The Mole

Ch.8

(8-1) Mole

• (mol): amt. of substance– # of atoms in 12g of carbon-12

• Avogadro’s constant: 6.02 x 1023 particles / mol– Atoms, molecules (covalent) ,

formula units (ionic)

Moles # of Atoms

• 2.66 mol x 6.02 x 1023 atoms = 1.60 x 1024 1 mol atoms

• # of Atoms Moles

• 2.54 x 1024 atoms x 1 mol = 4.22 mol

6.02 x 1023 atoms

Moles to Molecules

• 1.04 mol x 6.02 x 1023 molecules = 6.26 x 1023 1 mol molecules

• Formula units to moles3.49 x 1024 form.u. x 1 mol = 5.80 mol

6.02 x 1023 form.u.

Moles to Atoms Practice

How many atoms are in 2.5 mol of Si?1. List known

2.5 mol Si

2. Set up conv. factor (what you’re going to is on top, what needs to cancel is on bottom)

2.5 mol Si x 6.02 x 1023 atoms 1 mol Si

Moles to Atoms Practice

3. Cancel units, multiply top #’s & divide by bottom2.5 mol Si x 6.02 x 1023 atoms = 1.5 x 1024 atoms Si

1 mol Si

Atoms to Moles Practice

Convert 3.01 x 1023 atoms of Si to mols Si1. List known

3.01 x 1023 atoms Si

2. Set up conv. factor (what you’re going to is on top, what needs to cancel is on bottom)

3.01 x 1023 atoms Si x 1 mol Si = 6.02 x 1023 atoms

Atoms to Moles Practice

3. Cancel units, multiply top #’s & divide by bottom

3.01 x 1023 atoms Si x 1 mol Si = 0.500 mol Si 6.02 x 1023 atoms Si

Relative Atomic Mass

• Weighted avg. of each isotope’s mass– Need mass & % abundance

(62.94 amu)(.6917) + (64.93 amu)(.3083) = 63.55 amu

Isotope Mass (amu)

% Decimal

Copper-63 62.94 69.17 .6917

Copper-65 64.93 30.83 .3083

Molar Mass

• Sum of all the atomic masses of a substance

• Ex: CO2

C = 12.01 g/mol, O = 16 g/mol

12.01 g/mol + (2)(16 g/mol) = 44.01 g/mol

Molar Mass Practice

Calculate the molar mass of Ba(NO3)2

1. List known (at. masses from PT)Ba = 137.33 g/mol, N = 14.01 g/mol, O = 16 g/mol

2. Calculate # of each atom in the formula (outer subscript multiplies w/ inner subscripts)1 Ba1 N x 2 = 2 N3 O x 2 = 6 O

Molar Mass Practice

3. Multiply each at. mass by the # of atoms & add together

(1 Ba)(137.33 g/mol) + (2 N)(14.01 g/mol) +

(6 O)(16 g/mol) = 261.35 g/mol

Moles Mass

• 2.49 mol KF x 58.1 g KF = 145 g KF 1 mol KF

• Grams Moles

• 110 g KF x 1 mol KF = 1.89 mol KF 58.1 g KF

Moles to Mass Practice

What’s the mass in grams of 3.50 mol Cu?1. List known

3.50 mol Cu

2. Find at. mass or calculate molar mass

63.55 g Cu

1 mol Cu

Moles to Mass Practice

3. Set up conv. factor (what you’re going to is on top, what needs to cancel is on bottom)

3.50 mol Cu x 63.55 g Cu

1 mol Cu

4. Cancel units, multiply top #’s, & divide by bottom3.50 mol Cu x 63.55 g Cu = 222 g Cu

1 mol Cu

Mass to Moles Practice

Determine the # of moles in 237 g of Cu1. List known

237 g Cu

2. Find at. mass or calculate molar mass

63.55 g Cu

1 mol Cu

Mass to Moles Practice

3. Set up conv. factor (what you’re going to is on top, what needs to cancel is on bottom) 237 g Cu x 1 mol Cu =

63.55 g Cu

4. Cancel units, multiply top #’s, & divide by bottom237 g Cu x 1 mol Cu = 3.73 mol Cu

63.55 g Cu

Atoms to Grams

1.2 x 1024 atoms B x 1 mol B x 10.81 g B = 21.55 g B

6.02 x 1023 atoms 1 mol B

(8-2) Percentage Composition

• % by mass of each element in a cmpd– Ex: H2O = 88.7% O, 11.3% H

O

H

Determine % Comp. from Chemical Formula

• % = mass of component x 100

mass of whole

• To verify answers all components %, should = 100%

% Comp. Practice

Find the % Comp. of Cu2S

1. Find at. masses & multiply by # of atoms

Cu = (63.55 g/mol) (2) = 127.10 g/mol

S = (32.07 g/mol) (1) = 32.07 g/mol

% Comp. Practice

2. Divide each mass by total mass of cmpd & multiply by 100

127.10 g/mol Cu x 100 = 79.85% Cu

159.17 g/mol Cu2S

32.07 g/mol S x 100 = 20.15% S

159.17 g/mol Cu2S

3. Add % to verify it’s close to 100%

79.85% + 20.15% = 100%

Hydrates

• To determine % comp. of a hydrate, use same format as before

• % = mass water x 100

mass of whole

Hydrate % Comp. Practice

Determine the % water in Na2CO3•10H2O

1. Calculate molar mass of water (coef. is distributed)

(20)(1.01 g/mol) + (10)(16 g/mol) = 180.2 g/mol H2O

2. Divide water mass by the total hydrate mass 180.2 g/mol H2O x 100 = 62.96%H2O

286.2 g/mol Na2CO3•10H2O

Empirical Formula

• Empirical Formula: simplest ratio among the elements of a cmpd– Ex: CH2O: CH2O, C2H4O2, C6H12O6

Determining Emp. Formulas

63.0% Mn, 37.0% O by mass

1. Convert to g assuming you have 100 g63.0 g Mn, 37.0 g O

2. Convert g to mols using molar mass63.0 g Mn x 1 mol Mn = 1.15 mol Mn

54.94 g37.0 g O x 1 mol O = 2.31 mol O

16 g

Emp. Formula (cont.)

3. Divide by smallest amt (round to whole #)1.15 mol Mn = 1 mol Mn

1.15

2.31 mol O = 2 mol O

1.15

4. Write empirical formula MnO2

Molecular Formula

• Actual # of atoms in a cmpd

• Determined by emp.formula & molar mass

Determining a Molecular Formula

emp. formula = P2O5

molar mass of cmpd = 284 g/mol

1. Find emp. formula molar mass(2)(30.97 g/mol) + (5)(16 g/mol) = 141.94 g/mol

2. Solve for n (round to nearest whole #)n = molar mass of cmpd = 284 g/mol = 2

molar mass emp. 141.94 g/mol

Determining a Molecular Formula

3. Use n to get molecular formula

n (emp.form.) = molecular form.

2 (P2O5) = P4O10

4. To verify answer, calculate molar mass & compare to value given

(4)(30.97 g/mol) + (10)(16 g/mol) = 284 g/mol