Empirical formula:

1

Hydroquinone, used as a photographic developer is 65.4% C, 5.5% H, and 29.1% O, by mass. What is the empirical formula?

If we work with 100 grams, we would have 65.4 g C, 5.5 g H and 29.1 g O.

Then: Change g to moles and convert to whole numbers.

C: H: O: 65.4 g mol 12 .

0 g 5.5 g 29.1 g mol 1 .

01 g mol 16.0

g = 5.42 mol C = 5.45 mol H



1.82



2.97



1.82



2.99

= 1.82 mol O



1.82



1

C 3 H 3 O 1 Divide by smallest to get whole number

2

A compound was sent out for analysis and was found to contain 6.7% H, 39.9% C, and 53.4% O. It was also found to have a molecular mass of 60.0 g/mol.

Determine both the empircal and the molecular formulas.

If we work with 100 grams, we could call the %’s grams!

3

A compound was sent out for analysis and was found to contain 6.7% H, 39.9% C, and 53.4% O. It was also found to have a molecular mass of 60.0 g/mol.

Determine both the empirical and the molecular formulas.

H: 6.7 g mol 1 .

01 g = 6.64



3.32

= 2.00

C: 39.9 g mol 12 .

01 g O: 53.4g

mol 16 .

00 g = 3.32



3.32

= 1.00

= 3.33



3.32

= 1.00

empirical formula: CH 2 O empirical formula wt.: 30.03

molecular formula: 60 .

0



30.03



2.0

so: C 2 H 4 O 2 We need whole numbers

4

Stoichiometry: “working with ratios”

5

When N 2 O 5 is heated, it decomposes: 2 N 2 O 5 (g)



4 NO 2 (g) + O 2 (g) a. How many moles of NO 2 2 N 2 O 5 (g)



can be produced from 4.3 moles of N 2 O 5 ?

4 NO 2 (g) + O 2 (g) 4.3 mol ? mol 4.3 mol N 2 O 5 4 mol NO 2 2 mol N 2 O 5 = moles NO 2 b. How many moles of O 2 N 2 O 5 (g)



can be produced from 4.3 moles of N 2 O 5 ?

4 NO 2 (g) + O 2 (g) 4.3 mol ? mol 4.3 mol N 2 O 5 1 mol O 2 2mol N 2 O 5 = mole O 2

6

When N 2 O 5 is heated, it decomposes: 2 N 2 O 5 (g)



4 NO 2 (g) + O 2 (g) 210 g a. How many moles of N 2 O 5 were used if 210g of NO 2 were produced?

210 g NO 2 mol NO 2 46 .

0 g NO 2 2 mol N 2 O 5 4 mol NO 2 = moles N 2 O 5 b. How many grams of N 2 O 5 are needed to produce 75.0 grams of O 2 ?

75.0 g O 2 mol O 2 32.0

g O 2 2 mol N 2 O 5 1mol O 2 108 g N 2 O 5 mol N 2 O 5 = grams N 2 O 5

7

Limiting/excess/theoretical yield Reactant Problems: Potassium superoxide, KO 2 , is used in rebreathing gas masks to generate oxygen.

4KO 2 (s) + 2H 2 O(l)



4KOH(s) + 3O 2 (g) a. How many moles of O 2 can be produced from 0.15 mol KO 2 Determine the limiting reactant.

4KO 2 (s) + 2H 2 O(l)



4KOH(s) + 3O 2 and 0.10 mol H 2 O?

(g) First copy down the the BALANCED equation!

8

Limiting/excess/theoretical yield Reactant Problems: Potassium superoxide, KO 2 , is used in rebreathing gas masks to generate oxygen.

4KO 2 (s) + 2H 2 O(l)



4KOH(s) + 3O 2 (g) a. How many moles of O 2 can be produced from 0.15 mol KO 2 Determine the limiting reactant.

4KO 2 (s) + 2H 2 O(l)



4KOH(s) + 3O 2 and 0.10 mol H 2 O?

(g) Now place numerical the information below the compounds.

9

Limiting/excess/theoretical yield Reactant Problems: Potassium superoxide, KO 2 , is used in rebreathing gas masks to generate oxygen.

4KO 2 (s) + 2H 2 O(l)



4KOH(s) + 3O 2 (g) a. How many moles of O 2 can be produced from 0.15 mol KO 2 Determine the limiting reactant.

4KO 2 (s) + 2H 2 O(l)



4KOH(s) + 3O 2 and 0.10 mol H 2 O?

(g) 0.15 mol 0.10 mol ? moles Hide one Two starting amounts? Where do we start?

10

Limiting/excess/theoretical yield Reactant Problems: Potassium superoxide, KO 2 , is used in rebreathing gas masks to generate oxygen.

