Body parts Mole calculations: Moles level 2(continued)

`body’ parts example 2 : Weight moles first, then mol Mol

Octane = C8H18

How many moles of octane contain 24 grams of C (at. wt. = 12 g/mol)

24 g = 2 moles C12 g/mol C

Step 1) Convert grams C to moles C (all roads lead through moles divide up)

`body’ parts example 2 : (continued) Octane = C8H18

How many moles of octane contain 24 grams of C (at. wt. = 12 g/mol)

Step 2) 2 moles C => how many moles octane?

1 mol octane = x8 mol C 2

x= 2/8=0.25 moles octane

(body parts relationship ?)

`body’ parts example 3 : Weight moles , then mol Mol weight

Octane = C8H18 (MW=114)How many grams of H are in 10 grams of Octane ? (1 mole H=1 g H)

Step 1) Convert grams octane to moles octaneAll roads lead through moles; divide up

10 g = 0.0877 mol octane114 g/mol

`body’ parts example 3 (continued) : Octane = C8H18 (MW=114)

How many grams of H are in 10 grams of Octane ? (1 mole H=1 g H)

Step 2) Relate moles H to moles octane(body parts relationship ?)

14 mol H = x 1 octane 0.0877

x= 14*0.0877=1.228 mol H

`body’ parts example 3 (continued) : Octane = C8H18 (MW=114)How many grams of H are in 10 grams of Octane ? (1 mole H=1 g H)

Step 3) convert moles H to grams H(multiply down)

1.228 mol H * 1 g H/mol=1.228 mol H

Solving Mole body parts problems: A SUMMARY

Step 1: Convert any given weight or molecule count to moles (“all roads lead through moles” DIVIDE UP)

Step 2: Relate the computed moles above to desired moles (set up ratio and solve x)

If necessary –Step 3: convert desired moles to target grams or molecule count (MULTIPLY DOWN)

Isohexane has the formula C6H14. How many grams of H (at. Wt. = 1 g/mol) are combined

in 300 grams of C6H14 (MW=86 g/mol)

A. 36.84 g H

B. 1.32 g H

C. 0.249 g H

D. 0.020 g H

E. 48.84 g H

36.84 g H

1.32 g H

0.249 g H

0.020 g H

48.84 g H

0% 0% 0%0%0%

Isohexane has the formula C6H14. How many moles of isohexane are formed with 24 grams of C (1 mol C=12 g)

A. 0.333 moles isohexane

B. 2 moles isohexane

C. 0.1667 moles isohexane

D. 3 moles isohexane

E. Stoichiometry blows

F. None of the above

0.333 moles i

sohexa

ne

2 moles i

sohexa

ne

0.1667 moles i

sohexa

ne

3 moles i

sohexa

ne

Stoich

iometry blows

F. None of t

he above

69%

7%2%

9%9%4%

Mole Body Part Calculations level 2: extra problems for you to try at home if you want

1)mole to mole: how many moles of O are present in 0.1666 mole of C6 H12O6?

2)mole to mole how many moles of C6H12O6 can be made with 12 mole of O ?

1 mol O

2 moles C6H12O6

3) weight to moles how many moles of C6H12O6 in a sample containing 216 g C ?

3 mol

4) moles to weight how many grams of C6H12O6 are formed with 0.2666 mol H? = 4 g

Mole Calculations: part 2 (cont.)

5) moles to molecules how many molecules of O are present in 0.8333 mol of C6 H12O6?

=5*1023

6) molecules to moleshow many moles of C6 H12O6 are formed from 4.32*1025 atoms of H ? 6 moles

Mole Calculations: part 2 (cont.)

7) mass to molecules: how many molecules of C6 H12O6 form from 84 g of C ?

7 *1023 molecules of C6H12O6

8) atoms to mass: how many grams of H are combined with 2.4*1024 atoms of O in C6H12O 6 ?

8 grams H

Mole Calculations: part 2 (cont.)

• Basic mole-mass-count conversions ( divide up, multiply down)

• Mole ratio conversion within a compound (`body parts’ relationships)

The road trip through mole land so far…

• % composition problems and combustion analysis (pp. 94-103)

• Reaction balancing (pp. 105-108)

• Reaction stoichiometry predictions, limiting yields and % yields ( pp. 108-123): moles level 3

The mole road ahead...

Why bother with all of this ????

Sample % composition problem #1

A compound of N and O contains 63.63 wt % N and 36.36 wt % O. What is the empiric formula of the compound ?

% composition problems: using mole concept to convert weights to formulas

Definition:

Example: CH2O = empiric formula of glucose (sugar we metabolize)

empiric formula= mole ratio of elements in a compound expressed in lowest whole numbers

CH2O = empiric formula of glucose (sugar we metabolize)

(CH2O)6

Sugar =Carbo hydrate

Actual molecular formula (what it really has in atom count):

C6H12O6=

Empiric formulas (continued)

Which formulas below are empiric (= lowest common denominator form) ??

C3H6O3

NOT EMPIRIC

÷ 3

CH2OEMPIRIC

Which formulas below are empiric (continued) ??

N3F7EMPIRIC…3 & 7 have no shared factors

Which formulas below are empiric (continued) ??

H3P9O12S5EMPIRIC(3,9,12 divisible by 3…but 5 is not)

A compound of N and O contains 63.63% N and 36.36 %O. What is the empiric formula of the compound?

% composition Problem 1: empiric formula determination

let’s do it on the board…

Sample problem #1; SUMMARIZED

element

N

O

w(g)

63.63

36.36

14

16

63.63/14= 4.54536.36/16= 2.273

4.545 = 22.273

2.273 =12.273

AW=Atomic wt (g/mol)

n=w/AWmoles

nnmin

=>N2O=Laughing gas

OFFICIAL NAME…

Dinitrogen monoxide

http://www.youtube.com/watch?v=_Ha-ZrUPJ_E

Joseph Priestley 1772

An oxide of nitrogen contains 46.7 wt. % N and 53.3 % O. What is it’s empiric formula ?

A. NO3

B. NO2

C. N2O

D. N3O7

E. NO

F. None of the above

0% 0% 0%0%0%0%

A compound composed of C,H and O is 48.64% C and 8.16 % H and 43.2 % O by mass. What is the empiric formula ? Answer: C1.5H3O1 C3H6O2

% composition Problem 2: empiric formula determination with a twist

element Mass,g Atomic wt(g/mol)

n=mass Atomic wt

n/nmin

C 48.64

12

H 8.16 1

O 43.2 16

48.64/12=4.05

8.16/1=8.1643.2/16=2.70

4.05/2.7 =1.5 ?

8.16/2.7 =3.0

2.7/2.7= 1

x FACTOR1.5*2= 3

3.0*2= 6 1*2= 3