What Relationships Are Needed to Convert 1.5 Grams of Carbon to the Number of Carbon Atoms?

The Mole

Avogadro'south Number

  • Ascertain mole.
  • Define Avogadro's number.

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

Is there an easier way to load this truck?

When the weather is prissy, many people begin to 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 order up 20 1000000 grains of sand and have people really stare at you. You could society past the pound, merely that takes a lot of time weighing out. The all-time bet is to order by the yard, pregnant a cubic thousand. The loader can easily scoop upward what you need and put it direct in your truck.

Avogadro'south Number

It certainly is easy to count bananas or to count elephants (every bit long as yous stay out of their manner). Still, you would be counting grains of sugar 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 non only unwise, it is absolutely impossible. One drop of water contains about 10 22 molecules of h2o. If you counted 10 molecules every second for l years without stopping yous would accept counted merely 1.6 × 10 10 molecules. Put another fashion, at that counting rate, it would have you lot 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 called the mole. A mole (mol) is the amount of a substance that contains six.02 × ten 23 representative particles of that substance. The mole is the SI unit for amount of a substance.  Simply like the dozen and the gross, it is a name that stands for a number. There are therefore 6.02 × 10 23 water molecules in a mole of h2o molecules. In that location also would be 6.02 × 10 23 bananas in a mole of bananas, if such a huge number of bananas ever existed.

Portrait of Amedeo Avogadro

Figure ten.1

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

The number vi.02 × 10 23 is called 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 of measurement in which a substance naturally exists. For the majority of elements, the representative particle is the atom. Atomic number 26, carbon, and helium consist of iron atoms, carbon atoms, and helium atoms, respectively. Seven elements exist in nature equally diatomic molecules and they are H 2 , N ii , O ii , F ii , Cl 2 , Br 2 , and I 2 . The representative particle for these elements is the molecule. Likewise, all molecular compounds such equally H 2 O and CO 2 exist every bit molecules and so the molecule is their representative particle.  For ionic compounds such every bit NaCl and Ca(NO 3 ) 2 , the representative particle is the formula unit. A mole of whatever substance contains Avogadro'southward number (six.02 × 10 23 ) of representative particles.

The animal mole is very different from the unit mole

Figure 10.2

The creature mole is very different than the counting unit of measurement of the mole. Chemists nevertheless have adopted the mole as 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 any substance contains Avogadro'southward number (half-dozen.02 × 10 23 ) of representative particles.

Practice

Questions

Utilize the link beneath to answer the following questions:

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

  1. What was Avogadro'south hypothesis?
  2. Who first calculated this number?
  3. Who coined the term "Avogadro'due south 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 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 of measurement 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 piddling numbers?

Do yous hate to type subscripts and superscripts? Even with a good word-processing plan, having to click on an icon to get a superscript and then remembering to click off subsequently you type the number can be a real hassle.  If nosotros did not know almost moles and but knew near numbers of atoms or molecules (those big numbers that require lots of superscripts), life would be much more than complicated and we would make many more typing errors.

Conversions Between Moles and Atoms

Conversions Betwixt Moles and Number of Particles

Using our unit conversion techniques, we tin utilize the mole characterization to convert back and forth betwixt the number of particles and moles.

Sample Problem 1: Converting Number of Particles to Moles

The element carbon exists in two master forms: graphite and diamond.  How many moles of carbon atoms is 4.72 × 10 24 atoms of carbon?

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

Known

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

Unknown

  • iv.72 × x 24 = ? mol C

One conversion factor will allow us to catechumen from the number of C atoms to moles of C atoms.

Step 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 iii: Think well-nigh your outcome.

The given number of carbon atoms was greater than Avogadro'south number, so the number of moles of C atoms is greater than ane mole.  Since Avogadro'south number is a measured quantity with three significant figures, the result of the calculation is rounded to iii significant figures.

