Unit 6: The Mathematics of Chemical Formulas Chemistry Na 2 CO 3. 10 H 2 O p v MOL m Cu(OH) 2 SO 3.

Unit 6:The Mathematics ofChemical Formulas

Chemistry

Na 2CO 3

. 10 H 2O

p

v MOL

m

Cu(OH)2

SO3

# of H2O molecules

# of H atoms # of O atoms

1

2

3

100

6.02 x 1023

12

4

16.0 g

2

6 3

100200

2 (6.02 x 1023) 6.02 x 1023

2.0 g18.0 g

molar mass: the mass of one mole of a substance

PbO2

HNO3

ammonium phosphate

Pb: 1 (207.2 g) = 207.2 g

O: 2 (16.0 g) = 32.0 g 239.3 g

H: 1 (1.0 g) = 1.0 g

N: 1 (14.0 g) = 14.0 g 63.0 g

O: 3 (16.0 g) = 48.0 g

(NH4)3PO4

H: 12 (1.0 g) = 12.0 g

N: 3 (14.0 g) = 42.0 g

149.0 g P: 1 (31.0 g) = 31.0 g

NH41+ PO4

3–

O: 4 (16.0 g) = 64.0 g

percentage composition: the mass % of each element in a compound

Find % composition.

PbO2

(NH4)3PO4

% of element =g element

molar mass of compoundx 100

207.2 g Pb 239.2 g : = 86.6% Pb

32.0 g O 239.2 g = 13.4% O

31.2 g P 149.0 g = 20.8% P

64.0 g O 149.0 g = 43.0% O

42.0 g N 149.0 g = 28.2% N

12.0 g H 149.0 g = 8.1% H

:

::

::

(see calcs above)

zinc acetate

Zn2+ CH3COO1–

Zn(CH3COO)2

183.4 g = 3.3% H

= 34.9% O

= 35.7% Zn

= 26.2% C :

C: 4 (12.0 g) = 48.0 g

Zn: 1 (65.4 g) = 65.4 g

H: 6 (1.0 g) = 6.0 g

O: 4 (16.0 g) = 64.0 g

183.4 g

“What’s your flavorof ice cream?”

Finding an Empirical Formula from Experimental Data

1. Find # of g of each element.

2. Convert each g to mol.

3. Divide each “# of mol” by the smallest “# of mol.”4. Use ratio to find formula.

A compound is 45.5% yttrium and 54.5% chlorine.Find its empirical formula.

YCl3

Yg 45.5

Yg 88.9 Ymol 1

Ymol 0.512 0.512 1

Cl g 54.5

Cl g 35.5Cl mol 1

Cl mol 1.535 0.512 3

A ruthenium/sulfur compound is 67.7% Ru.Find its empirical formula.

RuS1.5 Ru2S3

Ru g 67.7

Ru g 101.1Ru mol 1

Ru mol 0.670 0.670 1

S g 32.3

S g 32.1S mol 1

S mol 1.006 0.670 1.5

A 17.40 g sample of a technetium/oxygen compoundcontains 11.07 g of Tc. Find the empirical formula.

TcO3.5 Tc2O7

Tc g 11.07

Tc g 98Tc mol 1

Tc mol 0.113 0.113 1

O g 6.33

O g 16.0O mol 1

O mol 0.396 0.113 3.5

A compound contains 4.63 g lead, 1.25 g nitrogen,and 2.87 g oxygen. Name the compound.

PbN4O8

Pb(NO2)4

Pb? 4 NO21–

lead (IV) nitrite

(plumbic nitrite)

?

Pb g 4.63

Pb g 207.2Pb mol 1

Pb mol 0.0223 0.0223 1

N g 1.25

N g 14.0N mol 1

N mol 0.0893 0.0223 4

O g 2.87

O g 16.0O mol 1

O mol 0.1794 0.0223 8

?

(How many empiricals “fit into” the molecular?)

To find molecular formula…

A. Find empirical formula.

B. Find molar mass of empirical formula.C. Find n = mm molecular mm empirical D. Multiply all parts of empirical formula by n. (“How many scoops?”)

(“What’s yourflavor?”)

A carbon/hydrogen compound is 7.7% H and has amolar mass of 78 g. Find its molecular formula.

emp. form. CH

mmemp = 13 g 78 g 13 g

= 6 C6H6

H g 7.7

H g 1.0H mol 1

H mol 7.7 7.69 1

C g 92.3

C g 12.0C mol 1

C mol 7.69 7.69 1

A compound has 26.33 g nitrogen, 60.20 g oxygen,and molar mass 92 g. Find molecular formula.

mmemp = 46 g = 2 N2O4 92 g 46 g

NO2

N g 26.33

N g 14.0N mol 1

N mol 1.881 1.881 1

O g 60.20

O g 16.0O mol 1

O mol 3.763 1.881 2

Mole Calculations

1 mol = 6.02 x 1023 particles

MOLE(mol)

Mass(g)

Particle(at. or m’c)

1 mol = molar mass (in g)

Volume(L or dm3)

1 mol = 22.4 L

1 mol = 22.4 dm3

New Points aboutIsland Diagram:

a. Diagram now has four islands.

b. “Mass Island” now for elements or compounds

c. “Particle Island” now for atoms or molecules

d. “Volume Island”: for gases only

1 mol @ STP = 22.4 L = 22.4 dm3

1 mol = 6.02 x 1023 particles

MOLE(mol)

Mass(g)

Particle(at. or m’c)

1 mol = molar mass (in g)

Volume(L or dm3)

1 mol = 22.4 L

1 mol = 22.4 dm3

1.29 mol

What mass is 1.29 mol ?

