problems.  Endeavor as many every bit you have time for:

http://chemistry.about.com/od/chemistry-test-questions/tp/Molecular-Formula-Exercise-Test-Questions.htm

Review

Questions

  1. What is the difference betwixt an empirical formula and a molecular formula?
  2. In addition to the elemental analysis, what practice you need to know to calculate the molecular formula?
  3. What does the empirical formula mass tell y'all?
  • 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.

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

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