Suppose that you wanted to know how many hydrogen atoms were in a mole of water molecules.  Commencement, you would need to know the chemical formula for water, which is H 2 O.  There are two atoms of hydrogen in each molecule of water.  How many atoms of hydrogen would there be in 2 water molecules?  There would be two × two = iv hydrogen atoms. How well-nigh in a dozen?  In that case a dozen is 12 and then 12 × 2 = 24 hydrogen atoms in a dozen water molecules.  To get the answers, (iv and 24) you had to multiply the given number of molecules past 2 atoms of hydrogen per molecule.  Then to observe 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 first 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 ten.three

Two water molecules contain four hydrogen atoms and ii oxygen atoms. A mole of h2o molecules contains 2 moles of hydrogen atoms and 1 mole of oxygen atoms.

Sample Problem two: Atoms, Molecules, and Moles

Sulfuric acid has the chemical formula H two So 4 .  A sure quantity of sulfuric acid contains four.89 × 10 25 atoms of oxygen.  How many moles of sulfuric acid is the sample?

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

Known

  • 4.89 × 10 25 = O atoms
  • 1 mole = six.02 × x 23 molecules H 2 SO 4

Unknown

  • mol of H ii SO 4 molecules

Two conversion factors will be used.  Starting time, convert atoms of oxygen to molecules of sulfuric acid.  Then, convert molecules of sulfuric acid to moles of sulfuric acid.

Step 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: Think nigh your result.

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

Summary

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

Exercise

Read the relevant portions of the post-obit article and practise problems 3, 5, 9, thirteen, and eighteen.  Do not worry nigh the calculations involving conversions dealing with tooth mass (that will come next).

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

Review

Questions

  1. What of import number practice nosotros 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 six.02 × 10 23 atoms/mole.  Is this correct?
  3. Why should I know the formula for a molecule in social club to summate the number of moles of 1 of the atoms?

Tooth Mass

  • Define tooth mass.
  • Perform calculations involving tooth mass.

Pile of potassium dichromate

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

I want to brand a solution that contains 1.8 moles of potassium dichromate.  I don't have a balance calibrated in molecules, just I do take one calibrated in grams.  If I know the relationship between moles and the number of grams in a mole, I can employ my balance to mensurate out the needed amount of material.

Molar Mass

Molar mass is defined as the mass of one mole of representative particles of a substance. By looking at a periodic tabular array, we can conclude that the tooth mass of lithium is 6.94 g, the molar mass of zinc is 65.38 yard, and the molar mass of gold is 196.97 g.  Each of these quantities contains 6.02 × 10 23 atoms of that detail chemical element.  The units for tooth mass are grams per mole or g/mol.

Molar Masses of Compounds

A molecular formula of the compound carbon dioxide is CO ii .  I molecule of carbon dioxide consists of i cantlet of carbon and 2 atoms of oxygen.  Nosotros tin calculate the mass of one molecule of carbon dioxide past adding together the masses of 1 atom of carbon and 2 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 any compound is the mass in grams of ane mole of that compound.  I mole of carbon dioxide molecules has a mass of 44.01 g, while i mole of sodium sulfide formula units has a mass of 78.04 grand.  The tooth masses are 44.01 1000/mol and 78.04 yard/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 ii Due south, it is the formula unit of measurement.

Sample Trouble:  Molar Mass of  a Compound

Calcium nitrate, Ca(NO iii ) 2 , is used equally 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 three ) 2
  • molar mass Ca = forty.08 g/mol
  • molar mass Due north = 14.01 g/mol
  • molar mass O = 16.00 g/mol

Unknown

  • molar mass Ca(NO 3 ) 2

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

Step 2:  Calculate.

Use the tooth 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 molar mass of an atom or a compound.

Practice

Read the material at the link beneath and work the issues at the terminate:

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

Review

Questions

  1. What is the tooth mass of Pb?
  2. Why practise we demand to include the units in our answer?
  3. I want to calculate 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 ) 3 PO 4 ?
  • molar mass : The mass of i mole of representative particles of a substance.
  • molecular mass : The mass of i molecule of that chemical 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 production?

