CHEMICAL COMPOUNDS AND THE MOLE

Chapter 7

Formula Mass

Mass of H2O? H 2(1.01)

+ O 16.00_

18.02 amu Formula Mass: mass of molecule, formula unit,

or ion is sum of masses of all atoms represented (amu)

Ca(NO3)2

Ca 40.08

N 2(14.01)

+ O 6 (16.00)

164.10 amu

Molar Mass

Definition: mass of 1 mole of compound – use molar masses of elements (g/mol)

MgCl224.31 g/mol + 2(35.45 g/mol) = 95.21 g/mol

(NH4)2CrO4

2(14.01 g/mol) + 8(1.01 g/mol) + 52.00 g/mol + 4(16.00 g/mol) = 152.10 g/mol

CuSO4 * 5H2O63.55 g/mol + 32.07 g/mol + 4(16.00 g/mol) + 5(18.02 g/mol) = 249.72 g/mol

Molar Mass in Conversions

Remember flow chart from chapter 3? What is mass (g) of 3.04 mol of ammonia

vapor, NH3? ? g = 3.04 mol X 17.04 g = 51.8 grams NH3

1 mol

How many moles of sodium chloride are present in 100.0 grams?

? Moles NaCl = 100.0 g X 1 mol = 1.711 mol NaCl

58.44 g

PRACTICE MOLAR MASS

Work as a group of 4. 1st group member: 1,5,9,13,17,21 GroupII:1 2nd group member: 2,6,10,14,18,22 Group

II:2 3rd group member: 3,7,11,15,19,23 Group II:3 4th group member:4,8,12,16,20,24 Group II:4

Show work for each problem you complete.Explain your work to the other group membersand write in their answers.HW: complete your set and 33-40.

“Super Mole” Conversions

How many molecules are in 4.15 x 10-5g C6H12O6?? Molec.= 4.15 x 10-5 g X 1 mol X 6.02 x 1023 molec. = 1.39 x 1017 molecules

180.18 g 1 mole

How many H atoms are in 7.1 moles of C6H12O6?

? Atoms = 7.1 mol X 12 mol H X NA = 5.1 x 1025 atoms H

1 mol C6H

12O

6 1 mol H

How many formula units are in 4.5 kg Ca(OH)2?

?f.un = 4.5 kg X 103 g X 1 mol X NA X 1 f. un = 3.7 x 1025 formula units

1 kg 74.10 g 1 mol 1 molecule

**NA = 6.02x1023

molecules

More “Super Mole” Conversions What is the mass of H2SO4, if you have 1.53 x

1023 sulfate ions your compound? 1.53x1023 ions x 1 mol SO4 x 1 mol H2SO4 x 98.09 g = 24.9 g

NA 1mol SO4 1 mol

How many water molecules are present in in a 5.00 g sample of copper (II) sulfate pentahydrate? 5.00 g CuSO4 * 5H2O x 1 mol x 5mol H2O x NA

249.72 g 1mol CuSO4 * 5H2O 1 mol H2O

= 6.03x1022 molecules H2O

Percent Composition

The percent by mass of each element in a compound.

% = mass due to 1 element x 100 mass of whole compound

Percent Composition

What is the percent composition by mass of each element in (NH4)2O? [MM (NH4)2O = 52.10 g/mol]

%N = 2(14.01) x 100 = 53.78% 52.10

% H = 8(1.01) x 100 = 15.5%52.10

% O = 16.00 x 100 = 30.71% 52.10 What percentage by mass of Al2(SO4)36H2O is water?

% H2O = 6(18.02) x 100 = 24.01%450.29

Given a 25.0 gram sample of aluminum sulfate hexahydrate, how much water (g) could be driven off?

g H2O= (%H2O) (total sample mass)(24.01%) (25.00 g) = 6.00g

Empirical Formula

Definition: formula showing smallest whole-number mole ratio of atoms in a compound

Ex: B2H6 Molecular Formula

BH3 Empirical Formula Given (CH2O)x as the empirical, determine

possible molecular formulas.x=2 x=1 x=6

C2H4O2 CH2O C6H12O6

Formaldehyde acetic acid glucose

(all have different molar masses)

Empirical Formula Calculation

Given the molecular formula: reduce Given % compostion data

Find grams of each element Find moles of each element Find mole ratio of atoms by dividing through

by the smallest number of moles

If the ratio yields a 0.33, 0.50 multiply the entire formula through to clear fractional mole amounts.

Finding Empirical Formula

Determine the empirical formula of the compound with 17.15% C, 1.44% H, and 81.41% F.

(CHF3)x

• Assume 100 g sample. C1.427H1.43F4.285

17.15g C x 1mol = 1.427mol C 1.427 1.427 1.427

12.01g

1.44g H x 1mol = 1.43mol H =(CHF3)x

1.01 g

81.41g F x 1mol = 4.285mol F

19.00g

Finding Empirical Formula

Find empirical formula of 26.56% K, 35.41% Cr, and rest O.

(K2Cr2O7)xAssume 100 g sample.

26.56 g K x 1 mol = .6793 moles K K.6793 Cr .6810 O 2.377

39.10 g .6793 .6793 .6793

35.41 g Cr x 1 mol = .6810 moles Cr

52.00 g = (KCrO3.5 ) x 2

38.03 g O x 1 mol = 2.377 moles O = (K2Cr2O7 )x

16.00 g

Finding Molecular Formula

x(empirical formula) = molecular formulax(emp.form mass) = molec.form mass

x = Molecular formula Mass Empirical formula Mass

Finding Molecular Formula

Determine molecular formula of compound with empirical formula CH and formula mass of 78.110 amu.x = 78.110 amu = 6 (CH)6 = C6H6

(12.01+1.01) amu

Finding Molecular Formula

Sample has formula mass of 34.00 amu has 0.44 g H and 6.92 g O. Find its molecular formula.%H = .44g x 100 = 6.0%

7.36g

%O = 6.92g x 100 = 94.0%

7.36 g

Assume 100 g sample.

6.0g H x 1 mol = 5.9 mol 94.0 g O x 1mol = 5.88 mol

1.01g 16.00g

H5.9 O5.88 = (HO)x x = 34.00 amu = 2 H2O2

5.88 5.88 (1.01 + 16.00) amu hydrogen peroxide

Combustion Analysis

A compound contains only carbon, hydrogen, and oxygen. Combustion of the compound yields .01068 grams of carbon dioxide and .00437 grams of water. The molar mass of the compound is 180.1 g/ mol. The sample has a total mass of .0100 grams. What are the empirical and molecular formulas of the compound?

CHO + O2 H20 + CO2

.01068g CO2 x 1 mol CO2 x 1 mol C = 2.427 x 10-4 mol C x 12.01 g = .002915 g C

44.01gCO2 1 mol CO2 1 mol

.00437g H20 x 1mol H20 x 2mol H = 4.85 x 10-4 mol H x 1.01 g = 4.90 x 10-4 g H

18.02 g 1mol H20 1 mol

C+H+O = CHO

.002915 g + 4.90 x 10-4 g + O = .0100 g

gO = .00660 g = 4.12 x 10-4 mol O

Combustion Analysis (continued)C 2.427 x 10-4 H 4.85 x 10-4 O 4.12 x 10-4 = (CH2O2)x

2.427 x 10 -4

X = molecular mass = 180.1 g = 4

empirical mass 46.03 g

(CH2O2)4 = C4H8O8