Mole Concept

 

 How you measure how much? You can measure mass, or volume, or you can count pieces. We measure mass in grams. We measure volume in liters. We count pieces in MOLES. 

 

For modern systems; 12C is taken as a standard. So; 12C is assumed to be 12 a.m.u. 1

C = 12 a.m.u.

12

 

1 a.m.u.=1/12 of the mass of23a single 12C atom =1/12 (12/6.02 X 10 )   =1/6.02 X 1023 grams   =mass of 1 1H atom Then by comparing the masses of each atom with 12

theeasily mass calculated of C atomic masses of other atoms can be “RELATIVE “RELATIVE ATOMIC MASSES”. Relative atomic masses easily calculated by MASS by  MASS SPECTROMETER. 1 16O = 16 a.m.u 1 40Ca = 40a.m.u. 1 1H = 1 a.m.u.

 

Moles Defined as the number of carbon atoms in exactly 12 grams of carbon-12.  1 mole contains 6.02 x 1023  particles. Treat it like a very large dozen 6.02 x 1023  is called Avogadro’s number.

 

Representative particles The smallest pieces of a substance. For a molecular compound it is a molecule. For an ionic compound it is a formula unit. For an element it is an atom.

 

Units of ameasurement withtogether equations. . equations We use mole conceptused to bring the concepts of counting numbers and atomic weights of elements. The mole is derived from the following information. Atomic weights are an average of the relative masses of all of the isotopes of the given element. The number of C-12 atoms in exactly 12.00 g of C-12 is 6.02 X 1023. This called Avogadro’s number.

 

An amount of a substance thations, contains Avogadro’s number of atoms, molecules, or any other chemical unit is MOLE.. called A MOLE

A mole of C-12 atoms is defined as having a mass of exactly 12.00 g, a mass that is equal to its atomic weight.

 

 Mole Calculations

1 mole (mol): Amount of matter that contains as many objects (atoms, molecules) as the number of atoms in 12

isotopically pure C. 6.022 x 1023 Also known as Avogadro’s number. 1 mol

12

C atoms = 6.02 x 1023 12C atoms

1 mol H2O molecules = 6.02 x 10 23  molecules 1 mole NO3- ions = 6.02 x 10 23 NO3- ions.

 

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The mole concept for (A) elements, (B) compounds, and (C) molecular substances. A mole contains 6.02 X 1023 particles. Since every mole contains same number ofthe particles, the ratio of the mass of any two moles is the same as the ratio of the masses of individual particles making up the two moles.

 

Molar Mass The atomic mass of any substance expressed in grams corresponds to 1 mol of the t he substance. Atomic mass of a substance expressed in grams is the molar mass. The molar mass of a diatomic substance is equal to twice twice its  its atomic mass.

 

Molar Mass The generic term for the mass of one mole. The same as gram molecular mass, gram formula mass, and gram atomic mass.

 

Gram Atomic Mass The mass of 1 mole of an element in grams. 12.01 grams of carbon has the same number of pieces as 1.008 grams of hydrogen and 55.85 grams of iron. We can right this as 12.01 g C = 1 mole We can count things by weighing them.

 

Mole Calculations Calculate the number of sodium atoms in 0.120 mol Na?

0.120 mol Na x 6.02 x 10 23 atoms Na = 7.22 x 10 22  mol Na

1

 

 Mole Calculations Calculate the number of moles of potassium potass ium in 1.25 x 1021 atoms of K.

1.25 x 1021 atoms K x 1 mol K = 2.08 x 10-3 mol K   

6.02 x 1023 atoms K 

 

 Mole Calculations What is the mass in grams of 2.01 x 1022 atoms of sulfur?

2.01 x 1022 atoms S x 1 mol S  

x

6.02 x 1023 atoms S = 1.07g S

32.07 g S 1 mol S

 

Mole Calculations How many O2 molecules are present in 0.470g of oxygen gas?

0.470 g O2 x 1mol O2 x 6.02 x 1023 molecules O2   32.00 g O2

1 mol O2

= 8.84 x 1021 molecules O2

 

What about compounds? In 1 mole of H2O molecules there are two moles of H atoms and 1 mole of O atoms To find the mass of one mole of a compound determine the moles of the elements they have Find out how much they would weigh add them up.