Chemical manufacturing plants are e'er seeking to ameliorate their processes.  One of the ways 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 tin can try different means of getting more production at less cost.

Conversions Between Moles and Mass

The molar mass of any 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 human relationship is frequently used in the laboratory.  Suppose that for a certain experiment you lot 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 mensurate the mass that is needed.  The molar mass of CaCl two is 110.98 grand/mol.  The conversion cistron that tin can be used is then based on the equality that i mol = 110.98 m CaCl ii .  Dimensional analysis will allow you to calculate the mass of CaCl two 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 lot mensurate the mass of 333 yard of CaCl 2 , y'all are measuring 3.00 moles of CaCl 2 .

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

Figure 10.4

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.  Discover 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 factor 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 iii: Think virtually your upshot.

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

A like conversion factor utilizing molar mass tin exist used to convert from the mass of an substance to moles.  In a laboratory situation, yous may perform a reaction and produce a certain amount of a product which can be massed.  It will ofttimes then exist necessary to decide the number of moles of the production that was formed.  The adjacent problem illustrates this state of affairs.

Sample Problem: Converting Mass to Moles

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

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

Known

  • mass = 2.81 g

Unknown

  • mol Cu(OH) 2

I conversion factor volition allow us to convert from mass to moles.

Step 2: Calculate.

Outset, information technology is necessary to calculate the molar 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 iii: Think well-nigh your result.

The relatively small mass of production formed results in a minor number of moles.

Summary

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

Practice

Read the material in the link below and piece of work the issues at the end.

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

Review

Questions

  1. Why would y'all desire to calculate the mass of a material?
  2. Why would you want to make up one's mind how many moles of material you produced in a reaction?
  3. You have 19.7 grams of a textile and wonder how many moles were formed. Your friend tells you to multiply the mass past 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 conditions independent the same number of particles.  Other researchers studied how many gas particles were in a specific book of gas. Eventually, scientists were able to develop the relationship between number of particles and mass using the thought of moles.

Conversions Between Mass and Number of Particles

In "Conversions between Moles and Mass", you learned how to convert back and forth between moles and the number of representative particles.  Now you 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 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.five

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

Sample Problem: Converting Mass to Particles

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

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

Known

  • molar mass Cl 2 = 70.90 1000/mol
  • 20.0 yard Cl two

Unknown

  • number of molecules of Cl 2

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

Step 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 trouble is done using two consecutive conversion factors. There is no demand to explicitly calculate the moles of Cl 2 .

Pace three: Think about your result.

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

Summary

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

Exercise

Read the cloth at the link beneath and then do practise problems on page 9 and the problem on page 17 (don't tiptop at the answers until yous have tried the bug).

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

Review

Questions

  1. Why can't we convert directly from number of particles to grams?
  2. How many atoms of chlorine are nowadays in the problem above?
  3. The periodic tabular array says the diminutive weight of chlorine is 35.v. Why can't I employ that value in my calculations?

Avogadro'southward Hypothesis and Molar Volume

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

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

How do scuba defined know if they will 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'southward back is equipped with gauges to tell how much gas is nowadays and what the pressure is.  A basic knowledge of gas behavior allows the diver to appraise how long to stay under water without developing problems.

Avogadro'southward Hypothesis and Molar Volume

Volume is a third way to measure the amount of matter, after item 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 close together with very little space in between the particles.  However, gases are largely composed of empty space between the bodily gas particles (meet Effigy beneath ).

Gas particles are small compared to the empty space between them

Figure x.six

Gas particles are very small compared to the large amounts of empty infinite betwixt them.

