Dimensional Analysis

• Conversion factors

• Units

• Cancel to solve

Counting Counting AtomsAtoms

Mg burns in air (OMg burns in air (O22) to ) to produce white magnesium produce white magnesium oxide, MgO. oxide, MgO.

How can we figure out how How can we figure out how much oxide is produced much oxide is produced from a given mass of Mg?from a given mass of Mg?

Counting AtomsCounting AtomsChemistry is a quantitative Chemistry is a quantitative

science—we need a science—we need a “counting unit.”“counting unit.”

1 mole is the amount of 1 mole is the amount of substance that contains as substance that contains as many particles (atoms, many particles (atoms, molecules) as there are in molecules) as there are in 12.0 g of 12.0 g of 1212C.C.

1 mole is the amount of 1 mole is the amount of substance that contains as substance that contains as many particles (atoms, many particles (atoms, molecules) as there are in molecules) as there are in 12.0 g of 12.0 g of 1212C.C.

MOLEMOLE

Particles in a Particles in a MoleMole

6.02214199 x 106.02214199 x 102323

Avogadro’s NumberAvogadro’s Number

There is Avogadro’s number of particles in a mole of any substance.There is Avogadro’s number of particles in a mole of any substance.

Amedeo AvogadroAmedeo Avogadro1776-18561776-1856

Converting particles to moles

moles = representative particles x 1 mole

6.02 x 1023 representative particles(atoms/molecules/formula units)

moles = 2.80 x 1024 atoms Si x 1 mole Si

6.02 x 1023 atoms Si(atoms/molecules/formula units)

= 4.65 moles Si

Converting moles to particles

Rep part = moles x

6.02 x 1023 representative particles

1 mole

Rep part = 1.14 mol SO3 x

6.02 x 1023 molecule

1 mole

= 2.75 x 1024 molecules SO3

The mass of moles

• The atomic mass of an element expressed in grams is the mass of a mole of the element.

• Molar Mass - the mass of a mole of an element.• Molar mass of a compound is the total mass of all

the atoms making up the compound.• Molar mass of a compound is calculated by adding

the masses of each element in the compound. – Take the subscript and multiply it by the molar mass of

that element then add all the elemental masses together.

Molar MassMolar Mass1 mol of 1 mol of 1212C C

= 12.00 g of C= 12.00 g of C = 6.022 x 10 = 6.022 x 102323 atoms atoms

of Cof C

12.00 g of 12.00 g of 1212C is its C is its

MOLAR MASSMOLAR MASS

Taking into account all of Taking into account all of

the isotopes of C, the the isotopes of C, the

molar mass of C is molar mass of C is

12.011 g/mol12.011 g/mol

One-mole AmountsOne-mole Amounts

PROBLEM: What amount of PROBLEM: What amount of Mg is represented by 0.200 g? Mg is represented by 0.200 g? How many atoms?How many atoms?

PROBLEM: What amount of PROBLEM: What amount of Mg is represented by 0.200 g? Mg is represented by 0.200 g? How many atoms?How many atoms?

Mg has a molar mass of 24.3050 g/mol.

0.200 g • 1 mol

24.31 g = 8.23 x 10-3 mol

8.23 x 10-3 mol • 6.022 x 1023 atoms

1 mol

= 4.95 x 10= 4.95 x 102121 atoms Mg atoms Mg

How many atoms in this piece of Mg?

Converting moles to mass

mass = number of moles x mass (grams)

1 mole

mass = 3.00 mol Na x 22.98977grams Na

1 mole Na

= 69.0 grams Na

Converting mass to moles

moles = mass in grams x 1 mole

mass (grams)

moles = 10.0 g Na2SO4 x 1 mole Na2SO4

142.1 grams Na2SO4

= 0.070 g Na2SO4

Mole to volume

• Avogadro’s hypothesis states equal volumes of gases at the same temperature and pressure contain equal numbers of particles.

• STP – 0oC, 273 K / 1 atm, 101.3 kPa

• Molar volume at STP – 22.4 L/mol

Converting mole to volume

volume = moles of gas x 22.4 L

1 mol at STP

= 8.40 L O2

volume = 0.375moles of O2 x 22.4 L

1 mol at STP

moles = 0.200 L H2 x 1 mol

22.4 L at STP

Converting volume to moles

moles of gas = volume x 1 mol

22.4 L at STP

= 8.93 x 10-3 L H2

Converting Molar Mass from Density

• Density is mass per unit volume (g/L)

• Molar mass grams per mol

• Need to cancel moles and insert liters

• Use molar volume – mol/liter

Grams = grams 22.4 LxMoles L 1 mol

Converting Molar Mass from Density

Grams = grams 22.4 LxMoles L 1 mol

= 43.99 g/mol

= 1.964 grams 22.4 Lx 1L 1 mol

molar mass(g/mol)

Molgrams

liters

atoms

molar mass

Avogadro’s number

molar volume

22.4 L/mol

6.02 x 1023 particles/mol

Periodic table

•• •

•

••

Percent Composition

• Relative amounts (mass) of elements in a compound.

