empirical formula:

1

Empirical formula:

2

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

g 0.12mol = 5.42 mol C

5.5 gg 01.1

mol= 5.45 mol H

29.1 gg 16.0

mol= 1.82 mol O

Divide by smallest to get whole number

2.97 1.82

2.99 1.82

1 1.82

C3H3O1

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 empircal and the molecular formulas.

If we work with 100 grams, we could call

the %’s grams!

4

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

C: 39.9 g

O: 53.4g

g01.1mol = 6.64

g01.12mol = 3.32

g00.16mol = 3.33

We need whole

numbers

3.32

3.32

3.32

= 1.00

= 1.00

= 2.00 empirical formula:CH2O

empirical formula wt.:

30.03

molecular formula:2.0 30.03 0.60

so: C2H4O2

5

Stoichiometry: “working with ratios”

6

When N2O5 is heated, it decomposes:

2N2O5(g) 4NO2(g) + O2(g)

a. How many moles of NO2 can be produced from 4.3 moles of N2O5?

= moles NO2

4.3 mol N2O5

52

2

ON mol2NO mol4

8.6

b. How many moles of O2 can be produced from 4.3 moles of N2O5?

= mole O2

4.3 mol N2O5

52

2

ON 2molO mol1

2.15

2N2O5(g) 4NO2(g) + O2(g)4.3 mol ? mol

2N2O5(g) 4NO2(g) + O2(g)4.3 mol ? mol

7

When N2O5 is heated, it decomposes:

2N2O5(g) 4NO2(g) + O2(g)

a. How many moles of N2O5 were used if 210g of NO2 were produced?

= moles N2O5

210 g NO2

2

52

NO mol4ON mol2

2.283

b. How many grams of N2O5 are needed to produce 75.0 grams of O2?

= grams N2O5

75.0 g O2

2

52

O 1molON mol2

506.25

2

2

NO g0.46NO mol

2

2

O g 32.0O mol

52

52

ON mol ON g108

210 g

8

Limiting/excess/theoretical yield Reactant Problems:Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen.

4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g)

a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?Determine the limiting reactant.

4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g)

First copy down the the BALANCED

equation!

9

Limiting/excess/theoretical yield Reactant Problems:

Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen.

4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g)


4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g)

Now place numerical the

information below the compounds.

10


4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g)


4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g) 0.15 mol 0.10 mol ? moles

Two starting amounts?

Where do we start?

Hide

one

11


4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g)


4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g) 0.15 mol 0.10 mol ? molesHide

Based on:KO2 = mol O2

0.15 mol KO2

2

2

KO 4molO mol3 0.1125

12

Limiting/excess /theoretical yield Reactant Problems:

Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen.

4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g)




0.15 mol KO2

2

2


Hide

Based on: H2O

= mol O20.10 mol H2O

OH 2molO mol3

2

2 0.150

13

Limiting/excess /theoretical yield Reactant Problems:Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen.

4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g) a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?

Determine the limiting reactant.



0.15 mol KO2

2

2


Based on: H2O

= mol O20.10 mol H2O

OH 2molO mol3

2

2 0.150

What is the theoretical yield? Hint: which

reactant was the limiting reactant?

it was limited!

H2O = XS reactant!

14

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 reactantActual yield = 12.3 g experimentally recovered

100x yield ltheoretica

yield actual yield %

yield 63.1% 100x 19.512.3 yield %

15

4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g) If a reaction vessel contains 120.0 g of KO2 and 47.0 g of H2O, how many grams of O2 can be produced?

4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g) 120.0 g 47.0 g ? gHide one

Based on:KO2

= g O2 120.0 g KO2

g1.71mol

2

2

KO 4molO mol3

2

2

O molO g0.32 40.51

16

4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g) If a reaction vessel contains 120.0 g of KO2 and 47.0 g of H2O, how many grams of O2 can be produced?

4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g) 120.0 g 47.0 g ? g

Based on:KO2

= g O2 120.0 g KO2

g1.71mol

2

2

KO 4molO mol3

2

2

O molO g0.32 40.51

Based on:H2O

= g O2

Question if only 35.2 g of O2 were recovered, what was the percent yield?

yield 86.9% 100x 51.402.35

Hide

47.0 g H2OOH g 02.18

OH mol

2

2

OH mol 2O mol 3

2

2

2

2

O molO g0.32 125.3

17

If a reaction vessel contains 120.0 g of KO2 and 47.0 g of H2O, how many grams of O2 can be produced?

4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g) 120.0 g 47.0 g ? g

Based on:KO2

= g O2 120.0 g KO2

g1.71mol

2

2

KO 4molO mol3

2

2

O molO g0.32 40.51

Based on:H2O

= g O247.0 g H2O

OH g 02.18OH mol

2

2

OH mol 2O mol 3

2

2

2

2

O molO g0.32 125.3

Determine how many grams of Water were left over.The Difference between the above amounts is directly RELATED to the XS H2O.

125.3 - 40.51 = 84.79 g of O2 that we have XS water to form.

= g XS H2O84.79 g O2

2

2

O g 32.0O mol

2

2

O mol 3OH mol 2

OH mol 1OH g 02.18

2

2 31.83

empirical formula:

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