In 1811, Amedeo Avogadro explained that the volumes of all gases tin can be easily determined. Avogadro'southward hypothesis states that equal volumes of all gases at the same temperature and pressure contain equal numbers of particles.  Since the total volume that a gas occupies is fabricated up primarily of the empty space between the particles, the bodily size of the particles themselves is nearly negligible.  A given volume of a gas with small low-cal particles such equally 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, significant that when put under loftier pressure, the particles are forced closer to 1 another.  This decreases the amount of empty infinite and reduces the volume of the gas.  Gas book is also afflicted past temperature.  When a gas is heated, its molecules movement faster and the gas expands.  Considering of the variation in gas book due to pressure and temperature changes, the comparison of gas volumes must be done at one standard temperature and pressure. Standard temperature and pressure level (STP) is defined as 0°C (273.xv Thou) and 1 atm force per unit area. The tooth volume of a gas is the book of ane mole of a gas at STP.  At STP, ane mole (half dozen.02 × ten 23 representative particles) of any gas occupies a book of 22.iv L ( Effigy below ).

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

Figure 10.7

A mole of whatsoever gas occupies 22.4 L at standard temperature and pressure (0°C and 1 atm).

The Effigy below illustrates how molar volume tin be seen when comparison different gases. Samples of helium (He), nitrogen (N 2 ), and methane (CH 4 ) are at STP.  Each contains ane mole or 6.02 × ten 23 particles.  Even so, the mass of each gas is different and corresponds to the molar mass of that gas: iv.00 g/mol for He, 28.0 m/mol for North 2 , and sixteen.0 m/mol for CH 4 .

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

Figure 10.eight

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

Summary

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

Exercise

Questions

Utilise the link below to answer the post-obit 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 between gas weight and molecular weight?

Review

Questions

  1. What practise we know about the space actually taken upwards by a gas?
  2. Why practise we need to do all our comparisons at the aforementioned temperature and pressure?
  3. How can we use this data?
  • Avogadro's hypothesis: Equal volumes of all gases at the aforementioned temperature and pressure contain equal numbers of particles.
  • molar volume: The volume of ane mole of a gas at STP.
  • standard temperature and pressure level (STP): 0°C (273.xv K) and one atm pressure.

Conversions between Moles and Gas Book

  • 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?

Pocket-sized gas tanks are often used to supply gases for chemistry reactions.  A gas estimate volition give some information about how much is in the tank, simply quantitative estimates are needed so the reaction will be able to proceed to completion.  Knowing how to calculate needed parameters for gases is very helpful to avoid running out besides early.

Conversions Betwixt 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 ane mole = 22.4 L is the basis for the conversion gene.

Sample Problem One: Converting Gas Volume to Moles

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

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

Known

  • 86.v L H 2
  • ane mol = 22.iv L

Unknown

  • moles of H 2

Utilise a conversion gene 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: Remember about your result.

The volume of gas produced is nearly 4 times larger than the molar volume.  The fact that the gas is hydrogen plays no part in the calculation.

Sample Problem Two: Converting Moles to Gas Volume

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

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

Known

  • 4.96 moles O 2
  • 1 mol = 22.4 L

Unknown

  • volume of O 2

Step 2:  Calculate.

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

Step 3: Call back nigh your result.

The book seems correct given the number of moles.

Sample Problem Three: Converting Book to Mass

If we know the volume of a gas sample at STP, nosotros tin can decide how much mass is present.  Assume we have 867 liters of Due north 2 at STP.  What is the mass of the nitrogen gas?

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

Known

  • 867 Fifty Due north 2
  • one mol = 22.iv L
  • tooth mass of Northward two = 28.02 g/mol

Unknown

  • mass of N two

Step 2: Summate.

We showtime by determining the number of moles of gas present. We know that 22.4 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}

We also know the molecular weight of Northward ii (28.0 grams/mole), and so we can then 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: Recall about your result.

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

Summary

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

Exercise

Work the exercise issues at the link below. Focus on conversions between volume and moles, just try some of the others:

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

Review

Questions

  1. Why practise the gases demand to exist 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 we run a reaction to produce a gas, we await it to rise into the air.  Many students take washed 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 manus, sinks when it is released.  Carbon dioxide has a density greater that air, so it volition not rise like these other gases would.

Gas Density

As y'all know, density is defined equally the mass per unit volume of a substance.  Since gases all occupy the same volume on a per mole basis, the density of a particular gas is dependent on its molar mass.  A gas with a small molar mass will have a lower density than a gas with a big tooth mass.  Gas densities are typically reported in one thousand/50.  Gas density tin be calculated from molar mass and tooth volume.