 

 Molar Mass

 

 The Mole Calculate the number of Magnesium and Chlorine ions present in 0.450 mol of MgCl 2.

# Mg2+ ions: 0.450 mol MgCl2 x 6.02 x 1023 formula units x 1 Mg2+  mole

1 formula unit # Mg2+ ions = 2.71 x 1023  # Cl- ions: 0.450 mol MgCl2 x 6.02 x 1023 formula units x 2 Cl-  1 mole

1 formula unit

1

# Cl- ions = 5.42 x 1023  

The mole concept applied to compounds The formula weight of a species is the sum of atomic masses (amu) of the atoms in a species. Molecular weight of NH3 =   For an ionic cmpd formula weight

MgF2 =

 

Mass of one mole of MgF2 is Mass of one formula unit of MgF 2 is Mass of 6.022 x 1023 formula units of MgF2 is

 

Gram Formula Mass The mass of one mole of an ionic compound. Calculated the same way. What is the GFM of Fe 2O3? 2 moles of Fe x 55.85 g = 111.70 g 3 moles of O x 16.00 g = 48.00 g The GFM = 111.70 g + 48.00 g = 159.70g

 

Molar Mass

 

What about compounds? What is the the mass mass of one mole of CH4?

1 mole of C = 12.01 g 4 mole of H x 1.01 g = 4.04g 1 mole CH4 = 12.01 + 4.04 = 16.05g The Gram Molecular mass of CH4 is 16.05g The mass of one mole of a molecular compound.

 

The gram atomic weight of weight of an element is the mass in grams of one mole of an element that is numerically equal to its atomic weight. weight of  of a compound The gram formula weight is the mass in grams of one mole of the compound that is numerically equal to its formula weight. The gram formula weight of a compound is the sum total of all the individual atomic weight in the formula. The gram molecular weight is weight is the gram formula weight of a molecular

compound.  

Molar Mass Calculate the molar mass of Ag and of magnesium nitrate, Mg(NO3)2. Ag is an element so its molar mass equals its atomic mass = 107.87g/mol. Mg(NO3)2 = 24.31 + 2(14.01 + 3 x 16.00)  

= 148.33 g/mol

 

For example How many moles is 5.69 g of NaOH?

   5.69 g  

1   mole  0.142 142 mol NaOH NaOH 40.00  g    = 0.



need to change grams to moles

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for NaOH

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1mole Na = 22.99g

1 mol O = 16.00 g 1 mole of H = 1.01 g

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1 mole NaOH = 40.00 g

 

 Mole Calculations Calculate the mass in grams of a single molecule of carbon dioxide, CO2.

44.01 g CO2 x 1mol CO2 = 7.31 x 10-23 g/molecule 23

  1 mol CO2 

6.02 x 10  molecules

 

Molar Mass The number of grams of 1 mole of atoms,ions,or molecules. We can make conversion factors from these. To change grams of a compound to moles of a compound.

 

Examples How many moles is 4.56 g of CO2 ? How many grams is 9.87 moles of H2O? How many molecules in 6.8 g of CH C H 4? 49 molecules of C6H12O6 weighs how much?

 

Calc the molar mass of Ca(NO 3)2.

Calc the molar mass of a compound if 0.372 mol of it has a mass of 152g.

 

How many grams0.100 of each areof  required to have mol A. NaOH B. H2SO4   C. C2H5OH  

D. Ca3(PO4)2

 

 How many moles are in 50.0 g of  A. CS2  

B. Al2(CO3)3

  C. Sr(OH)2 D. LiNO3

 

Calc the no. of C, H, and O atoms atoms in 1.50 g of glucose (C6H12O6). ).  

What is the average mass of one C 3H8  molecule?

What is the mass of 5.00 x 1024 molecules of NH3?

 

Types of questions How many molecules of CO2 are the in 4.56 moles of CO2 ? How many moles of water is 5.87 x 10 22  molecules? How many there in 1.23 many atoms of carbon are there moles of C6H12O6 ? How many moles is 7.78 x 1024 formula units

of MgCl2?  

