Mullis 1

Relationship between mass, moles and molecules in a compound

Mass(g)

Amount(moles)

# moleculesor

Formula units

X molar mass

(__g__ mole)

X 6.022 x 1023

(units mole)

Grams moles = moles gram

moles units = units or molecules mole

moles grams = grams mole

Mullis 2

Molar mass• Molar mass of a substance = mass in grams of one

mole of the substance.• A compound’s molar mass is NUMERICALLY equal to its

formula mass.2 mol H x 1.01 g H = 2.02 g H

1 mol H1 mol O x 16.00 g O = 16.00 g O

1 mol O molar mass H2O =18.02 g/mol

• Formula mass H2O = 18.02 amu• Molar mass H2O = 18.02 g/mol

Mullis 3

Molar Mass Example

What is the molar mass of K2SO4?2 mol K x 39.10 g K = 78.20 g K

1 mol K1 mol S x 32.10 g S = 32.07 g S

1 mol S4 mol O x 16.00 g O = 64.00 g O

1 mol Omolar mass K2SO4 =

174.27 g/mol

How many moles of each element are present in this compound?

2 mol K, 1 mol S, 4 mol O

Mullis 4

What is the molar mass of C6H12O6?6 mol C x 12.01 g C = 72.06 g C

1 mol C12 mol H x 1.01 g H = 12.12 g H

1 mol H6 mol O x 16.00 g O = 96.00 g O

1 mol Omolar mass C6H12O6 =

180.18 g/mol

How many moles of each element are present in this compound?

6 mol C, 12 mol H, 6 mol O

Mullis 5

Converting to grams from molesHow many moles of glucose are in 4.15x10-3 g C6H12O6?

4.15x10-3 g x 1 mol C6H12O6 = 2.30 x 10-5 mol C6H12O6

180.18 g

How many molecules of glucose are in 4.15x10-3 g C6H12O6?

2.30 x 10-5 mol C6H12O6 x 6.022 x 10 23 molecules =1 mol

(2.30 x 6.022)(10(-5+23)) = 13.90 x 10 –18 molecules = 1.39 x 10 –19 molecules

Mullis 6

What is the mass in grams of 6.25 moles copper (II) nitrate?

Cu 2+ NO3 - : formula is Cu(NO3)2

Find molar mass of Cu(NO3)2 first.1 mol Cu x 63.55 g Cu = 63.55 g Cu

1 mol Cu2 mol N x 14.01 g N = 28.02 g N

1 mol N6 mol O x 16.00 g O = 96.00 g O

1 mol O

molar mass Cu(NO3)2 = 187.57 g/molNow find mass in grams of 6.25 moles:

6.25 moles x 187.57 g = 1172 g Ans. 1170 g Cu(NO3)2 1 mol

Mullis 7

Atoms and Ions Within Compounds• How many carbon atoms are in one mole of C2H6?

1 mole C2H6 | 2 moles C | 6.022 x 1023 atoms = 1.204 x 1024 atoms

|1 mole C2H6 | 1 mole C

• How many MOLES of carbon atoms are in one mole of C2H6?

1 mole C2H6 | 2 moles C = 2 moles C atoms

|1 mole C2H6

• How many moles of hydroxide ions are in one mole of calcium hydroxide? How many moles of Ca2+?

1 mole Ca(OH)2 | 2 moles OH-- =2 moles hydroxide ions

|1 mole Ca(OH)2

1 mole Ca(OH)2 | 1 mole Ca2+ =1 mole calcium ions

|1 mole Ca(OH)2

Mullis 8

Percentage Composition % Composition is the % by mass of each element in a compound.Find the percentage composition of sodium chloride.Na+ Cl - : formula is NaCl

1 mol Na x 22.99 g Na = 22.99 g Na 1 mol Na

1 mol Cl x 35.45 g Cl = 35.45 g Cl

1 mol Cl molar mass NaCl = 58.44 g/mol

22.99 g Na x 100 = 39.34 % Na58.44 g NaCl

35.45 g Cl x 100 = 60.66 % Cl58.44 g NaCl

Mullis 9

Find the percentage composition of sodium nitrate.

Na+ NO3 - : formula is NaNO3

1 mol Na x 22.99 g Na = 22.99 g Na 1 mol Na

1 mol N x 14.01 g N = 14.01 g N 1 mol N

3 mol O x 16.00 g O = 48.00 g O

1 mol O molar mass NaNO3 = 85.00 g/mol22.99 g Na x 100 = 27.05 % Na85.00 g NaNO3

14.01 g N x 100 = 16.48 % N85.00 g NaNO3

48.00 g O x 100 = 56.47% O85.00 g NaNO3

Mullis 10

Percentage Composition

• Why mass instead of moles?• Isn’t 2/3 of the water molecule

hydrogen?• Moles indicate the amounts of each

atom needed to make the molecule stable from an electron standpoint.

2.02 g H x 100 = 11.21 % H18.02 g H2O H H

O

Mullis 11

Empirical FormulaUse % composition to convert to original formula:1. Assume 100 g sample, so % = grams2. Convert grams to moles for each element3. Divide the number of moles for each element by the smallest

number of moles4. The result for each type of element is its subscript in the

empirical formula.5. The order of elements is usually:

Organics: C,H,O,NInorganics: Metal, nonmetal, oxygen

6. Keep 4 decimal places when dividing numbers. If the result has a decimal between .2 and .8, may need to multiply all numbers by the number needed to get a whole number. Ex: 3.5 should be multiplied by 2 to get 4. Then multiply all other elements by the same number.

Mullis 12

Example: Empirical and Molecular formula

What is the empirical formula of a compound with 54.82% C, 5.624% H, 32.45% O, 7.104% N?

54.82 g C| 1 mole C = 4.568 mole C / .5074 = 9 C | 12 g C

5.624 g H| 1 mole H =5.624 mole H / .5074 = 11 H

| 1 g H C9H11O4N32.45 g O| 1 mole O = 2.028 mole O / .5074 = 4 O

| 16 g O7.104 g N| 1 mole N =0.5074 mole N / .5074 = 1 N

| 14 g N

If a compound has this same composition but its molecular weight is 394 g/mol, what is its molecular formula?

MW. C9H11O4N = 197 g/mol

394/197 = 2 so molecular formula is C18H22O8N2

Mullis 13

Oxidation Numbers• Used to indicate the general distribution of electrons

among the bonded atoms in molecular compounds or polyatomic ions.

• Analogous to charges in ionic compounds.• An oxidation number is assigned to each element.• Assign the ones you know 1st.• Find the others based on the numbers it takes to make the

charge equal to the charge of the ion or compound. (A compound has a charge of zero.)

Mullis 14

Oxidation Numbers: Rules1. Pure element = 02. F = -13. O = -2 (except in peroxides and bonds with halogens)

4. H = +1 (except in bonds with metals)

5. The more electronegative element = same (-) charge as its anion

6. The less electonegative element = same (+) charge as its cation

7. The sum of a compound or polyatomic ion’s oxidation numbers is equal to its charge.