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

Effigy 10.ix

Balloons filled with helium gas float in air considering 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?

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

Known

  • N 2 = 28.02 k/mol
  • 1 mol = 22.4 Fifty

Unknown

  • density = ? g/L

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

Step two: 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 set up with a conversion gene, the mol unit cancels, leaving g/L as the unit in the upshot.

Stride three: Think near your outcome.

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

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

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

Known

  • N 2 = 28.02 g/mol
  • 1 mol = 22.4 L

Unknown

  • tooth mass = ? g/L

Molar mass is equal to density multiplied past molar volume.

Step 2: Calculate.

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

Footstep three: Remember well-nigh your upshot.

Because the density of the gas is less than 1 g/L, the tooth mass is less than 22.4.

Summary

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

Do

Questions

Use the link below to reply 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 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 book of 3.5 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 become from here to there?

If I desire to visit the boondocks of Manteo, North Carolina, out on the coast, I volition need a map of how to become there.  I may have a printed map or I may download directions from the internet, only I need something to go me going in the correct direction.  Chemistry road maps serve the same purpose.  How do I handle a certain blazon of adding? There is a process and a set of directions to assistance.

Mole Road Map

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

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

Figure 10.10

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

The mole is at the center of whatsoever adding involving amount of a substance.  The sample problem below is one of many unlike issues that can be solved using the mole route map.

Sample Problem One:  Mole Route Map

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

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

Known

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

Unknown

  • volume = ? L

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

Step 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 three:  Retrieve nigh your consequence.

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

Summary

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

Practice

Employ the link below to bear out some practise calculations.  Do problems 1, 2, and 5 (you tin endeavour the others if you lot are feeling especially brave):

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

Review

Questions

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

Percent Composition

  • Ascertain pct composition.
  • Perform pct limerick calculations.

Nutritional information labels can let you know the percent composition

Is at that place anything good for you in this jar?

Packaged foods that you eat typically have nutritional information provided on the label.  The characterization on a jar of peanut butter (shown above) reveals that 1 serving size is considered to exist 32 grand.  The label also gives the masses of diverse types of compounds that are present in each serving.  One serving contains 7 g of protein, xv g of fatty, and three k of sugar.  By calculating the fraction of protein, fat, or sugar in ane serving of size of peanut butter and converting to per centum values, we tin can determine the composition of the peanut butter on a per centum by mass ground.

Percent Composition

Chemists frequently need to know what elements are present in a compound and in what percent.  The percent composition is the percent by mass of each element in a compound.  It is calculated in a similar way that nosotros but 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 percent limerick of a compound based on mass data.

Sample Problem One: Pct Composition from Mass

A sure newly synthesized chemical compound is known to comprise the elements zinc and oxygen.  When a 20.00 g sample of the sample is decomposed, xvi.07 1000 of zinc remains.  Make up one's mind the percentage composition of the chemical compound.

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

Known

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

Unknown

  • percent Zn = ? %
  • pct O = ? %

Subtract to observe the mass of oxygen in the compound.  Divide each element'due south mass by the mass of the compound to discover the percent by mass.

Pace ii: Summate.

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} \

Stride 3: Retrieve nigh your issue.

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

Percent Limerick from a Chemic Formula

The per centum composition of a compound can also be determined from the formula of the chemical compound.  The subscripts in the formula are first used to summate 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 percent composition of a given chemical compound is ever the aforementioned equally long equally the compound is pure.

Sample Problem Two: Percentage Composition from Chemic Formula

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

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

Known

  • mass of Cl in 1 mol Cl two O 7 = 70.xc g
  • mass of O in ane mol Cl 2 O 7 = 112.00 g
  • molar mass of Cl 2 O 7 = 182.90 g/mol

Unknown

  • percentage Cl = ? %
  • per centum O = ? %

Calculate the percent by mass of each chemical element by dividing the mass of that element in 1 mole of the compound by the tooth mass of the compound and multiplying past 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 effect.