4KO 2 (s) + 2H 2 O(l)



4KOH(s) + 3O 2 (g) a. How many moles of O 2 can be produced from 0.15 mol KO 2 Determine the limiting reactant.

4KO 2 (s) + 2H 2 O(l)



4KOH(s) + 3O 2 and 0.10 mol H 2 O?

(g) 0.15 mol ? moles Based on: KO 2 0.15 mol KO 2 3 mol O 2 4mol KO 2 = mol O 2

11

Limiting/excess /theoretical yield Reactant Problems: Potassium superoxide, KO 2 , is used in rebreathing gas masks to generate oxygen.

4KO 2 (s) + 2H 2 O(l)



4KOH(s) + 3O 2 (g) a. How many moles of O 2 can be produced from 0.15 mol KO 2 Determine the limiting reactant.

4KO 2 (s) + 2H 2 O(l)



4KOH(s) + 3O 2 and 0.10 mol H 2 O?

(g) 0.10 mol ? moles Based on: KO 2 0.15 mol KO 2 3 mol O 2 4mol KO 2 = mol O 2 Based on: H 2 O 0.10 mol H 2 O 3 mol O 2 2mol H 2 O = mol O 2

12

Limiting/excess /theoretical yield Reactant Problems: Potassium superoxide, KO 2 , is used in rebreathing gas masks to generate oxygen.

4KO 2 (s) + 2H 2 O(l)



4KOH(s) + 3O 2 (g) a. How many moles of O 2 can be produced from 0.15 mol KO 2 Determine the limiting reactant.

4KO 2 (s) + 2H 2 O(l)



4KOH(s) + 3O 2 and 0.10 mol H 2 O?

(g) 0.15 mol 0.10 mol ? moles Based on: KO 2 0.15 mol KO 2 3 mol O 2 4mol KO 2 = mol O 2 it was limited!

Based on: 0.10 mol H 2 O 3 mol O 2 H 2 O 2mol H 2 O H 2 O = XS reactant!

What is the theoretical = mol O 2 yield? Hint: which reactant was the limiting reactant?

13

Theoretical yield vs.

Actual yield Suppose the theoretical yield for an experiment was calculated to be 19.5 grams, and the experiment was performed, but only 12.3 grams of product were recovered. Determine the % yield.

Theoretical yield = 19.5 g based on limiting reactant Actual yield = 12.3 g experimentally recovered

% yield



actual yield theoretica l yield x 100

% yield



12.3

19.5

x 100



63.1% yield

14

4KO 2 (s) + 2H 2 O(l)



4KOH(s) + 3O 2 (g) If a reaction vessel contains 120.0 g of KO 2 how many grams of O 2 can be produced?

4KO 2 (s) + 2H 2 O(l)



120.0 g and 47.0 g of H 2 O, 4KOH(s) + 3O ? g 2 (g) Based on: KO 2 120.0 g KO 2 mol 71 .

1 g 3 mol O 2 4mol KO 2 32 .

0 g O 2 mol O 2 = g O 2

15

4KO 2 (s) + 2H 2 O(l)



4KOH(s) + 3O 2 (g) If a reaction vessel contains 120.0 g of KO 2 how many grams of O 2 can be produced?

4KO 2 (s) + 2H 2 O(l)



47.0 g and 47.0 g of H 2 O, 4KOH(s) + 3O ? g 2 (g) Based on: KO 2 120.0 g KO 2 mol 71 .

1 g 3 mol O 2 4mol KO 2 32 .

0 g O 2 mol O 2 Based on: H 2 O 47.0 g H 2 O mol H 2 O 18 .

02 g H 2 O 3 mol O 2 2 mol H 2 O 32 .

0 g O 2 mol O 2 = g O 2 = g O 2 Question if only 35.2 g of O 2 were recovered, what was the percent yield?

35 .

2 40 .

51 x 100



86.9% yield

16

If a reaction vessel contains 120.0 g of KO 2 how many grams of O 2 can be produced?

4KO 2 (s) + 2H 2 O(l)



120.0 g 47.0 g and 47.0 g of H 2 O, 4KOH(s) + 3O ? g 2 (g) Based on: KO 2 120.0 g KO 2 mol 71 .

1 g 3 mol O 2 4mol KO 2 32 .

0 g O 2 mol O 2 = g O 2 Based on: H 2 O 47.0 g H 2 O mol H 2 O 18 .

02 g H 2 O 3 mol O 2 2 mol H 2 O 32 .

0 g O 2 mol O 2 = g O 2 Determine how many grams of Water were left over.

The Difference between the above amounts is directly RELATED to the XS H 2 O.

125.3 - 40.51 = 84.79 g of O 2 that we have XS water to form.

84.79 g O 2 mol O 2 32.0

g O 2 2 mol H 2 O 3 mol O 2 18 .

02 g H 2 O 1 mol H 2 O 31.83

2 O

17