How many molecules is 415 L at STP?sulfur dioxide

Fe2+ NO31– Fe(NO3)2

ferrous nitrateferrous nitrate

( )1 mol179.8 g

= 232 g

sulfur dioxideSO2

415 L ( )1 mol22.4 L ( )1 mol

6.02 x 1023 m’c

= 1.12 x 1025 m’c

22.4 L

( )1 mol

87.3 L

What mass is 6.29 x 1024 m’cules ? Al3+ SO4

2– Al2(SO4)3

aluminum sulfate aluminum sulfate

= 3580 g

( )1 mol

342.3 g( )1 mol6.02 x 1023 m’c

6.29 x 1024 m’c

At STP, how many g is 87.3 dm3 of nitrogen gas?N2

( )1 mol28.0 g = 109 g

But there are 9 atoms per molecule, so…

How many m’cules is 315 g of iron (III) hydroxide?Fe3+ OH1–

Fe(OH)3

315 g( )1 mol106.8 g ( )1 mol

6.02 x 1023 m’c

= 1.78 x 1024 m’c

How many atoms are in 145 L of CH3CH2OH at STP?

145 L ( )1 mol22.4 L ( )1 mol

6.02 x 1023 m’c

= 3.90 x 1024 m’c

9 (3.90 x 1024) = 3.51 x 1025 atoms

Hydrates and Anhydrous Salts

anhydrous salt: an ionic compound (i.e., a salt) that attracts water molecules and forms loose chemical bonds with them; symbolized by MN

“anhydrous” = “without water”

Uses:

hydrate: an anhydrous salt with the water attached

-- symbolized by MN . ? H2O

Examples:

“desiccants” in leathergoods, electronics, vitamins

CuSO4 . 5 H2O BaCl2 . 2 H2ONa2CO3 . 10 H2O FeCl3 . 6 H2O

ENERGY

ENERGY

+

+

MNH2O

H2OH2O

H2OH2O H2O

MNH2O

H2O

H2OH2OH2O H2O

HEAT+

hydrate anhydrous salt water

Finding the Formulaof a Hydrate

1. Find the # of g of MN and # of g of H2O.

2. Convert g to mol.

3. Divide each “# of mol” by the smallest “# of mol.”

4. Use the ratio to find the hydrate’s formula.

Find formula of hydrate for each problem.

sample’s mass beforeheating = 4.38 g

sample’s mass afterheating = 1.93 g

molar mass of anhydrous salt = 85 g

MN . ? H2O

MN

(hydrate)

(anhydrous salt)

MN g 1.93

MN g 85MN mol 1

MN mol 0.0227 0.0227 1

OH g 2.45 2

OH g 18OH mol 1

2

2 OH mol 0.1361 2 0.0227 6

MN . 6 H2O

beaker +salt + water

A. beaker = 46.82 g

B. beaker + sample beforeheating = 54.35 g

C. beaker + sample afterheating = 50.39 g

molar mass of anhydrous salt = 129.9 g

beaker + salt

MN g 3.57

MN g 129.9MN mol 1

MN mol 0.0275 0.0275 1

OH g 3.96 2

OH g 18OH mol 1

2

2 OH mol 0.22 2 0.0275 8

MN . 8 H2O

beaker +salt + water

A. beaker = 47.28 g

B. beaker + sample beforeheating = 53.84 g

C. beaker + sample afterheating = 51.48 g

molar mass of anhydrous salt = 128 g

beaker + salt

MN g 4.20

MN g 128MN mol 1

MN mol 0.0328 0.0328 1

OH g 2.36 2

OH g 18OH mol 1

2

2 OH mol 0.1311 2 0.0328 4

MN . 4 H2O

or…

For previous problem, find % water and% anhydrous salt (by mass).

g 128 ) g (18 4O)H g (18 4

OH % 22

g 4.20 g 2.36OH g 2.36

OH % 22

hydrate of mass molarformula in OH g

OH % 22

OH 36.0% 2MN 64.0%

OH 36.0% 2MN 64.0%

Review Problems

Find % comp. of

Fe3+ Cl1–

FeCl3

iron (III) chloride.iron (III) chloride.

Fe: 1 (55.8 g) = 55.8 g

Cl: 3 (35.5 g) = 106.5 g

162.3 g

162.3 g 34.4% Fe

65.6% Cl :

A compound contains 70.35 g C and 14.65 g H.Its molar mass is 58 g. Find its molecular formula.

emp. form. C2H5

mmemp = 29 g 58 g 29 g

= 2 C4H10

H g 14.65

H g 1.0H mol 1

H mol 14.65 5.86 2.5

C g 70.35

C g 12.0C mol 1

C mol 5.86 5.86 1

At STP, how many g is548 L of chlorine gas?

22.4 L548 L ( )1 mol

Cl2

( )1 mol71.0 g = 1740 g

beaker +salt + water

A. beaker = 65.2 g

B. beaker + sample beforeheating = 187.9 g

C. beaker + sample afterheating = 138.2 g beaker + salt

MN g 73.0

MN g 158.6MN mol 1

MN mol 0.46 0.46 1

OH g 49.7 2

OH g 18OH mol 1

2

2 OH mol 2.76 2 0.46 6

SrCl2 . 6 H2O

is an anhydrous salt on which thefollowing data were collected. Find formula of hydrate. Strontium chlorideStrontium chloride

Sr2+ Cl1– SrCl2

0 0

0 0