The percentages add together up to 100%.

Percent composition can also be used to make up one's mind the mass of a certain chemical element that is contained in whatever mass of a compound.  In the previous sample problem, it 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 g sample of dichlorineheptoxide.  Y'all tin set up a conversion factor based on the percentage by mass of each chemical 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 1000, the mass of the sample size.

Summary

  • Processes are described for calculating the percent composition of a textile based on mass or on chemic composition.

Do

Apply the link below to review cloth and practice calculations. Read both parts of the lesson and exercise equally many calculations as y'all accept time for.

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

Review

Questions

  1. What is the formula for calculating percent limerick?
  2. What data do you need to calculate percent composition past mass?
  3. What do subscripts in a chemical formula tell y'all?
  • per centum composition: The per centum by mass of each chemical element in a compound.

Per centum of Water in a Hydrate

  • Define hydrate.
  • Summate the percent water in hydrate when requite relevant data.

Copper sulfate changes color when hydrated

Why does the color change?

If yous wait at a typical bottle of copper sulfate, information technology will be a bluish-green.  If someone tells you that copper sulfate is white, yous won't believe them.  Y'all are both correct; it just depends on the copper sulfate.  Your blue-green copper sulfate has several water molecules attached to it 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 color.  When the water is removed, the electron configuration changes and the color disappears.

Pct of Water in a Hydrate

Many ionic compounds naturally contain water equally office of the crystal lattice construction.  A hydrate is a compound that has one or more water molecules bound to each formula unit of measurement.  Ionic compounds that contain a transition metal are often highly colored.  Interestingly, it is mutual for the hydrated form of a chemical compound to exist of a different color than the anhydrous form, which has no water in its structure.  A hydrate can usually be converted to the anhydrous compound past heating.  For instance, 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.xi

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

The hydrated course of cobalt(II) chloride contains six h2o molecules in each formula unit.  The name of the compound is cobalt(Ii) chloride hexahydrate and its formula is CoCl ii •6H 2 O.  The formula for water is set autonomously at the terminate 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 per centum of water contained within a hydrate.  The sample problem below demonstrates the procedure.

Sample Problem One: Pct of Water in a Hydrate

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

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

The mass of h2o in the hydrate is the coefficient (six) multiplied past the molar mass of H 2 O.  The tooth mass of the hydrate is the molar mass of the CoCl 2 plus the mass of water.

Known

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

Unknown

  • percent H 2 O = ? %

Calculate the percent by mass of water by dividing the mass of H 2 O in 1 mole of the hydrate by the tooth mass of the hydrate and multiplying past 100%.

Pace 2: 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: Think nearly your result.

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

Summary

  • The process of calculating the pct water in a hydrate is described.

Practice

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

http://world wide web.sd84.k12.id.us/shs/departments/science/martz/2007_ssem2/Chemistry/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 compound that has 1 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 picture?

In the early days of chemistry, in that location were few tools for the detailed written 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 not being able to characterize a compound completely.  The relative amounts of elements could be determined, but then many of these materials had carbon, hydrogen, oxygen, and maybe 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 lowest whole-number ratio of the elements in a compound.  Because the construction of ionic compounds is an extended three-dimensional network of positive and negative ions, all formulas of ionic compounds are empirical.  However, we tin can also consider the empirical formula of a molecular compound.  Ethene is a small hydrocarbon compound with the formula C 2 H 4 (see Figure below ).  While C 2 H 4 is its molecular formula and represents its truthful molecular construction, it has an empirical formula of CH 2 .  The simplest ratio of carbon to hydrogen in ethene is 1:2.  There are 2 ways to view that ratio.  Because one molecule of ethene, the ratio is one carbon atom for every two atoms of hydrogen.  Because one mole of ethene, the ratio is i mole of carbon for every 2 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 10.12

Ball-and-stick model of ethene, C 2 H 4 .