Gases Many of the chemicals we deal with are gases. They are difficult to weigh. Need to know how many moles of gas we have. Two things effect the volume of a gas Temperature and pressure Compare at the same temp. and pressure.

 

Standard Temperature and Pressure 0ºC and 1 atm pressure abbreviated STP At STP 1 mole of gas occupies 22.4 L Called the molar volume Avogadro’s Hypothesis Hypothesis  - at the same temperature and pressure equal volumes of gas have the same number of particles.

 

 Molar Volume Avogadro's Theory: Two gases containing equal numbers of molecules occupy equal volumes under similar conditions. Standard Temperature and Pressure: 0OC and 1 atm. Molar Volume at STP = 22.4 L

 

Examples What is the volume of 4.59 mole of CO2 gas at STP? How many moles is 5.67 L of O2 at STP? What is the volume of 8.8g of CH4 gas at STP?

 

Molar Volume Density of gas = molar mass in grams   molar volume in liters Calculate the density of ammonia gas, NH 3, at STP. Density of NH3 = molar mass NH3 = 17.04g   molar volume NH3 22.4 L  = 0.761 g/L

 

Density of a gas  D = m /V for a gas the units wil willl be g / L • We can determine the density of any gas at STP if we know its formula. • To find the density we need the mass and the volume. If you assume you have 1 mole then the mass is the molar mass (PT) At STP the volume is 22.4 L.

 

Examples Find the density of CO2 at STP. Find the density of CH4 at STP.

 

Volume 22.4 L

PT

Mass

Moles 23

6.02 x 10   Representative   Particles Atoms

Ions

 

 Mole Calculations  What is the mass of 3.36 L of ozone gas, O3, at STP?

 

3.36 L O3 x 1 mol O3 x 48.00g O3 = 7.20 g O3 22.4 L O3 1 mol O3

 

 Mole Calculations How many molecules of hydrogen gas, H2, occupy 0.500 L at STP? 0.500 L H2 x 1 mol x

 

6.02 x 1023 molecules H2

22.4 L

1 mol 22

= 1.34 x 10  molecules H2

 

 Mole Calculations

 

 Percent Composition Percentage Composition: Percentage by mass contributed by each element in a substance. % Element (E) = (# atoms of Element)(AW of E) x 100%  Formula Weight of Compound

According to the law of definite composition the elements in a compound are always present in the same proportion by mass.

 

Percentage composition by mass of a compound

It shows us how many grams of each element exist in 100 gram of a compound.

 

Percent Composition Like all percents  Part

x

100 %

whole

Find the mass of each component, divide by the total mass.

 

 Percent Composition

Calculate the percentage of Nitrogen, by mass in Ca(NO3)2:

 

Formula Weight = 1 x Ca + 2 x N + 6 x O = 1 x 40.1 + 2 x 14.0 + 6 x 16.0   = 164.1 amu

% N = (2)(14.0) = 17.1% 164.1 

 

 Percent Composition

Calculate the percent composition of trinitrotoluene (TNT), C7H5(NO2)3.

Assume that there is 1 mol of o f TNT. 1 mol of TNT contains: 7 mol carbon atoms = 84.07 g 5 mol hydrogen atoms = 5.05 g 3 mol of nitrogen atoms = 42.03 g 6 mol oxygen atoms = 96.00 g

 

 Percent Composition Calculate the molar mass of TNT: Molar mass = 227.15 g/mol. Compare the mass of each element to the molar mass of the compound:

  84.07 g C x 100 = 37.01% C   227.15 g C7H5(NO2)3

 

Percent Composition 5.05g H x 100 = 2.22% H 227.15 g C7H5(NO2)3 42.03 g N x 100 = 18.50% N 227.15 g C7H5(NO2)3 96.00 g 0 x 100 = 42.26% O 227.15 g C7H5(NO2)3

 

 Empirical Formula Empirical formula gives: the ratio of atoms in a molecule: H2O. the ratio of moles of atoms on the molar level. The molar ratio of the elements in a compound allows the calculation of the subscripts in the empirical formula of a compound.