• Percent by mass.• Mass of the element divided by the mass of the

compound times 100.• Use the subscripts to determine the mass of each

element and the mass of the compound if the question calls for a percent composition and only the formula is given.

Percent CompositionPercent CompositionPercent CompositionPercent CompositionConsider NOConsider NO22, Molar mass = ?, Molar mass = ?

What is the weight percent of N and What is the weight percent of N and of O?of O?

Wt. % O 2 (16 .0 g O per mole )46 .0 g

x 100 % 69 .6%Wt. % O 2 (16 .0 g O per mole )46 .0 g

x 100 % 69 .6%

Wt. % N = 14.0 g N

46.0 g NO2 • 100% = 30.4 %Wt. % N =

14.0 g N46.0 g NO2

• 100% = 30.4 %

What are the weight percentages of What are the weight percentages of N and O in NO?N and O in NO?

Determining Determining FormulasFormulas

Determining Determining FormulasFormulas

In In chemical analysischemical analysis we determine the % by we determine the % by weight of each element in a given amount of pure weight of each element in a given amount of pure compound and derive the compound and derive the EMPIRICALEMPIRICAL or or

SIMPLESTSIMPLEST formula. formula. Remember Remember formulas formulas are mole to mole ratios.are mole to mole ratios.

PROBLEMPROBLEM: A compound of B and H : A compound of B and H is 81.10% B. What is its empirical is 81.10% B. What is its empirical formula?formula?

• Because it contains only B and H, it Because it contains only B and H, it must contain 18.90% H.must contain 18.90% H.

• Assume a 100 gram sample.Assume a 100 gram sample.• In 100.0 g of the compound there In 100.0 g of the compound there

are 81.10 g of B and 18.90 g of H.are 81.10 g of B and 18.90 g of H.• Calculate the number of moles of Calculate the number of moles of

each constituents.each constituents.

A compound of B and H is 81.10% A compound of B and H is 81.10% B. What is its empirical formula?B. What is its empirical formula?A compound of B and H is 81.10% A compound of B and H is 81.10% B. What is its empirical formula?B. What is its empirical formula?

Calculate the number of moles of Calculate the number of moles of each element in 100.0 g of sample.each element in 100.0 g of sample.

81.10 g B • 1 mol

10.81 g = 7.502 mol B81.10 g B •

1 mol10.81 g

= 7.502 mol B

18.90 g H • 1 mol

1.008 g = 18.75 mol H18.90 g H •

1 mol1.008 g

= 18.75 mol H

A compound of B and H is 81.10% A compound of B and H is 81.10% B. What is its empirical formula?B. What is its empirical formula?A compound of B and H is 81.10% A compound of B and H is 81.10% B. What is its empirical formula?B. What is its empirical formula?

Now, recognize that Now, recognize that atoms combine in atoms combine in the ratio of small whole numbers.the ratio of small whole numbers.

1 atom B + 3 atoms H --> 1 molecule BH1 atom B + 3 atoms H --> 1 molecule BH33

oror

1 mol B atoms + 3 mol H atoms ---> 1 mol B atoms + 3 mol H atoms ---> 1 mol BH1 mol BH33 molecules molecules

Find the ratio of moles of elements Find the ratio of moles of elements in the compound.in the compound.

A compound of B and H is 81.10% A compound of B and H is 81.10% B. What is its empirical formula?B. What is its empirical formula?A compound of B and H is 81.10% A compound of B and H is 81.10% B. What is its empirical formula?B. What is its empirical formula?

But we need a whole number ratio. But we need a whole number ratio.

2.5 mol H/1.0 mol B = 5 mol H to 2 mol B2.5 mol H/1.0 mol B = 5 mol H to 2 mol B

EMPIRICAL FORMULA = BEMPIRICAL FORMULA = B22HH55

Take the ratio of moles of B and H. Take the ratio of moles of B and H. AlwaysAlwaysdivide by the smaller number.divide by the smaller number.