In a process called elemental analysis , an unknown compound can be analyzed in the laboratory in lodge to determine the percentages of each chemical 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 exist taken are outlined below.

  1. Assume a 100 g sample of the chemical compound so that the given percentages tin can be direct 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, divide the moles of each chemical element by whichever of the moles from step ii 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, ane or more of the moles calculated in step 3 volition not be whole numbers.  Multiply each of the moles past the smallest whole number that will convert each into a whole number.  Write the empirical formula.

Sample Trouble I: Determining the Empirical Formula of a Compound

A chemical compound of iron and oxygen is analyzed and institute to contain 69.94% atomic number 26 and thirty.06% oxygen.  Find the empirical formula of the compound.

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

Known

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

Unknown

  • Empirical formula = Atomic number 26 ? O ?

Steps to follow are outlined in the text.

Footstep 2:  Summate.

1. Assume a 100 yard sample.

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

2. Catechumen 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. Divide both moles by 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}

4/5. Since the moles of O, is still non a whole number, both moles can be multiplied by ii, 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 Atomic number 26 ii O 3 .

Stride iii: Think almost your outcome.

The subscripts are whole numbers and represent 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 percent composition of that compound.

Practice

Apply the link beneath to read about calculating empirical formulas and practise 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?
  3. What does it not tell you?
  • 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 requite the empirical formula and the molar mass of the compound.

Fischer projection of glucose

Structure of sucrose

How can you lot decide the differences between these 2 molecules?

Above we see 2 carbohydrates: glucose and sucrose.  Sucrose is nearly exactly twice the size of glucose, although their empirical formulas are very similar.  Some people could distinguish them on the basis of gustatory modality, just it's not a skillful idea to become around tasting chemicals. The best way is to determine the molecular weights – this approach allows y'all 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 chemical compound.  In many cases, the molecular formula is the same every bit the empirical formula.  The molecular formula of marsh gas is CH iv and because it contains but i carbon cantlet, that is too its empirical formula.  Sometimes, yet, the molecular formula is a simple whole-number multiple of the empirical formula.  Acerb acid  is an organic acid that is the main component of vinegar.  Its molecular formula is C 2 H 4 O 2 .  Glucose is  a uncomplicated sugar that cells utilize as a chief source of energy.  Its molecular formula is C half dozen H 12 O half dozen .  The structures of both molecules are shown in the effigy below.  They are very different compounds, yet both have the aforementioned empirical formula of CH 2 O.

Acetic acid and glucose both the same empirical formula

Figure ten.xiii

Acetic acrid (left) has a molecular formula of C 2 H 4 O two , while glucose (correct) has a molecular formula of C six H 12 O 6 . Both accept the empirical formula CH 2 O.

Empirical formulas can exist determined from the percent composition of a chemical compound.  In order to determine its molecular formula, it is necessary to know the molar mass of the chemical compound.  Chemists use an instrument called a mass spectrometer to decide the molar mass of compounds.  In gild to get 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. Carve up the molar mass of the compound by the empirical formula mass. The event should be a whole number or very shut to a whole number.
  3. Multiply all the subscripts in the empirical formula by the whole number constitute in step two. The result is the molecular formula.

Sample Problem One: 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 thou/mol.  Determine the molecular formula of the chemical compound.

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

Known

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

Unknown

  • molecular formula = ?

Steps to follow are outlined in the text.

Footstep 2: Summate.

one. The empirical formula mass (EFM) = xiii.84 chiliad/mol

2. 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 chemical compound is B 2 H 6 .

Stride three: Remember nigh your result.

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

Summary

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

Practice

Use the link below to access exercise issues.  Try every bit many equally you have time for:

http://chemical science.about.com/od/chemistry-test-questions/tp/Molecular-Formula-Practise-Examination-Questions.htm

Review

Questions

  1. What is the difference between an empirical formula and a molecular formula?
  2. In improver to the elemental analysis, what do you demand to know to calculate the molecular formula?
  3. What does the empirical formula mass tell you?
  • empirical formula mass (EFM): The tooth mass represented past 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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