 

 Empirical Formula Percent composition - the mass percent of each element present in a compound. 1. Calculate the chemical chemical formula from percent composition. 2.Find the relative numbers of moles of each element in the compound. 3.Use molar masses of the elements as conversion factors.

 

 Empirical Formula Find the ratio of the numbers of moles by dividing the larger number of moles by the smaller number. Multiply the subscripts by small integers in a trial-and-error procedure until whole numbers are found. 

 

Empirical• Formula The flow diagram summarizes the •

 relationship mass percent,among moles, mole ratios and subscripts in the formula of a compound used in the determination of empirical formula from percent composition.

 

Calculating Empirical Just find the lowest whole number ratio C6H12O6  CH2O It is not just the ratio of atoms, it is also the ratio of moles of atoms. In 1 mole of CO2 there is 1 mole of carbon and 2 moles of oxygen.

In one molecule of CO 2 there is 1 atom of C and  

Calculating Empirical Means we can get ratio from percent composition. Assume you have a 100 g. The percentages become grams. Can turn grams to moles. Find lowest whole number ratio by dividing by the smallest.

 

Example Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N.

Assume 100 g so 38.67 g C x 1mol C 12.01 gC 16.22 g H x 1mol H 1.01 gH 45.11 g N x 1mol N 14.01 gN

= 3.220 mole C

= 16.09 mole H = 3.219 mole N

 

Example The ratio is 3.220 mol C = 1 mol C 3.219 mol N 1 mol N The ratio is 16.09 mol H = 5 mol H 3.219 mol N 1 mol N C1H5 N1 • A compound is 43.64 % P and 56.36 % O. What is the empirical formula? • Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. What is its empirical formula?

 

 Empirical Formula Q. The percent composition of a solid is known to Q.  be 68.4% Ba, 10.3% P, and 21.3% O O.. What is the empirical formula of the compound? A. Assuming  Assuming a 100 g sample gives us 68.4 g Ba, A. 10.3 g P, and 21.3 g O. Convert these masses to numbers of moles: 0.498 moles Ba, 0.332 mol P, 1.33 mol O

 

 Empirical Formula Knowing the relative numbers of moles, find the ratio by dividing the two larger numbers by the smaller number. Answer = 1.5 The O:Ba:P ratio of 4:1.5:1 gives an empirical formula of Ba1.5PO4. However, we must multiply the subscripts by a small integer (in this case 2) to find whole numbers for the formula. Ba3P2O8, or Ba3(PO4)2.

 

 Empirical Formula Q. What is the percent composition of sodium hydrogen carbonate?

A. The formula for sodium hydrogen hydro gen carbonate is  NaHCO3. The Na:H:C:O mole ratio is 1:1:1:3. Convert this mole ratio into a mass ratio r atio by assuming there is a 1 mole sample present. Answer = 23g Na, 1.0g H, 12g C, 48 g O

 

 Empirical Formula To determine the percent composition, divide the mass of each element present by the total mass of the compound and multiply by 100. Total mass of 1 mole of NaHCO3 = 84 g Answer = 27% Na, 1.2% H, 14% C, 57% O 

 

 Empirical Formula 0.500 g of scandium was heated and allowed to react with oxygen. The resulting product oxide had a mass of 0.767g. What is the empirical formula for scandium oxide? Calculate the number of moles of scandium:   0.500 g Sc x 1 mol Sc = 0.0111 mol Sc   44.96 g Sc

 

 Empirical Formula Calculate the moles of oxygen: Mass of oxygen = 0.767g– 0.500g O = 0.267 g O  Number of moles of oxygen: oxygen: 0.267g x 1 mol 0 = 0.0167 mol O   16.00 g Mole ratio of Sc:O is 0.0111:0.0167 Divide by smallest number: 1:1.50

 

Empirical Formula Sc1.00O1.50 Using the law of multiple proportions we need a whole number ratio: Sc2.00O3.00

 

Empirical Formula

Calculate the empirical formula of glycine given that it contains 32.0% carbon, 6.7% hydrogen, 18.7% nitrogen and 42.6% oxygen.