18.75 mol H7.502 mol B

= 2.499 mol H1.000 mol B

= 2.5 mol H1.0 mol B

18.75 mol H7.502 mol B

= 2.499 mol H1.000 mol B

= 2.5 mol H1.0 mol B

A compound of B and H is 81.10% A compound of B and H is 81.10% B. What is its empirical formula?B. What is its empirical formula?A compound of B and H is 81.10% A compound of B and H is 81.10% B. What is its empirical formula?B. What is its empirical formula?

A compound of B and H is 81.10% B. A compound of B and H is 81.10% B. Its Its empirical formulaempirical formula is B is B22HH55. What . What

is its is its molecular formulamolecular formula??

Is the molecular formula BIs the molecular formula B22HH55, B, B44HH1010, B, B66HH1515, ,

BB88HH2020, etc.? , etc.?

BB22HH66 is one example of this class of compounds. is one example of this class of compounds.

B2H6

Molecular formulas

• Molecular formulas are either the empirical formula or it is a whole number multiple of its empirical formula.

• Determine the empirical formula first, to determine the molar mass of the compound.

• Divide the experimental molar mass by the empirical molar mass. This give you the multiplier to convert the empirical formula to the molecular formula.

A compound of B and H is 81.10% B. Its empirical A compound of B and H is 81.10% B. Its empirical

formula is Bformula is B22HH55. What is its molecular formula. What is its molecular formula??A compound of B and H is 81.10% B. Its empirical A compound of B and H is 81.10% B. Its empirical

formula is Bformula is B22HH55. What is its molecular formula. What is its molecular formula??

We need to do an We need to do an EXPERIMENTEXPERIMENT to find the to find the MOLAR MASSMOLAR MASS..

This experiment gives This experiment gives 53.3 g/mol53.3 g/molCompare with the mass of BCompare with the mass of B22HH55

= = 26.66 g/unit26.66 g/unit

Find the ratio of these masses.Find the ratio of these masses.53.3 g/mol

26.66 g/unit of B2H5 =

2 units of B2H51 mol

53.3 g/mol26.66 g/unit of B2H5

= 2 units of B2H5

1 mol

Molecular formula = BMolecular formula = B44HH1010

Determine the formula of a Determine the formula of a compound of Sn and I using the compound of Sn and I using the following data.following data.

Determine the formula of a Determine the formula of a compound of Sn and I using the compound of Sn and I using the following data.following data.

• Reaction of Sn and IReaction of Sn and I22 is done using excess Sn. is done using excess Sn.• Mass of Sn in the beginning = 1.056 gMass of Sn in the beginning = 1.056 g• Mass of iodine (IMass of iodine (I22) used ) used = 1.947 g = 1.947 g• Mass of Sn remaining Mass of Sn remaining = 0.601 g= 0.601 g• See p. 104See p. 104

Find the mass of Sn that combined Find the mass of Sn that combined with 1.947 g Iwith 1.947 g I22..

Mass of Sn initially = 1.056 gMass of Sn initially = 1.056 g

Mass of Sn recovered = 0.601 gMass of Sn recovered = 0.601 g

Mass of Sn used = 0.455 gMass of Sn used = 0.455 g

Find moles of Sn used:Find moles of Sn used:0.455 g Sn • 1 mol

118.7 g = 3.83 x 10-3 mol Sn0.455 g Sn •

1 mol118.7 g

= 3.83 x 10-3 mol Sn

Tin and Iodine Tin and Iodine CompoundCompound

Tin and Iodine Tin and Iodine CompoundCompound

Now find the number of moles of INow find the number of moles of I22 that that combined with 3.83 x 10combined with 3.83 x 10-3-3 mol Sn. Mass of I mol Sn. Mass of I22 used was 1.947 g.used was 1.947 g.

1.947 g I2 • 1 mol

253.81 g = 7.671 x 10-3 mol I21.947 g I2 •

1 mol253.81 g

= 7.671 x 10-3 mol I2

How many mol of How many mol of iodine atomsiodine atoms? ?

= 1.534 x 10-2 mol I atoms= 1.534 x 10-2 mol I atoms

7.671 x 10-3 mol I2 2 mol I atoms

1 mol I2

7.671 x 10-3 mol I2 2 mol I atoms

1 mol I2

Tin and Iodine Tin and Iodine CompoundCompound

Now find the ratio of number of moles of Now find the ratio of number of moles of moles of I and Sn that combined.moles of I and Sn that combined.

1.534 x 10-2 mol I

3.83 x 10-3 mol Sn =

4.01 mol I1.00 mol Sn

1.534 x 10-2 mol I

3.83 x 10-3 mol Sn =

4.01 mol I1.00 mol Sn

Empirical formula is Empirical formula is

SnISnI44