Assume a 100g sample. Then the percent composition equals number of grams.

 

 Empirical Formula 32.0 g C, 6.7g H, 18.7g N, 42.6g O. Convert grams into moles:   32.0 g C x 1 mol = 2.66 mol C    

12.01 g C   6.7 g H x 1 mol H = 6.6 mol H 1.01 g H

 

 Empirical Formula 18.7 g N x 1 mol N = 1.33 mol 14.01 g N 42.6 g O x 1 mol O = 2.66 mol O 16.00 g O Divide the mole ratios by the smallest number: 1.33 C2H5 N1O2

 

The Empirical Formula The lowest whole number ratio of elements in a compound. The molecular formula the actual ration of elements in a compound. The two can be the same. CH2 empirical formula C2H4 molecular formula

C3H6 molecular formula H2O both  

Molecular Formula The formula obtained from percentage compositions is always the empirical formula. Molecular formula - gives the actual numbers of atoms in a molecule.   1. May be the same as the empirical formula. 2. May be a multiple of the empirical formula. a. Multiple = Molecular mass   Empirical Formula

 

Empirical to molecular Since the empirical formula is the lowest ratio the actual molecule would weigh more. By a whole number multiple. Divide the actual molar mass by the the mass of one mole of the empirical formula.

 

 Molecular Formula The empirical formula for fructose is CH2O. If the molar mass of fructose is 180 g/mol, what is the molecular formula?

Molar mass of the empirical formula = 30 g/mol.

 

 Molecular Formula Multiple = molar mass of fructose  

molar mass of empirical formula Multiple = 180 g/mol = 6  

30 g/mol

Multiply the empirical formula by 6: C6H12O6

 

Molecular Formula

Q. Determine the molecular formula of ethyl butyrate. The molecular mass of the compound is 116 g/mol and the empirical formula is C3H6O. A. The molecular formula of the compound may be the same as the empirical formula or it may be a multiple of the empirical formula. For ethyl butyrate: Multiple = 116 =2 58 The subscripts in the empirical formula are

multiplied by 2 giving a molecular formula of C 6H12O2.  

Some problems How many atoms are there in 5.10 moles of sulfur? What’s the mass of 5.10 moles of S? How many moles of calcium atoms are in 1.16 x 1024 atoms of Ca? How many grams?

 

Which of the following has more atoms: 1.10g of hydrogen atoms or 14.7 g of chromium atoms? How many moles are in 0.040 kg Na?

 

Example A compound is known to be composed of 71.65 % Cl, 24.27% C and 4.07% H. Its molar mass is known (from gas density) is known to be 98.96 g. What is its molecular formula?

 

STRUCTURAL FORMULA The atoms in a molecule are connected or chemically bonded in a precise way. A SF. Shows how the atoms in a molecule are arranged. For ex: H2O C2H6

H-O-H

CH3 

H-C- C- H H

H

 

Empirical formula Empirical formula •The simplest whole number ratio of atoms of elements in a compound, described with the use of subscripts. •Ionic compounds are always shown as empirical formulas. formulas.

Molecular Formula The actual numbers of atoms in a molecule. molecule.

Structural Formula Show the relative arrangements of atoms

in a molecule

 

•If you know the name of an ingredient, you can write a chemical formula, and the percent composition of a particular substance can be calculated from the

formula. This can be useful information for consumer decisions.

 

HYDRATES Solids which are found in combined form with water in definite proportion are called HYDRATES.  HYDRATES. When hydrates areasheated, H2O  evaporates, and only solid is obtained in amorphous . (w/o a certain geometric structure, generally in powdered form. H2O

molecules surround ionic substances with certain amounts.  

WATER OF HYDRATION : HYDRATION : Water molecules of a hydrate.

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Na2CO3.10H2O

Na2CO3(s) + 10H2O(g)

DEHYDRATION:  Evaporation of water of hydration.  DEHYDRATION: 

 Na2CO3.10H2O,

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4 2 CaSO .2H O,  CuSO .5H O 4 2